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Linear Algebra for Machine Learning — Transcript

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0:00this in-depth course provides a
0:02comprehensive exploration of all
0:03critical linear algebra Concepts
0:06necessary for machine learning you'll
0:09learn the mathematical foundations to
0:11excel in AI tdiv from lunar Tech
0:15developed this course she has created
0:18many popular machine learning courses
0:20machine learning is at the Forefront of
0:22the Innovation powering the most
0:24advanced and transformative systems for
0:27the companies like apple Tesla Netflix
0:30Amazon open Ai and many others it
0:33enables the creation of the intelligent
0:35systems that can predict Trends
0:38personalized user experience and
0:40automate complex tasks to develop these
0:44practical applications a deep
0:46understanding of the underlying
0:47mechanics is important this requires a
0:50solid grasp of mathematics behind the
0:53machine learning so all these technical
0:55details with a particular focus on
0:58linear algebra
1:00this all-encompassing course explores
1:03the linear algebra in an interactive and
1:06machine learning Focus manner welcome to
1:08the linear algebra for machine learning
1:10course you will acquire the critical
1:13principles needed to build optimize and
1:16analyze sophisticated machine learning
1:18models from designing customer
1:20algorithms to enhancing curent
1:23Technologies this course provides the
1:25mathematical foundations with vital
1:28interest for the those for pioneering
1:31advancements in machine learning for
1:33those dedicated to mastering the
1:35mathematical aspect and the technical
1:37details behind machine learning our
1:40extensive 26 plus hour course of
1:43fundamentals of machine learning within
1:45the mathematics boot camp as well as a
1:47separate course offers an in-depth
1:50exploration this extensive program
1:53includes certification and is tailored
1:55for individuals that are serious about
1:58advancing their career in the field of
2:00machine learning Andi engineering this
2:03crash course in mathematics will serve
2:06you as a great starting point by
2:08establishing a robust foundation in
2:11linear algebra you will be well prepared
2:13to excel as machine learning
2:15practitioner equipped with the
2:17mathematical knowledge that drives the
2:20Innovation and efficiency in this field
2:23so if you're ready I'm really excited
2:26and without further Ado let's get
2:28started welcome to the course on the
2:31fundamentals of linear algebra presented
2:34by Lun Tech Academy my name is D Vasan
2:38and today we are going to start with
2:39some basic concepts that are important
2:42for understanding linear algebra linear
2:45algebra is one of the most applicable
2:46areas of mathematics it is used by pure
2:49mathematicians that you will see in a
2:52universities doing research publishing
2:54research papers but also by the
2:56mathematically trained scientists of all
2:58disciplines this is really one of those
3:01areas in mathematics that you will see
3:03time and time again appearing in your
3:06professional life if you want to become
3:09a job ready data scientist or you want
3:11to do some handson machine learning deep
3:15learning and AI stuff but also linear
3:17algebra is used in cryptology it is used
3:20in cyber security and in many other
3:22areas of computer science and artificial
3:25intelligence so if you want to become
3:28this well-rounded professional you want
3:30to go beyond using libraries and you
3:33want to truly understand the uh
3:36mathematics and the technical side of
3:38these different machine learning
3:39algorithms from very basic ones like
3:41linear regression to most complex ones
3:44coming from Deep learning like
3:45architectures in neural network how the
3:47optimization algorithms work how the
3:50gradient descent works and all these
3:52other different methods and models then
3:54you are in the right place because you
3:56must know linear algebra such that you
3:59will understand these different concepts
4:01from very basic ones to most advanced
4:04ones in data science machine learning
4:07deep learning artificial intelligence
4:09data analytics but also in many other
4:13applied science
4:14disciplines so before starting this
4:17comprehensive course that will give you
4:19everything that you need to know about
4:21linear algebra first I'm going to tell
4:23you what we assume that you already know
4:26because linear algebra it comes from
4:28about third PA of Bachelors of different
4:31highly technical studies and here we are
4:35assuming that you already know certain
4:38Concepts so to ensure that this course
4:41stays really on the topic of linear
4:43algebra and that you understand all
4:46these Concepts really well for that we
4:49need to be able to know different topics
4:53so before we dive into this Concepts
4:55let's familiarize ourselves with the
4:57basic prerequisites and notation used
5:00throughout this course and you will
5:02really need to know this in order to
5:05understand these Concepts really well
5:06such that instead of memorizing you'll
5:09actually just hear me once or maybe
5:11twice and then every time you hear later
5:14on or you see it in the papers or in
5:16some algorithms you will recognize this
5:18is something that we already
5:20learned so some key prerequisites
5:23overview is here first of all to fully
5:26grasp the upcoming material you should
5:28be familiar with some basic concept like
5:30real numbers Vector spaces so you don't
5:33need to know this idea of vectors though
5:36you already most likely are familiar
5:39with this given that you know how to
5:41plot different lines you know the idea
5:44of exes and wise and how to plot these
5:47different graphs but here we are going
5:50to touch base on this every time when we
5:52come close to these Concepts I will
5:53refresh you your memory and we will go
5:56through this numbers the idea of norms
5:59and distance measures because when it
6:01comes to the vectors when it comes to
6:03the magnitude and all these different
6:05topics that we are going to discuss as
6:07part of linear algebra knowing the what
6:10Norm is and what is the definition of
6:13distance what is the length between two
6:16points when we plot it into
6:18two-dimensional space or
6:19three-dimensional space those are all
6:22very basic concept that usually use as
6:24part of a basic pre-algebra or just
6:27common algebra courses and lessons in
6:30order to truly understand what the new
6:33algebra is about to understand the
6:35direction of vectors the angle and then
6:38the dimensionality reduction how linear
6:41algebra is applied for instance in
6:42different algorithms in machine learning
6:44deep learning data science statistics
6:47you really need to understand this
6:48Cartesian coordinate system so this is
6:51not only important for linear algebra
6:54but I assume you already know it given
6:56that you have passed those other courses
6:58like calcul or usually they are covered
7:01as part of pre-algebra or algebra so the
7:04cartisian coordinate system I mean here
7:07understanding what is for instance the
7:09the common description of them for
7:11instance when you when we write like X
7:14and then y on the vertical axis and then
7:16we can we have here zero and then we can
7:20always plot this different plots you
7:22know we have a clear understanding what
7:24is this Y is equal to X line we
7:27understand how by knowing certain points
7:29we can plot different plots for instance
7:32that this is the Y is equal to X line
7:34that here it means that if we have here
7:37one then this is just one two this is
7:39two so we understand when we have the
7:42function of the line and we have a
7:44certain value at is our y coordinate or
7:47x coordinate then the corresponding
7:49coordinate can be found then you also
7:52need to know some basic things that I
7:55just didn't mention right now so for
7:57instance that the numbers here can be
7:59like 1 2 three up to Infinity so you
8:02understand this concepts of infinity and
8:04then here the same story then here we
8:07have minus one you know minus two uh and
8:11then this is then used later on and we
8:15will be pouch basing this one we will be
8:18describing our vectors and how we can
8:21visualize our vectors either two
8:23dimensional space like we have here
8:25because this is two dimensional so we
8:26have X and Y but we can also of course
8:29visualize it in three-dimensional
8:32Etc so this idea of basic coordinate
8:35system is really important usually
8:37covered as part of algebra if not
8:40pre-algebra then we have basic triog
8:43genetry which means that you need to
8:45have a clear understanding what sinus is
8:47what cosine is what tangent is and their
8:49reciprocals and here I mean that you
8:52know for instance what is cosine
8:54function what is s function you know
8:57that you have an understanding
9:00for instance that what is this line you
9:02know whether it's a sinus line or cosine
9:05line you have also an understanding what
9:08this Pi is one thing that I didn't
9:11mention but it it just goes around all
9:13these topics some basic things that you
9:16understand what is X what is y why we
9:19use them and this idea of
9:22variables and also you need to
9:25understand this idea of square or you
9:28know 90
9:30degree angle and then Pythagoras Theorem
9:33here we have the same so what is this
9:35relationship between different sides of
9:37a triangle that is a very unique
9:40triangle and that has one of the angles
9:43as 90° and this idea of you know the
9:48sides how this relates to the sinus
9:50cosinus tangent cotangent and also how
9:53the Pythagorean Pythagorean theorem
9:55applies when we have triangular but it
9:59is is no longer with angle that is 90°
10:02what is the sum of all the angles of
10:05triangle so those are basic stuff that
10:07are com commonly covered as part of
10:10trigonometric lessons or part of General
10:15geometry then another prerequisite is
10:18this understanding of identities and
10:21equations in triog genometric lessons
10:24something part of which I already
10:25covered and this is goes around of basic
10:29having a basic un uh understanding of
10:31algebra and geometry those are super
10:33important to understand more Advanced
10:35Techniques from linear algebra then we
10:38have finally this idea of orthogonality
10:40perpendicularity in vectors so this also
10:43comes from geometry and from a
10:46trigonometric lessons so you understand
10:49that if we have for instance the two
10:51lines that don't have any intersections
10:53then we are talking about two orthogonal
10:56lines and otherwise for instance if we
10:58have and the two lines like this then we
11:01are talking about perpendicular vectors
11:05when you have two lines that are
11:06actually parallel so they don't have any
11:09intersection and you won't find any
11:11point that is common for the two so when
11:14it comes to this R so as part of real
11:17numbers and Vector spaces R represents
11:20the set of all real numbers so you can
11:22be dealing with for instance an integers
11:25like 1 2 three this can Al this will
11:27also cover all the negative numbers like
11:29-1 - 2 - 3 but also the floting numbers
11:34like 1. 223 and all the other numbers
11:39that you can think of those are the set
11:41of all real
11:43numbers so this is in one dimensional
11:45space right so you can see that I'm
11:47writing just one number you know two
11:50three and other numeric numbers then we
11:53have the idea of R2 R3 up to RN where
11:57now all these numbers they represent
11:59represent in this case the N it
12:01represents the N dimensional aidian
12:04space so when it comes to this idea of n
12:08dimensional numbers so for instance
12:12R2 here we just mean 2D plane so I'm
12:16pretty sure you are familiar with this
12:17idea of for instance
12:19xais and Y AIS here we are dealing with
12:23two dimensional plane so for every point
12:27that we can find here we can describe
12:29them by assigning them a value X so
12:32coordinate X and a coordinate y that's
12:36exactly what we mean by saying that the
12:39number can be represented in a 2d plane
12:42so here we are dealing with this two
12:44dimensional space this is our two
12:47dimensional Elan space and every number
12:50in
12:51here that is part of this R2 can be
12:55pictured here can be represented in this
12:57visualization so for instance if I have
13:00this number and let's assume that the
13:02value on the x-axis is two and we can
13:07see here that the corresponding Y is
13:08zero I can describe this number which I
13:11will call a I can describe this by
13:13writing down first the x coordinate
13:16which is two and then the y-coordinate
13:18which is zero so I'm then saying that a
13:21which is a point with x coordinate 2 and
13:24y coordinate Z it is part of my R2
13:29and it's part of my two dimensional
13:31alian space when it comes to R3 similar
13:35thing we can do with that only in that
13:36case we need not just x axis and y AIS
13:40but we need to add our third
13:42dimension so here for instance when it
13:45comes to the r
13:48Tre then we need to do y AIS we need to
13:53have xaxis but also we need to have some
13:57Z axis so
13:59such that every time every point in the
14:03space we can then describe by x y and
14:08Zed
14:09coordinates so if we write it in terms
14:12of the vector something that we will see
14:14very soon as part of our first unit of
14:17this course we will then need to
14:19represent every number in this
14:21three-dimensional Alan Space by writing
14:23down first the x coordinate let's say
14:25one and then y coordinate let's say
14:28another one and then Z coordinate which
14:30is one or even better even easier let's
14:33use 0 0 0 which means that we are
14:36dealing with this initial number which
14:38is the center of this three-dimensional
14:40Alan space when it comes to the N
14:44dimensional or the higher dimensional
14:46spaces it's much harder to visualize
14:49therefore usually when it comes to
14:52visualizations we do usually we usually
14:54only visualize the onedimensional two
14:57dimensional and thre dimensional spaces
14:59above then it just no longer does make
15:01sense to visualize it but we definitely
15:04deal with them and they are part of
15:07Applied linear
15:10algebra so understanding this spaces is
15:13very important for analyzing vectors for
15:16their interactions and this holds not
15:19just for this two-dimensional and
15:20three-dimensional but really for
15:23multidimensional
15:27spaces let's now quickly Define this
15:29idea of Norm so the norm of a vector
15:32denoted by this V which you can see kind
15:35of like similar to the absolute value
15:38from
15:39pre-algebra you can see here that we
15:41have this double straight lines like
15:44from absolute value then we have the
15:47name of the vector or the variable name
15:49that we are assigning to our vector and
15:52then you might notice here on the top of
15:54this this Arrow this basically says that
15:58we are deing not with just a variable
16:01but really we are dealing with a vector
16:04this is really important because you can
16:05see that there makes a huge difference
16:08if we have for instance just V or V1 I
16:11have to say or just V those are really
16:14important and things that you need to
16:16keep in mind when it comes to Leading
16:18your algebra and trying to differentiate
16:20vectors from a
16:22point you will notice that when it comes
16:24to Norm we can represented it either by
16:29this notation or this usually it's a
16:32common notation in machine learning or
16:34in data science with this two bars and
16:38when we do this we automatically also
16:41know that we are dealing with aladine
16:43distance we call it also L2 norm and
16:47this is something very common and
16:49usually used as part of
16:52retrogression which is an application of
16:55linear algebra and it's used in
16:57regularization so we are regularizing
17:00our machine learning algorithms so when
17:02you get into machine learning you will
17:04see time and time again this notation so
17:07next time when you see this then you
17:08know automatically that you are dealing
17:10with L2 norm and L2 Norm which is also
17:13used a lot in machine learning it is
17:16referring to the usage of L2 Norm to uh
17:20in the retrogression and retrogression
17:23or L2 regularization is a very popular
17:27regularization techniques as part of
17:29machine learning so right now even you
17:32can see this intersection or linear
17:34algebra or this idea of norms in machine
17:39learning all right so now let's see why
17:42we call it actually L2 Norm or often
17:46referred as Eline distance so Eline
17:50distance you can see here which is also
17:52the in this case this V which describes
17:56the norm of the vector v is equ Al to
17:59square roof and then we have all these
18:01coordinates assuming that the vector
18:03comes from an N dimensional space so you
18:05can see here the RN the V Vector the
18:09idian distance or the norm of this
18:12vector v is equal to square roof and
18:14then V1 2 plus vs2 S Plus and all this
18:18in between numbers plus VN squ so here
18:22basically it means take square root of
18:25V1 2 V2 S Plus plus V3 2 blah blah blah
18:31plus VN s so basically take all the
18:36units that form this vector and then so
18:39are on this vector and use them Square
18:43them and then add them and then take the
18:45square root of that that's the distance
18:48or I have to say the norm of this Vector
18:52so why this is important this idea of
18:54norms and equity in distance beside of
18:56being used in machine learning and why
18:58is it used
18:59so Norms they provide a way to measure
19:01the size or the length of a vector in
19:03Vector spaces which means that when we
19:06want to measure a distance a similarity
19:10relationship between for instance
19:12vectors then it becomes much easier to
19:15use this idea an Elan distance is not
19:18only used in regularization techniques
19:21like L2 regularization or retrogression
19:24but it's also used in other machine
19:27learning or deep learning Al items as a
19:29way to measure the distance or the
19:31relationship or the similarity between
19:34two different entities those can be
19:36variables those can be two people that
19:38we want to compare in our algorithm or
19:41two entities um for instance the Norms
19:45or the Al and distance they are also
19:47used as part of K me algorithm something
19:50that you might have heard and if you
19:51follow later on the machine learning and
19:53the clustering section of machine
19:55learning you will see that Alan distance
19:57is used as part of C's algorithm that
20:00aims to Cluster observations into
20:02different groups so this also yet
20:04another highly applicable uh topic that
20:07you must know in order to understand
20:09different linear algebra top topics but
20:12also machine learning topics let's now
20:14talk about simple topic that we must
20:16know about and refresh our memory very
20:18quickly before moving forward to our
20:21next topic that is a prerequisite for
20:23this course so the cartisian coordinate
20:26system is just a fancy word of
20:28describing this idea of X and Y or XY Z
20:32when we just want to visualize them and
20:35showcase this numbers related to the
20:38space so we just learned the and I just
20:42quickly was talking about this idea of X
20:44and and Y and how we can visualize that
20:47in plain so the cial coordinate system
20:49is a framework for specifying points in
20:52a plane or a space using ordered list of
20:55numbers so we know for instance when we
20:58plot this then here we need to put X and
21:02Y in our two dimensional space R2 and we
21:06know that here in the middle we have
21:08zero and here we have 1 2 three four and
21:12the same here one two and then three
21:15four which means that everyone that is
21:18in the industry whether it's in
21:20mathematics in physics in data science
21:22or ml or AI we all universally agree on
21:27this system we know this is this ordered
21:30list of numbers and we know that if we
21:32have for instance a point here then for
21:35this point we know that the xaxis and Y
21:38AIS is definitely positive even if we
21:40know don't know the corresponding
21:42numbers and then once we have more
21:44General lines here so not General but
21:47specific lines then we even know the
21:50exact coordinates and values here and we
21:52definitely know that this number should
21:54be so the x coordinate should be between
21:56two and three so first we have the two
21:59and then tree and not the other way
22:00around so this ordered nature helps us
22:03to understand how we can put all these
22:06different numbers and organize them in
22:08our two dimensional space and we also
22:10know the corresponding y so we know that
22:13for instance our Y is not minus three
22:16because it's lying in here in this part
22:19of our coordinate system and not
22:20somewhere here where the y axis are
22:24negative and why do we know that because
22:27it's an ordered list of numbers that we
22:29can visualize in this 2D plane and here
22:33you also need to keep in mind and we
22:35need to remind ourselves about this idea
22:38of these four different parts that we
22:40got so we have our here the first part
22:42the second part the third part and then
22:45the four you know part of our coordinate
22:48system and here we we are dealing with a
22:51two dimensional plane but if we were to
22:54deal with the three-dimensional plane we
22:57no longer have just x-axis and y axis
23:00where X AIS were on the horizontal and
23:03y- axis on the vertical but we have our
23:05third line which is the Z so we have now
23:10three different dimensions so X Y and Z
23:15and we are basically extending our
23:17two-dimensional plane to
23:19three-dimensional so this system is
23:21fundamental for visualizing and working
23:24with vectors geometrically so then we
23:26can just use this two dimension
23:29uh plane in order to visualize this
23:32Vector for instance knowing what are all
23:34these points that appear on this Vector
23:37what is its direction where is it headed
23:40you know what is the beginning and then
23:43we can also find out all the so the
23:46relationship of these vectors with all
23:48the other vectors for instance if we
23:49have an other Vector here then we can
23:52use the coordinates of them and
23:54information about vectors to understand
23:56that we are dealing with two parallel
23:58vectors that don't have anything in
24:00common so no intersection points where
24:02to say if we have another Vector like
24:04this and we know that here we are
24:06dealing with perpendicular you know
24:09orthogonal
24:11vectors so this is why those this
24:13coordinates Cartesian coordinate system
24:15is important and it's not just important
24:17for linear algebra but just in general
24:20for mathematics and for data science and
24:22for AI and you will see this coordinate
24:25system time and time again in different
24:26visualizations even when you want to
24:28visualize the mean of your data or you
24:31want to visualize the probability
24:33distribution function describing your
24:35population from statistics or from data
24:38science you want to visualize for
24:41instance how your optimization is
24:43working or you want to visualize how
24:45your model is performing in terms of its
24:48evaluation Matrix for all these cases
24:51and for any visualizations this idea of
24:54the Cartesian coordinate system is going
24:56to become very handy let's now talk
24:58about this idea of angles and the idea
25:00of circles radian the pi as well as this
25:05degree sign so this comes usually from
25:09geometry or
25:10tonometry and this is very important
25:13when it comes to the vectors because
25:14when we have two different
25:17vectors then we want to understand their
25:20relationship do they form this less than
25:2390° or so are we dealing with sharp
25:26corner sharp angle or with we are
25:28dealing with 90° angle so we are dealing
25:31with this type of vectors where we have
25:34you know 90° or we are dealing with um
25:39this type of vectors when the angle is
25:43180° which is by the way uh something
25:47that we are referring as
25:49Pi and here is one thing that is
25:52important here is that it's not just Pi
25:54but it's Pi
25:56radians why because in mathematics we
25:59also have this idea of Pi which is
26:01usually a number that is 3.14 so we
26:04should not confuse this Pi with pi
26:06radians so the relationship between the
26:08two is something that we have also seen
26:10as part of our pre-algebra and algebra
26:12courses so if it's something that you
26:14want to just refresh your memory on this
26:16will be super helpful to check our very
26:19initial course on um all these Basics so
26:22pre-algebra so this number comes from
26:24per algebra and then this idea of P
26:26radians and just in general all this
26:28information about what is 180° what is
26:31this angle what is 360° and all the
26:34information that comes from triogen
26:36metry and geometry can be found in our
26:39corresponding course so the next topic
26:41is the unit circle unit circle is highly
26:45related to this idea of radians degrees
26:47cosine sign but also understanding the
26:50Cartesian coordinate system will help
26:52you to understand the unit circle so
26:54this also comes from theog gometry and
26:57geometry and it's basically a fancy way
27:00of saying we have x-axis we have y AIS
27:03we have here zero so our common
27:06Cartesian coordinate system only we are
27:09trying to focus on this part of the
27:12system where we have here one we have
27:13here one so on the x-axis we have one
27:16and then here minus one here minus one
27:18for y AIS and here y the Y is equal to 1
27:22so we have here all these points and
27:24then we have the circle with the radius
27:26of one so here is this you know this is
27:29the radius and here we plot this circle
27:33and this will help us to understand this
27:36concepts of sinus cosinus you know the
27:38Theta is just variable that we use to
27:41describe the angle and for instance here
27:44we are dealing with
27:4545° this angle is
27:4890° this entire thing is
27:51360° and half of it so this part only is
27:56180° so those are all important part of
28:01understanding this idea of unit circle
28:03so you might have already guessed that
28:06unit circle refers to this idea that we
28:08have here one unit here one unit one
28:10unit one unit forming this entire circle
28:12so with the radius that is equal to
28:15one all right so this is something that
28:18is very easy and this comes from the
28:21geometry and pre and triog
28:23gometry uh you also need to understand
28:26this concept of the sinus and and
28:28cosinus and how sinus and cosinus are
28:30related to this what do we refer by the
28:34sinus and cosine you know what is this
28:37what are these points so 1 Z for
28:39instance we understand that here the x
28:42is equal to one and Y is equal to zero
28:45so here this point is simply 1 Z so this
28:50point and then we have 2 p radians so
28:54what is this idea of P so we know that a
28:57p Radian
28:58so P radians is simply the 180° which
29:03means that you also need to understand
29:05this concept of P2 which is simply the
29:0890° so you can see here one thing that I
29:11forgot to mention you need to understand
29:12this concept the relationship between
29:14the pi and so Pi radians and radians and
29:18this unit circle you need to know that
29:20here the pi ided two is simply this
29:24angle and then the entire Pi is this
29:27angle
29:28and then this entire thing the entire
29:31angle with
29:35360° is equal to 2 pi so 2 pi radians is
29:39simply this entire thing so those are
29:42very easy Concepts that come from
29:44geometry and trometry and if you want to
29:47refresh them then head towards those
29:49sores because this will help you to
29:52understand all this concept from scratch
29:55let's now continue our refreshment when
29:57it comes to so genometric identities and
30:00we just spoke about this unit circle we
30:02talked about the sinus cosinus it's
30:05really important to relate this back to
30:07bit more advanced topics coming from the
30:10same do domain and from the same area of
30:14mathematics and here we we need to know
30:17this concept before learning linear
30:19algebra few other things that um would
30:22be really great if you know but it's
30:24actually not a must to understand all
30:26these different topics it is the idea of
30:29Pythagorean identity so don't confuse
30:31this with Pythagoras Theorem this is the
30:34Pagan identity so this one that the
30:37square of the S of an angle plus the
30:39cosine squared is equal to one and all
30:42these different rules that go around the
30:44S and cosine and also the what is for
30:47instance the S 2 Theta which is equal to
30:502 s of theta and cosine of theta you
30:54know those are all different rules that
30:57would be handed to know and if you are
30:59so far I assume that you also know
31:02geometry and fundamentals to triogen
31:05ometry which means that you also know
31:07these trues but this might be just a
31:09great time to go ahead and quickly
31:11refresh your memory on these Concepts
31:13because those might become handy in your
31:17applied linear algebra and Applied
31:19Mathematics Journey but for now I would
31:21say this is not one of the most
31:24important things to know to learn this
31:26and to go through this course
31:28but just something to keep in mind so
31:31when it comes to the triog genometric
31:33equations uh this can become very handy
31:36later on when we want to prove something
31:38in linear algebra so to follow along
31:41it's actually a good idea to know for
31:43instance what is how you can solve this
31:45different equations and this will go
31:48back and refer to the unit circle that
31:50we just saw for instance if the sinus
31:53Theta is equal to 1 / 2 then you will
31:56need to quickly remember what is that
31:58angle for which the sinus is equal to 1
32:01/ 2 then you realize that is actually
32:05the angle where you take the p and
32:08remember that Pi is equal to
32:11180° and that is the one corresponding
32:14to and then Pi / to 6 is simply 180 / to
32:196 so this is basically the 30 degree so
32:23those are things that you can do when
32:25you know for instance all these
32:27different sinus and cosinus so you have
32:29memorized for these different angles so
32:32what is the sinus and cosinus for 30°
32:34for
32:3560° um let me actually remove this to
32:39make it easier so this type of problems
32:42is very easy to solve when we keep in
32:44mind and we memorize what are these
32:47different values for sinus and cosinus
32:49when it comes to different angles for
32:52instance for the angle equal to zero let
32:55me actually remove this and clean this
32:59part for better understanding so if we
33:01have for instance 0 degrees then we know
33:04that the sinus for this is zero and the
33:06cosine of this is one so we are
33:09basically dealing so if I plot a unit
33:14circle we are dealing with this number
33:17so remember that sinus and cosinus those
33:21refer to the Y and X on our unit circle
33:26so keep this one in mind
33:28so if the cosine Theta is then equal to
33:311 and the sinus so Y is equal to Z we
33:33are dealing automatically this number
33:35with this number and you can see that
33:37here the angle is also zero so here we
33:40are dealing with one and zero coordinate
33:44so this is
33:45our
33:48cosine of zero angle and this is then
33:52our sinus of zero angle so we
33:55automatically even from this graph can
33:57see very easy easily that the S of 0° is
34:00equal to 0 and the cosine is equal to 1
34:04all right so let's quickly also refresh
34:07our memory on few other degrees so for
34:11the
34:1230° which is simply the Pi / to 6 so
34:17this is
34:1930° then the sinus or the Y AIS is equal
34:23to 1 / 2 and the cosine or the X x value
34:28x coordinate is equal to square root of
34:303 2 so we are dealing with this this
34:33corner or angle so
34:3630° so even from here you can see that
34:39the coordinates make sense make sense
34:41then we have the pi for another famous
34:44value which is corresponding to the 45°
34:47it's simply this angle and for this
34:50angle the X AIS which is the coine so
34:54this number is equal to 1 / to 2
34:58and then for the sinus the so the y
35:01coordinate is equal to 1 / 2 square root
35:05of 2 as you might have guessed because
35:07in this number the x-axis and y axis is
35:09equal to is the same so you can see that
35:12this distance and this distance is the
35:14same because we are dealing with this
35:17type of figure so here we have 45° here
35:21we have 45° so this values are the same
35:24and this is something that you would
35:25know knowing the pag Ian Pagan
35:29theorem so then you can go ahead and
35:32refresh your memory for the
35:3460° so here I'm referring to the Pi / to
35:39three and then the 90° which is the very
35:43easy case this one obviously the x-axis
35:45is equal to zero so here you should have
35:47zero and the y axis is equal to one so
35:50here you should have one and so
35:53on all right so we went into quite
35:55detailed here but I think this is a very
35:57important topic knowing this idea of a
36:01trigonometric equations identities this
36:03idea of unit circle are super important
36:06because they are highly applicable to
36:09different fields in artificial
36:10intelligence data science machine
36:12learning and will definitely set you
36:15apart all right let's now talk about the
36:18law of signs and cosiness those are
36:20things that I won't be going on into too
36:23much details I just wanted to quickly
36:24showcase to you if you want to get the
36:27proof of those definitely check out our
36:29corresponding courses but for here I'm
36:32assuming that you already know so you
36:34know the law of signs which means that
36:36if you have this triangle you know you
36:39have this different sides so you have an
36:41angle a the corresponding side is a and
36:44then you have angle B corresponding side
36:46is B and then here C and the
36:48corresponding side is C then you know
36:50that a / to sinus of that angle is equal
36:53to B / to the sign of that angle and
36:56then is equal to C divided to the sign
36:58of that angle so basically take this
37:01value divide it to the sinus of this
37:04angle you know right in front of it is
37:08equal to taking this value and then
37:10dividing into the sinus of this angle so
37:13the proof of this low is outside so out
37:16of the scope of this course but knowing
37:18this will help you to understand
37:19different concepts and then the law of
37:22cosine is simply saying take the side of
37:26a Target angle so in our triangular we
37:30have here a we have here angle B and the
37:33C and if we go and look into this
37:36specific angle so angle C just randomly
37:39picking one of the three angles then the
37:42side right in front of that angle so the
37:45C c^ squ is equal to if we take this you
37:49know the other two sides forming that
37:51angle so A and B is equal to a s so this
37:55is just a constant a distance of this
37:58side a 2 + b 2 so this side squar minus
38:032 * a * B times the cosine of that angle
38:09this is what we are referring as the law
38:11of cosin quite easy we are not going to
38:13prove it again if you want to get the
38:15proofs make sure to check our other
38:18courses on the geometry and triog
38:20genetry we're almost done with the
38:22prerequisites just a quick refreshment
38:24we saw already the norm here is just a
38:27not EX exle what Norm is and on a
38:30specific two dimensional Vector when we
38:32have for instance that a vector is equal
38:35to three and four which means for the
38:37First Dimension let's say on xaxis we
38:39have three and then on Y axis is equal
38:41to four then the norm or the Alid
38:43distance so this is equal to we take the
38:47x value so three and then we Square it
38:50so V you can see here this is the case
38:53when n is equal to 2 this is simply
38:56equal to square Ro of v1^2 + v2^ 2 and
39:00as V1 is equal to 3 so this is our maybe
39:05I can make this just V1 and this is my
39:09V2 then the norm or the equan distance
39:12for this Vector so this thing is equal
39:15to V1 2 + v2^ 2 which is equal to 3^ 2 +
39:214 S and this value is square root of 25
39:25and it's equal to 5 so let's now see the
39:28difference between aladine distance and
39:30the norm so you could see here the norm
39:33here we have just one vector like here
39:38and this Norm it has just two
39:41corresponding values into two
39:42dimensional space you see here we have
39:45just three and then four so this is V1
39:47and V2 when it comes to the Alan
39:50distance this is kind of the
39:51generalization of this idea of Norm so
39:54the aladine distance between two points
39:57a and B in RN so in the N dimensional
40:00space is the norm of the vector
40:03connecting a to B so we see that the
40:06norm and the elidan distance are highly
40:09related to each other only we are
40:12talking about the norm when it comes to
40:14one vector but when we have this Vector
40:18a and the vector
40:21B this is simply the Alan distance so
40:27for the Aline distance we know already
40:31this idea of distance how we can measure
40:32it and you can see that this comes very
40:35similar to what we see here notation and
40:38here we are saying well we have this
40:41vector and then it has the two
40:43coordinates in N is equal to 2 in two
40:45dimensional space when it comes to the
40:47Alan distance Alan distance helps you
40:50understand what is this distance between
40:53two points in an N dimensional space so
40:57the aladan distance between two points
40:59let's say A and B in N dimensional space
41:03is the norm of the vector connecting a
41:07to B so for instance if we have a point
41:09a and we have a point B we are
41:13connecting this and this is the vector
41:15connecting these two points then the
41:17aladan distance is simply the norm of
41:21this Vector so this is the aladan
41:24distance so we can see that nor
41:27and the distance they are highly related
41:30to each other in the Alan distance we
41:32are using this idea of norm and
41:34specifically the norm two as I mentioned
41:38before so here you can see that the
41:40definition of aladine distance so the
41:43distance between A and B the two point
41:45is equal to square root of A1 - B1 2 + a
41:50and then here we have basically A2 - b
41:532^ 2 and then plus A3 - B 3 squ those
41:58are things that we cover as part of this
42:00dot dot dot and then plus up to the last
42:02point when we have a n minus BN 2 so
42:05here what we mean basically is that if
42:08we
42:10have two points here is a and here's B
42:13and this s vector and we know all these
42:16different points so A1 B1 A2 B2 A3 B3
42:23blah blah blah and then here a n BN we
42:26know all these points lying here in this
42:29distance then we are taking them and
42:31using them to calculate the line
42:33distance so here for instance if we have
42:37point A and B so in this example let's
42:42do quick one specific example when we
42:45have a point a which has coordinates 1
42:47and two so this is basically A1 A2 and
42:51then point B with points in it like B1
42:55B2 you can notice that the da AB so the
42:58distance or the Eid in distance of these
43:00two points which is equal to the norm of
43:04this
43:07vector or here this is a and this is B
43:10and this is this Vector this is equal to
43:13Square < t of 4 - 1 so it takes the B1
43:17so this is B1 and this is
43:21A1 takes the square and then says plus
43:25B2 minus H ^ 2 takes the square root of
43:30that and says this equal to 5 now you
43:33might be wondering but hey why do we do
43:35then instead of 1 - B1 2 we do B1 - A1 2
43:39and the answer to this question lies in
43:42the uh properties that we learn as part
43:44of prealgebra because it doesn't matter
43:47when we take A1 - B1 squ or B1 - A1
43:51squared because this squared ensures
43:53that it doesn't matter which one we take
43:55first and subtract the other now the
43:57proof of that is outside of the scope of
43:59this of course is this is part of
44:01pre-algebra but I just wanted to put
44:03this out there to ensure that you are
44:06seeing what we are seeing here because
44:08here it says A1 minus B1 but in this
44:10example we are taking instead depth B1
44:13and we are subtracting A1 this is a
44:15common thing that we do in prealgebra
44:18and just in general in different cting
44:21distance or distance related cases so I
44:23just wanted to put this here to ensure
44:25that later on this is something that can
44:28be clear from the first view right and
44:30in here we will quickly refresh our
44:32memory on the Pythagorean theorem which
44:34basically says in the right angle
44:36triangle so if we have this type of
44:41triangle so here we have
44:4490° this is a right angle
44:47triangle the square of the length of the
44:51the side opposite to the right angle so
44:53this side this we over refer C and this
44:56as B uh and then a those two are not
44:59very important but this is commonly
45:01referred by C so the the side opposite
45:05to the right
45:06angle then we know that the square of
45:09the C so c^ s is equal to a 2 + b^2 this
45:13is super important theorem and a
45:16fundamental principle for defining the
45:18Norms the distances in equity and spaces
45:21in and in many other
45:24applications so the angles play Cru Ro
45:27in understanding the direction of the
45:28vectors and you know how they can be
45:30measured in degrees or in radians we saw
45:33also the pi radian this idea of you know
45:36that the P radian is equal to 180° those
45:39are all very important when it comes to
45:41linear algebra and just in general
45:43application of mathematics in machine
45:46learning in Ai and other applications
45:48the relationships between this angle
45:50measurements and the triog genometric
45:53functions is foundational in solving
45:56different problems
45:57that are about these vectors and their
46:01orientations for instance this angle of
46:03s cosine you know what is this idea of
46:06tangent they are very important just to
46:09give you an idea the um uh Tang tangent
46:14is specifically used as part of the
46:16activation functions we call it tank
46:18activation function and knowing this
46:20tank will help you to understand the
46:22activation functions that I use as part
46:24of deep learning which are more advanced
46:26machine learning type of models and they
46:30are fundamentals in all these different
46:32new and Cutting Edge techniques large
46:35like large large language models
46:37Transformers encoder and decoder based
46:39algorithms Etc they're also important in
46:42this idea of computing dot products so
46:45very important and must know when it
46:48comes to linear algebra so this is just
46:51an simple example when it comes to this
46:53right angle triangle and Pythagorean
46:56theorem and how is applied I will skip
46:58this for now it's also important to
47:00understand this idea of orthogonality so
47:03the two vectors let's say A and B they
47:06are orthogonal to each other if their
47:10dotproduct is zero so later on as part
47:13of the vectors when we will talk about
47:15dot product we will see what we mean
47:18when we say that the dot product is
47:20equal to zero and here you can even see
47:22that if the a norm if the a vector so
47:27you see here and B Vector if those
47:30vectors if we multiply them to each
47:32other their dot product is equal to zero
47:35it means they are orthogonal so this
47:38angle that they form is equal to 90°
47:41orthogonality implies that the vectors
47:43from the from a right angle with each
47:46other they are in you know we we are
47:48dealing with that in R2 in R Tre so they
47:51are super important when it comes also
47:54to visualizing them correctly
47:57this concept is visually represented all
48:00this you know Vector a and then Vector B
48:03and they are perpendicular in the 2D uh
48:06coordinate
48:08system all right so when it comes to the
48:11applications of orthogonality
48:12orthogonality plays a crucial role in
48:14various aspect of linear algebra it's
48:16fundamental in defining Vector spaces
48:19subspaces in solving a systems of linear
48:22equation later on when we pass the
48:24vector ideas and we go on to the
48:27matrices solving linear systems so
48:30equations with many unknowns and then we
48:32use this idea of reductions or gausian
48:35reductions we will see how this idea of
48:37orthogonality can be important and how
48:40also it relates back to the norm of two
48:43vectors so it's fundamental in defining
48:46all these different identities and
48:48solving system of linear equations and
48:50also orthogonal vectors are used in
48:52finding the shortest distance from a
48:55point to the plane um something that is
48:58important when it comes to the
49:00optimizations and here you can see an
49:03example the vector a which is equal to 2
49:05three and then Vector B which is equal
49:07to minus 3 and 2 you can see that when
49:10we multiply 2 by minus 3 so we obtain
49:13basically the dot product by the way
49:15this is something that we are going to
49:16cover also as part of this course but
49:18for now you can see that if we take this
49:21number we multiply with this so 2 * - 3
49:25we take this number multiply with this
49:27so we take three and multiply with two
49:30you can see that this equal to minus 6
49:32this is equal to 6 so - 6 + 6 is equal
49:35to zero so you can see that the
49:38dotproduct of these two vectors is
49:41simply equal to zero and this is what we
49:43are referring as orthogonality this
49:45means that these two vectors form a
49:48right angle where we see here this angle
49:51is equal to 90° why this prerequisites
49:55matter and why I meant those
49:57understanding this concept is very
49:59crucial they underpin this geometric
50:01interpretation of linear algebra they
50:03will help you to better understand these
50:05Concepts and not just to memorize them
50:07but really understand and later on when
50:10you go into your machine learning and AI
50:12journey and in your data science Journey
50:14seeing these Concepts will help you to
50:18better understand those different alori
50:20this optimization techniques what we
50:22mean when we say we want our
50:24optimization algorithm to move towards
50:27local minimum Global minimum but this
50:29idea of movement this idea of vectors
50:32later on will you will also understand
50:34this different concepts in deep learning
50:37how these models work how the neural
50:38networks work and those are essential
50:42Concepts that you need for solving
50:44different systems of linear equation a
50:46core part of this course they also help
50:50you in visualizing vectors spaces which
50:53are critical to understand this concept
50:55of linear algebra the applications of
50:57linear algebra when it comes to the real
51:00world applications so those are things
51:03that you can definitely must by
51:06following some of our other courses but
51:08for this course I assume that you are
51:10already familiar with this Concepts
51:13right so now we are ready to actually
51:15begin and with this prerequisites in
51:17mind you are prepared to start your
51:20linear arbra Journey we are going to
51:23learn everything in the most efficient
51:25way in such a way that you will learn
51:27the theory you are going to see many
51:29examples we are going to learn
51:31everything in detail but at the same
51:33time you're going to learn the must know
51:35Concepts and I'm not going to overwhelm
51:37you with this most difficult concept
51:39that you will not be seeing in your
51:41career I'm going to give you this bare
51:44minimum when it comes to really knowing
51:47and the must know for linear algebra
51:49such that you will be ready to apply
51:52linear algebra in your professional
51:54Journey whether you want to get into
51:56machine learning deep learning
51:58artificial intelligence data science
52:00knowing these different concepts in
52:02linear algebra you will be a pro in your
52:05field going to give you everything that
52:07you need the theory examples
52:10implementations everything in detail but
52:13at the same time you will be doing that
52:15in the most efficient and time-saving
52:18way so without further Ado let's get
52:23started let's Now quickly Define this
52:25idea of norm so the normal of a vector
52:27denoted by this uh uh V which you can
52:31see kind of like similar to the absolute
52:34value from
52:35pre-algebra you can see here that we
52:37have this double straight lines like
52:41from absolute value then we have the
52:43name of the vector or the variable name
52:46that we are assigning to our vector and
52:48then you might notice here on the top of
52:51this this Arrow this basically says that
52:55we are dealing not with just a variable
52:57but really we are dealing with a vector
53:00this is really important because you can
53:02see that there makes a huge difference
53:05if we have for instance just V or V1 I
53:08have to say or just V those are really
53:11important and things that you need to
53:13keep in mind when it comes to linear
53:15algebra and trying to differentiate
53:17vectors from a
53:18point you will notice that when it comes
53:21to Norm we can uh represented it either
53:25by this not ation or this usually it's a
53:28common um notation uh in machine
53:31learning or in data science um with this
53:34uh two bars and um when we do this we
53:39automatically also know L2 norm and this
53:42is something very common and uh usually
53:45used as part of um
53:47retrogression which is an application of
53:51um linear algebra uh and it's used in uh
53:54regularization so we are regularizing
53:57our machine learning algorithms so when
53:59you get into machine learning you will
54:00see time and time again this um notation
54:03so uh next time when you see this then
54:05you know automatically that you are
54:07dealing with L2 norm and L2 Norm which
54:10is also used a lot in machine learning
54:13it is referring to the usage of L2 Norm
54:16to uh in the uh regression and
54:19regression or L2 regularization is a
54:23very popular regularization techniques
54:25as part of machine learning so right now
54:28even you can see this uh intersection or
54:31linear algebra or um this uh idea of
54:34norms in machine learning the norm of
54:37this vector v is equal to square roof
54:40and then V1 S Plus V2 squ plus and all
54:43this in between numbers plus VN squ so
54:47here basically it means take square root
54:50of V1 squ then V2 squar plus V3 squ blah
54:56blah BL plus VN 2 so basically take all
55:01the units that form this vector and then
55:04so are on this vector and use them
55:08Square them and then add them and then
55:11take the square root of that that's the
55:13distance or I have to say the norm of
55:16this Vector we saw already the norm here
55:19is just a not example what Norm is in um
55:22on a specific two dimensional Vector
55:25when we have for instance that the
55:27vector is equal to three and four which
55:29means for the First Dimension let's say
55:31on xaxis we have three and then on Y
55:34axis is equal to four then the norm or
55:36the AL in distance so this is equal to
55:39we take the x value so three and then we
55:42Square it so V you can see here this is
55:46the case when n is equal to 2 this is
55:49simply equal to square root of V1 2 +
55:52v2^ 2 and as V1 is equal to 3
55:56so this is our maybe I can make this
55:59just V1 and this is my V2 then the norm
56:04or the in distance for this Vector so
56:07this thing is equal
56:09to V1 2 + V2 s which is equal to 3^ 2 +
56:154^ 2 and this value is square root of 25
56:19and it's equal to five so let's now see
56:21the difference between aladine distance
56:24and the norm so you you could see here
56:26the norm here we have just one vector
56:30like here and this Norm it has just two
56:35corresponding values into two
56:36dimensional space you see here we have
56:38just three and then four so this is V1
56:41and V2 when it comes to the Alan
56:43distance this is kind of the
56:45generalization of this idea of Norm so
56:48the Alan distance between two points A
56:51and B in RN so in the N dimensional
56:54space is the norm of the the vector
56:57connecting a to B so we see that the
57:00norm and the elidan distance are highly
57:03related to each other only we are
57:06talking about the norm when it comes to
57:08one vector but when we have this Vector
57:12a and the vector
57:15B this is simply the Alan distance so
57:21for the Aline distance we know already
57:24this idea of distance how we can measure
57:26it and you can see that this comes very
57:29similar to what we see here notation and
57:32here we are saying well we have this
57:35vector and then it has this two
57:37coordinates in N is equal to two in two
57:39dimensional space when it comes to the
57:41AQ in distance Eline distance helps you
57:44understand what is this distance between
57:47two points in an N dimensional space so
57:51the aladan distance between two points
57:53let's say A and B in n dimens space is
57:57the norm of the vector connecting a to B
58:01so for instance if we have a point a and
58:04we have a point B we are connecting this
58:07and this is the vector connecting these
58:09two points then the aladan distance is
58:12simply the norm of this Vector so this
58:17is the aladine distance so we can see
58:20that the norm and the distance they are
58:23highly related to each other in the Alan
58:25distance where using this idea of norm
58:28and specifically the norm two as I
58:31mentioned before so here you can see
58:34that the definition of alodine distance
58:36so the distance between A and B the two
58:39point is equal to square root of A1
58:41minus B1 2qu plus a and then here we
58:44have basically A2 minus b 2 2ar and then
58:48plus A3 minus B3 2 those are things that
58:52we cover as part of this dot dot dot and
58:55then plus up to the last point when we
58:56have a n minus bn^ 2 so here what we
59:00mean basically is that if we
59:03have two points here is a and here is B
59:07and this s vector and we know all these
59:10different points so A1 B1 A2 B2 A3 B3
59:17blah blah blah and then here a n BN we
59:20know all these points lie in here in
59:22this distance then we are taking them
59:25and using them to calculate the Lan
59:27distance so here for instance if we have
59:31um point A and B so in this example
59:36let's do a quick one specific example
59:38when we have a point a which has
59:40coordinates 1 and two so this is
59:42basically A1 A2 and then point B with u
59:47points in it like B1 B2 you can notice
59:51that the da AB so the distance or the
59:53equid distance of these two points which
59:56which is equal to the norm of this um
1:00:02vector or here this is a and this is B
1:00:05and this is this Vector this is equal to
1:00:08square root of 4 - 1 so it takes the B1
1:00:12so this is B1 and this is
1:00:16A1 takes the square and then says plus
1:00:21B2 - A2 2 takes the square root of that
1:00:25and says this equal to 5 now you might
1:00:28be wondering but hey why do we do then
1:00:31instead of 1 - B1 2 we do B1 - A1 2 and
1:00:35the answer to this question lies in the
1:00:37um uh properties that we learn as part
1:00:40of pre-algebra because it doesn't matter
1:00:43when we take uh A1 - B1 squ or B1 - A1
1:00:47squ because this squared ensures that it
1:00:49doesn't matter which one we take first
1:00:51and subtract the other now the proof of
1:00:54that is outside of the scope of this um
1:00:57course is this is part of pre-algebra
1:00:59but I just wanted to put this out there
1:01:00to ensure that uh you are uh seeing what
1:01:04we are seeing here because here it says
1:01:06A1 minus B1 but in this example we are
1:01:08taking instead depth uh B1 and we are
1:01:11subtracting A1 this is a common thing
1:01:14that we do in um pre-algebra and just in
1:01:17general uh in different um eling
1:01:20distance or distance related cases so I
1:01:22just wanted to put this here to ensure
1:01:24that uh later on this is something that
1:01:27can be clear um from the first view why
1:01:29this is important this idea of norms and
1:01:32Al IND distance beside of being used in
1:01:34machine learning and why is it used so
1:01:36Norms they provide a way to measure the
1:01:38size or the length of a vector in Vector
1:01:41spaces which means that when we want to
1:01:44measure a distance a similarity a
1:01:47relationship between for instance
1:01:49vectors then it becomes much easier to
1:01:52use this idea an Alan distance is not
1:01:55only used in regularization techniques
1:01:58like L2 regularization or retrogression
1:02:02but it's also used in other machine
1:02:04learning or deep learning algorithms as
1:02:06a way to measure the distance or the
1:02:09relationship or the similarity between
1:02:12two different entities those can be
1:02:14variables those can be two people that
1:02:16we want to compare in our algorithm or
1:02:18two entities um for instance the um
1:02:23Norms or the Al and distance they are
1:02:25also used as part of K algorithm
1:02:28something that you might have heard and
1:02:29if you follow later on the machine
1:02:31learning and the clustering section of
1:02:33machine learning you will see that Elin
1:02:35distance is used as part of C's
1:02:37algorithm that aims to Cluster
1:02:39observations into different groups so
1:02:42this is also yet another highly
1:02:44applicable uh topic that you must know
1:02:46in order to understand different linear
1:02:48algebra top topics but also machine
1:02:51learning topics welcome to the course on
1:02:53the fundamentals of linear arbra
1:02:56my name is D Vasan and today we are
1:02:58going to start with some basic concepts
1:03:01that are important for understanding
1:03:02linear algebra linear algebra is one of
1:03:05the most applicable areas of mathematics
1:03:08it is used by pure mathematicians that
1:03:10you will see in universities doing
1:03:13research publishing research papers but
1:03:15also by the mathematically trained
1:03:17scientists of all disciplines this is
1:03:20really one of those areas in mathematics
1:03:22that you will see time and time again
1:03:24appearing in your professional life if
1:03:27you want to become a job ready uh data
1:03:31scientist or you want to do some handson
1:03:33machine learning deep learning and AI
1:03:36stuff but also linear algebra is used in
1:03:38cryptology it is used in cyber security
1:03:41and in many other areas of computer
1:03:43science and artificial
1:03:45intelligence so if you want to become
1:03:48this well-rounded professional you want
1:03:50to go beyond using libraries and you
1:03:53want to truly understand the uh
1:03:56mathematics and the technical side of
1:03:58this different machine learning
1:03:59algorithms from very basic was like
1:04:02linear regression to most complex ones
1:04:04coming from Deep learning like
1:04:06architectures in neural network how the
1:04:08optimization algorithms work how the
1:04:10gradient descent works and all these
1:04:12other uh different methods and models
1:04:15then you are in the right place because
1:04:17you must know linear algebra such that
1:04:19you will understand these different
1:04:21concepts from very basic ones to most
1:04:24advanced ones in the data science
1:04:26machine learning deep learning
1:04:28artificial intelligence data analytics
1:04:31but also in many other applied science
1:04:35disciplines so before starting this
1:04:38comprehensive course that will give you
1:04:40everything that you need to know about
1:04:41linear algebra first I'm going to tell
1:04:44you what we assume that you already know
1:04:47because linear algebra it comes from
1:04:49about third uh year of bachelors's um of
1:04:53different uh highly technical studies
1:04:56and um here um we are assuming that you
1:04:59already know certain Concepts so uh to
1:05:03ensure that this course Tes really on
1:05:05the topic of linear algebra and that you
1:05:08uh understand all these Concepts really
1:05:11well for that we need to uh be able to
1:05:14know different topics so before we dive
1:05:18into this Concepts uh let's familiarize
1:05:21ourselves with the basic prerequisites
1:05:23and notations used throughout this
1:05:25course and you will really need to know
1:05:27this in order to understand these
1:05:29Concepts really well such that instead
1:05:31of memorizing you will actually just
1:05:34hear me once or maybe twice and then
1:05:37every time you hear later on or you see
1:05:39it in the papers or in some algorithms
1:05:41you will recognize ah this is something
1:05:43that we already
1:05:45learned so uh some key prerequisites
1:05:48overview is here um first of all to
1:05:51fully grasp the upcoming material you
1:05:53should be familiar with some basic
1:05:55concept like real numbers Vector spaces
1:05:58so you don't need to know this idea of
1:06:00vectors though you uh already most
1:06:04likely are familiar with this given that
1:06:06you know how to plot different uh lines
1:06:09you know the idea of x's and y's and how
1:06:13to plot these different graphs but um
1:06:16here we are going to touch base on this
1:06:18every time when we come close to this
1:06:20Concepts I will refresh you uh your
1:06:22memory and we will go through this
1:06:24numbers the idea of norms and distance
1:06:27measures because when it comes to the
1:06:30vectors when it comes to the magnitude
1:06:32and all these different uh topics that
1:06:34we are going to discuss as part of
1:06:36linear algebra knowing the what Norm is
1:06:39and um what is the definition of
1:06:41distance what is the length between uh
1:06:45two points when we plot it in the
1:06:47two-dimensional space or
1:06:48three-dimensional space those are all
1:06:50very basic concept that usually you see
1:06:53as part of a basic pre-algebra or is uh
1:06:56common algebra ques and um lessons in
1:07:01order to truly understand what the your
1:07:03algebra is about to understand the
1:07:05direction of vectors the angle and then
1:07:09um the uh dimensionality reduction how
1:07:11linear algebra is applied for instance
1:07:13in different algorithms in machine
1:07:15learning deep learning data science
1:07:17statistics you really need to understand
1:07:19this Cartesian coordinate system so uh
1:07:22this is not only important for linear
1:07:25algebra but I assume you already know it
1:07:27given that you have passed those um uh
1:07:29other courses like calculus or usually
1:07:33they are covered as part of pre-algebra
1:07:34or algebra so the cartisian coordinate
1:07:38system I mean here understanding uh what
1:07:41is for instance the the common um
1:07:44description of them for instance when
1:07:45you when we write like X and then y on
1:07:48the vertical axis and then we can uh we
1:07:51have here zero and um then uh we can
1:07:54always PL this different plots you know
1:07:57we we have a clear understanding what is
1:07:59this um Y is equal to X line we
1:08:02understand how by knowing certain points
1:08:05we can plot different plots for instance
1:08:07that this is the Y is equal to X line
1:08:10that here it means that if we have here
1:08:12one then this is just one two this is
1:08:15two so we understand when we have the
1:08:17function of the line and we have a
1:08:19certain value where is our y coordinate
1:08:22or x coordinate then the corresponding
1:08:24uh coordinate can be found then um you
1:08:28also need to know um some basic things
1:08:31that I just didn't mention uh right now
1:08:34so for instance that the numbers here
1:08:36can be like 1 2 three up to Infinity so
1:08:39you understand this concepts of infinity
1:08:42and then here the same uh story then
1:08:45here we have minus one you know minus
1:08:48two uh and then this is then used later
1:08:52on and we will be uh touch basing this
1:08:55is when we will be describing our
1:08:57vectors and how uh we can visualize our
1:09:00vectors either two dimensional space
1:09:02like we have here because this is two
1:09:04dimensional so we have X and Y but we
1:09:06can also of course visualize it in
1:09:09three-dimensional
1:09:10Etc so this idea of basic coordinate
1:09:13system is really important um usually
1:09:16covered as part of algebra if not
1:09:19pre-algebra then we have basic triog
1:09:22genetry which means that you need to
1:09:24have a clear understanding what sinus is
1:09:26what cosine is what tangent is and their
1:09:29reciprocals and here I mean uh that you
1:09:31know for instance um what is cosine
1:09:34function what is s function um you know
1:09:38that you have an
1:09:40understanding for instance that um uh
1:09:43what is this line you know um whether
1:09:46it's a sinus line or cosine line you
1:09:49have also an understanding what this Pi
1:09:51is um one thing that I didn't mention
1:09:54but it it just goes um around all these
1:09:58topics some basic things that you
1:10:00understand what is X what is y why we uh
1:10:03use them and this idea of uh
1:10:07variables uh and also uh you need to
1:10:10understand this idea of a square uh or
1:10:14you know a
1:10:1690° uh angle and then uh Pythagoras
1:10:20Theorem here we have the same so what is
1:10:23this relationship between different
1:10:24sides of the triangle uh that is a very
1:10:27unique triangle and that has one of the
1:10:30uh angles as 90° um and uh this idea of
1:10:36um you know the sides how this relates
1:10:38to the sinus cosinus tangent cotangent
1:10:41um and also um how the Pythagorean um
1:10:45Pythagorean theorem applies when we have
1:10:48uh a triangular but it is no longer with
1:10:52a angle that is 90° what is the sum of
1:10:55all the angles of triangle so those are
1:10:58basic stuff that are com commonly
1:11:01covered as part of uh trigonometric uh
1:11:04lessons or part of General
1:11:07geometry then another prerequisite um is
1:11:11this uh understanding of uh identities
1:11:15and equations in triog genometric um
1:11:18lessons something part of which I
1:11:20already covered and this is goes around
1:11:23of basic having a basic
1:11:25understanding of algebra and geometry
1:11:27those are super important to understand
1:11:29more Advanced Techniques uh from linear
1:11:32algebra then we have finally this idea
1:11:34of orthogonality perpendicularity in
1:11:37vectors for instance if we have um the
1:11:41two lines like this then we are talking
1:11:43about uh perpendicular vectors when you
1:11:46have two lines that are actually
1:11:48parallel so they don't have any
1:11:50intersection and you won't find any
1:11:52point that is common for the tube hi
1:11:55there so let's get started with our
1:11:57first module which is foundations of
1:11:58vectors in this module we are going to
1:12:00talk about fundamentals of linear
1:12:02algebra vectors we are going to make a
1:12:04differentiation with between scalers and
1:12:07vectors we are going to Define them so
1:12:09first we will learn the theory then we
1:12:11will Implement them into practice by
1:12:13plotting them by looking into different
1:12:15examples then we will look into this
1:12:17representation of vectors by looking
1:12:20into the magnitude and the direction of
1:12:22it and the representation of them just
1:12:25in general we are going to plot them in
1:12:27our coordinate system then we are going
1:12:30to see the common notational vectors and
1:12:32indexing of them vectors are super
1:12:36important when it comes to linear
1:12:37algebra and application of it and uh
1:12:40they matter not only in mathematics but
1:12:43beyond so uh vectors help us in many
1:12:46ways from figuring out how objects move
1:12:49to solving math problems in science and
1:12:52just in general in technology including
1:12:54in data science
1:12:55machine learning artificial intelligence
1:12:57Etc they are super useful tool so uh
1:13:02let's start our journey with looking
1:13:04into scalers so scalers they are just
1:13:07plain numbers and by definition a scaler
1:13:11is a single numeric volume often
1:13:13representing magnitude or
1:13:15quantity for example uh scalers can be
1:13:20describing um the temperature outside
1:13:23for instance the temperature
1:13:25of um a
1:13:2722° uh can be represented by a scaler or
1:13:31a height of a person can be represented
1:13:33it's a scaler so let's assume we have a
1:13:37scaler that we will Define by a letter s
1:13:39it's just a variable this scaler is then
1:13:42equal to 22 for instance and we are
1:13:45measuring it in degrees so it means that
1:13:49uh if this s measures a room temperature
1:13:52then the scaler s which is equal to
1:13:5522° which represents the room
1:13:57temperature it can be for instance 18°
1:14:00or 9° if it's very called uh it just
1:14:04measures a single volume it represents
1:14:07just a single number or it can be for
1:14:10instance 17 100
1:14:132.22 so all these they are just scalers
1:14:17they represent a single numeric volume
1:14:21they often represent a magnitude or a
1:14:23quantity very we will see that scalers
1:14:26they are a value that represent the
1:14:28magnitude of a
1:14:30vector so uh now when we are clear on
1:14:33this very basic concept of scalers let's
1:14:37actually move to this idea of vectors so
1:14:40by definition a vector is an ordered
1:14:42array of numbers which can represent
1:14:45both magnitude and direction in space so
1:14:49uh vectors they are bit more they
1:14:51represent bit more than scalers there
1:14:53are numbers that also show show
1:14:55direction like a car spitting down the
1:14:58highway or a bow uh being
1:15:01thrown for instance uh when it comes to
1:15:04our previous example we were using this
1:15:06uh uh room temperature as a way to uh
1:15:10think about the scaler a scaler for
1:15:14instance scaler that we just saw was
1:15:17this room temperature room temperature
1:15:20which was
1:15:2222° when it comes to the vector
1:15:25Vector is different for Vector for
1:15:28instance we can have an example when a
1:15:31bird for instance bird it
1:15:35flies
1:15:37flies at 10
1:15:41kilomet per
1:15:44hour and I also add here another
1:15:47information which will make this as a
1:15:49vector which is that it flies
1:15:53South so here as you can see what I'm
1:15:56doing is that I'm not just oh let me
1:16:00actually remove this part to make it
1:16:02easier to
1:16:04understand okay so uh in this example
1:16:08let me write it down that the
1:16:12example
1:16:14bird
1:16:16FES
1:16:19s at 10 kilomet per hour so you can see
1:16:26that I'm not just adding the scaler
1:16:29which is in this case the
1:16:32magnitude we will see very soon the
1:16:34formal definition of it so I'm writing
1:16:37down the speed I'm defining the speed
1:16:40but also the direction so I'm
1:16:43saying I know that the bird is flying
1:16:46south that's the direction and I know
1:16:49also the speed of it which is the
1:16:51magnitude so 10 kilomet per hour so here
1:16:54in the vector I have much more
1:16:56information than in the scaler because
1:16:59in the scaler I just got temperature
1:17:01room temperature single volue but in
1:17:03case of a vector I not only have um
1:17:07magnitude or speed like 10 kilm per hour
1:17:10but I have extra information which is
1:17:12the direction of it for instance flying
1:17:14to the South so let's now look into some
1:17:17real examples and plotting them to make
1:17:20more sense out of this idea of vectors
1:17:23and what is this magnitude what is the
1:17:25direction so let's assume we have a 2d
1:17:29plane so we have xaxis we have y AIS
1:17:33here like usual we have our z0 Center
1:17:37and we want to plot a simple Vector so
1:17:42uh usually the way we represent Vector
1:17:44in tutorials or just writing down is by
1:17:47writing the name of the vector this can
1:17:49be just a a random name let's assume
1:17:52that it's a v letter v and then on the
1:17:56top we are always adding this Arrow so
1:17:59this Arrow it says and it tells the
1:18:02person who is reading that we are
1:18:04dealing with the vector arrow on the top
1:18:06is that reference so let's assume this
1:18:10uh vector v it starts from the center of
1:18:14our coordinate system and it goes to
1:18:17this point so let's say in here this is
1:18:21our vector v
1:18:25so let's assume that this point in here
1:18:28is equal to 4 which means that the x
1:18:32coordinate is four and the y-coordinate
1:18:33is zero as the um uh Arrow it just as
1:18:38the point in here it has a a y value of
1:18:42zero so you can see that it goes
1:18:43straight from zero to this one to this
1:18:47point okay so what tells this Vector uh
1:18:51to us is that we have a value that
1:18:54describes the length of the vector so it
1:18:57goes from 0 to 4 which means that the
1:19:00length is equal to unit four so it's
1:19:04equal to
1:19:05four um
1:19:08and we have just learned and we were
1:19:10just talking about that the magnitude is
1:19:14the length in this case so the length
1:19:16describes the magnitude in this
1:19:20case so this means that the magnit ude
1:19:25of this Vector is equal to 4 and then um
1:19:29what else we can see here we can see the
1:19:31direction of the vector which means that
1:19:33the direction is also something that we
1:19:36can see here this is the direction of
1:19:39the vector so this going straight from
1:19:42this point to this point in a horizontal
1:19:46way
1:19:48so independent whether I plot this
1:19:51Vector from 0 to 4 in here or in here
1:19:54here or in here or in here or in here in
1:19:59all cases as long as the length is this
1:20:02I'm dealing with the same Vector because
1:20:04I am basically in this entire
1:20:09R2 space I have exactly the same Vector
1:20:12all I care is about the magnitude and
1:20:15the Direction Where will this Vector
1:20:19start and where will it
1:20:20end I am not interested I'm interested
1:20:23that the uh that the magnitude in this
1:20:26case the length is equal to the
1:20:28direction of the vector so let's now
1:20:30look into another example where we go a
1:20:33bit more difficult on our coordinates
1:20:35and on our Vector we already saw that we
1:20:38had this Vector where we went let me
1:20:41change the color so this was our vector
1:20:45v and it went from zero till 4 so this
1:20:48point to be more specific is so this
1:20:51Vector it goes the vector B it goes from
1:20:550 0
1:20:58to 40 so the coordinate X was 4 and the
1:21:03Y was Zero now let's plot another one um
1:21:08where the direction is no longer
1:21:09horizontal for this Vector let's call it
1:21:12Vector
1:21:13W and for this Vector w we will again
1:21:16start with Z 0 so we will start again in
1:21:18here but this time we will go bit like
1:21:23this so let's say we go all the way to
1:21:26this
1:21:27point so this
1:21:30point has a value for an x- axis of
1:21:35three and for y axis it has a value of
1:21:38four which means it goes from this point
1:21:41to this point and this is the direction
1:21:45of our vector v so it goes to
1:21:4934 because this point is 3 0
1:21:55and this point is 04 so xaxis is 0o x
1:22:00coordinate and y coordinate is 4 so now
1:22:03you can see that the direction of this
1:22:07Vector is like
1:22:09this while the direction of the vector v
1:22:12was like
1:22:14this and like in case of vector v i
1:22:18again no longer care about where exactly
1:22:21my Vector W stars and ends but all I
1:22:24care is about its magnitude so the
1:22:27length and the direction so for
1:22:30instance I can have the same Vector in
1:22:33here the same Vector in
1:22:36here as long as the length the magnitude
1:22:40is the same and the direction I am
1:22:43dealing with the same Vector that's all
1:22:44I care so the magnitude and the
1:22:46direction is all that you care about all
1:22:50right so now about the length um that's
1:22:53uh something that you can see very
1:22:55easily from this specific example
1:22:57because by using the Pythagoras Theorem
1:23:00or P Pythagorean theorem we can see very
1:23:03quickly that as the length of this side
1:23:06of our uh right angle
1:23:1030° so right triangle we can see that
1:23:12this side is three this side is
1:23:15four which means that this side is 5
1:23:19because 4 2 + 3^ 2 then we take the
1:23:23square root of that square root of 25
1:23:25and it's equal to 5 so the length or the
1:23:29magnitude of this vector v is simply
1:23:31equal to
1:23:335 all right this was about this uh
1:23:36specific vectors let's now look into the
1:23:38uh common representation of the vectors
1:23:41so we always use the magnitude as well
1:23:43as the direction you know to represent
1:23:45the vectors and they commonly are
1:23:47represented by two different uh ways
1:23:50let's now look into the first way that
1:23:52the vectors can be represented and then
1:23:54we will move on to the next one so when
1:23:56it comes to the vector v so we saw that
1:23:59vector v was moving from
1:24:010 till uh to the point of 40 so we can
1:24:07represent the vector B by 4 and
1:24:11zero when it comes to the vector w we
1:24:15can represent that uh Vector so Vector w
1:24:19we can again do the parenthesis and we
1:24:21can say that it's equal to 3 four so by
1:24:25using the coordinates from the
1:24:26coordinate system we can then represent
1:24:30our uh vectors so this is just one way
1:24:33of representing a vector another way of
1:24:36representing these vectors is by using
1:24:38this Square braces given that we are in
1:24:41a two dimensional space first we will
1:24:43mention here the four then we will
1:24:45mention the zero in here twoo so we can
1:24:50say three and four this is yet another
1:24:53way of represented the vectors in a two
1:24:56dimensional
1:24:57space so if we were to have a
1:25:00threedimensional space so let me
1:25:02actually show it on a new page so if we
1:25:05were to um if we were um to have vectors
1:25:09in three dimensional space so we are
1:25:11dealing with
1:25:13R3 so we have points that can be
1:25:16described by X Y and Z
1:25:20so coordinate space like this so X and
1:25:24and the Y and then the Z then every
1:25:27point so let's say we have this Vector
1:25:30then we had to represent it by a value
1:25:33let's say x uh X1 y1 and Z1 or um better
1:25:39let me actually use a different
1:25:42letters a b and c and this would be my
1:25:47vector v and I could also represent this
1:25:50Vector
1:25:51B is the same so vector v can be
1:25:55represented as a b and c so one thing
1:26:00that you can notice is that unlike the
1:26:02R2 now I have three different entries
1:26:06what we are also referring as rows and
1:26:09we just got one column so um we can uh
1:26:13often represent and usually that's a
1:26:15common way of representing vectors by
1:26:18using this um columns so columns help us
1:26:22to represent our vectors and you can see
1:26:26very clearly then when it comes to the
1:26:28two dimensional space so when we have
1:26:32R2 so then our vectors have just two
1:26:37rows so three and four four zero like in
1:26:40here when it comes to three dimensional
1:26:42space we have three entries and so on so
1:26:45the same holds of course also for for
1:26:47instance R5 then for R5 um our vectors
1:26:51so coordinate space can be for instance
1:26:53x y Zed and then GMA and then let's say
1:26:57Delta and then the coordinates uh of a
1:27:01vector in that space can be V and then
1:27:04arrow is equals sh and then we would
1:27:06have uh let's say A B C D E you get the
1:27:11idea so depending on the space the
1:27:13coordinate space and the dimension of
1:27:15that space then the corresponding
1:27:17vectors can be represented accordingly
1:27:24so the vectors are quantities that have
1:27:27both magnitude and direction as we just
1:27:29so distinguishing them from scalers
1:27:32which only have magnitude so we saw that
1:27:34the scalers got only magnitude while in
1:27:37case of vectors we saw both for the
1:27:39vector v and for the vector w we didn't
1:27:42we didn't only have the magnitude so the
1:27:44length of the vector but also the
1:27:46corresponding
1:27:48Direction so uh when it comes to the um
1:27:52vectors so the this is exactly what we
1:27:55just saw in our example a vector in a
1:27:57two dimensional space so in
1:28:01R2 uh can be represented by using this
1:28:04Square braces and the corresponding
1:28:07entries for X and Y where X is basically
1:28:09the x coordinate in our coordinate
1:28:12system so in our X and Y system whenever
1:28:16you have this x and y
1:28:19coordinate then uh this x coordinate
1:28:22Will then describe your magnitude and
1:28:25the y coordinate Will then describe your
1:28:27second entry that you need to put when
1:28:29representing your
1:28:32vectors so here the X and Y indicate the
1:28:36movement in the horizontal and in the
1:28:38vertical Dimensions respectively so for
1:28:41X's it's always the x coordinate so how
1:28:44far you move towards the horizontal
1:28:46Direction in here in here or independent
1:28:49in here so always take the x coordinate
1:28:52that is the value that you need to put
1:28:54first and then the Y need to be put it
1:28:56in here so indexing in vectors when it
1:29:00comes to the um indexing the standard
1:29:02mathematical notation uh indices in the
1:29:05N vectors goes from I is equal to 1 to I
1:29:09is equal to n so the um notation here
1:29:13can be bit ambiguous so AI uh could mean
1:29:16the E element of AI uh the a vector or
1:29:19the each Vector in a collection so let's
1:29:22start with a simple one and then move
1:29:24move on to this next part so what this
1:29:27means and what this means we will look
1:29:29into now so uh usually uh when we have a
1:29:35um n dimensional space we are having
1:29:38hard time visualizing it therefore we
1:29:40use this two dimensional space or
1:29:42maximum three-dimensional space in order
1:29:44to get an understanding of what these
1:29:46vectors are so we just s examples of
1:29:49them uh when uh creating our vectors in
1:29:52um V and V uh and W in uh R2 and also in
1:29:58R3 but we can have similar vectors also
1:30:02in R4 in
1:30:04R5 or all the way down to RN where n can
1:30:09be 100 200 500 any number as large as
1:30:13you want the thing is is that
1:30:16visualizing R 4 R5 RN is very hard but
1:30:20we can still benefit from this great
1:30:22properties of the vectors metrices and
1:30:25in general linear algebra in order to
1:30:27describe different things that have more
1:30:29than three dimensions therefore we have
1:30:32this a bit more ambiguous notation where
1:30:36we use r n and this n can be any real
1:30:40number and it can be all the way to
1:30:42Infinity so very large
1:30:44number and uh let's say we have a vector
1:30:47in this RN then this Vector is usually
1:30:51described by using similar Square uh
1:30:55brackets like before only with uh more
1:30:58entries so like before we got just one
1:31:01column so that's something that we
1:31:03didn't uh change but here we have
1:31:06instead of just two entries or three
1:31:08entries like in the two dimensional or
1:31:09three dimensional spaces now we have A1
1:31:12A2 A3 all the way down to a n minus one
1:31:18and a n so we got in total n elements in
1:31:23our column and this describes our uh
1:31:26single Vector so this Vector in an N
1:31:29dimensional space this we can call also
1:31:33a so one thing that we just saw is that
1:31:36it was saying in our definition and
1:31:38notation that uh we might also be
1:31:40dealing with the E Vector in a
1:31:43collection which means that sometimes
1:31:46you will
1:31:47see this while here the A1 A2 they are
1:31:51vector themselves so here
1:31:54we saw that these are just entries so A1
1:31:57is a number A2 is a number A3 is a
1:31:59number a n is just a number but it's
1:32:02also possible uh when you have a much
1:32:05more difficult and complicated case that
1:32:08you got an
1:32:09A let's write it down with a capital
1:32:12letter A which is equal to
1:32:17A1 or let's actually remove
1:32:21this so we got
1:32:24let's say
1:32:26A1
1:32:28A2 A3 all the way down to a n minus one
1:32:34and a
1:32:37n where you can already see what is
1:32:39going on so instead of having just a
1:32:41number as an entries instead we have
1:32:44vectors in here so our first element is
1:32:48actually Vector our second element is
1:32:50actually Vector so A2 Arrow A3 Arrow all
1:32:52the way down to a n arrow so while here
1:32:57this can be for instance some numbers
1:32:59let's say one one one all the way down
1:33:01to one
1:33:02one here we have a vector vector another
1:33:07vector and all the way down here yet
1:33:09another Vector where for instance let me
1:33:13remove this
1:33:16part where for
1:33:19instance A1 arrow is actually equal to
1:33:25A1 1 A1 2 A1 3 all the way down to A1 n
1:33:34one thing that you will notice here is
1:33:36that unlike in
1:33:39here here I got double indices so I got
1:33:44here a11 and then A1 2 and then a13 all
1:33:49the way to A1 n so the first index it
1:33:53doesn't change as I have here a one so
1:33:57I'm writing down the index corresponding
1:33:59to this Vector but the second index it
1:34:04changes per entry indicating which
1:34:07element specifically in the vector I'm
1:34:09talking about so from the first index
1:34:12you can identify the vector that I'm
1:34:15referring to which is A1 and from the
1:34:17second index you can see the
1:34:20corresponding um entry or the volume you
1:34:24that that um element is positioned in
1:34:28this Vector so you can see that this
1:34:30values for instance in the um Vector one
1:34:34so A1 to be more specific but then it is
1:34:37in the first position this is in the
1:34:39second position in the third position
1:34:41all the way down to the end position so
1:34:45this is something that is really
1:34:47important to understand well because
1:34:49this notation is going to appear time
1:34:52and time again across various
1:34:54applications of matrices and vectors so
1:34:57is really important to understand well
1:34:59therefore I want to go one more time
1:35:02through this to make sure that we are
1:35:04clear on what this indexes represent so
1:35:07whenever we have an index uh an a vector
1:35:11that we want to uh represent and it's um
1:35:14it has just um it is just a vector which
1:35:17means that it's not a nested vector
1:35:20vector in a vector then um we can Define
1:35:24it by let's say a and then on top an
1:35:26array and it's equal to and here we can
1:35:29have A1 A2 all the way down to a n so
1:35:35you can see what we are also referring
1:35:37as dimension of this Vector is equal to
1:35:41n by 1 so I got n entries and just one
1:35:46column so n by
1:35:49one which means that this already gives
1:35:51me an indication that most likely this
1:35:54A1 is a number this A2 is a number this
1:35:57a and is a number so let's say this
1:35:58equal to 1 2 uh three blah blah blah and
1:36:03then here I have let's say
1:36:07100 but if I'm dealing with the nested
1:36:11Vector later we will see that this can
1:36:13be represented by a matrix then um I can
1:36:18also Define
1:36:20this by capital letter A
1:36:24which is a common way to refer to either
1:36:26matrices or nested vectors and then this
1:36:30is equal to A1 Arrow A2 Arrow A3 Arrow
1:36:37this already sends a message to the
1:36:39reader that we are dealing with no
1:36:41longer U constants within a vector but
1:36:46rather vectors in a vector and uh what
1:36:50can we see here is that the dimension of
1:36:53this nested vector or which we can also
1:36:56refer to as a matrix here the number of
1:37:00rows so the number of entries this
1:37:02elements we can see it's equal to n but
1:37:05then this time the number of values that
1:37:08form these vectors is no longer one
1:37:11because we are not dealing with just a
1:37:13constant this is not some constant but
1:37:16rather this is yet another Vector so
1:37:19let's assume this Vector has a length of
1:37:21M so let's say this has a length of M
1:37:26then the dimension of this Matrix a is
1:37:29equal to M so something that we will see
1:37:32also when talking about
1:37:34matrices so let me actually clarify this
1:37:37bit more for better understanding let's
1:37:40say we look into one of those um one uh
1:37:44one other example of an entry so let's
1:37:47say we look into this specific Vector
1:37:49which is in the uh the third uh vector
1:37:53within this Vector capital A so this
1:37:58A3
1:38:02Vector so one thing to see here already
1:38:06is that I assumed that these vectors
1:38:09they got M elements and keep in mind
1:38:12that all these vectors they should be of
1:38:13the same size so it means that I already
1:38:16know that this specific Vector A3 has M
1:38:21elements so m elements so I'm
1:38:25representing this uh A3 Vector from here
1:38:29I'm taking this out from this entire uh
1:38:32nested a vector and I just want to
1:38:34represent this and now unlike this
1:38:38elements that got an arrow on the top
1:38:41this time I will have uh constants
1:38:45forming the A3 Vector so I no longer
1:38:47have vectors but I have elements in it
1:38:51so in here I will have
1:38:54a a let me actually write down all the
1:38:57A's but to refer and to make sure that I
1:39:00recognize that I'm dealing with the
1:39:02third a vector so a Tre Arrow here I
1:39:06will put three Tre all the way here Tre
1:39:09so they all come from the same third A3
1:39:11Vector but then their positions is
1:39:14different because this is let's say uh
1:39:17one two and then all the way down
1:39:20to Ed position
1:39:24so this indices help us to keep track
1:39:28what are the um position that these
1:39:32values are taking part in the vector A3
1:39:37errow this might seem bit complicated at
1:39:39the moment but once we move on onto bit
1:39:42more complex material like uh matrices
1:39:45it will make much more sense this is bit
1:39:47of an extra I just wanted to Showcase
1:39:50this but this is what uh is at its core
1:39:53and what you need to uh understand at
1:39:55the moment to understand this concept of
1:39:57vectors so you need to know that vectors
1:40:00can be represented by this arrow on the
1:40:02top so let's say Vector a and it has
1:40:05let's say n elements then you can write
1:40:07the square brackets and then you will
1:40:09need to mention A1 A2 all the way to a n
1:40:12which means that you have n different
1:40:14entries describing your vector so you
1:40:17have A1 which is the first element in
1:40:19your vector A2 the second element all
1:40:21the way to a n which is the end element
1:40:23where here you can see for instance so
1:40:26if I had here A3 that uh A1 is simply
1:40:30equal to one A2 is equal to 2 A3 is
1:40:33equal to 3 all the way to a n is equal
1:40:36to 100 so this numbers I'm basically
1:40:40taking and I'm representing them I'm
1:40:43putting them in here within Square
1:40:44braces in order to get a representation
1:40:47of my Vector so my Vector a has all
1:40:50these different entries and different
1:40:52entries and it starts with one and it
1:40:54ends with 100 this is a vector and then
1:40:57when it comes to the vectors within
1:40:59vectors here we need to be a bit more
1:41:02careful CU here we not just have uh
1:41:05constant values forming a vector but we
1:41:08have vectors that form yet not vectors
1:41:11so our Vector a our nested Vector a
1:41:15which we uh later will refer as Matrix a
1:41:20has actually entries that also are
1:41:22vectors so we have a 1 Vector A2 Vector
1:41:25A3 Vector they are not just constants
1:41:27but only own they are vectors so here
1:41:30for instance we have defined also an
1:41:32example of it we have said let's look
1:41:34into this third specific Vector that is
1:41:37part of a which is A3 uh vector and uh
1:41:41that one has M different
1:41:45elements here we have then the index
1:41:47referring to the which Vector from the
1:41:50nested Vector a it is which is the third
1:41:52one because we have taken it from here
1:41:55but then on its own this Vector has
1:41:57different members and different members
1:41:59to be more specific therefore we have
1:42:01also an index to keep track of the
1:42:04position of this value one to up to M
1:42:07and this can be yet another uh this time
1:42:10it can contain some elements an example
1:42:12of which is for instance Z 1 2 all the
1:42:16way to let's say 500 and this can be
1:42:20different numbers it doesn't need to be
1:42:22ordered it doesn't need to have a
1:42:24specific pattern they can be just random
1:42:26numbers describing this A3 Vector so
1:42:30hopefully this makes sense if it doesn't
1:42:32don't worry because we are going to see
1:42:34this time and time again I just wanted
1:42:36to give you a brief of an intro such
1:42:39that you can uh remember this when we
1:42:42come uh back to bit more uh complex
1:42:46topics like uh indexing in matrices so
1:42:50now let's talk about special vectors and
1:42:52operation
1:42:53here we are going to talk about zero
1:42:55vectors unit vectors the concept of
1:42:57sparcity in vectors as well as vectors
1:43:00in higher Dimensions like we just saw
1:43:02about this n dimensional space we will
1:43:05also talk about different operations we
1:43:07can apply when it comes to vectors like
1:43:09uh addition subtraction and then later
1:43:12on in the next module we will also talk
1:43:15about multiplication we will also be
1:43:17looking into the properties of vector
1:43:19addition after we have looked into some
1:43:21detailed examples when it comes to
1:43:23operations on
1:43:25vectors all right so let's start with
1:43:28the zero vectors and unit vectors when
1:43:30it comes to zero vectors you can see
1:43:32here already that um the zero and arrow
1:43:36on the top it basically refers to the
1:43:39vector like we saw before only with the
1:43:42difference that all its members are zero
1:43:45so you can see here that we have zero
1:43:49and then an arrow and then underneath
1:43:51here we have some number tree and then
1:43:53this is described by this common
1:43:56representation with the square braces
1:43:58and then three different members z0 0 so
1:44:02all zero and then it says in R
1:44:06Tre okay so why are we doing this well
1:44:10uh when it comes to uh different linear
1:44:13Lal operation sometimes we just need to
1:44:16add zero vectors or we just want to
1:44:18create zero vectors it's just easier to
1:44:21work with you we want to uh just create
1:44:24an empty uh Vector we want we know the
1:44:27length but we want to keep it empty such
1:44:30Laton we can add something on the top or
1:44:33knowing that when we add a zero on a
1:44:36number the number stays the same we can
1:44:38make use of this property to uh do
1:44:41different um uh tricks when it comes to
1:44:45programming in Python in SCAR or in C++
1:44:47Etc so therefore this idea of zero
1:44:50vectors can become very handy now one
1:44:53thing that you need to notice here is
1:44:55that we are not just writing down this
1:44:57zero to emphasize we are dealing with
1:45:00the vector but like uh before we have
1:45:02this error on the top emphasizing that
1:45:05we are dealing with a vector then what
1:45:08we are doing is that we are also adding
1:45:11the dimension of this Vector so in what
1:45:15dimension in what space are we um uh
1:45:18creating this zero Vector that this
1:45:20Vector is located is it in r R 2 in RN
1:45:24in R3 in this specific case you can see
1:45:26that in this example the uh index that
1:45:29we got here is three which basically
1:45:31indicates we are dealing with a zero
1:45:34Vector in threedimensional
1:45:36space so in the
1:45:39R3 uh in general we would just note this
1:45:42by n keeping the uh notation general
1:45:46which means that we are dealing with 0 0
1:45:49all the way down to zero so it has n one
1:45:53dimension in r
1:45:57n all right so this is about zero
1:46:00vectors it is just a way to uh make our
1:46:03programming life easier also to use it
1:46:05in different uh algorithms when it comes
1:46:08to bit more advanced
1:46:09algebra uh the next type of special
1:46:12vectors that we will look into is this
1:46:14unit vectors so vectors with a single
1:46:18element equal to one and all the others
1:46:21zero denoted as EI for the E unit Vector
1:46:25in N dimensions are referred by unit
1:46:28vectors
1:46:30so uh what we mean here when it comes to
1:46:34the unit
1:46:35vectors um if we have for instance E1 it
1:46:39means that we have a vector where the e
1:46:43in this case the first element is equal
1:46:45to one so you can see that E1 is equal
1:46:48to 1 0 0 so in the first element we got
1:46:53one and the remaining is zero and this
1:46:55is really important that we are dealing
1:46:57with vectors that contain only elements
1:47:00of zeros and ones and the only member
1:47:05that is equal to the only element in
1:47:06that Vector that is equal to one is the
1:47:08E element in the entire Vector all the
1:47:11remaining ones are zero and you can see
1:47:13here that the dimension is no longer
1:47:16specified but just the um index of the
1:47:19entry where the um uh the uh one is
1:47:24located so let's look at another example
1:47:27in here for instance when it comes to
1:47:29the um uh unit Vector yet another unit
1:47:33Vector is E2 which basically means that
1:47:36in the second element so in the second
1:47:40place uh the uh Vector contains one and
1:47:44all the other members are zero so here
1:47:46you can see Zero here it can see Zero
1:47:48only in the second element we have one
1:47:51and then in the E3 what we have here is
1:47:54that the third element is one and all
1:47:57the other ones are zero so let's
1:47:59actually look into uh one um bigger
1:48:04Vector uh in higher Dimension to make it
1:48:07even more sense so first I will Define
1:48:10and assume that we are dealing with a
1:48:12vector in RN so in an N dimensional
1:48:15space this gives me an idea that we are
1:48:17dealing with um so we are not dealing
1:48:20with nested Vector we are dealing with a
1:48:21simple and dimensional Vector so it has
1:48:24n rows and one column so using the
1:48:27square braces I'm going to represent my
1:48:30Vector so I have all these different
1:48:32members n members C so e let's say it is
1:48:40E5 so what does this mean it means that
1:48:43I is equal to 5 and this I element so
1:48:46the fifth element is equal to one and
1:48:49all the other entries the elements in
1:48:51this Vector are zeros so let's look into
1:48:54this is
1:48:56z0 0 Z I'm approaching the fifth element
1:49:01in my Vector so it's this one this is
1:49:04one and the remaining all zeros so this
1:49:09is a unit Vector in an N dimensional
1:49:12space and I'm defining it by
1:49:15E5 because my fifth element is equal to
1:49:18one now those are very handy when it
1:49:22comes to some other uh techniques in
1:49:24linear algebra and just in general think
1:49:27about techniques like um uh row etum
1:49:31form solving linear equation something
1:49:34that we will see as part of the next
1:49:36unit so many things um we can do by
1:49:40using unit vectors unit vectors are
1:49:42super important so you need to
1:49:45understand this concept uh very well
1:49:47such that later on you will understand
1:49:49uh more advanced concepts in linear
1:49:52algebra let now look into the topic of
1:49:54sparsity in vectors so by definition a
1:49:58sparse Vector is characterized by having
1:50:00many of its entries as zero so its
1:50:03parity pattern indicates the position of
1:50:06a nonzero
1:50:08entries so uh what we are basically
1:50:10saying is that if we are dealing with a
1:50:12vector that contains too many zeros we
1:50:16are dealing with the sparse Vector so uh
1:50:19this sparsity pattern indicates uh all
1:50:22Al the positions of a nonzero elements
1:50:26so um if we have um unit Vector it means
1:50:31that we are already dealing with a
1:50:33sparse uh Vector this is a concept that
1:50:37is super important when it comes to
1:50:38linear algebra but also in general data
1:50:41science machine learning and AI because
1:50:44having a spity in your vector it means
1:50:46that you don't have much of an
1:50:48information usually a value zero it
1:50:50means you don't know much about that
1:50:53specific volum and if you got just too
1:50:56many of zeros and too few numbers which
1:50:59do um provide information it means that
1:51:02you are dealing with a vector that
1:51:04doesn't provide you much information and
1:51:06there's always a problem when it comes
1:51:08to data science machine learning and AI
1:51:11so sparcity is something that you need
1:51:13to be aware of you need to know how to
1:51:15recognize it and you also need to know
1:51:17whe there's a problem in your specific
1:51:19case or not so let's look into an
1:51:22example let's say we are dealing with
1:51:25this Vector X that has five different
1:51:28elements so X is a vector coming from um
1:51:32five dimensional space so we have for
1:51:36instance an element of three the first
1:51:38entry then we have z0 in the second and
1:51:40third uh entries then we have an entry
1:51:43um four which coincident also contains
1:51:46value four and then the last element in
1:51:49our five dimensional Vector X is equal
1:51:52to zero
1:51:53now what do we see here we see that the
1:51:55majority of elements of a vector X is
1:51:59equal to zero because we got in total
1:52:01five
1:52:02elements and then we got three of it
1:52:06actually uh being equal to zero and only
1:52:09two of them containing information like
1:52:11equal to three and four so only two
1:52:14elements that are not zero so non-zero
1:52:17elements it means that 3 / to 4 which is
1:52:20basically 60% 60% of all the entries in
1:52:25the vector X are equal to zero so the
1:52:3060% it means that is above half so above
1:52:3450% 60% of all the information in this
1:52:38Vector um the majority is simply equal
1:52:41to zero this type of vectors we are uh
1:52:44calling sparse vectors and sparcity is
1:52:47really important concept uh that we need
1:52:49to keep in mind later on
1:52:53so uh while we can visualize vectors in
1:52:55two and three dimensions in linear
1:52:57algebra like we just saw in case of this
1:53:00n dimensional vectors
1:53:02visualizing uh the this type of higher
1:53:05dimensional vectors becomes very
1:53:07difficult so uh this mathematical
1:53:10flexibility uh to work with uh this type
1:53:14of uh information so when we can
1:53:17represent uh information many with many
1:53:20entries we can represent it by vector s
1:53:23which we can actually not visualize
1:53:25becomes very handy for complex data
1:53:28structures for different simulations in
1:53:30physics and much more so uh we just saw
1:53:33in couple of examples uh how we can
1:53:36represent vectors in a high dimensional
1:53:38space using this Square braces and this
1:53:41common Vector notation representation we
1:53:44saw that in an N dimensional space we
1:53:46could uh very easily represent this uh
1:53:50very large Matrix or
1:53:52vectors uh by just um using this Vector
1:53:56representation for instance if we got a
1:53:58vector that had any different entries
1:54:01where n is for instance thousand so
1:54:03let's say we have thousand then uh we
1:54:06can
1:54:07represent uh this uh vector or this
1:54:11information by using common Vector
1:54:13notation so A1 A2 all the way to a th000
1:54:18so of course we cannot visualize this it
1:54:21just doesn't make sense we can visualize
1:54:23two dimensional vectors we can visualize
1:54:25three dimensional vectors but we cannot
1:54:28uh visualize thousand dimensional
1:54:31vectors so Vector that comes from uh
1:54:35r, but what we can do is still make use
1:54:39of this very useful
1:54:41information in order to uh do different
1:54:44operations when and later on we will see
1:54:47that uh this property and specifically
1:54:50this part of linear algebra it helps us
1:54:53to work with vectors in any number of
1:54:55Dimensions whether thousands million
1:54:57billions this mathematical flexibility
1:55:00is super important for more complex data
1:55:03structures uh for metrix multiplications
1:55:07when for instance we are doing different
1:55:09uh algorithms including how we can
1:55:12represent uh very large matrices very
1:55:15large feature spaces all this different
1:55:18information we can represent just by
1:55:21making use of
1:55:22vectors coming from this specific uh
1:55:26part of linear
1:55:28algebra let's now finish of this module
1:55:31by looking into some applications of
1:55:33vectors so one common application of
1:55:36making use of vectors is uh when we are
1:55:39performing different operations while
1:55:41having words and we want to count those
1:55:44words so this is a super common
1:55:46application of vectors and we can
1:55:49account this words and you can even plot
1:55:50a histogram over how often each of these
1:55:53words appear in a document so a vector
1:55:57of a length n for instance can represent
1:56:00the number of times each of these words
1:56:02in a dictionary of n words appears in a
1:56:05document so uh just for the sake of
1:56:08Simplicity let's assume that we got um
1:56:11dictionary that contains only three
1:56:13words of course in the reality uh the um
1:56:17dictionary what we also refer often as
1:56:20Corpus it contains much many much more
1:56:22many words but for the Simplicity we
1:56:25will assume that we just got three
1:56:26different words in our dictionary so
1:56:28that's a total now let's assume that we
1:56:31got a document uh with these different
1:56:33words and we want to count how many
1:56:36times each of those words that we got in
1:56:38dictionary actually appear in our
1:56:40document so uh if our document is
1:56:44described by this Vector so it contains
1:56:47an entry of 25 2 and zero it means that
1:56:51in our our document we got
1:56:5625 word one in now from our dictionary
1:57:00so in the position one two * word two
1:57:06and zero *
1:57:09word three so basically we have a
1:57:13predetermined set of words in our
1:57:15dictionary in this case three words word
1:57:18one word two and word three and they
1:57:21have a specific IND specific position in
1:57:23our vector and when we are putting these
1:57:27values in here then the machine or the
1:57:31uh computer the program will understand
1:57:33that if we have 25 in the first position
1:57:36then the word one in the dictionary
1:57:40appeared 25 times in our document
1:57:43whereas the second word appeared only
1:57:45two times and the last word for three
1:57:47didn't appear at all so zero times in
1:57:50the entire document
1:57:52so let's look into a practical example
1:57:55actually to make even more
1:57:59sense so um this is by the way a common
1:58:03practice to count different variations
1:58:05of a word there are common application
1:58:07in engrams large language models
1:58:10Transformers they are just the
1:58:11Cornerstone of many language models when
1:58:14we want to count the words in the
1:58:18document to understand how often the
1:58:20word appears because this gives us a
1:58:22idea what this document is about knowing
1:58:24how many times the same word appears in
1:58:26that uh document it gives us an
1:58:29indication of the topic of uh the do
1:58:32document also we can make use of a to do
1:58:35sentiment analysis to understand what
1:58:37this document is about not only in terms
1:58:39of the topic but also is it a positive
1:58:42is it a natural or a negative uh
1:58:45document so to say so uh for instance if
1:58:50we got uh the following words uh that
1:58:53correspond to our dictionary and in our
1:58:56dictionary we got just um let's say 6
1:59:00different words then what we can do is
1:59:03that we can say 3 2 1 let's say Zer 4
1:59:09two and the corresponding words are
1:59:12word
1:59:14row
1:59:16[Music]
1:59:17number
1:59:20horse is
1:59:23and then
1:59:24document what this means is that we have
1:59:28a text what we refer as a document that
1:59:31contains three times the word
1:59:33word that contains two time the word row
1:59:37contains one time the word number zero
1:59:40times the word horse and four times the
1:59:42word eel and two times the word
1:59:47document so uh this is basically a
1:59:50common way represent presenting the uh
1:59:53frequency of the words in the
1:59:55document let me actually give you uh
1:59:58another example
2:00:01and in here I want to emphasize another
2:00:05thing the concept of stop words so uh
2:00:09let's say I make
2:00:12this 10 and then
2:00:16here I say there is a three times the
2:00:20word I
2:00:23two times the
2:00:25word uh
2:00:29reading two * the word
2:00:33library four times the word
2:00:36book 0er * the word
2:00:40shower and 10 times the word
2:00:45uh so uh you can see a that in here we
2:00:50are dealing with the document that
2:00:52contains 10 times the word uh which is
2:00:56what something that we refer as a stop
2:00:58word so those are things that actually
2:01:00don't give us too much information about
2:01:03what the document is about because uh
2:01:05it's just used everywhere but it is
2:01:07appearing too often so you can see 10
2:01:10times the most frequently appearing word
2:01:13this is what we refer as a stop word and
2:01:16then another thing that we can observe
2:01:18the second thing we can observe is that
2:01:20we are dealing most like ly with a
2:01:22document that describes library reading
2:01:25uh because you see the words like
2:01:26reading you see the word like book
2:01:29library but another word shower that is
2:01:32totally unrelated to reading book or
2:01:34library is appearing zero times so even
2:01:37by looking at discounts we can already
2:01:39get an idea what a topic of this
2:01:41document is about so uh you can see
2:01:45already know from this very basic
2:01:47example where I made too many
2:01:49assumptions regarding how small the the
2:01:51uh dictionary should be uh you can even
2:01:54see now how we can use discounts in our
2:01:58dictionary from our text in order to get
2:02:01idea about the topic of the document or
2:02:04topic of the conversation it can be
2:02:06topic of the uh tweets if you have a
2:02:08tweet data it can be topic uh regarding
2:02:12book if you have many book um uh book
2:02:16text it can be for instance the topic of
2:02:19the review if you got a reviews from uh
2:02:23Amazon for instance using this count can
2:02:26help you to get a topic regarding topic
2:02:31from that text then you can also use it
2:02:34to remove the stop wordss because
2:02:35usually the stop words are the most
2:02:37frequently P words it can also give you
2:02:40an idea about the sentiment for instance
2:02:42here we are dealing with natural
2:02:43sentiment it's not positive it's not
2:02:45negative it's just reading a book in
2:02:47library that kind of topic so all this
2:02:50can be super helpful when it comes to
2:02:52natural language processing that's a
2:02:54field where this uh text processing text
2:02:57cing and then using that for modeling
2:03:00purposes is what uh what plays a central
2:03:03role it also plays a super important
2:03:05role in the large language models in the
2:03:07Transformer models and uh in simple
2:03:10matters like uh back of words or uh in
2:03:15the uh TF IDF all these they are based
2:03:17on this idea of counting words and how
2:03:21we can use it information and you can
2:03:23see how vectors come into play in the
2:03:26different applications of linear algebra
2:03:29in data science natural language
2:03:31processing in artificial intelligence in
2:03:34machine learning so they are super
2:03:39important another application of vectors
2:03:42can be representing customer purchases
2:03:45for example an N Vector P so let's say p
2:03:51can record a customer purchases over
2:03:53time with pi being the quantity or
2:03:56dollar value of an item I now what does
2:03:59this mean so let's say we have Vector P
2:04:03that represents the customer purchases
2:04:06and we are dealing with a single
2:04:07customer and we are just saving over
2:04:10time that information how many time this
2:04:12customer has made purchases over
2:04:16time so the quantity is in the um
2:04:20dollars so the dollar value of item I
2:04:23purchase so we are basically keeping
2:04:25track of uh what is the value of the
2:04:29item I that the customer has purchased
2:04:34so what we can
2:04:36do is we can assume that in here
2:04:40actually it already Mak that assumption
2:04:42it says n Vector which means that the
2:04:45number of rows or number of um items
2:04:49that the customer purchases is
2:04:53n now what the um the problem says that
2:04:59it represents is that in each
2:05:03entry and here we have in total n
2:05:06entries we got a dollar value of item I
2:05:09which means that here if I have P1 P2
2:05:14all the way to PN and here somewhere in
2:05:16the middle I got Pi in the East position
2:05:20it means p Pi represents the
2:05:25value of
2:05:28item I so for example if I'm dealing
2:05:33with a
2:05:34customer that buys um let's say uh
2:05:40courses and uh the first item that the
2:05:44customer is buying is a mathematics
2:05:46course so I'm writing
2:05:49mathematics course and this is the first
2:05:52course that it buys e is by the way just
2:05:57a um way to refer to the E purchase so
2:06:02let's say um here somewhere in the
2:06:04middle the um customer decides to buy a
2:06:08deep learning course deep
2:06:12learning
2:06:14learning
2:06:17course and then it continues buying uh
2:06:20the customer continues buying courses
2:06:22and the last course that a customer buys
2:06:24is let's say um career
2:06:29coaching
2:06:31course
2:06:33now let's say the mathematics course
2:06:36costs uh around
2:06:40$1,000 let's say the uh deep learning
2:06:43course costs
2:06:46$33,000 and then let's say the career
2:06:48coaching service which is usually one of
2:06:50the most apply and personalized one can
2:06:53cost all the way to
2:06:56$5,000 now we see that in the each
2:07:00position this is the East position let
2:07:02me change the color by the
2:07:04way so let's say this is the East
2:07:07position this is the first position and
2:07:09this is the last position so those are
2:07:11just
2:07:13indices we can see that in the E
2:07:15position we got the
2:07:173,000 which means that the p e is equal
2:07:21equal to
2:07:24$3,000 so this indicates that in the
2:07:27East
2:07:28purchase the customer purchased deep
2:07:31learning course and the value of that
2:07:34item was equal to
2:07:40$3,000 all right so now we are ready to
2:07:43go on to next major topic which is about
2:07:46vector addition and subtraction so we
2:07:48are going to do some operations and
2:07:50apply this operations two vectors so
2:07:53let's first formally Define this ideal
2:07:55of vector addition so uh two vectors of
2:07:59the same size are added by adding their
2:08:01corresponding elements the result is a
2:08:04vector of the same size so uh let's
2:08:08unpack this it says two vectors of the
2:08:12same size are added by their
2:08:14corresponding elements so here it refers
2:08:18to two different vectors let's say
2:08:20vector v and Vector W and it says let's
2:08:24add them what we refer as vector
2:08:27addition and says for that what we need
2:08:30to do is to take all the elements of v
2:08:33and then all the elements of w and using
2:08:37their corresponding elements so
2:08:40indices that helps us to understand
2:08:43where those elements are located we are
2:08:45using in order to add each
2:08:47element in the vector v to the element
2:08:51of the vector W in the same position and
2:08:55do note that in the second part it says
2:08:58the result is a vector of the same size
2:09:01because we are adding two different
2:09:03vectors of the same size it's mentioning
2:09:07here it means if we add two different
2:09:09vectors to the same uh that have the
2:09:11same size we are going to end up with a
2:09:14vector that has the same
2:09:17size now once I go into the examples it
2:09:20will make much more per let's quickly
2:09:23also look into this concept of
2:09:25substraction so on its own uh
2:09:27substraction is very similar to this
2:09:29idea of addition so if we have a
2:09:31substraction let's say we have vector v
2:09:34We substract Vector W then we are doing
2:09:37basically uh what we just did to the
2:09:39addition only instead of uh doing add we
2:09:43are doing subtract so again we are just
2:09:46uh we are just subtracting from vector v
2:09:49Vector W they have the same size so we
2:09:52end up having the result which is a
2:09:55vector of the same size only one thing
2:09:59that you can see is that this can be
2:10:01also written as V Vector plus and then
2:10:05minus W so we basically can represent
2:10:09subtraction um on its own as a way of
2:10:13adding only we take the negative so the
2:10:16um opposite directed Vector so this will
2:10:20make even much more
2:10:22uh once we go on to the examples so
2:10:24let's look into our first operation
2:10:27example where we are adding two
2:10:29different vectors this a basic example
2:10:31we got just two dimensional two vectors
2:10:34we got Vector a that has entries two
2:10:37three and Vector B that has entries one
2:10:40form and what we are doing is that we
2:10:42are adding Vector a to Vector B we just
2:10:45learned that a we need to have the same
2:10:48size of vectors so you can see that
2:10:50Vector a has a dimension 2 by one vector
2:10:53B has a dimension of 2 by one so their
2:10:56sizes is the same both they got two
2:10:59entries only two
2:11:02elements and at the same time we just
2:11:04learned that what we need to do is to
2:11:06take their corresponding elements and
2:11:09add them to each other now what does
2:11:11this mean it means that we take from a
2:11:16the first element
2:11:19two and then we take the first element
2:11:22of the second Vector which is the B so
2:11:25we take the two from here and one from
2:11:28here the first element of a and the
2:11:30first element of B and then we are
2:11:32adding them to each other 2 + 1 is equal
2:11:35to three and then the same holds for the
2:11:38second and Tre so three which is the
2:11:40second element of vector a and then four
2:11:43which is the second element of vector B
2:11:45we are saying 3 + 4 is = to 7 so let me
2:11:49write it down even in a simpler manner
2:11:51such that it will make much more sense
2:11:54so Vector a has elements 2 three in the
2:11:57first element we got two in the second
2:11:59element we got three so a then we want
2:12:04to add B which has in the first element
2:12:07element equal to 1 and the second
2:12:09element is equal to 4 this means that if
2:12:12we want to add these vectors 2
2:12:153+ 1 4 this is equal to we need to take
2:12:20two we need to add one so this element
2:12:24and this element and then we need to
2:12:25take three we need to add to four so
2:12:28this one and this one which is equal to
2:12:312 + 1 is equal to 3 3 + 4 is equal to 7
2:12:34so we got uh Vector
2:12:3837 do you note that this Vector the
2:12:41result Vector contains again two
2:12:43elements and just one column so 2 by 1
2:12:47so you notice that the sign that the
2:12:49size is the same of this result
2:12:53Vector now let's actually generalize
2:12:55this concept before moving on to the
2:12:57next example so if we got let's say
2:13:03Vector
2:13:05a that contains n elements A1 A2 all the
2:13:10way down to a
2:13:13n and it is
2:13:16from n dimensional
2:13:19space and we got Vector
2:13:24B that also has n elements so remember
2:13:27that they both need to have the same
2:13:29size so B1 B2 all the way to b
2:13:34n so they come also so B comes also from
2:13:39n dimensional
2:13:44space so then when we add a to B this is
2:13:50equal to
2:13:55A1 A2 all the way to a
2:14:00n
2:14:02plus B1 B2 all the way
2:14:07to
2:14:10BN so n by one n by one the sizes this
2:14:14is equal
2:14:16to let me actually use this color to
2:14:19make it even more visible so I for the
2:14:21first entry for my result factor I will
2:14:26get A1 + B1 then A2 + B2 so all the way
2:14:32down onto the end element which is a n
2:14:40plus then me use a different color A1
2:14:44B1
2:14:45B2 b
2:14:48n so you can notice is now in general
2:14:53terms what we are doing here so we are
2:14:55taking the A1 coming from the vector a
2:15:00we are adding in the same
2:15:03uh position the value that comes from
2:15:07Vector B which is B1 we are saying take
2:15:10the A1 Plus B1 this is the uh first
2:15:14element so the position stays the same
2:15:16and then in the result Vector so we take
2:15:19all the corresponding values that are
2:15:22have the same position in the
2:15:24corresponding Vector first from Vector a
2:15:26and then Vector B we are adding them and
2:15:28this forms our new vector and this new
2:15:31Vector will again have a size n by one
2:15:35so you can see that a the sizes of the
2:15:38two vectors are the same both have n
2:15:40elements and then we are using their
2:15:43corresponding elements to add them to
2:15:45each other element wise and then we are
2:15:47getting the result that has the same
2:15:49size so n by 1 so this is a more General
2:15:53description of how you can add two
2:15:58vectors let's now look into this
2:16:00specific example so we have a vector
2:16:03with the entry 073 so this comes from R
2:16:07three you can see so three dimensional
2:16:09vectors the second Vector is 1 2 0 and
2:16:13then the final result is 1 193 so how we
2:16:16got this we took zero we added 1 7 we
2:16:20added two and then three we added zero
2:16:23so you can see all these
2:16:25elements element Y and then this is
2:16:28equal to 0 + 1 is 1 7 + 2 is 9 and then
2:16:313 + 0 is 3 exactly what we got here so
2:16:35again the same sizes and the result is
2:16:38from the same
2:16:39size so quite
2:16:41straightforward now when it comes to the
2:16:43vector substruction what are we doing
2:16:46that um so what are we doing here so we
2:16:50are doing kind of very similar thing we
2:16:53are taking this element one we are
2:16:56subtracting the other one in this first
2:16:59element then we are taking the nine in
2:17:01the second position and subtracting this
2:17:05again from the second position of the
2:17:06second vector and we are putting in here
2:17:091 and then 1 - 1 is = to 0 9 y - 1 is =
2:17:13to 8 so we get result Factor 08 like in
2:17:17here and you can see that the sizes stay
2:17:20the same so also in this case let's
2:17:22write more General um this idea of
2:17:25subtraction if we got a vector
2:17:29a
2:17:31from RN so n dimensional space and it
2:17:34can be represented by A1 A2 all the way
2:17:38down to a n so it has n elements n by
2:17:43one and then we got
2:17:47B also from RN so coming from the n
2:17:51dimensional space which means that it
2:17:53got n elements so B1 B2 all the way down
2:17:57to BN again with the same size n by one
2:18:02then a minus B is simply equal
2:18:08to a A1 let me actually use the same
2:18:13colors to make it easier to
2:18:18follow so let me first draw my Square
2:18:23races and then here I will use blue for
2:18:28a and then red for the uh color for
2:18:32second Vector which is
2:18:34B here I will use black
2:18:44minus then given that the same size
2:18:48should be for the result Vector I I
2:18:50already know that I expect n different
2:18:53elements for this and then
2:18:56here I'm taking
2:19:01this first element that comes from
2:19:04Vector
2:19:05a I substracting from this the first
2:19:09element that comes from Vector B so
2:19:11element wise subtraction B1 and I'm
2:19:15already getting the result for the first
2:19:18element in my result Vector so you can
2:19:22see A1 minus B1 I'm taking this element
2:19:26and this element and subtracting them
2:19:28from each other to get A1 minus B1 and
2:19:31then the same holds for all the other
2:19:34values only coming from different
2:19:37elements from Vector
2:19:40a subtracting from this the
2:19:44corresponding values element Wise from
2:19:47the vector B so B2 B3 all the way to a n
2:19:52so you can see that in my result
2:19:55Vector a vector minus B Vector in the
2:19:58first element I get A1 minus B1 then A2
2:20:02- B2 then A3 - B3 in the third element
2:20:05all the way down to the end element
2:20:07which is equal to oh this should be b a
2:20:11n minus
2:20:12BN so um this already should makes uh
2:20:16much more sense so every time we take
2:20:18the element in the same position from
2:20:20one vector than the other we subtract
2:20:23from each other in order to get the
2:20:24corresponding element in the final
2:20:29Vector all right so let's now uh before
2:20:33moving on to the
2:20:34properties um I want to to show you um
2:20:39this only in a coordinate space so what
2:20:43this means in terms of visualization in
2:20:45a coordinate
2:20:47space so uh let's say we have a
2:20:52coordinate
2:20:58space this is my Y
2:21:05axis this is my x axis
2:21:09so this is X and the Y and this is my
2:21:14Center so 0
2:21:160 and what I'm doing here
2:21:21is
2:21:22simply I want to have Vector a let's say
2:21:27this is just um Vector a simple one with
2:21:31the coordinates um let's say four and
2:21:35minus
2:21:372 and I got Vector
2:21:40B let me use a different
2:21:44color Vector
2:21:47B that has coordinates
2:21:51let's
2:21:53say
2:21:56minus 4 and
2:22:00four so let's actually visualize them
2:22:03let's first start with D Vector a uh
2:22:07which has a x value of
2:22:10four three
2:22:12four one 2 three and
2:22:15four and the Y value minus 2 so this
2:22:25is my Vector
2:22:30a and let's now visualize the vector B
2:22:35so minus 4 and
2:22:374 which means
2:22:43that let me
2:22:45actually extend
2:22:48this this is minus 4 so the x coordinate
2:22:52is min - 4 so it should be here and then
2:22:56the y coordinate is four so 1 2 3 and 4
2:23:01it's this one which means that my Vector
2:23:06B is this
2:23:13one all
2:23:16right so you can see now that the vector
2:23:20a is in here and the vector B is in
2:23:23here now what I want to do is to add
2:23:27these two vectors to each other so what
2:23:30I want to do is to take this Vector
2:23:34a and add to
2:23:37this the vector B which
2:23:42is is equal to 4 - 4 = 0 and then - 2 +
2:23:484 is = 2 so zero and then two it
2:23:57is z and
2:24:03two
2:24:11two so this is my result Vector so now
2:24:18when we are clear on how we can in
2:24:20vector s how we can perform these
2:24:21different operations and what it means
2:24:24in practice when it comes to looking at
2:24:26the vectors in a cordan space and adding
2:24:29them or subtracting them we are ready to
2:24:31look into the properties of vector
2:24:35additions so this is something that will
2:24:37definitely seem familiar to you uh from
2:24:40pre-algebra where we are basically using
2:24:43all these properties that we already
2:24:45know that holds for uh numeric values
2:24:48for the scalers that being transferred
2:24:50to this Vector space so we are going to
2:24:54talk about this four different
2:24:56properties that a vectors have the first
2:24:58one is the cumulative property which
2:25:01says that if we add a vector a to Vector
2:25:05B then this is the same as adding a
2:25:08vector B to Vector a so basically the
2:25:11order of the vectors doesn't really
2:25:13matter when it comes down to adding them
2:25:16so formally A + B is equal to B+ a for
2:25:19any vectors A and B of the same size
2:25:23then we have associative property which
2:25:26says A + B + C is equal to a + b + C we
2:25:33can write both As A + B + C now what
2:25:37does this mean we know from pre-algebra
2:25:39that this parenthesis means first do
2:25:42this addition and then do the the rest
2:25:45of operations in here it basically says
2:25:47if you add a to the B first and then you
2:25:51add the C is the same as first you add B
2:25:55to the C and then on the top of that you
2:25:57add D Vector a so then the third
2:26:02property is addition of zero vectors
2:26:04which says if we add a zero Vector to
2:26:06Vector a then this is equal to adding a
2:26:10vector zero to a and this is equal to
2:26:13Vector a so adding the zero Vector has
2:26:16basically no impact on the vector
2:26:19whatsoever
2:26:21then the final property is subtracting a
2:26:23vector from itself which means if we
2:26:25take the vector we substract the same
2:26:27Vector from itself so a minus a and we
2:26:30get a zero Vector so a minus a is equal
2:26:34to zero vector and this heals the zero
2:26:37Vector now let's look into each of those
2:26:39properties one by one and let's uh look
2:26:42into specific examples uh in some cases
2:26:45we will prove this on the example that
2:26:47we have to make this Concepts much more
2:26:50clear so let's start with this
2:26:52cumulative property of vector additions
2:26:56so we want to see whether A+ B is equal
2:26:58to B + a so let's say we have a vector
2:27:03a that has coordinates or magnitude and
2:27:08direction that is equal to one and two
2:27:12then we have a vector uh let's say
2:27:18B that has a magnitude and direction of
2:27:22- 2 and
2:27:253 so the first thing that we want to
2:27:28check is indeed whether the A + B is
2:27:32equal to B +
2:27:34a so therefore let's first calculate
2:27:37this part and then we will calculate
2:27:39this part that I will Define by one and
2:27:41two and we will see whether we are
2:27:43indeed having the same value the same
2:27:46vector or not so let's see so we have
2:27:48here a
2:27:50so A + B which is the first value that
2:27:55we want to calculate a plus b is = to 1
2:28:012+ -
2:28:0323 and we learned before that this is
2:28:06simply equal 2 take this value and then
2:28:09add this one so 1 + - 2 and then 2 +
2:28:153 so this gives us a vector 1 - 2 is =
2:28:20to - 1 and 2 + 3 is = 5 so we get that A
2:28:25+ B is = to -1 5 this Vector now let's
2:28:30look at the second quantity so B Vector
2:28:33B plus Vector a this is equal to - 2 3 +
2:28:391 2 and this is equal to - 2 + 1 and
2:28:44then 3 + 2 this gives us - 2 + 1 is =
2:28:49to- 1 and 3 + 2 is equal to 5 so we can
2:28:53already see from here that the quantity
2:28:57one is indeed equal to quantity 2 which
2:29:01proves that indeed the A + B is equal to
2:29:05B+ a what this basically means is that
2:29:09adding two different vectors the
2:29:11direction or the order is not important
2:29:14whether you add a on the top of the b or
2:29:16B to a it doesn't matter at the end is
2:29:19the same and actually you can also see
2:29:21it if you uh combine this or if you do
2:29:25this in more general terms so let's say
2:29:28if we
2:29:29have a vector a which is equal to in an
2:29:33N dimensional space A1 A2 up to a n so
2:29:39it has n by one dimension and you have a
2:29:42vector B with the same size from the
2:29:46same RN
2:29:48Dimension and it has element B1 B2 up to
2:29:52BN and the dimension is equal to M by1
2:29:56then if we calculate first
2:29:59A+
2:30:01B and this is equal to Simply A1 + B1 A2
2:30:07+ B2 up to a n +
2:30:12BN and if you calculate the second uh
2:30:16amount which is B+ a
2:30:20this is equal to B1 + A1 B2 + A2 up to
2:30:27BN + a n you can see that A1 + B1 is
2:30:35equal to B1 + A1 simply from prealgebra
2:30:39you know that if those are all constants
2:30:41for instance 2 + 3 is equal 3 + 2 in the
2:30:45same way A2 + B2 is = to B2 + A2 and
2:30:50then here up to a n + BN is equal to BN
2:30:55+ a n what this means is that all these
2:30:58elements they are basically the same
2:31:01which means that we already have a proof
2:31:04so we get this proof and we can see that
2:31:07even for the general term independent
2:31:10what this Vector a is what this Vector B
2:31:12is that a + b is equal to B + a
2:31:20this is exactly what we saw before in
2:31:23the first property which is called
2:31:24commutative property of the vectors that
2:31:27a plus b is equal to B+ a now let's move
2:31:30on to the other property which is called
2:31:32associative property of the vectors now
2:31:35what this property does and says is that
2:31:38a plus b so first we do this plus C is
2:31:41equal to a + b + C and this is then
2:31:46equal to a + b + C now let's then see um
2:31:50this specific property on an actual
2:31:54example so what this basically says is
2:31:57that if we have this example where a is
2:32:00equal to actually I had this before let
2:32:03me simply just remove this part let's
2:32:07then add our third Vector which is C and
2:32:11let's call it let's say it has a
2:32:14representation of four and
2:32:17five then the IDE behind this property
2:32:21is that what we need to prove here that
2:32:25A + B within the parenthesis plus C is
2:32:29equal to
2:32:34a
2:32:43plus B+
2:32:46C and then this is equal to a plus b
2:32:52plus C so let's see actually whether
2:32:55this is indeed true for this specific
2:32:57case now this should come very intuitive
2:33:01so I'm going to do it very
2:33:02quickly so first we have this quantity
2:33:05this one then we have this one and the
2:33:07third one let's do it very quickly so A
2:33:11+
2:33:12B plus
2:33:15C is equal to
2:33:20one Tu
2:33:24plus and then we had C so it is
2:33:30simply 4
2:33:32five and then this is equal to we saw
2:33:36before when doing this that we were
2:33:38getting
2:33:391 - 2 2 + 3 and then we add this four
2:33:46five this is simply equal to 1 - 2 is
2:33:52-1 and 2 + 3 is
2:33:565 + 4
2:34:005 now given that it doesn't really
2:34:02matter no longer that we have uh here
2:34:05parenthesis or not this basically means
2:34:09that this volue is simply equal
2:34:14to -1 + 4 so here -1 + 4 here 5 +
2:34:215 so this is then equal to three and
2:34:26then
2:34:2610 all
2:34:29right let's then now quickly do the
2:34:33second amount which says first add the
2:34:36vector B to Vector
2:34:39C and only then add the vector a on the
2:34:42top what this means is that we need to
2:34:45take one two this is Vector a and we
2:34:47will only add this once we have added D
2:34:50minus 23 the vector B plus to the Vector
2:34:5645 okay
2:34:58so we can see that we are just leaving
2:35:01this in here let's first add this two
2:35:04minus
2:35:072 + 4 3 +
2:35:125 so this gives us 1
2:35:162+ - 2 + 4 is uh 2 and then 3 + 5 is 8
2:35:22so this gives
2:35:24us let me remove this
2:35:30calculations so this gives us 1 + 2 is =
2:35:34to 3 and then 2 + 8 is equal to 10 okay
2:35:40great so now we got already the quantity
2:35:441 being equal to quantity 2 let's check
2:35:47whether this is all equal to this one it
2:35:50should already be um something that you
2:35:53see now given that um we know just from
2:35:56mathematics that parentheses doesn't
2:35:58really matter when it comes to the
2:35:59scalers and adding two vectors is
2:36:02basically very close to this idea of
2:36:04addited property um of the edited
2:36:07property of the scalers but just let's
2:36:10quickly do it to be 100% sure so when we
2:36:13take this Vector a to the B and to the C
2:36:16we had all this this is equal to one
2:36:19want to added to minus 2 three and then
2:36:24added this to four and five now what
2:36:27this is equal
2:36:29to let me actually write this in bit
2:36:32shorter way such that it can be all fit
2:36:35in in the small place so 1 2 + - 2 3 + 4
2:36:445 this is equal to basically 1 - 2 + + 4
2:36:50and then 2 + 3 +
2:36:525 now what is this
2:36:59number 1 - 2 + 4 is simply equal to 1 -
2:37:032 is = to -1 and then + 4 is equal to 3
2:37:07so first element is three 2 + 3 + 5 is
2:37:10equal to 5 + 5 which is equal to 10
2:37:14perfect so now we get the confirmation
2:37:16that indeed a + b plus c C is = to A + B
2:37:21+ C is = to A + B +
2:37:26C so let's quickly also look into this
2:37:29addition of zero vector and the
2:37:31subtracting a vector from itself
2:37:32properties and uh the detailed
2:37:35explanation of this or example of this I
2:37:37will leave it to you so when it comes to
2:37:39this A+ um 0 is equal to 0 + a is equal
2:37:43to a so this property let's say if a is
2:37:47equal to this 23 and then we are adding
2:37:51on this a plus some zero Vector which
2:37:55basically means take two three and then
2:37:58added the same size of zero Vector you
2:38:01can see that this is the same as adding
2:38:04this zeros on these values now what do
2:38:08we get we get that this is equal to 2 +
2:38:120 is 2 and then 3 + 0 is
2:38:15three there we go so we already see very
2:38:18quickly that it doesn't really matter
2:38:21whether we add a zero Vector to this
2:38:23original a vector or not we in all cases
2:38:27it just adding a zero Vector has no
2:38:29effect and seeing from the commutative
2:38:32property that a plus b is equal to B+ a
2:38:35we already know that if um a + 0 is
2:38:38equal to uh a and is equal to this then
2:38:42also 0 + a will be the same and we can
2:38:47see indeed that we just saw that a a + 0
2:38:50is simply equal to a so we basically
2:38:52have quickly proven all
2:38:54this now when it comes to the
2:38:56subtracting Vector from itself I think
2:38:59this is a very nice one just to see how
2:39:01we um uh take the same vector and
2:39:04subtract from that value and we get zero
2:39:07and this is very similar to working with
2:39:09just real numbers in the same way as 3
2:39:12minus 3 is equal to Z also when we have
2:39:15a vector consisting of the scalers like
2:39:18a is equal 2
2:39:2123 in the same manner if we take this a
2:39:25and we subtract it from itself so A Min
2:39:27- A then what we will get is 23 - 23 and
2:39:33this will give us 2 - 2 is 0 and then 3
2:39:36- 3 is zero so we get a vector zero so
2:39:39zero
2:39:41Vector so now when we are clear on how
2:39:44we can perform different operations on
2:39:45our vectors and also we know uh what are
2:39:48the prop properties of uh adding and
2:39:51subtracting uh different vectors we are
2:39:54ready to move on to a bit more advanced
2:39:56topics so uh in this module we are going
2:39:58to discuss this idea of scalar
2:40:01multiplication we're going to look into
2:40:03the example how uh what happens and how
2:40:06we can do the uh Vector multiplication
2:40:09with the scalar then we are going to uh
2:40:11look into the span of vectors what it
2:40:14means to have a sp of vectors uh what is
2:40:17this IDE of linear combination and the
2:40:19relationship between the span and linear
2:40:21combination and the unit vectors then we
2:40:24are going to look into the application
2:40:26of scalar Vector multiplication in audio
2:40:29scaling uh example and then finally we
2:40:32are going to finish off this module by
2:40:34looking into the length of a vector and
2:40:36a DOT product and we are going to uh go
2:40:39back to this idea of distance
2:40:42understanding vector magnitude and
2:40:44understanding Vector
2:40:46l so let's get started now before we
2:40:49look into this idea of span and linear
2:40:51combination I quickly wanted to look
2:40:54into this idea of scalar multiplication
2:40:56and the um specific definition of it so
2:41:00formally the scalar multiplication
2:41:02involves multiplying each component of a
2:41:05vector by scalar value effectively
2:41:07scaling the vector's magnitude so what
2:41:10do I mean here let's say we have a
2:41:13vector and I will write it in the
2:41:16general terms to keep everything General
2:41:18so let's see we have a vector a let me
2:41:21pick up my pen a and this Vector a is
2:41:26from n dimensional space so it is from
2:41:30RN and it can be represented by A1 A2 up
2:41:35to a
2:41:36n and I have this magnitude um of a
2:41:41vector and now I want to scale this uh
2:41:45Vector for which I know the magnitude
2:41:47and the direction I want to scale it
2:41:49with a scaler and we learned before that
2:41:51the scaler is just a number so um scaler
2:41:55in this case I will be uh referring it
2:41:58to uh by C so c will be my scaler and uh
2:42:02this comes from R which means that it's
2:42:05a real
2:42:06number let me actually use a different
2:42:10color to make it easier to
2:42:13follow okay so my scaler will be with
2:42:16the color uh red so C and C comes from
2:42:23R
2:42:25so what do I mean by scalar
2:42:27multiplication I mean that I want to
2:42:29find what is
2:42:32this c
2:42:35times
2:42:37a this is what we mean by scalar
2:42:41multiplying with Vector now what does
2:42:44this definition say it says when we are
2:42:47multiplying scalar we Vector so the
2:42:50scalar multiplication meaning
2:42:51multiplying Vector with the scalar it
2:42:54involves multiplying each component of a
2:42:57vector by a scalar volume so if we
2:43:00translate it to this specific example it
2:43:03means
2:43:04that this
2:43:06amount so this amount is equal
2:43:16to taking C and multiply find it with
2:43:20each
2:43:21element of this Vector so each component
2:43:25of vector and what are the components of
2:43:27my Vector the A1 A2 a up to the point of
2:43:31a n so all these
2:43:33components so that means that the first
2:43:35element of this new Vector the scalar
2:43:39multiplication result will be C *
2:43:44A1 C * A2
2:43:49dot dot dot so all this middle elements
2:43:51and at the end again c times and then a
2:43:57n and then in both cases of course the
2:44:01number of elements doesn't change so the
2:44:03so the number of rows of my Vector
2:44:06doesn't change it's n so here also n and
2:44:09then number of columns is the same so
2:44:11it's just a column Vector so one
2:44:15column so what we see here is that we go
2:44:18from a 1 to C * A1 we go from A2 to C C
2:44:23* A2 up to the a n transforms into C * a
2:44:29n so we see very easily that I keep all
2:44:33the elements from this Vector I take
2:44:35them in here and instead what I'm doing
2:44:37is that I'm multiplying every element
2:44:40from this vector by the scaler
2:44:44C so this is exactly what this
2:44:46definition says and let's actually go
2:44:50ahead and do a Hands-On example with
2:44:54some real numbers to have this um method
2:44:58and to have this uh definition very
2:45:00clear in our mind because we are going
2:45:02to make use of this fundamental
2:45:04operation scalar multiplication on and
2:45:07on in the upcoming lectures and just in
2:45:09general in your journey in any applied
2:45:13sciences so this is an example of scalar
2:45:16multiplication uh here what we are doing
2:45:18is that we want to multiply this Vector
2:45:21C so in this case the vector is defined
2:45:25by a letter C and then on the top we can
2:45:27see the arrow indicating that this is
2:45:29the vector now and here we refer the
2:45:32scalar by a letter K we are saying we
2:45:36want to perform scalar multiplication
2:45:38which means that we want to
2:45:40multiply the uh a vector C by the scaler
2:45:45K so how we can do that so what we want
2:45:49is to multiply K by C and we just
2:45:52learned that for that what we need to do
2:45:56let me write this
2:45:57over so this equal to minus 2 multiply
2:46:04it by 4 - 3 this is my Vector so this is
2:46:09the K and this is the C this is equal to
2:46:13so I take my
2:46:15scaler and I multiply it with the each
2:46:18of the ele element of the C so - 2 * 4
2:46:21and then - 2 *
2:46:25-3 so - 2 * 4 is = to - 8 and then - 2 *
2:46:32- 3 so- minus it goes away it becomes a
2:46:35plus and 2 * 3 is 6 so my end result the
2:46:39K * C is equal to - 8 6 this is my final
2:46:45result so let's quickly also do yet
2:46:49another example and this one is a unique
2:46:52one because it's relating to this idea
2:46:54of U multiplying something with a zero
2:46:59uh which is something that we also uh
2:47:01know from a high school that when we
2:47:03multiply number let's say seven by zero
2:47:06we are getting zero and here in this
2:47:09example the uh problem is describe the
2:47:13effect of a scalar multiplication by
2:47:15zero on any Vector which means
2:47:19what we are doing is that in this
2:47:21example is we want to know what is this
2:47:24result
2:47:25of any Vector let's say Vector uh C so
2:47:29we will use the same example C only this
2:47:32time instead of multiplying it with
2:47:34scalar k equal to minus 2 our scalar
2:47:36will be zero which means that c is equal
2:47:39to 4 - 3 and then K is now equal to zero
2:47:43and we want to find out what is this K *
2:47:47C let me actually write down the K with
2:47:50a different
2:47:58color k is equal to zero so what we want
2:48:02to find out is K * and then
2:48:05C and this is that equal
2:48:12to0 so I'm taking the
2:48:15k0 times then I'm taking each of the
2:48:19elements of C which is four and then
2:48:22minus 3 and I know that when multiplying
2:48:25the number with is 0 it gives me 0o
2:48:28which means that I end up with 0 here 0
2:48:31* 4 is 0 0 * - 3 is also 0 so I end up
2:48:35with a zero Vector now this gives me an
2:48:40idea already that I can make a general
2:48:43conclusion that independent of the type
2:48:47of vector that I have independ and what
2:48:49are this values in my C uh if I have any
2:48:53Vector C and I'm multiplying it with
2:48:58zero then this will always give me a
2:49:01vector of zero because all the members
2:49:05of this final Vector will be just zeros
2:49:10so if for instance the C comes from uh
2:49:15let's say r n so it has n different
2:49:18elements it comes from n dimensional
2:49:20space then my final result of 0 * C so
2:49:27this zero Vector this one so zero that
2:49:31this one will come also from RN so you
2:49:35will be having a vector so 0
2:49:40* c will then be equal to z0 blah blah
2:49:45blah blah zero so n time zeros
2:49:49so this is then the idea of multiplying
2:49:52so scaling a vector with zero and this
2:49:56is our example two all right so let's
2:50:01now move on on to our application of
2:50:04scalar vectal multiplication and then
2:50:06after this we will go back to this idea
2:50:08of linear combinations and
2:50:10dispense so in this specific application
2:50:13we have a scalar Vector multiplication
2:50:15and we are looking into application of
2:50:17audio scaling
2:50:19so the scalar Vector multiplication
2:50:21audio processing uh this can change the
2:50:23volume for instance of an audio signal
2:50:25without altering its content so um you
2:50:29might have noticed that um when uh when
2:50:33you are listening to video you can
2:50:35simply increase the volume of that video
2:50:38or decrease it but you will notice that
2:50:40the content doesn't change you are just
2:50:42increasing the volume or decreasing it
2:50:44even on the TV when you are watching a
2:50:46show you are increasing The Voice or
2:50:48decreasing
2:50:49now what you're basically doing behind
2:50:51and this is super interesting is that
2:50:53behind the scenes what is happening is
2:50:55that there is simply um audio that um
2:51:00contains that show and the audio of that
2:51:03show is being multiplied with a scaler
2:51:06and that scale is simply the volume
2:51:08scale if you scale it in such way that
2:51:12you want to decrease the volume so the
2:51:15audio will then have a lower volume then
2:51:18you are simply multiplying it uh your
2:51:21vector containing the audio information
2:51:24in such way that those newer volume
2:51:27indications they will be they will be
2:51:30containing lower
2:51:31numbers hope this makes sense let's look
2:51:33into the example this make uh this will
2:51:36definitely clear this out so um let's
2:51:39assume we have an a vector a that
2:51:42represents the audio signal and we want
2:51:46to multiply Vector a a by scalar B to
2:51:50adjust the volume so B is some sort of
2:51:53number it can be so B comes from R so is
2:51:59a real number while
2:52:01a is simply a vector given that it
2:52:04doesn't mentioning here I'm assuming
2:52:06that a comes from RN so it comes from r
2:52:10n dimensional space so imagine of a as
2:52:14this Vector A1 A2 blah blah blah blah to
2:52:19a n and each of these values it
2:52:22basically describes uh an uh the audio
2:52:25signal so it represents um uh an amount
2:52:29so it contains an amount that represents
2:52:31the audio signal of your uh video or uh
2:52:35your uh
2:52:37show and then the b in this case for
2:52:40instance in this example you can see
2:52:42that the B is then uh equal to for
2:52:45instance 1.2 1 / 2 or B is equal to Min
2:52:49- 1 / 2 so you can see that b is equal
2:52:53to 1 / 2 which basically is a fensive of
2:52:56saying that b is equal to 0.5 or B can
2:53:01be equal to minus1 / 2 which is minus
2:53:060.5 now then it says then the B * a
2:53:10which basically means multiplying our um
2:53:15scalar
2:53:16beta by the Vector containing the audio
2:53:20signal a so this B * a is perceived as
2:53:24the same audio signal but at the lower
2:53:27volume now why lower because you can see
2:53:30that b is equal to 0.5 or minus 0.5 it
2:53:34means that once you take all these
2:53:37elements of your a and you multiply it
2:53:40with a number that is smaller than one
2:53:42in this case 0.5 then all these numbers
2:53:45will decrease which means that also your
2:53:48audio volume will
2:53:51decrease so let me actually uh show you
2:53:55an
2:53:56example so let's say our talk show is
2:54:00very short and you know the audio
2:54:02variation is very low you have a vector
2:54:06a that is quite small it comes from a
2:54:10three dimensional space so R Tre and it
2:54:13has numbers like three uh six and then
2:54:16five so 3x1 vector and then we have our
2:54:22audio adjustment scalar beta which is
2:54:25equal to
2:54:260.5 now when we take the beta we're
2:54:29multiply it by our audio
2:54:31signal then what we
2:54:34do
2:54:36times is clear so times what we are
2:54:40doing is that we are simply taking all
2:54:44the elements of our a so Three 6 and
2:54:47five and what we are doing is that we
2:54:50are multiplying it by
2:54:520.5 0.5 and 0.5 or you can also say 1 /
2:54:582 so what this is equal is that 3 * 0.5
2:55:02is
2:55:031.5 6 * 0.5 is 3 and then 5 * 0.5 is
2:55:102.5 and you can see that all this
2:55:12numbers 1.5 3 and 2.5 they are smaller
2:55:17and specifically two times times less
2:55:19than all the original values in the um
2:55:22original audio so original audio is a
2:55:27which was 3 6 and 5 and the new audio
2:55:32the the scaled one is so audio scaled so
2:55:38B * a is equal to
2:55:411.5 3 and
2:55:432.5 so you can clearly see this
2:55:46transformation where this element three
2:55:49is larger than 1.5 6 is larger than
2:55:52three and then the last element five is
2:55:54larger than 2.5 which means that this
2:55:57audio
2:55:59audio is much at a higher volume so the
2:56:05volume two times
2:56:08higher than this
2:56:16audio so this is basically the idea of
2:56:20uh applying scalar multiplication to our
2:56:23audio pre-processing I will leave the
2:56:25other example to you that will show that
2:56:27when your scaler is equal to minus 0.5
2:56:31you again will end up with the lower
2:56:34volume only that time the volume will be
2:56:36much much lower than the original one so
2:56:39now that we know how we can perform
2:56:41scale multiplication in theory as well
2:56:44as we have looked into an example how we
2:56:46can do it in terms of the numbers and
2:56:47multiply apping them and we have also
2:56:50seen uh applying SK multiplication in
2:56:53practice uh so we have seen in this
2:56:56audio processing stage the uh
2:56:59multiplication process we are ready to
2:57:02look into the visualization of it this
2:57:04will help us to get a better
2:57:07understanding on uh what exactly happens
2:57:10when we are scaling different vectors
2:57:13let's look actually in the following
2:57:14example so let's assume we have a vector
2:57:18oh let me remove
2:57:22that so let's usum we have a vector and
2:57:27the vector is let me get a color this
2:57:31one for instance a vector a and this
2:57:35Vector a consists of elements one and
2:57:38two so where does this Vector Li the
2:57:43vector is with um one so here in our
2:57:47coord system this is our xaxis this is
2:57:50our y AIS and here we got uh let me
2:57:53actually pick another color let's say
2:57:56black
2:57:59one and then we got one and then two
2:58:04right this is two this is one so it is
2:58:06this one so the line that we get here it
2:58:10is this one so this is our Vector
2:58:14a now let's assume I want to multiply
2:58:18my Vector a so I want to scale my Vector
2:58:21a by a constant Tree by scalar tree so I
2:58:25have a scaler let's say I call K and
2:58:28this
2:58:29k a different number let's say k is
2:58:33equal to
2:58:35three so what I wanted to do is to
2:58:38perform a scale of multiplication so I
2:58:40want to obtain K multiplied by a and we
2:58:45learned that this is simply equal to
2:58:49three
2:58:51times and
2:58:53then one
2:58:55two and then this is equal
2:59:01to 3 * 1 3 * 2 which is equal to 3 and
2:59:08then
2:59:09six so let's also visualize this scaled
2:59:13uh
2:59:14Vector so let me pick this yellow color
2:59:18this will be our scaled Vector so we
2:59:21have done scale multiplication and we
2:59:23are going to visualize that so we have
2:59:26three and six so this is three 1 2 three
2:59:30and this is
2:59:32six so we have this
2:59:34point so you should already see what is
2:59:37going on
2:59:38here
2:59:40so we
2:59:43got 3A
2:59:47here so you can see that this part is
2:59:50our Vector a and this longer one is 3 a
2:59:54and even visually you can see that this
2:59:57longer Vector is simply the three times
3:00:00of the shorter Vector so we got this and
3:00:05then if you add on the top of this the
3:00:07same three times you will then end up
3:00:11with
3:00:13the
3:00:15original so scaled version of that
3:00:22so basically this is a this is a this is
3:00:25a we combine three different so we scale
3:00:28a three times and we simply get a three
3:00:32times longer version with the same
3:00:35direction so you can see that when we
3:00:37are scaling even visually it makes sense
3:00:40so we are scaling our Vector a three
3:00:43times and we are just getting that
3:00:45Vector so we are transforming
3:00:48oh let me remove
3:00:55this so
3:00:58basically we are taking this vector and
3:01:02we are scaling it up to this
3:01:06point if I would do it only two times
3:01:10then it would be something like
3:01:14this or one and a half times it would be
3:01:18something like this so only half of
3:01:24it so now this should make much more
3:01:27sense let us actually do yet another
3:01:29example to uh make sure that we are
3:01:32clear on this visualizations because we
3:01:34are going to make use of it when uh
3:01:36looking into this idea of linear
3:01:38combination in a span so let's say we
3:01:42have a vector B and this Vector B has
3:01:46elements zero and three
3:01:48so let's visualize and uh plot this
3:01:51Vector so it contains elements Z zero
3:01:55and three so zero and three so this is
3:01:59the X element and the Y element on the Y
3:02:02AIS we can see this is three which means
3:02:04that our Vector B is this
3:02:08Vector all right perfect so this is our
3:02:11B let's now multiply so scale our Vector
3:02:15B by scaler t Q so let's say we want to
3:02:20get 2
3:02:22*
3:02:24B so what is this amount this is equal
3:02:28to 2 * 2 times and I'm simply taking
3:02:31each of those elements zero and then
3:02:34three
3:02:37so this is then equal to 2 * 0 is 0 and
3:02:40then 2 * 3 is equal to
3:02:436 so this is my new scaled Vector 2 *
3:02:47time B Vector this one so let's
3:02:50visualize this the xais value is zero so
3:02:54we are still here and then the y- axis
3:02:57value is six so what is sixth this thing
3:03:02all right so you already should see that
3:03:05this is very similar what we had before
3:03:08so this is 2
3:03:11B all right so this all uh should make
3:03:15sense uh also we learned as part of the
3:03:18um High School when visualizing
3:03:20different plots so this is quite similar
3:03:22to this idea of having Y is equal to X
3:03:25and then scaling it getting like Y is
3:03:27equal to 2x so in this case only we know
3:03:31exactly where the vector starts and ends
3:03:34uh so we have a much more specific
3:03:38definition instead of having all this
3:03:40infinite number of points on the
3:03:42line but the idea stays the same so we
3:03:45are taking this vector and we are then
3:03:48scaling it two times so we get 2 B
3:03:51vector and I could do the same only
3:03:55instead what I could also do is I could
3:03:57do like uh 0.5 or 1 / 2 * B so I take
3:04:02the half of it which means I would get
3:04:04this
3:04:06vector or I could multiply it with minus
3:04:09one so
3:04:12minus - 1 * B so I was scale with minus1
3:04:17and and then I will simply get the
3:04:19negative
3:04:21version of
3:04:23my original Vector so this thing this
3:04:27would be minus b or min-1 * B so this is
3:04:32basically the idea of uh scaling
3:04:35multiplication when visualizing it in
3:04:37our coordinate system cartisian
3:04:40coordinate system and now when we know
3:04:42all this we are ready to move on on this
3:04:45idea of linear combin
3:04:48and now when we know all this we are
3:04:50ready to move on on this idea of linear
3:04:54combination so let's now formally Define
3:04:56this ideal linear combinations a linear
3:04:59combination of vectors A1 up to a m
3:05:02using scalers B1 up to BM or what we
3:05:05also refer as beta 1 to Beta m is the
3:05:09vector beta 1 * A1 Plus up to Beta M * a
3:05:14m and the scalers are called the
3:05:17coefficient
3:05:18of linear
3:05:19combination and any Vector B in N
3:05:23Dimensions can be expressed as a linear
3:05:26combination of the standard unit vectors
3:05:28E1 up to n the coefficients in this
3:05:31combination are then the entries of B
3:05:34itself well this is whole bunch of
3:05:37information uh let's unpack them one by
3:05:40one firstly um I want to mention about
3:05:43this m so far we have seen this idea of
3:05:45n so I just wanted P to experiment with
3:05:49a different one just to ensure that we
3:05:51are clear that you can use any source of
3:05:54identifier to describe the size of your
3:05:58um uh number of vectors that you got and
3:06:02uh in this case we got M different
3:06:05vectors because so far we were using
3:06:07this n in order to describe the size of
3:06:10a vector and now we are no longer
3:06:12talking about the size of a vector but
3:06:14the number of vectors therefore I
3:06:16specifically didn't use use the letter N
3:06:18so here m is simply the number of
3:06:24vectors so don't confuse this with this
3:06:27thing where we were plotting this and we
3:06:29were saying this A1 A2 up to a n because
3:06:33in here we basically mean that we are
3:06:35dealing with some Vector a and this has
3:06:39n different elements whereas in here we
3:06:42are already moving from this idea of one
3:06:45vector and now we are talking about mve
3:06:47multiple vectors so we have M different
3:06:50vectors they all look like kind of this
3:06:53only with bit more complex indexing that
3:06:56we also saw
3:06:58before all right but we will learn this
3:07:01um that's not an issue I just wanted to
3:07:04mention this to ensure we are at the
3:07:06same page so then let's move on to this
3:07:09idea of using scalers beta 1 till beta M
3:07:14so it's a common uh practice in linear
3:07:17algebra in just in general in
3:07:19mathematics but also definitely in data
3:07:22science statistics and in artificial
3:07:23intelligence to use beta 1 as a way to
3:07:27describe the coefficient so what do you
3:07:30mean by coefficient it is just a scalar
3:07:32so it's just a constant or a number so
3:07:35in this case for instance this beta 1
3:07:38can be 0.5 beta 1 can be uh two bet one
3:07:43can be let's say 100 it just describes
3:07:46how much we are multiplying scaling this
3:07:49Vector A1 so so far we have done a lot
3:07:53of scal and multiplication already lot
3:07:56of details there and we have seen
3:07:58different times different scalers that
3:07:59we use we can use um zero as a scaler we
3:08:02can use any other number as long as it's
3:08:05a real number so this beta 1 should
3:08:09belong uh in the a real number space so
3:08:13it's a real number and of course the
3:08:15same holds for uh all the betas so we
3:08:19have M different vectors which means we
3:08:20are going to have M different scalers
3:08:23because each of those vectors we are
3:08:25going to multiply with their
3:08:27corresponding or respective scalers so
3:08:31beta one is basically the scaler uh or
3:08:34the um uh coefficient that we are using
3:08:38to
3:08:40scale a one maybe I can actually write
3:08:44this down on a new page such that we can
3:08:46save this as a SL Light page for you
3:08:49let's write it down so what do we have
3:08:51as this idea of linear combination so a
3:08:54linear combination simply involves
3:08:57taking several vectors uh to go from
3:09:00this uh formal definition to more
3:09:02practical uh terms so we got this
3:09:06A1 A2 up to
3:09:10a and what we want to do is to take the
3:09:13linear combination of this m different
3:09:16vectors so we got m is the number of
3:09:21vectors and to get a linear combination
3:09:25we need to uh scale each of those
3:09:28vectors which means that we need to have
3:09:32this different scalers let's say beta
3:09:361 for
3:09:38A1 and then plus beta 2 for A2 so each
3:09:43time we are scaling each of those
3:09:45vectors where beta 1 is the uh scaler or
3:09:49the coefficient of the vector A1 and we
3:09:53are multiplying we are performing scalar
3:09:56multiplication of our scalar beta 1 with
3:09:59the vector A1 and then we are adding to
3:10:02this our beta 2 which is the coefficient
3:10:06corresponding to the vector A2 and then
3:10:08adding beta 3 * A3 and then dot dot dot
3:10:12so all these different uh vectors up to
3:10:15the point of beta
3:10:17M time a
3:10:21m and all this so A1 A2 up to a those
3:10:27are all vectors belonging to the
3:10:31m space so those are all vectors coming
3:10:35from the um M dimensional
3:10:39space so um in
3:10:43here this is the linear combination of
3:10:47our
3:10:47M different vectors and the uh beta
3:10:531 beta 2 up to Beta M those are all
3:11:01constants so those are scalers or real
3:11:05numbers that belong to R so those are
3:11:10real
3:11:12numbers all right so now when we are
3:11:14clear on that let's also unpack this
3:11:16idea of coefficients so the scalers are
3:11:19called the coefficients of linear
3:11:20combination so basically all this
3:11:24members so beta 1 beta 2 of two beta M
3:11:31that belong to real number
3:11:34space they are
3:11:38called
3:11:42coefficients this is what we are
3:11:43referring as coefficients and this
3:11:45coefficients this IDE and name is super
3:11:48important because you will see this time
3:11:50and time again appearing in your uh very
3:11:52basic machine learning models or some
3:11:55other applications of linear algebra
3:11:57because the end goal is always to find
3:11:59these coefficients so these coefficients
3:12:02those are numbers that we are using to
3:12:05scale these different vectors and uh the
3:12:08idea of coefficients is very Central
3:12:11because those are numbers that Define
3:12:14how exactly we are combin ining all
3:12:18these different vectors because this
3:12:20beta 1 beta 2 Beta 3 they can be
3:12:22different numbers real numbers and every
3:12:25time when we are choosing these
3:12:27coefficients or these betas we will then
3:12:30end up with a different combination of
3:12:33these vectors so we are basically mixing
3:12:37all these different vectors and the way
3:12:39we mix it and how we will mix it it will
3:12:42depend on the values of this beta 1 beta
3:12:452 Beta 3 up to Beta m so these
3:12:48coefficients so therefore coefficients
3:12:50are super important and they Define the
3:12:53end results from our linear
3:12:56combination so any Vector B in N
3:13:00Dimensions can be expressed as a linear
3:13:03combination of standard unit vectors E1
3:13:06up to n so when looking into this um
3:13:09idea of unit vectors uh we saw already
3:13:14what this E1 is what is E2 is up to e n
3:13:19and we saw that E1 is for instance if
3:13:22it's from an N dimensional
3:13:24space and it says from n
3:13:27dimensions then E1 simply means 1 0 0
3:13:32dot dot dot dot Z then E2 means 0 1 Z
3:13:37dot dot dot dot zero so we already saw
3:13:39this this is not something new that we
3:13:41are seeing so 0 Z blah blah blah and
3:13:44then one at the end and what this
3:13:47definition basically says is that any
3:13:51Vector b as long as the B comes from n
3:13:55dimensional space we can represent this
3:13:59by using this uh unit vectors and by
3:14:04linearly combining
3:14:05them so this is yet another part of this
3:14:08definition and we are going to by the
3:14:11way um go through each of the parts of
3:14:13this definition one by one going to each
3:14:16of the examp examples as well as
3:14:18visualizing them so now I just want to
3:14:20quickly unpack all the parts in this
3:14:23definition before moving on to step by
3:14:25step examples and
3:14:27explanation so this is about this linear
3:14:30combination of any n dimensional uh
3:14:33Vector B that we can uh create by using
3:14:37a linear combination of these unit
3:14:39vectors I will come to this in a bit so
3:14:42then the final part of this definition
3:14:45is that the coefficient in this
3:14:48combination are the entries of B
3:14:51itself so it says that the coefficients
3:14:56so beta 1 up to Beta m in this linear
3:14:59combination that we can create are the
3:15:02entries of B
3:15:05itself so we will come to this section
3:15:08once we are done with the first part so
3:15:11first let's have a good understanding of
3:15:13what this linear combination is and also
3:15:17touch base and we will also formally
3:15:19Define the idea of span and after that
3:15:22we will move on on uh representing and
3:15:26expressing any Vector B in N Dimension
3:15:29Space by using standard unit vectors E1
3:15:32up to e n and this idea of coefficients
3:15:35and then entries of B so let's start
3:15:38with the first one so let's assume we
3:15:41have two different
3:15:43vectors we have Vector a
3:15:47and this Vector a is equal to one
3:15:512 so let's plot
3:15:54this one and two in our coordinate space
3:15:58that is this one which means that our
3:16:00Vector a is this
3:16:03one and let's assume that we have a
3:16:07vector
3:16:10B and this Vector
3:16:13B is equal to Z
3:16:1803 so 0 is here and then three is here
3:16:22which means that our Vector B is this
3:16:25one this is Vector
3:16:27B now I want to create a linear
3:16:31combination of this Vector a and Vector
3:16:35B so we just learned from the formal
3:16:38definition that in order to do so I need
3:16:41a beta
3:16:431 to
3:16:45multiply the vector
3:16:58a and then I need beta 2 which is the
3:17:01coefficient corresponding to to my
3:17:03second Vector in order to multiply the
3:17:06second Vector which is
3:17:08B Vector B okay so I'm getting the
3:17:11linear combination of A and B by taking
3:17:15any beta 1 and beta 2 which are real
3:17:18numbers so beta 1 and beta 2 belong to R
3:17:24so they are real numbers and then I'm
3:17:26getting a linear combination of the two
3:17:29so let's look into a few examples of a
3:17:31linear combination of vector A and B
3:17:35depending on the different choice of the
3:17:37coefficients like beta 1 and beta 2 so
3:17:41example one is that beta 1 is equal to
3:17:45zero and then beta beta 2 is equal to Z
3:17:49Now what is the linear combination of A
3:17:52and B when my coefficients beta 1 and
3:17:54beta 2 both are zero it just means that
3:17:57I'm
3:17:59getting 0
3:18:05time
3:18:08A
3:18:12Plus 0 times
3:18:19B which is of
3:18:27course 0 * 1 0 *
3:18:332
3:18:35plus and then multiplying Vector B with
3:18:38a scaler zero which is 0 * 0 0 * 3 so
3:18:43let's quickly do this what this value is
3:18:46this is equal to 0 * 1 is 0 0 * 2 is
3:18:520 0 * 0 is equal to 0 0 * 3 is equal to
3:18:570 and this is then equal to 0 + 0 0 0 +
3:19:030 is0 so I'm basically getting a vector
3:19:08zero all right so this is then equal to
3:19:14zero so this equal to vector is
3:19:20zero so I can also say that this vector
3:19:26or it's actually a point so this
3:19:30point is simply a linear combination of
3:19:34these two
3:19:35vectors now this is a super basic case
3:19:38let's look at another case when our in
3:19:41our second example the beta 1 and beta 2
3:19:43so our coefficients they are actually
3:19:46not zero there are some other nonzero
3:19:49real
3:19:58numbers so in this example I will then
3:20:02take beta 1 = to 3 and then beta 2 is =
3:20:06to
3:20:092 and then what I will do is that I will
3:20:13take actually I will take the um minus
3:20:16two
3:20:18then I can also get rid of one of the
3:20:20elements and I can get actually a zero
3:20:22for one of the elements I will show you
3:20:23in a bit so then the linear combination
3:20:28of A and B using these coefficients beta
3:20:311 and beta 2 where beta 1 isal to 3 and
3:20:33beta 2 is equal to minus 2 is then equal
3:20:37to so this
3:20:42amount this
3:20:45amount is equal =
3:20:482 3
3:20:51* 1
3:20:542 and then
3:20:57plus we got - 2
3:21:04* 0 and three now what does this give
3:21:10us 3 * 1 is = 3 3 * 2 is = 6 plus and
3:21:17then - 2 * 0 is = to
3:21:200 and then - 2 * 3 is = - 6 so you might
3:21:25have already noticed why I picked the
3:21:27beta 2 equal to minus 2 I wanted these
3:21:29two numbers to actually cancel each
3:21:32other so you see because 6 + - 6 is
3:21:36equal to Z so what do I get in my final
3:21:38result as a linear combination of these
3:21:40two vectors I get 3 + 0 so 3 + 0
3:21:51so 3 + 0 and then 6 + -
3:21:576 and this gives me 3 + 0 is 3 6 + - 6
3:22:02is zero there we go so this is my linear
3:22:06combination of the vector A and B when
3:22:10using the coefficients equal to 3 and
3:22:12minus 2 respectively so this value is
3:22:16actually = to three and zero in this
3:22:20case all right so let me actually clean
3:22:23this up because I also want to visualize
3:22:26this idea and then we will go uh back to
3:22:29this uh linear combination let just
3:22:32summarize uh what we got before moving
3:22:35on to the plotting part so if we simply
3:22:37take a and we add to this B so this is
3:22:41the first case so this is as you might
3:22:44have already guessed this is also l your
3:22:46combination here we are saying take 1 *
3:22:49a and take a 1 * B and this is yet in
3:22:54our linear combination here the beta 1
3:22:57is equal to 1 and then beta 2 is equal
3:23:00to 1 so this linear combination gives us
3:23:04a vector that is 1 + 0 is = to 1 and
3:23:08then 2 + 3 is = 5 this is our first
3:23:13linear combination when the beta 1 and
3:23:15beta 2 is equal to one this is a basic
3:23:18case so doesn't require too much
3:23:20explanation here we have seen already
3:23:22this let's now look into the other
3:23:25example that we saw when we use uh the
3:23:28zeros as our coefficient so that is 0 *
3:23:33A+ 0 * B then this gave
3:23:38us 0 0 this was our second linear
3:23:43combination when beta 1 and beta 2 were
3:23:46both equal to
3:23:48zero and then the third linear
3:23:50combination that we saw was that 3 *
3:23:55A+ - 2 * B this gave
3:24:01us 3 and zero this was our third linear
3:24:05combination where beta 1 was three and
3:24:07then beta 2 was minus
3:24:102 so so then the linear combination of
3:24:14these two vectors is basically
3:24:17all the possible combinations of these
3:24:19two vectors that I can get when scaling
3:24:23or when multiplying these two different
3:24:26vectors by different sorts of uh vector
3:24:29by different sorts of scalars so in all
3:24:33these different cases what I'm simply
3:24:34doing is I'm taking different sorts of
3:24:36coefficients beta 1 and beta 2 and then
3:24:40I'm getting the linear combination of
3:24:42these two vectors we saw that in the
3:24:44simple case when we take a and we had to
3:24:47B so basically the coefficients are
3:24:48equal to 1 so 1 * a + 1 * B then the
3:24:52corresponding linear combination is
3:24:54equal to one and five it means that we
3:24:57are getting this vectors so one and five
3:24:59is in here which means that we are
3:25:01getting this one this Vector if we get
3:25:04if we take the zero as a scal so beta 1
3:25:06and beta 2 are both equal to zero then
3:25:09the linear combination of these two
3:25:10vectors is simply the vector zero which
3:25:13means that it is this point then if if
3:25:16we take the linear combination using
3:25:18three and minus 2 as coefficients then
3:25:20we are getting this so 0o and three so
3:25:24one two and three this is three then
3:25:26this is our linear
3:25:28combination I can also take any other uh
3:25:33like scaled version of my B and of my a
3:25:37and then I will get entirely different
3:25:39sort of vector so let me actually show
3:25:43you a few more times um a couple of
3:25:45other examples so let's say I keep my a
3:25:48so I just take the beta 1 equal to 1 but
3:25:51instead I scale my Vector B two times so
3:25:55this was at three I'm taking two times
3:25:58of my Beta which means that I'm
3:26:01here then I can take this I can add this
3:26:05to my a so this is 2
3:26:07B this will give me another linear
3:26:11combination of these two different
3:26:15vectors I can also you might recall that
3:26:18we said that the starting point and the
3:26:20end point doesn't really matter for for
3:26:22us what matters is that we uh have the
3:26:26same magnitude and the same direction
3:26:28for our vectors so this means that for
3:26:32me the vector being here and the vector
3:26:35being here doesn't matter when I scale
3:26:37it with two I can be here with three I
3:26:40can be here so this is the same as my B
3:26:43only 3 * B this is basically basically
3:26:46scaling B with three and this in here
3:26:51means that my Beta 2 is simply equal to
3:26:54tree and then this means that I can
3:26:57combine this with my a which was in here
3:27:02you remember so this here this means
3:27:06that I get yet another linear
3:27:08combination of these vectors which means
3:27:11that I'm
3:27:12taking
3:27:143B and I'm on this my a so 1 * a my beta
3:27:211 is equal to 1 my Beta 2 is equal to 3
3:27:24which means that the linear combination
3:27:26of this is equal to one two plus and
3:27:29then 3 * B is equal to 0 and then 3 * 3
3:27:32is 9 this is then the new linear
3:27:35combination which is 1 and 11 so the new
3:27:38linear combination is equal to 1 and
3:27:4211 so this thing
3:27:48which is the same
3:27:50as this
3:27:53thing and then you can go on and on you
3:27:57can also calculate the same with a
3:27:59negative B so you can take B and then
3:28:01you can scale it with minus one so this
3:28:04is minus b or you can go in here in here
3:28:09the same holds for a so you can scale it
3:28:11all the way to here or in the negative
3:28:14side so so this already uh give us the
3:28:19idea that we will go into to the next
3:28:21point which is the span so when it comes
3:28:24to the linear combination and in this
3:28:26specific case when we have these two
3:28:28vectors we can combine these two vectors
3:28:31in anyway and uh we can mix them up by
3:28:35using different sorts of coefficients of
3:28:37beta 1 and beta 2 and we will can we can
3:28:40get any Vector in our R2 so this means
3:28:45that any vector in our R2 we can
3:28:47represent by using only these two
3:28:50vectors and this is not always the case
3:28:53for this specific case we are dealing
3:28:55with two vectors that we can use to
3:28:57represent any Vector in our r
3:29:03two so what I mean here is that let me
3:29:07clean this
3:29:09up so independent what kind of vector
3:29:13you will give me in the R2 so it has two
3:29:17different elements it is 2x one I can
3:29:21use a linear combination of A and B so a
3:29:25linear combination of A and B is beta 1
3:29:28* a
3:29:30plus beta 2 * B in order to
3:29:35represent this
3:29:37Vector X1 and
3:29:40X2 therefore we are saying and we will
3:29:43come to this um in the next slide too
3:29:46that the
3:29:48spend of the vectors A and B so this is
3:29:52the set of all possible combinations of
3:29:55these two vectors is equal to R2 because
3:29:59any Vector in R2 can be represented as a
3:30:04linear combination of these two vectors
3:30:08so we have a linear combination of A and
3:30:10B and I'm saying that I can represent
3:30:13any Vector so here vector X and this
3:30:17Vector X I'm representing by X1 and X2
3:30:20and X1 and X2 can be any real numbers so
3:30:24X1 and X2 they belong to R and X is
3:30:30simply a two-dimensional Vector so X1
3:30:34and X2 those can be any numbers 0 1 2us
3:30:38100 anything and I'm saying any number
3:30:42in this two dimensional space so whether
3:30:45it is this one any Vector this one this
3:30:47one or this vector or this one any
3:30:51Vector that you give me in two
3:30:52dimensional space I can find a linear
3:30:56combination of this A and B that is
3:30:59equal to that Vector so I can represent
3:31:01that Vector as a linear combination of
3:31:03vector A and B that we saw before so
3:31:06let's actually prove that so I'm going
3:31:09to represent this uh X1 and X2 by a
3:31:13linear combination of this Vector A and
3:31:15B and how we can do that so we have beta
3:31:171 * a plus beta 2 * B it is equal to X1
3:31:21and X2 where beta 1 and beta 2 so beta 1
3:31:24and beta 2 they are constants so they
3:31:27are also real
3:31:30numbers so let's unpack this which is
3:31:34beta 1 * 1 2 + beta 2 *
3:31:4003 and this should be equal to X1 and X2
3:31:46now
3:31:48this is
3:31:51equivalent of so beta 1 * 1 beta 1 * 2
3:31:58plus beta 2 * 0 and then beta 2 * 3 and
3:32:04this should be equal to X1
3:32:08X2 so this is my beta 1 a this is my
3:32:12Beta 2 B and this is my
3:32:21X all right so now what we get is that
3:32:25and this is
3:32:31equivalent beta 1 *
3:32:341 is equal to beta 1 beta 1 * 2 is 2
3:32:39beta
3:32:401
3:32:41plus then here beta 2 * 0 is 0 and then
3:32:45beta 2 * 3 is 3 beta
3:32:562 so we have learned U from the uh
3:32:59operations on the vectors that beta one
3:33:02uh so in this case when we are adding
3:33:04two vectors so beta 1 +0 is the uh
3:33:08amount that we need to put as our first
3:33:10element so when we are adding two
3:33:11vectors we just need to take their
3:33:12corresponding elements we need to add
3:33:14them up so equal to beta 1 + 0 and then
3:33:182 beta 1 + 3 beta 2 this is the result
3:33:23and this should be equal to X1 and X2 at
3:33:26least this is what I'm
3:33:34claiming so this zero doesn't matter so
3:33:37we what we are getting from here is that
3:33:39beta 1 is = to X1 and then 2 beta 1 + 3
3:33:46b. 2 is equal to X2 this is the two
3:33:49expressions that we are getting based on
3:33:51all these different calculations so let
3:33:53me actually remove all
3:34:05this so we have beta 1 is equal to X1
3:34:10and 2 beta 1 + 3 beta 2 is equal to
3:34:18X2 so here given that we have already
3:34:21that beta 1 is equal to X1 and here we
3:34:23have two unknowns I'm going to fill in
3:34:27the value for beta 1 in here so I'm
3:34:30going to take
3:34:33this and I'm going to fill in it in here
3:34:37so for this
3:34:40volue so remember that beta 1 and beta 2
3:34:43are two unknowns and each one and next
3:34:45to are just uh numbers that we will get
3:34:49when we uh know exactly the vector and
3:34:51we just want to represent the vector as
3:34:54a linear combination of two vectors so
3:34:58when I take this uh value for beta 1
3:35:01which is equal to X1 and I'm going to
3:35:03fill that in in here it means that I'm
3:35:06going to get from here that beta 1 is
3:35:10equal to X1 and 2 *
3:35:17X1 because beta 1 is equal to X1 and
3:35:20here I got beta 1 I'm just filling in
3:35:22that value for beta 1 which is equal to
3:35:24X1 so 2 *
3:35:27X1 and then the rest I'm just taking
3:35:29over 3 beta 2 is equal to
3:35:34X2 let me remove this and from here what
3:35:38we are getting is that beta 1 is equal
3:35:41to X1 and I will solve this equation for
3:35:44the unknown which is equal beta 2 so I
3:35:48will just take the three beta 2 from
3:35:51left hand side I will leave it there and
3:35:53I will take this and I will take it over
3:35:54to the right so I'm taking X2 over and
3:35:58this two X1 so this part I'm just taking
3:36:02to the right two of the equation so Min
3:36:05- 2
3:36:09X1 which then on its turn is equal to so
3:36:14it goes to B 1 is = to X1 and then beta
3:36:192 is = to X2 - 2 X1 / 2
3:36:26three perfect so what do we get
3:36:30here what is our end result and why is
3:36:33it
3:36:36significant so what we are getting here
3:36:39is that based on all this information
3:36:42without knowing beta 1 and beta 2 we got
3:36:46that beta 1 should be equal to X1 and
3:36:49beta 2 should be equal to X2 -
3:36:532x1 / to three this means that
3:36:58independent what kind of x's you will
3:37:02give me so what kind of vector we have
3:37:05in our R2 so this X1 and X2 they are
3:37:09just real numbers we can always find
3:37:12beta 1 and beta 2 that we can use to
3:37:16represent that X1 X2 so our X Vector as
3:37:21a linear combination of these two
3:37:24vectors let me actually give you an
3:37:27example so let's remove
3:37:40this so let's assume we have a vector X
3:37:44and this x is is equal to four and let's
3:37:51say
3:37:53three so if we got this Vector X and we
3:37:57are saying we can use this Vector A and
3:37:59B to represent X as a linear combination
3:38:03of vector A and B which means that I can
3:38:05find I can find real number beta 1 and
3:38:09beta 2 that I can use to multiply the
3:38:12vector A and B respectively combine them
3:38:14together so their linear combination
3:38:17that will be equal to this Vector X so
3:38:20this is my
3:38:21X1 this is my
3:38:24X2 so this is equal to 4 and three Now
3:38:30using
3:38:31this let's actually see whether that is
3:38:35true so based on this example my beta 1
3:38:40should be equal to X1 which is four my
3:38:43Beta 2 should be equal to X2 which is 3
3:38:46so beta 2 should be equal to X2 which is
3:38:503 - 2 * X1 which is 4 / to three and
3:38:56what's this number this means that my
3:38:59beta 1 should be equal to 4 and my Beta
3:39:012 should be equal to 3 - 8 so 3 - 8 / to
3:39:063 and this is equal to Minus 5 / to
3:39:123 so this means that I use a
3:39:16coefficients beta 1 is equal to 4 and
3:39:18beta 2 = to - 5 / to 3 to represent my
3:39:22Vector X as a linear combination of
3:39:26vector a and Vector B so let's actually
3:39:30prove that too as a final
3:39:37step
3:39:39so let's see where the four
3:39:42times Vector a which is 1
3:39:462 + - 5 / to 3 whether this is indeed
3:39:52equal to Vector X so my Vector B is
3:39:5803 so this is the first part
3:40:16and I want to prove that this is indeed
3:40:18equal to X and we already know what x is
3:40:24so this is equal to 4 * 1 4 * 2 plus and
3:40:31then here we got - 5 / 3 * 0 and then -
3:40:355 / to 3 *
3:40:373 this is equal
3:40:41to 4 * 1 is = to 4 4 * 2 is = to 8
3:40:46and then here we need to subtract minus
3:40:4953 5 5 / 3 * 0 is equal to 0 so this one
3:40:54is zero and then minus 5 / to 3 so 5/3 *
3:40:593 this ones are canceling out and we got
3:41:028 + - 5 so here the plus and here minus
3:41:06just to make sure we got everything
3:41:08right and this is equal to 4 and then 8
3:41:12+ - 5 is equal to 3
3:41:15so you can see already that this amount
3:41:18that we got here is equal to X which was
3:41:21equal to 4 / to3 and this helps us to uh
3:41:27verify and to know for sure that indeed
3:41:32while given any Vector in a two
3:41:35dimensional space in R2 X independent
3:41:39what this X1 is or X2 is we can always
3:41:42find a pair of beta 1 and beta 2 that
3:41:46will ensure that the beta 1 a plus beta
3:41:502 B is actually equal to this
3:41:54x where X A and B they are part of
3:42:00R2 and a is equal to 1 2 and then B is
3:42:06equal to
3:42:0803 so we can represent any Vector in our
3:42:12two-dimensional space as a linear comp
3:42:15combination of this Vector a with
3:42:16elements 1 2 and um Vector B with
3:42:20elements 03 and that's exactly what we
3:42:22saw here because we could find any
3:42:25vector and we can represent this Vector
3:42:27as a linear combination of this Vector A
3:42:29and B this Vector as a linear
3:42:31combination of this A and B this Vector
3:42:33as a linear combination in any vector or
3:42:36a point in this plane we can represent
3:42:38as a linear combination of this Vector a
3:42:42and Vector B and in this specific case
3:42:45with this Vector a and Vector B we are
3:42:47saying that Vector a and Vector B they
3:42:50spin
3:42:51R2 so Vector
3:42:56A and
3:42:58B
3:43:00span R 2 now we will come to these
3:43:05definitions of the span and uh just in
3:43:08general for different sorts of vectors
3:43:11we will see what this IDE of span is but
3:43:13for now given that we just proved that
3:43:16we can represent any Vector in R2 as a
3:43:19linear combination of these two vectors
3:43:22A and B therefore we can say and we
3:43:25usually say it in linear algebra that
3:43:28the vector a and Vector B they spend R2
3:43:32before moving on onto this concept of
3:43:34Spence that we just touched upon in our
3:43:37example I wanted to quickly go back to
3:43:40this example that I promised to discuss
3:43:43uh which was part of the definition of
3:43:45the linear combinations and unit vectors
3:43:48because we saw in our definition and let
3:43:51me just show you that uh the uh
3:43:53definition was providing these two
3:43:56highlights these two bullet points and
3:43:57was saying any Vector B in N Dimensions
3:44:00can be expressed as a linear combination
3:44:02of the standard unit vectors E1 to up to
3:44:05e n and the coefficients in this
3:44:07combination are the entries of B itself
3:44:11so let's look into the example and see
3:44:13what we mean by that in this specific
3:44:16example we have this Vector B it is
3:44:19coming from the three dimensional space
3:44:21which we can see given that we have
3:44:22three different uh three entries so
3:44:24three uh elements in our Vector so it's
3:44:273 by one and this means that b belongs
3:44:32to
3:44:34R3 and in here we can see that B can be
3:44:38written as a linear combination of these
3:44:42three vectors so you can see that b e is
3:44:45equal to -1 * this Vector 1 0 0 so this
3:44:52one then we have + 3 * 0 1 0 Vector so
3:44:59this
3:45:00one and plus 5 * this third Vector which
3:45:05is 0
3:45:0601 now we already know from the unit
3:45:10vectors that
3:45:13E1 is equal to
3:45:151 0
3:45:170 assuming that we are in three
3:45:19dimensional
3:45:22space E2 is equal to 0 1
3:45:260 and E3 is equal to 0 01 you can notice
3:45:33that that's exactly what we got here
3:45:35this Vector is E1 this Vector is E2 and
3:45:39this Vector is E3 where E1 E2 and E3
3:45:44belong to three-dimensional
3:45:47space okay so another thing that we can
3:45:50see is that here we got coefficients
3:45:54minus one here three and here five so
3:45:57this is basically how beta 1 beta 2 and
3:46:00beta 3 using the common conventions that
3:46:02we saw before when describing the linear
3:46:05combination so let's actually check that
3:46:08and then we will comment on these values
3:46:16so let's check whether -1 * E1 + 3 * E2
3:46:21+ 5 * E3 is indeed equal to this B so
3:46:25this is equal to -1 * this Vector gives
3:46:29us -1 0
3:46:310 three times this E2 gives us 0 3 and
3:46:37zero and then 5 * E3 gives us 0 0 5 and
3:46:44this is equal to
3:46:45-1 + 0 + 0 is = -1 0 + 3 + 0 is = 3 and
3:46:52then 0 + 0 + 5 is equal to
3:46:555 now what do we get here we see that
3:46:58this which is equal to this it is equal
3:47:02to this Vector B indeed okay so now when
3:47:06we have indeed checked that B can be
3:47:10represented as a linear combination of
3:47:14this three vectors this unit vectors E1
3:47:19E2 E3 another thing that we can notice
3:47:22and I'm sure you already
3:47:24did is that those coefficients they are
3:47:27not just randomly picked coefficients
3:47:30those are the entries of this Vector B
3:47:34so this is exactly what that definition
3:47:37was about it was saying that any Vector
3:47:39B including this example in in this case
3:47:42threedimensional space can be Express as
3:47:45a linear combination of the standard
3:47:48unit vectors E1 A2 E3 Etc so this
3:47:53coefficients in this combination so you
3:47:55can see that the beta 1 beta 2 and beta
3:47:583 which are our coefficients in our
3:47:59linear combination they are the entries
3:48:02so this
3:48:04values of the B
3:48:06itself so the same will hold for
3:48:09four-dimensional case five dimensional
3:48:12case n dimensional case so this means
3:48:15that if we write this down for General
3:48:19case just to ensure that we are clear on
3:48:23this part of the
3:48:26definition so if we got
3:48:30B Vector in N dimensional space so it
3:48:34got B1 B2 up to BN as the elements of it
3:48:39comes from
3:48:41RN then we can represent this B
3:48:45as a linear combination of unit vectors
3:48:49coming from the N dimensional space so
3:48:52we got E1 E2 up to e n that belong to n
3:48:58dimensional space and we can represent
3:49:00this B as a linear combination of these
3:49:03unit vectors so by using beta 1
3:49:08time so this this is a common Convention
3:49:10of the coefficient as you Rec called
3:49:12time C1 then B beta 2 * E2 blah blah
3:49:17blah plus beta n * e n and what is
3:49:22important here is that this beta 1 beta
3:49:252 and beta
3:49:28n those are not just some coefficients
3:49:31but we already know what these
3:49:32coefficients are
3:49:35because we can then represent this
3:49:40beta by taking the values so those are
3:49:44the entries the elements of the vector B
3:49:47itself so it is B1
3:49:52*
3:49:55B1
3:49:57plus b2 time E2 dot dot dot
3:50:04plus BN
3:50:07times e
3:50:10n where B1 B2 up Q BN they are all real
3:50:18numbers so basically knowing what these
3:50:22Vector is these elements of this
3:50:25Vector we can always describe and
3:50:28express it as a linear combination of
3:50:31the standard unit vectors and if you're
3:50:33wondering why is this important in some
3:50:36cases when performing different
3:50:38operations or working on different
3:50:40algorithms it just becomes handy to
3:50:43represent your vector as a linear
3:50:45combination of multiple
3:50:47vectors and in those cases exactly you
3:50:50can make use of this property of linear
3:50:52combinations to express your n
3:50:54dimensional Vector b as a linear
3:50:56combination of the standard unit vectors
3:50:59because everything is then down to you
3:51:01by having this Vector B you will already
3:51:03know what are the entries that you can
3:51:05use as your coefficients in this case
3:51:08beta 1 beta 2 so those are all these
3:51:10values coming from your vector itself
3:51:12and then the remaining is also none
3:51:14because you know exactly what these unit
3:51:16vectors are and how they are represented
3:51:19so here for instance the E1 is basically
3:51:22one 0 0 blah blah blah blah 0 and then
3:51:25this is n by1 Vector here the E2 is
3:51:28equal to 0 1 Z blah blah blah and then
3:51:31zero here so n * 1 again up to the point
3:51:35where you have the N where you have all
3:51:38the zeros only the last element is one
3:51:41again n by one vector so this is the
3:51:44idea behind this second part of this
3:51:46definition which says that any Vector
3:51:49being in N dimensional space can be
3:51:51expressed as a linear combination of the
3:51:53standard unit vectors E1 up to e n let's
3:51:58now talk about other concept which is
3:52:00also super important which is the span
3:52:03of vectors so by definition the span of
3:52:06a set of vectors is a set of all
3:52:08possible linear combinations of these
3:52:10vectors so if V is equal to V1 V2 up to
3:52:14V K and is a set of vectors then the
3:52:17span of V is written as a span V and it
3:52:21includes any vectors that can be
3:52:23expressed as C1 V1 up to C2 V2 up to CK
3:52:29VK so basically it is a common uh
3:52:34notation uh to say that if we got for
3:52:37instance vectors V1 V2 up to VN so we
3:52:43have n different vectors then we say
3:52:46that the
3:52:49span
3:52:51of
3:52:53V1 V2 up to
3:52:58VN that this is simply the notation that
3:53:02we use in order to describe the span of
3:53:04these vectors and we briefly spoke about
3:53:07this concept of span when we were
3:53:10looking into our example that we saw
3:53:12before so you might recall vectors A and
3:53:15B that we had and we saw and we said
3:53:19that the span of a and b is the entire
3:53:23space in the two dimensional uh real
3:53:25number space so we said that span of a
3:53:28and b is equal to R2 where our Vector a
3:53:33was simply equal to one 2 and B was
3:53:39equal to
3:53:4103 so we proved that the span
3:53:45of one two and
3:53:4903 was the
3:53:52entire R2 and how we knew that because
3:53:56we proved that any Vector in R2 could be
3:54:00represented as a linear combination of
3:54:03these uh two vectors so you might recall
3:54:05that we solved this equations we saw
3:54:08that in depend what kind of X1 and X2 uh
3:54:11one will give us we can always use the
3:54:14uh
3:54:15um we we found this amount let me see
3:54:18where I can find it back I no longer
3:54:20have this so we saw that for uh specific
3:54:23values of um beta 1 and beta 2 we can
3:54:28always get a linear combination of this
3:54:31A and B in order to get our desired
3:54:34factor
3:54:35x so beta 1 * a plus beta 2 * e will
3:54:39always then be equal to X1 and X2 if our
3:54:42Vector a and Vector B are those but of
3:54:45course this doesn't hold for all the
3:54:47vectors so not for all two-dimensional A
3:54:51and B uh we can say that the span of
3:54:54these vectors is the entire R2 therefore
3:54:57to better understand this concept of
3:54:59span and this concept of span of vectors
3:55:01I wanted to distinguish five different
3:55:04cases one of which we already spoke
3:55:06about and that is the case when we had
3:55:08this Vector a and Vector B and we said
3:55:10that the span of a and b is the entire
3:55:13R2 but we will also look into the case
3:55:16when we for instance have a span of the
3:55:18zero Vector the span of a single vector
3:55:21and the span of perpendicular vectors we
3:55:24might also look if there is time left we
3:55:26will also look into the span of parallel
3:55:29vectors so let's now look into this
3:55:32cases one by
3:55:36one so let's say we have a vector of
3:55:42zero so we have a zero vector so this is
3:55:45a very simple case we will start with
3:55:46the simplest case and we will move on B
3:55:48to two more advanced cases if we have a
3:55:51vector a that is a zero
3:55:58Vector 0
3:56:000 then
3:56:02independent what kind of scaler we will
3:56:05use to scale this so let's say um we
3:56:08Define it by C so C * Z independent what
3:56:13kind of scale we will use this will
3:56:16always end up being equal to 0 0 so if C
3:56:19is equal
3:56:20to0 C * 0 will be equal to
3:56:250 if C is equal to 1 C * 0 will be equal
3:56:30to
3:56:32Z or C is equal to 100 C * 0 will still
3:56:38be 0 0 so independent what kind of scal
3:56:42we will be using what kind of lead
3:56:44linear combination we will create from
3:56:47our Vector
3:56:50a this will always stay in here so the
3:56:54point the vector will always stay in
3:56:57here in our two Dimension space so this
3:57:00is completely different from what we saw
3:57:02before when we could create and we could
3:57:04take any Vector in our R2 and we could
3:57:06represent it as a linear combination of
3:57:08these two vectors that we saw in the
3:57:10previous example so in this specific
3:57:13case
3:57:14um scaling the zero with independent of
3:57:19any scalers we use this will not change
3:57:21the magnitude nor it will change the
3:57:23direction of our Vector so no matter how
3:57:26we scale it we still get zero this means
3:57:30that
3:57:31the
3:57:33span of zero
3:57:40Vector is just the zero Vector itself
3:57:45so you can see that independent what I
3:57:47scale the zero Vector I always end up
3:57:49with the same zero Vector so therefore
3:57:52this span of the zero Vector is equal to
3:57:54zero because by definition this Spen of
3:57:57set of vectors is the collection of all
3:58:00possible vectors that I can reach by
3:58:02performing linear combination and in
3:58:04this case I will always Reach This Z 0
3:58:08Vector so all possible collections of
3:58:11these vectors are the vector 0 0 which
3:58:13is single vector and the same as the
3:58:17input so this is the basic case now
3:58:20let's move on onto bit more uh Advanced
3:58:23case so bit more complicated than this
3:58:26one but itself also very easy which is
3:58:29when we got a single Vector
3:58:32a so let's say a is equal
3:58:40to one and two now I want to know what
3:58:45is the span of
3:58:48a in order to know what is the span of a
3:58:52we simply need to understand what are
3:58:54all these possible collections of
3:58:56vectors that I can get when I'm uh
3:58:59combining um a I'm multiplying a with
3:59:02different coefficients so what are the
3:59:05all possible linear combinations of this
3:59:09Vector because I got just single Vector
3:59:11a so a is one one two which means one in
3:59:15here and then two here my a is this
3:59:21vector and let's look into uh different
3:59:24uh scalar multiplications of this Vector
3:59:27so let's say I want to calculate C
3:59:30* a so the scalar multiplication of this
3:59:35where C is equal
3:59:37to C is equal to 2 C is equal to 3 C is
3:59:43equal to uh Min -1 C is = to - 3 and of
3:59:48course C is = to
3:59:501 so in all these cases when C is equal
3:59:54to
4:00:031 then the linear combination in this
4:00:07case just the scale multiplication of
4:00:09this single Vector a so 1 * a is simply
4:00:14equal
4:00:16to one and two so the same Vector
4:00:21a c is equal to 2 this will give me 2
4:00:26and
4:00:27four C is equal three this will give me
4:00:303 and
4:00:326 C is = to min-1 will give me -1 - 2
4:00:36for my a and then C is equal to -3 will
4:00:40give me -3 and then - 6
4:00:45so let's plot each of
4:00:47those so if we got for instance C is
4:00:50equal to one case you can see that we
4:00:53already got that Vector in here so it is
4:00:55this
4:00:56Vector when C is equal to two then we
4:00:59got this one so two and four so where is
4:01:02that it is in
4:01:07here let me use another color it is in
4:01:12here when when we got C is equal to
4:01:17three then we got so this one three and
4:01:22six so this is three and this is
4:01:26six so it gives me this
4:01:35Vector in the next example so in the
4:01:37next linear
4:01:39combination we have C is equal to minus
4:01:42one so we got Min -1 and Min
4:01:45-2 so where is min -1 it is in here
4:01:48where is min-2 it is in here so I'm
4:01:50getting this
4:01:53vector and then
4:01:56finally when I have let me change the
4:01:59color when I have C is equal to minus 3
4:02:02so this case then I got minus 3 and 6
4:02:05which means that here is my minus 3 here
4:02:08is my minus
4:02:106 so we got this thing
4:02:14so you already should see what is going
4:02:16on here when we got just the single
4:02:19Vector for which we need to know what is
4:02:21a linear combination and that Vector is
4:02:23not a zero Vector it has nonzero
4:02:25elements but um it's still it is just a
4:02:29single Vector then all its linear
4:02:32combinations given that it is simply a
4:02:34scaled multiplication of
4:02:37it we are all getting them on the same
4:02:41line so you can see all the linear
4:02:44combinations of this single Vector is
4:02:47just a scaled version of it and it lies
4:02:49on the same
4:02:51line so what this tells us is that
4:02:53essentially you can move along the line
4:02:56defined by this Vector a but you cannot
4:02:58leave it so you cannot get a vector that
4:03:01is in here that is in here that is in
4:03:03here in here so you cannot leave this uh
4:03:07line you will always stay on this line
4:03:10so this line essentially spend of
4:03:14a so when we got a single vector and
4:03:19that Vector is not equal to Z Vector
4:03:22then the span of a is equal to and this
4:03:25can be expressed as C * a given that the
4:03:30C is a real
4:03:32number so we already saw this
4:03:34independent of what kind of scalar we
4:03:36will take any linear combination of it
4:03:38will end up simply the C * a so
4:03:42therefore we are generally izing this
4:03:44and we are seeing that the span of a so
4:03:46to set of all possible linear
4:03:48combinations of this a is simply equal
4:03:51to C * a given that the C is a real
4:03:55number this is basically the spend of a
4:03:58real uh uh Vector in a two dimensional
4:04:02space let's now look into the next case
4:04:04the next example when we will calculate
4:04:07or we will Define the span of a
4:04:11perpendicular vectors so let's look in
4:04:13into another example when we are looking
4:04:15for a case when the um when we want to
4:04:19find out the span of perpendicular
4:04:21vectors so imagine we have these two
4:04:23vectors Vector a and Vector B where a is
4:04:26equal to 1 0 and then B is equal to 0 1
4:04:29so we are still in our lovely uh
4:04:32two-dimensional space so let's first
4:04:35visualize the vector a it's quite basic
4:04:37it is this one and then Vector B it is
4:04:40simply this one so we can already see
4:04:43why they are perpendicular so you can
4:04:45see that they are forming this um 90°
4:04:48angle so right angle
4:04:50here and then we know that the span so
4:04:55that's exactly what we want to find out
4:04:57so the span of a and b and this is what
4:05:03we want to find
4:05:05out and we know that the span of two
4:05:09vectors is the set of all possible
4:05:12linear combination of these vectors so
4:05:16we want to see what are these all
4:05:18possible inar combinations of
4:05:23C1 so all the possible outcomes that we
4:05:26will get when we get a linear
4:05:28combinations of these two vectors so
4:05:30basically C1 *
4:05:35a
4:05:37plus C2
4:05:40* B
4:05:44because those are all the linear
4:05:47combinations of these two
4:05:50vectors C1 * a + C2 *
4:05:56B nothing thing that we can see here
4:05:58quickly is that C1 * a so this part
4:06:03those are all the scaled versions of a
4:06:05so scaling multiplications of a and this
4:06:10second term in the linear combination
4:06:12those are all the scal
4:06:14variations so scalar multiplications of
4:06:17vector B which means that and we already
4:06:21have seen this time and time uh again
4:06:24that when it comes to Vector a all its
4:06:27linear combinations they will lie on the
4:06:29same line so let me take this color so
4:06:34if I do 2 a so C1 is equal C2 then I
4:06:38will be in here if C1 is equal to three
4:06:41then I will be here C1 is equal to 4 I
4:06:44will be here C1 is equal to 10 I will be
4:06:47in here and then the opposite holds as
4:06:51well if C1 is equal to for instance
4:06:52minus uh 2 then I will be in here if
4:06:55it's equal to Minus 5 I will be uh my
4:06:58Vector will look like this and so on so
4:07:01this means that all the scaled
4:07:04multiplications of vector a will lie on
4:07:07this
4:07:09line so I can also say that the span of
4:07:13see uh the span of a so span of
4:07:20a is simply equal
4:07:25to
4:07:28C1
4:07:31a so you can
4:07:34see in here so on this line basically so
4:07:37this
4:07:38is
4:07:41C1 a so this basically means independent
4:07:44what kind of C1 I will take with is 1 2
4:07:473 0 - 5 - 100 I will always end up on
4:07:51this line so this
4:07:54line so this about the uh scaled
4:07:58multiplication of a but of course to
4:08:01create this linear combination of A and
4:08:02B we also have the second element which
4:08:05is the all possible scaled
4:08:07multiplications with a vector B so C2 B2
4:08:11so let's see what that looks like
4:08:13so if I for instance take C2 is equal to
4:08:17Z I will be in here if I take C2 is
4:08:20equal to uh 2 I will be in here C2 is
4:08:24equal to 5 I will be in here C2 is equal
4:08:28to Minus 5 I will be here so you are
4:08:30already seeing what is happening here so
4:08:33all the possible scaled
4:08:36multiplications with Vector B will be on
4:08:39this line so now we are then getting
4:08:42that
4:08:45the
4:08:47span of B will then be equal
4:08:52to
4:08:54C2 and then
4:08:57B and here I'm not uh using formal
4:09:00notation I'm just trying to um I'm just
4:09:03trying to uh draft the idea of the spin
4:09:06of vector a and uh span of vector B
4:09:09because we are not uh done yet we still
4:09:12need to combine the two in order to find
4:09:14the span of vectors A and B when they
4:09:17are perpendicular so when this angle is
4:09:20simply
4:09:2190° okay so let's also add this on our
4:09:26plot so this is C2 and
4:09:30then
4:09:34B so this already gives us an idea that
4:09:40all the
4:09:41possible combination of the two so when
4:09:44we add these two elements to each
4:09:46other the outcome will always lie on
4:09:49these two
4:09:52lines but there is no way that we can
4:09:56find any other coefficient for C1 or C2
4:10:00that can help us to get a value that
4:10:02will be so a vector that will be in here
4:10:05or in here or in here or in here that's
4:10:07just not possible so just you can try to
4:10:11go ahead and solve that equations like
4:10:12we did before before and you will see
4:10:14that there there is no way that you can
4:10:18pick here a line and you can represent
4:10:21it as a linear combination of these two
4:10:24vectors it just not possible and later
4:10:27on we will see why but just keep in mind
4:10:31for now that once we have this this type
4:10:34of vectors when two vectors are
4:10:37perpendicular then um we cannot find a
4:10:41line a vector that is outside of the
4:10:45two lines so here you can see the xaxis
4:10:48and the Y AIS but it can also be like
4:10:51this it can also be like this but then
4:10:54you cannot find any other line that lies
4:10:56outside of this area that you can uh
4:11:00create a linear combination of these two
4:11:03different vectors and then you say then
4:11:05you cannot say that you can create a
4:11:07linear combination of these two vectors
4:11:10A and B so therefore
4:11:13when it comes to defining the span of
4:11:16the two
4:11:18perpendicular line we say that the
4:11:21span of a and
4:11:26b given
4:11:28that A and B are perpendicular but also
4:11:32given that these values in this case you
4:11:34know a is equal to 1 is z b is equal to
4:11:370 and one then their spend you might
4:11:40have already guessed is equal to
4:11:44C1 a plus C2 B given that C1 and C2 are
4:11:52of course real
4:11:54numbers so in this case C1 and and C2 as
4:11:58expected are just scalar so they are
4:12:00just some real numbers coming from R and
4:12:03uh this A and B those are vectors that
4:12:06are being spent and in this case
4:12:08specifically the vector a is equal to
4:12:11this one zero and Vector B is equal to 0
4:12:131 and this expression that we see here
4:12:16this pen this simply describes the set
4:12:19of all possible vectors that can be
4:12:21formed by adding the scaled versions of
4:12:23this A and B so C1 a plus C2 B in order
4:12:27to form this linear combination so this
4:12:30set this set of C1 a plus c2b which is a
4:12:33linear combination all possible linear
4:12:36combinations of the two
4:12:37vectors so um it effectively covers the
4:12:41entire plane illustrating that any point
4:12:44in 2B space can be reached by some
4:12:48combination of A and B let's now move
4:12:52towards our final example that we saw
4:12:54also as part of our definition for the
4:12:56span of vectors in order to check and to
4:12:59learn how we can usually check uh
4:13:02whether the two vectors they really
4:13:04spend the entire space so in this case
4:13:07we got two vectors we got Vector uh V1
4:13:10which is equal to one two and Vector V2
4:13:12which is it's equal to three4 so in here
4:13:16and also um we uh have in our example
4:13:20that it says the span of V1 and V2 is
4:13:23all over the R2 because any Vector in R2
4:13:27can be expressed as a linear combination
4:13:29of V1 and V2 so the example basically is
4:13:32saying that if we know that um we can
4:13:37express any Vector in R2 as a linear
4:13:41combination of V1 and V2 2 then we say
4:13:44that the span of V1 and V2 is the entire
4:13:47R2 so let's actually go ahead and prove
4:13:50that from our example so we have X which
4:13:53we can represent as X1 and X2 and X1 and
4:13:56X2 are just real numbers and we got V1
4:14:00which is 1 2 V2 which is 3 4 and we got
4:14:04in our example that uh we need to prove
4:14:08that the
4:14:10span of V1
4:14:14and
4:14:15V2 is the entire
4:14:18R2 so for that what we need to do is we
4:14:22need to prove that we can express our
4:14:26coefficients C1 and C2 in such way using
4:14:29X1 and X2 that independent of what these
4:14:33X1 and X2 are so what kind of X Vector
4:14:36we have whether this is like one two or
4:14:40this is 04 or this is th000 and uh 5,000
4:14:46independent what kind of vector we get
4:14:49uh we give here so X1 and X2 values as
4:14:52long as those are real uh numbers we can
4:14:56always find a set of C1 and C2 that we
4:15:00can use as coefficients in order to
4:15:02create a linear combination from vectors
4:15:05V1 and V2 and in that case we say then
4:15:09the span of V1 and V2 is the in par
4:15:14R2 okay so let's go ahead and actually
4:15:17prove that using our previous knowledge
4:15:20that we already gained so keeping in
4:15:22mind that C1 and C2 are unknown numbers
4:15:26for us whereas X1 and X2 are just a way
4:15:29to describe those elements in our Vector
4:15:32X that will be provided to us so X1 and
4:15:35X2 will be basically know and C1 and C2
4:15:38are the unknowns that we are chasing so
4:15:41for that the first thing that I'm going
4:15:42to do is to describe this linear
4:15:44combination that we got here C1 V1 plus
4:15:47C2 V2 with actual equations unknown
4:15:51equations and the way I'm going to do it
4:15:53is by simply filling in this Vector V1
4:15:56and Vector V2 um
4:15:59values so we have C1 and then C2 here
4:16:05and then here I got one two
4:16:09plus and then three and four here and
4:16:13what is this amount so this is equal
4:16:20to let me actually go on to the next
4:16:25row so we can create um set of equations
4:16:31for this so this is equal to X
4:16:43and we get
4:16:46C1 *
4:16:501+ C2 * 3 is =
4:16:552 and remember that this is X and we
4:16:59said that the x is equal to X1 and X2 so
4:17:01this basically equal to we can B right
4:17:05here is equal to X1 and
4:17:09X2 so this is then equal 2
4:17:15let's not skip all the
4:17:17steps X1 and
4:17:21X2 so here then the second elements need
4:17:25to be added so C1 *
4:17:302 and C2 *
4:17:384 which is then the same as
4:17:44C1 + 3 C2 and then 2 C1 + 4 C2 and
4:17:52then this we are saying this Vector is
4:17:54equal to X1 and
4:17:57X2 so this is what we have here and
4:18:00let's move from the vectors to equations
4:18:04so given that we have this we are
4:18:07allowed to say that this gives us
4:18:08actually two equations this means that
4:18:11this element this element from this part
4:18:14should be equal to this and this element
4:18:17should be equal to this now let's write
4:18:19it down we see that
4:18:22c1+ 3 C2 should be equal to
4:18:26X1 2
4:18:28c1+ 4
4:18:30C2 is equal to X2 this is all that we
4:18:34see in
4:18:35here let's remove this to keep the space
4:18:40clean now what this means is
4:18:43that we have two equations with two
4:18:46unknowns C1 and C2 and X1 and X2 are the
4:18:50numbers that will be provided to us as
4:18:52part of our Vector so what we want to
4:18:55prove is that we can describe and we can
4:18:58express C1 and C2 which are our
4:19:01unknowns using X1 and X2 so you see here
4:19:05this is C1 C2 those are our nouns and we
4:19:10want to describe them by using X1 x 1
4:19:12and
4:19:14X2 and very soon we will also see why so
4:19:18for now let's try to express those two
4:19:21unknowns using our nouns like X1 and X2
4:19:25so here I already see that C1 is alone
4:19:28so there is no scaler so I will make use
4:19:31of that opportunity to keep the C1 on
4:19:33the left hand side and I will take this
4:19:36this amount to the right so I will say
4:19:38C1 is equal to X1 minus 3 C2
4:19:43two okay slightly better so I have C1 at
4:19:46the left I do have X1 in the right but I
4:19:49also have three C2 in here but another
4:19:53thing that you will notice is that in my
4:19:55second expression here I got 2 C1 + 4 C2
4:19:58plus X2 I want to have the C2 only
4:20:04because then I will have an expression
4:20:06of my C2 only using X1 and X2 um so
4:20:12numbers that are that will be provided
4:20:14to me that are n so for that what I'm
4:20:17going to do is basically trying to solve
4:20:20uh two equations with two unknowns
4:20:22exactly the same um process so I'm going
4:20:25to take this C1 from the first equation
4:20:28and I'm going to fill in in the second
4:20:30equation so I am going to say two times
4:20:35and here I'm going to fill in that C1
4:20:38expression from here so X1 - 3 C2
4:20:442 so this is my C1 plus just taking over
4:20:48this part so 4 C2 is equal to X2 okay
4:20:57perfect so now what I end up with is C1
4:21:02is = to X1 - 3 C2 just taking it over
4:21:06and then here I'm opening parenthesis
4:21:08which is 2 X1 - 6
4:21:11c2+ 4 C2 is equal to here I forgot an X2
4:21:16is equal to X2 okay one step
4:21:20closer why because I in my second
4:21:23equation I no longer have a C1 I only
4:21:27have a C2 which is great which means
4:21:29that this gives me an
4:21:31indication that I can rewrite the C2
4:21:35which is unknown with nouns with X1 and
4:21:38X2 so let's make use of that opportunity
4:21:41the first equation I would just take
4:21:43over so C1 is equal to and then
4:21:47X1 - 3
4:21:51C2 and
4:21:53then here I will do 2 X1 and then here
4:21:57we got 2 * c2s which means we can
4:22:00combine them so Min - c - 6 * C2 + 4 C2
4:22:05it gives me - 2 *
4:22:09C2 and this is equal to
4:22:13X2 all right let's now solve that
4:22:18part so what I want to have is just the
4:22:21C2 in the left hand side which means I
4:22:23need to bring all this to the right and
4:22:25I need to get rid of them such that I
4:22:27can leave the C2 in the left entirely
4:22:30alone so this is what I'm basically
4:22:34chasing for that I'm going to once again
4:22:38rewrite C1 is equal to X1 - 3 C2
4:22:43and this time I'm going to take them
4:22:46minus 2 C2 here I'm going to leave that
4:22:49in the left but then this one I'm going
4:22:51to bring to the right so X2 - 2
4:22:56X1 here I need to be very careful to not
4:22:59make a mistake CU that will mess up my
4:23:01entire
4:23:05calculation all right so now we are one
4:23:07step closer just taking over the first
4:23:10equation again so C1 is equal to X1 - 3
4:23:14C2 and here what I need to do to get rid
4:23:17of this minus two is to divide the two
4:23:20sides so both 2 minus 2 CU that will
4:23:25help me to keep the C2 only in the left
4:23:30alone without any scaler so the C2 is
4:23:33then equal to X2 minus 2
4:23:37X1 / 2 - 2 so this is what I end up
4:23:43with
4:23:45perfect so we are very close stay with
4:23:48me so uh here what we are getting is
4:23:52that C2 is equal to this amount we see
4:23:55that now we no longer have any other C
4:23:59in here which is great and remember that
4:24:01X1 and X2 will be numbers that it will
4:24:03be provided to us I just wanted to give
4:24:05everything General and then uh another
4:24:08thing that I want to fix is this C2
4:24:11because this C2 is an unknown and I want
4:24:14to fill in uh this value of C2 in here
4:24:18such that for the C1 I will have a
4:24:21similar picture so in the left hand side
4:24:23I will have C1 in the right hand side I
4:24:25will Express the C1 with no number so X1
4:24:28and X2 but not the C2 or others all
4:24:32right so let's then go ahead and do that
4:24:37first I will write C2 in a simpler way
4:24:39so C2 is equal to here I got a min I
4:24:42will just write here minus so I will
4:24:45take the minus over here and then I will
4:24:47write X2 - 2 X1 to U be super careful
4:24:51with this minus therefore I'm using
4:24:53parenthesis so now I'm going to use this
4:24:56C2 and I'm going to fill that in in here
4:24:58so C1 is equal to
4:24:59X1 minus 3 * I can also make it plus
4:25:04because minus of
4:25:06here so minus of here and minus of here
4:25:10will cancel out therefore I will
4:25:14do plus three times and then X2 - 3
4:25:22X1 /
4:25:2522 and this then gives me
4:25:29C1 is equal to X1
4:25:33+ 3 / 2 * X2 - 3 sorry 2 almost made a
4:25:41mistake 2
4:25:43X1 and then C2 is = to X2 - 2
4:25:48X1 /
4:25:502 2 and here we
4:25:55got
4:25:57minus
4:25:58Perfect all right awesome so now we have
4:26:03expressed X1 and
4:26:07X2 well careful with this X1 and X2 we
4:26:12only noun numbers now what I'm going to
4:26:15do is that I'm going to prove that
4:26:18independent what kind of X we will be
4:26:20taking here we will end up getting the
4:26:23C1 and C2 using this what we just found
4:26:27here that will give us a linear
4:26:29combination of these two vectors that
4:26:31will be equal to that eight so for that
4:26:34so to prove that this pen of V1 and vs2
4:26:37is the entire R2 I need to prove that
4:26:40independent what kind of X I will take
4:26:42so X1 and X2 I can always find the C1
4:26:46and C2 that I um just calculate in here
4:26:49using that X1 and X2 that I can then use
4:26:53to combine with my V1 and V2 to find the
4:26:56linear combination of these two vectors
4:26:59with that C1 and C2 which will be equal
4:27:02to this
4:27:03x so for that what I need to do first is
4:27:07to take such a uh random X so let's say
4:27:11my X is equal to 0 and 4 this means that
4:27:16my X1 is equal to 0 and X2 is equal to 4
4:27:20what this means is that this gives me C1
4:27:22which is equal to and here X1 so I'm
4:27:26basically filling these two values for
4:27:28here to obtain my C1 so C1 corresponding
4:27:31to this specific Vector X so X1 is equal
4:27:35to 0 which means I end up C1 is = 0 + 3
4:27:39/ 2 * X2 is = 4
4:27:42so 4 minus and then 2 * X1 is equal to -
4:27:462 * 0 which is
4:27:480 and then C2 is = to minus and then X2
4:27:54is equal to 4 so 4 and then minus 2 * XY
4:27:59is = to 0 0 and then this divided to two
4:28:03now what are those
4:28:05numbers
4:28:07so C1 is equal
4:28:10to 3 / 2 * 4 which is 3 * 2 so 6 and
4:28:16then C2 is equal
4:28:18to- 4 and then minus so this is zero
4:28:22this cancels out which means 4 / 2 is 2
4:28:25and then C2 is equal to minus
4:28:282 so basically I have calculated the
4:28:33coefficients C1 and C2 by just knowing
4:28:38what is this Vector so knowing X the
4:28:41provide X1 and X2 I have calculated my
4:28:45C1 and C2 using my
4:28:49derivations in here so let's now get rid
4:28:52of
4:28:56this this calculations to clear some
4:29:00space and to do the final part which is
4:29:03compute the linear combination of vector
4:29:06V1 and V2 for this specific coefficients
4:29:08well knowing what this given Vector now
4:29:11is example random Vector so the C1 is
4:29:17equal to 6 which means 6 * and then
4:29:20Vector V1 is one 2 so this is first part
4:29:24of my linear combination
4:29:27plus and then C2 is equal to - 2
4:29:33times then here 3 4
4:29:46what is this this is equal
4:29:50to 6 and then 6 * 2 is
4:29:5512
4:29:57plus now let's calculate the second part
4:30:00- 2 * 3 is -
4:30:046 and - 2 * 4 is
4:30:09-8 so what does this give
4:30:14us 6 - 6 and 12 - 8 this gives
4:30:21us zero and
4:30:25four nice so this confirms that we have
4:30:29done everything also correctly which is
4:30:31great because we have seen that using
4:30:35this C1 and C2 that we have just
4:30:38calculated we have successfully uh
4:30:42computed the 6
4:30:45V1 so linear combination of this uh two
4:30:50vectors V1
4:30:55plus -
4:31:002 minus 2 and then
4:31:05V2 and we have seen that this linear
4:31:07combination is equal to 04 which is
4:31:09exactly our 8 so in this way we have
4:31:13proven that independent what kind of
4:31:15vector we will pick what kind of X we
4:31:18will pick here we can always find and
4:31:21calculate the corresponding coefficients
4:31:24C1 and C2 in the same way as I just did
4:31:27and then by using those when we
4:31:30calculate the linear combination of
4:31:32these two vectors with this specific
4:31:35coefficient this will be exactly equal
4:31:37to X and this proves that independent
4:31:41what kind of vector we have in our R2 we
4:31:44can always express that as a linear
4:31:47combination of the vector V1 and V2 and
4:31:50this proves and this concludes our proof
4:31:52that
4:31:54span of V1 and V2 is the entire
4:32:02R2 all right so we are very close to
4:32:05finishing up this unit so the next topic
4:32:08we are going to talk about is a linear
4:32:10Independence and all this important
4:32:13stuff that we learned as part of the
4:32:15previous modules are going to become
4:32:17super handy as part of this specific
4:32:20concept so we just spoke about the idea
4:32:22of span we have plotted a lot of vectors
4:32:25we have seen the linear combination of
4:32:27that and how we can find out whether the
4:32:30span of multiple vectors is the entire
4:32:32space uh for instance the R2 or it is
4:32:35just the line or it's maybe the zero
4:32:38Vector we have seen many examples and
4:32:40many operations we have we have also
4:32:42seen this idea of unit vectors and we
4:32:44are finally ready to come to this very
4:32:46important concept which is a concept of
4:32:49linear
4:32:51Independence so by definition linear
4:32:54Independence says that the set of
4:32:56vectors is linearly independent if no
4:32:59Vector in a set can be written as a
4:33:02linear combination of the others
4:33:05otherwise they are linearly
4:33:08dependent so vectors V1 V2 up to VN are
4:33:13linearly independent if and only
4:33:17if the only solution to the equation C1
4:33:21V1 + C2 V2 plus CN VN is equal to zero
4:33:27is C1 is equal to C2 up to CN is equal
4:33:31to Zer in other words in a l linearly
4:33:35independent set the equation C1 V1 + C2
4:33:39V2 plus CN VN is put zero has only the
4:33:43trial solution where all CIS are
4:33:48zeros so now what do we mean here there
4:33:51is a ton of information in this
4:33:53definition so let's unpack them firstly
4:33:55it's really important to uh keep in mind
4:33:58this idea of
4:34:00Independence and dependence Independence
4:34:03and dependence there are things that we
4:34:05commonly use in data science in
4:34:07artificial intelligence in statistics so
4:34:10those are really important so we
4:34:13basically have linear independent
4:34:17condition so there is a certain
4:34:19condition that our vectors should
4:34:20satisfy vectors in our set in our Vector
4:34:24space for them to be named as linearly
4:34:27independent and otherwise we are calling
4:34:30them linearly dependent and you can see
4:34:32that here there are a couple of Parts as
4:34:35part of this definition first it talks
4:34:38about um being unable to create a vector
4:34:42in the vector set while using the
4:34:45remaining vectors in our set so it says
4:34:48if you can use the remaining vectors in
4:34:51your vector space and linear create a
4:34:55linear combination of them so linearly
4:34:57combine them and we have already seen
4:35:00the definition of linear combination so
4:35:03if we cannot create such linear
4:35:05combination from the remaining vectors
4:35:08to get our Target vector
4:35:12then we are saying that we have a
4:35:14linearly independent vectors so if we
4:35:17want to say that all our vectors in our
4:35:20Vector set they are linearly independent
4:35:23it means that each of those vectors we
4:35:26should not be able to recreate out of
4:35:29the remaining vectors so we should not
4:35:31be able to find coefficients to create
4:35:34linear combination using the remaining
4:35:36vectors in order to get our Target
4:35:40vector now what do I mean by this target
4:35:43Vector what do I mean by this linear
4:35:45combination uh I will come to this in a
4:35:47bit for now let's just try to unpack
4:35:49this definition cuz uh with examples uh
4:35:53we will definitely go through this step
4:35:55by step in detail such that this ideal
4:35:58linear Independence and dependence is
4:36:00super clear so in the second part of the
4:36:03definition it says vectors V1 V2 up to
4:36:06VN are linearly independent if and only
4:36:10if the only solution to the equation and
4:36:13we have here in the left hand side you
4:36:15might recognize the linear combination
4:36:17of our vectors V1 up to VN so in in the
4:36:21right hand side you have zero so you are
4:36:22saying our linear combination of vectors
4:36:24is equal to zero if and only if C1 C2 up
4:36:29to CN is equal to zero so linear
4:36:33Independence basically claims that we
4:36:36will have linearly independent vectors
4:36:39only if and only in the condition when
4:36:44um the only way we can create linear
4:36:47combination of these vectors equal to
4:36:50zero only if those coefficients are zero
4:36:54there is no other way that we can get a
4:36:56linear combination that is equal to zero
4:36:59while those coefficients are not zero so
4:37:03the only way that we can get a linear
4:37:05combination out of all our vectors equal
4:37:07zero is only when all of the
4:37:10coefficients C1 1 C2 up to CN is equal
4:37:13to zero that's something that we will
4:37:15come later to this again this something
4:37:18also that we are going to come back in
4:37:20our next module and the next one so um
4:37:24this one will be also super clear once
4:37:26we go through those modules but for now
4:37:29keep in mind that the uh linear
4:37:31combination of all these vectors can
4:37:33only be zero in case when all these
4:37:36coefficients are equal to zero so and
4:37:40then we have the third part in our
4:37:42definition which says that in other
4:37:45words in a linearly independent set the
4:37:49equation C1 V1 plus C2 V2 up to CN VN is
4:37:54equal to zero has only the trivial
4:37:57solution where all CIS are zero so this
4:38:00explanation is basically what we just
4:38:02spoke about as part of this second part
4:38:04where we said that only in case the
4:38:07coefficients are all zero we can have a
4:38:09linear combination of our vector V1 V2
4:38:12up to VN which is equal to
4:38:16Zer and why we would like this linear
4:38:18combination to be equal to zero because
4:38:21it's a common way to find solution to
4:38:23our linear system so this is something
4:38:26that we will also see as part of the
4:38:28next module when we'll be discussing the
4:38:31idea of solving linear systems we will
4:38:33go into more uh Advanced topics but for
4:38:36now in order to understand this idea of
4:38:38linear Independence we should just keep
4:38:40in mind that we cannot find any CIS so
4:38:45C1 C2 so any coefficients that is not
4:38:48equal to zero and then expect that the
4:38:50linear
4:38:51combination of these uh linearly
4:38:54independent vectors is equal to zero so
4:38:56that's the uh if and only uh if and only
4:38:59uh if part which means that this holds
4:39:01from both sides on one hand we have V1
4:39:04V2 up to VN which are linearly
4:39:07independent only if the linear equation
4:39:11so the linear combination of all these
4:39:13vectors is equal to zero if all these
4:39:14coefficients are zero but also the other
4:39:17way around holds as well so if we have a
4:39:19linear combination that is equal to zero
4:39:22only if those coefficients are zero that
4:39:24means that we are dealing with a
4:39:26linearly independent vectors this is the
4:39:30if and only if part which means that we
4:39:33have this uh conditions from both sides
4:39:35if one holds the other one holds but
4:39:37also the other way
4:39:39around all right so let's now look into
4:39:42specific examples that will make our
4:39:44journey in understanding linear
4:39:46dependence much more convenient so let's
4:39:49say we have our coordinate system and we
4:39:52have these two different vectors so we
4:39:55have Vector let's
4:40:06say 2 and three which is our Vector a
4:40:11and we have a vector
4:40:14B that is equal
4:40:162 6 and
4:40:20N so those two are our vectors and what
4:40:24we want to understand is where those two
4:40:26vectors are linearly independent or
4:40:29linearly
4:40:31dependent so one thing that you can
4:40:34quickly notice is that b looks quite
4:40:38similar to a in terms of its scal so
4:40:42there is a way that we can recreate
4:40:45Vector B by using Vector a now you can
4:40:50see that if I take Vector
4:40:55a which is equal to
4:40:5923 if I take Vector a and I multiply it
4:41:03by three so three * Vector a this is a
4:41:07scale
4:41:08multiplication then what I can get is
4:41:14three times and then I have here two
4:41:18three and this is then equal
4:41:21to 3 * 2 is 6 3 * 3 is 9 this gives me 6
4:41:27and 9 which is our uh Vector now another
4:41:31thing that you can notice that that is
4:41:33exactly my B so you can see that those
4:41:37two are similar which means that three *
4:41:43a is equal to
4:41:47B now what this means is that I can
4:41:50recreate Vector B by using Vector a so
4:41:56in our definition we saw that a set of
4:41:59vectors is linearly independent if no
4:42:03Vector in the set can be written as a
4:42:06linear combination of the
4:42:08others so here I can take this 3A as a
4:42:13way to write down a linear
4:42:17combination so 3A + 0 *
4:42:22B is then equal to
4:42:27B which is basically saying 3 a is equal
4:42:30to B so by using these two vectors in a
4:42:34set I can then create a linear
4:42:36combination of the two and actually even
4:42:40basic way of writing this is saying I
4:42:44can use the vector a to write a linear
4:42:48combination from this so 3A is a linear
4:42:51combination so just a scaled
4:42:53multiplication in this case of course
4:42:55but if we have just two vectors our
4:42:58Target Vector is B and I want to write
4:43:00this uh I want to see whether I can
4:43:02rewrite the vector b as a linear
4:43:05combination of the remaining vectors
4:43:07which is Vector a so I can then write
4:43:10Vector B as a linear combination of
4:43:13vector a because I can say that 3 * a is
4:43:16equal to Vector
4:43:21B so this means that Vector a and Vector
4:43:27B they are
4:43:31linearly
4:43:34dependent this means that I can use
4:43:37Vector a to recreate Vector B and of
4:43:40course I can also do the other way
4:43:42around right what I can do is that I can
4:43:45just
4:43:46take Vector B so I can take Vector
4:43:53B I can multiply it by one ided to
4:43:58three 1 / 3 is real number so I'm just
4:44:01performing a linear combination using B
4:44:05and this will give me 6 / to 3 is 2 9 /
4:44:10to 3 is Tre and I'm getting exactly what
4:44:13I have under a so I can then also
4:44:16rewrite Vector a by using Vector B so I
4:44:21created a linear
4:44:23combination using Vector B in order to
4:44:26get a vector a and that's exactly the
4:44:29opposite what we have learned here
4:44:31because we should not be able to write
4:44:34this a vectors using the other ones in
4:44:36our set cuz otherwise we have a linearly
4:44:40dependent set
4:44:41therefore we are saying that Vector a
4:44:43and Vector B they are not a set that is
4:44:46linearly independent but they are
4:44:49linearly dependent before moving on to
4:44:52another example I also wanted to
4:44:54visualize these vectors just to see what
4:44:56is going on with this pan and uh how the
4:44:59two linearly dependent vectors look like
4:45:03in R2 so this is our r t we have a
4:45:06vector a which has two tree elements so
4:45:09we know already the magnitude and the
4:45:11direction this is two this is three
4:45:13which means here let me actually use
4:45:17another
4:45:18color so 2 three which means this is my
4:45:22Vector a and then my Vector B is simply
4:45:27six and N so it is this
4:45:32one so you can already see what is going
4:45:35on so this is Vector a and this entire
4:45:39thing is Vector B
4:45:47and you can see that those two vectors
4:45:52no matter how I combine them I can I
4:45:55will always get the combination so
4:45:58linear combination of the two on this
4:46:01line if I want to
4:46:05get um Vector that is for instance in
4:46:09here I can never
4:46:11find a scalers of
4:46:14C1 and C2 in such way that these
4:46:21vectors so A and B they can form a
4:46:25linear combination that will give me
4:46:27this Vector there is no way that I can
4:46:29do that and that's why uh we say that
4:46:33this span of this two
4:46:36vectors so span
4:46:40of A and B with this A and B is this
4:46:47line and we cannot express any of these
4:46:51other vectors like this one or this one
4:46:54using a linear combination of these
4:46:56vectors A and B the only linear
4:46:58combinations that we can recreate using
4:47:00these vectors A and B are on this line
4:47:04so you can see that even if I have two
4:47:07different vectors I actually just got um
4:47:10single Vector because I have two Tre and
4:47:13both of these vectors they are actually
4:47:16um uh scaled multiplication of the other
4:47:19one so B is equal
4:47:22to I'm missing here something 1 / 3 so B
4:47:27is simply equal to 3 * a and then a is
4:47:30equal to 1 / to 3 * B so in both cases
4:47:35they are simply a version of scaled
4:47:37multiplication of this Vector Q3
4:47:41so a is simply equal
4:47:46to
4:47:49B * 13 and then B is equal to 3 *
4:47:57a and both of them they are actually
4:47:59based on this Vector 2 Tre on this
4:48:02Vector
4:48:05a so therefore they both actually form
4:48:09and they span or round this single
4:48:16line and they are
4:48:18both linear we also call it collinear
4:48:21and they are linearly
4:48:25dependent okay so let's now move on to
4:48:28the next uh example where we will have
4:48:30bit more interesting case and we will
4:48:32look into this example when we have
4:48:35linear Independence
4:48:41look into another example bit more
4:48:43interesting one as we want to see
4:48:44whether those two are linearly
4:48:46independent or not so the first Vector
4:48:49that we got is the vector a the vector a
4:48:53is equal to 6 and Z so it is this
4:49:00Vector this is Vector a the vector
4:49:05B it is this
4:49:08one and it contains element of Z 0 and
4:49:137 so it is
4:49:16this Vector this is the vector B now in
4:49:21our definition of linearly independent
4:49:25vectors we saw that the idea of linear
4:49:28Independence is that the two vectors can
4:49:31only be linear independent if we cannot
4:49:34rewrite one of them by using the other
4:49:37so this means that we cannot rewrite a
4:49:41in terms of B and we cannot rewrite B in
4:49:44terms of a so there is no way that we
4:49:47can scale the vector a to get Vector B
4:49:50and there is no way that we can scale
4:49:51Vector B with vect with some uh scaler
4:49:54in order to get the vector a so there is
4:49:57no way that we can create a linear
4:49:59combination of this one vector to get
4:50:02the other one and the other way around
4:50:04so let's see whether this is the case
4:50:06just from uh trial and error we have a
4:50:10vector a
4:50:11which contains elements 6 and zero for
4:50:14us to go from A to
4:50:18B that has elements from so we need to
4:50:23go from 6 to zero in this case and we
4:50:26need to go from 0 to 7 now we can
4:50:30automatically already see from the
4:50:31second element that there is no way that
4:50:33we can go from 0 to 7 you cannot find
4:50:37any scaler
4:50:39C that you can multiply with zero in
4:50:43order to
4:50:45get seven there is no way that you can
4:50:48do that because any number any real
4:50:54number that is a real number if you
4:50:58multiply it with zero it will never
4:51:00become
4:51:02seven and of course another thing that
4:51:05you can notice here also very quickly is
4:51:07the other way around right so here if
4:51:10you go from this zero to six there is no
4:51:13way you can go from this Z to six
4:51:15because there is no such C that you can
4:51:19take this zero and multiplying it with
4:51:22that so here our
4:51:25scaler and you get this equal to six
4:51:29this is just not
4:51:33possible so what we are seeing here is
4:51:36that there is no way that we can somehow
4:51:39change this vector so there is no way
4:51:42that we can scale them in such
4:51:48way so this is minus
4:51:51B so all the scales scaled version of
4:51:56this or all the um scaled
4:51:58multiplications of vector B they will
4:52:00always be on this
4:52:02line and then the same holds for a as
4:52:05well so all the scaled multiplications
4:52:08of a will be on this line
4:52:11so then one thing we can quickly see
4:52:13here is
4:52:15that given that those two are
4:52:19perpendicular this
4:52:22pen of A and B is the entire
4:52:28R2 so we can see that by using those two
4:52:32lines we can recreate any other line in
4:52:35this
4:52:38R2 and this is highly related to this
4:52:41idea of linear
4:52:42Independence and given that we cannot
4:52:44come up with a linear combination using
4:52:47the a vectors to recreate the other one
4:52:50in this case given that we cannot
4:52:52recreate a using B and we cannot
4:52:55recreate B using a so no linear
4:52:58combination that exist that we can use
4:53:00to recreate Bay using a and the other
4:53:03way around we are saying that Vector a
4:53:08and Vector B are are
4:53:13linearly
4:53:20independent let's now look into another
4:53:22example that will uh clarify this linear
4:53:25Independence concept so we have three
4:53:28different vectors and the first Vector
4:53:30is Vector a 1 0 0 Vector B uh 0 1 0 our
4:53:34second vector and the third Vector 0 0 1
4:53:37you can notice that we are in R Tree
4:53:41and then the example goes on and it says
4:53:43that those three vectors are linearly
4:53:46independent and as an explanation we
4:53:48have that there is no way to add these
4:53:51vectors together with any scalar
4:53:53multiples to equal the zero Vector
4:53:56unless all scalers are zero now before
4:53:59even going on to next part it's actually
4:54:02very quickly um uh provable that those
4:54:06three vectors are linearly independent
4:54:08and you cannot create a linear
4:54:10combination of one using the remaining
4:54:12of the two let's look into this example
4:54:15in more detail so we have three vectors
4:54:24A1
4:54:26A2 sorry
4:54:29B so we got
4:54:33a B and
4:54:38C which are
4:54:411 0
4:54:430 0 1
4:54:460 and 0
4:54:5001 now you can quickly see that if we
4:54:53are in
4:54:54R3 and this is actually our unit Vector
4:54:57E1 this is our unit Vector E2 and this
4:55:00is our unit Vector E3 because in that
4:55:03positions we got our ones and the
4:55:05remaining they are all zero and this is
4:55:09actually very similar to the previous
4:55:10example because we can quickly see how
4:55:13we are we will not be able to recreate
4:55:16one vector using the other ones by even
4:55:19looking at the positions of the zeros so
4:55:22for us to recreate Vector a which is
4:55:27equal to 1 0 0 it means that we should
4:55:32be
4:55:34able to
4:55:36find a linear combination
4:55:41C1 C2 and then using these vectors this
4:55:45is the vector
4:55:48B 0 1
4:55:510 plus C2 * Vector C which is 0 0
4:55:591
4:56:02so in here
4:56:04basically we are already seeing a
4:56:08problem because we have here here an
4:56:12element
4:56:14one and we somehow need to be able to
4:56:18find
4:56:21C1 and
4:56:23C2 in such way that 1 is equal
4:56:27to C1
4:56:30* 0
4:56:32+
4:56:36C2 time Z but we know that there is no
4:56:40C1 and C2 that we can find such this uh
4:56:43expression actually is true because C1
4:56:47and C2 they should be real numbers and
4:56:50there are no real numbers that we can
4:56:51find to multiply with zero such that
4:56:54this will end up to one because this is
4:56:57always equal to zero and we basically
4:57:00get 1 is equal to Z which is not
4:57:04true and of course the same holds the
4:57:07other way around you can prove that B
4:57:09can never be um recreated by using the
4:57:12linear combination of a and c and also
4:57:15the C can never be recreated by using a
4:57:17linear combination of A and B therefore
4:57:21we are
4:57:26saying given that
4:57:30a can't be
4:57:34written as linear combination
4:57:43of B and
4:57:48C
4:57:51B can be
4:57:57written so the same only this time A and
4:58:03C and then
4:58:05C
4:58:07hunt B
4:58:11written as linear combination of A and
4:58:18B those
4:58:22vectors A B and
4:58:26C they
4:58:28are
4:58:30linearly
4:58:38independent and if stronger you can
4:58:41actually go ahead and prove that this
4:58:44Spen of these three vectors is the r Tre
4:58:48but that's outside of the scope of this
4:58:50example so we will just pass but I will
4:58:52leave that um to you to
4:58:55prove all right so now when we are done
4:58:58with that let's actually move on to the
4:59:01last module which is the dot product and
4:59:04its
4:59:05applications so uh the length of a
4:59:08vector and Dot product is a con cep that
4:59:10um we um are familiar from the high
4:59:13school so the length of a vector is
4:59:15deeply related to this do product idea
4:59:18the dotproduct of a vector v WID itself
4:59:21gives this a square of the length of V
4:59:25what basically um it means is that this
4:59:29dotproduct of vector v so this thing
4:59:33which means take the vector v and
4:59:36multiplying it with the with the other
4:59:38vector v is simply equal
4:59:42to the square of a length of B
4:59:47so this is way to express the length of
4:59:52the uh of the vector B and once we
4:59:55square that that is the dot product so
4:59:58that's basically this definition what is
5:00:00about so we know what this definition of
5:00:03the distance is and we Define it by this
5:00:07and then we take the square
5:00:10of that distance and there is our DOT
5:00:14product and we are going to see this IDE
5:00:17of dot product a lot especially when it
5:00:19comes to uh matrix multiplication Vector
5:00:23multiplications also in many
5:00:25applications of linear algebra you will
5:00:27see this idea of that product coming
5:00:29again uh and coming back to us so uh
5:00:32this is a concept that we really need to
5:00:35understand so in the two dimensional
5:00:37space let's say we have a vector B which
5:00:39is um consisting of the two elements X
5:00:42and Y then the dot product and the link
5:00:45are related by V by V this is the way we
5:00:48denote the dot product so we just simply
5:00:51use the dot and the name also makes
5:00:54sense because we are saying we are using
5:00:57the dot to perform dot product so we are
5:00:59multiplying to two we are creating the
5:01:01product of this Vector with itself and
5:01:04this is equal to x² + Y 2 which is equal
5:01:08to the uh um squared of the distance of
5:01:12this Vector now you might recall from
5:01:16the high school that we have learned
5:01:18this idea of distance so if we have
5:01:20x-axis Y axis then we basically use
5:01:24this uh x² + y sare to uh get the uh you
5:01:29know the formula for from our Circle and
5:01:32then uh we have the x square + y Square
5:01:35we take the square root of it and then
5:01:37this is our distance so once we take the
5:01:40square root of that square of that from
5:01:43this uh square root of x square + y
5:01:45Square then we are simply getting this
5:01:48two cancel out which is equal to x² +
5:01:51y^2 So This is highly related to this
5:01:54idea because we are again talking about
5:01:57distances and we are simply taking the
5:02:00distance we are squaring them up and
5:02:02then we are getting the dot
5:02:05product so this the double uh straight
5:02:09line
5:02:10this is just a notation that we use and
5:02:14we spoke about this also before this
5:02:16comes um from the
5:02:18pre-algebra and this um this is highly
5:02:22important related to this idea of
5:02:24pythagore theorem and how we compute the
5:02:26distances so for instance when we have
5:02:29this uh Square triangular so we have
5:02:33this um uh rectangle here and we have
5:02:37here the 90 uh
5:02:40uh great so here we have the right uh
5:02:43right um angle and here we have our C
5:02:47which is uh the side right in front of
5:02:49this uh 90° angle and here we have the A
5:02:53and the B and we say that the c² is
5:02:56equal to a sare + b
5:03:00sare and if I were to actually write
5:03:04this in terms of X and
5:03:07Y so if this side is X and this side is
5:03:11y and this is my Z let's say then z s
5:03:15would be equal to x² +
5:03:19y² and this is something that we can see
5:03:22here too and the two terms are highly
5:03:25related so the Z
5:03:27squ is equal to x² + Y 2 and this is
5:03:32simply equal to Z * Z right and this is
5:03:36something that we know from High School
5:03:55welcome to the module one of this new
5:03:57unit when we are going to talk about
5:04:00about matrices as well as linear systems
5:04:02so those are all fundamental concepts
5:04:05that you will see time and time again
5:04:06when applying linear algebra not only in
5:04:09mathematics techs but also in applied
5:04:11sciences like data science artificial
5:04:13intelligence when training different
5:04:15machine learning models and trying to
5:04:17see what is this mathematics behind
5:04:19machine learning models different
5:04:21optimization techniques when you want to
5:04:23solve different problems using linear
5:04:27algebra so in this first module as part
5:04:29of foundations of linear systems and
5:04:32matrices we're going to introduce this
5:04:34concept of linear systems and then we
5:04:37are going to talk about the general
5:04:38linear systems we are going to uh see
5:04:41this common labeling of the coefficients
5:04:43this idea of indices that refer to the
5:04:46rows and the columns we are going to see
5:04:48what is this differentiation between
5:04:50homogeneous and nonhomogeneous systems
5:04:54so without further Ado let's get started
5:04:57so uh the linear systems form the uh
5:05:01bedr of linear algebra modeling this
5:05:04array of problems thanks to this
5:05:07advancements in these linear systems and
5:05:09Sol in it in Computing we can now solve
5:05:12a large amount of problems in a very
5:05:15efficient and a fast
5:05:18way so uh the general linear systems can
5:05:21be represented by this uh set of M
5:05:24equations with n
5:05:26unknown in the previous unit when we
5:05:29were looking into this uh linear
5:05:32combination of vectors we saw this
5:05:35notation which was A1 and then we we
5:05:40had C1 multiplied or rather let me keep
5:05:44me uh let me keep the same notation so
5:05:47we had this linear combination of
5:05:48vectors so we had beta 1 and then we had
5:05:53A1 Plus beta 2 and then A2 and those are
5:05:57all vectors plus A3 so beta 3 * A3 dot
5:06:04dot dot and then beta m
5:06:09times a m this is the notation that we
5:06:13saw before and we said we want to come
5:06:16up we wanted to come up with the linear
5:06:18combination of these different vectors
5:06:20A1 A2 A3 up to a and then we use that in
5:06:24order to get a sense of whether we are
5:06:26dealing with linearly independent
5:06:28variables vectors or linearly dependent
5:06:31vectors and then we also commented on
5:06:33the span that these vectors
5:06:36take now when it comes to um the uh
5:06:40vectors and just in general linear
5:06:42systems we can represent what we had
5:06:45before now in terms of with a bigger
5:06:48system so in terms of M equations and
5:06:52with n unknowns so here what you can see
5:06:55here is that we have M different
5:06:59equations so we have beta B1 B2 up to
5:07:05BM so you can see it in here and then
5:07:09each of these equations it contains n
5:07:12unknowns so you can see that the
5:07:14unknowns stays the
5:07:16same so the unknowns are those X1 X2 up
5:07:21to xn so X1 X2 up to xn are the set of
5:07:28all n
5:07:33unknowns and then M equations that you
5:07:36can see in here are all these equations
5:07:39so a11 X1 + a12 X2 dot dot dot and then
5:07:43a1n and then xn is equal to
5:07:46B1 and here one thing that is really
5:07:50important to keep in mind is that the
5:07:54indexing is what we need to focus on so
5:07:58we need to keep this one in mind this a
5:08:01i
5:08:02j and this
5:08:04XI so this is something that we also
5:08:08spoke about when uh discussing the
5:08:10linear combination of vectors we
5:08:13slightly uh touched upon on this topic
5:08:17so let's now dive into this this
5:08:18indexing and how do we indexes a i j
5:08:23what are this A's what are this JS and
5:08:27here you can see that we have a11 and
5:08:30then a12 and then up to the a1n and this
5:08:34is in our equation one and then we have
5:08:39in our equation
5:08:41two A1
5:08:43two let may actually write this with
5:08:45different color so in our equation two
5:08:48we got a 21 a 23 up to
5:08:53a2n
5:08:54and this A's that you see here those are
5:08:58just real numbers so a11 can be 1 A1 2
5:09:02can be three A1 n can be 100 and then
5:09:06the same also holds for this B1 for this
5:09:09B2 and for this BM and all these values
5:09:12A's and B's they are just real numbers
5:09:16the only unknowns that we got here are
5:09:18those so the X1 X2 up to
5:09:31xn all right so what about the indexing
5:09:34now so we got a i j
5:09:39and as you can see in this
5:09:43case the first thing that we can see
5:09:45here it stays everywhere the same which
5:09:47is the one so we got here one we got
5:09:51here one and up to the point we got here
5:09:53one whereas the second
5:09:55Index this one it does change it grows
5:09:59gradually with
5:10:02one and it becomes it goes from 1 to two
5:10:06and up to n so you can see here that the
5:10:11first
5:10:15index first
5:10:19index or index
5:10:23I it
5:10:25goes from one it doesn't change it's
5:10:29just one so it is one one and one so
5:10:32here in all cases for this equation I is
5:10:36equal to 1 but another thing that you
5:10:39can notice here is that the index 2
5:10:43unlike index I so the second index which
5:10:46is the J so you see here that the second
5:10:49index is referred as J this is a general
5:10:52way of defining the indexes so here J is
5:10:56equal to 1 2 dot dot dot and then
5:11:01n so
5:11:05basically the I doesn't change in the
5:11:09same row but the G
5:11:12changes and then of course we have
5:11:15slightly different in terms of I but
5:11:17then the same for J for our second
5:11:19equation so here I is equal to 2 and
5:11:22then J is again equal to one and then
5:11:27two dot dot dot and then
5:11:29n and then here up
5:11:32to for the last equation our I is equal
5:11:36to M and then our J is again equal to
5:11:42one till two dot dot
5:11:45dot so you might notice that I was
5:11:48looking at this from the row perspective
5:11:51so I was saying pair equation or pair
5:11:56Row the I doesn't change but then the J
5:12:02stays the same and then it is either one
5:12:05two up to n but the set is the same so
5:12:08it is it contains all these different
5:12:10elements here so one one two and then n
5:12:13but it contains all these different real
5:12:16numbers going from one till n because we
5:12:18are combining and we are creating this
5:12:21combination the sum of all these values
5:12:24a11 and then X1 a12 X2 A1 n
5:12:29xn and another thing that you can also
5:12:31notice here is
5:12:34that here with the second
5:12:39index so with this J J is equal to one
5:12:44then here the X's corresponding index is
5:12:46also one when the J is equal to two then
5:12:50the ex's corresponding index is also two
5:12:54and then here the same story and you
5:12:56will notice that while the coefficient
5:12:58contains two indices 1 one one 2 or 1 n
5:13:03which are the two indices for the
5:13:06coefficients for the unknowns we got but
5:13:09just single index which goes from one
5:13:13till n so basically 4 a for the
5:13:17coefficients so
5:13:20I let me write with the right
5:13:23color so I can be one 2 all the way to
5:13:30M whereas in case of
5:13:36J it can be one to all the way to
5:13:41n
5:13:43and
5:13:46the indices are basically used to help
5:13:49us to keep track of in which row we are
5:13:53and what is the um
5:13:58variable that the coefficient belongs to
5:14:01because knowing this second Index this
5:14:06helps us to understand that we are
5:14:07dealing with a coefficient that
5:14:09corresponds to this
5:14:10first unknown the first variable X1 and
5:14:14then the same holds in here as you can
5:14:18see in here and in here we are dealing
5:14:20with the same variable X1 therefore the
5:14:25second index the index J is then the
5:14:28same both in the first equation and in
5:14:31the second one in both cases it's equal
5:14:33to
5:14:35one okay so now when we are clear on
5:14:38that let's
5:14:41also understand this high level concept
5:14:44because you will see this system of
5:14:47linear systems this m equations and N
5:14:49unknowns appearing a lot not only in
5:14:52terms of calculating and finding the
5:14:54solution to this linear system but this
5:14:57actually has a very common application
5:15:00when it comes to um running regression
5:15:03linear regression
5:15:05specifically and one thing that you can
5:15:08notice here is
5:15:10that here we got also this B1 B2 up to
5:15:14BM and you will notice that here the
5:15:17index also uh goes from one but then
5:15:20this time to M so when it comes to the
5:15:25rows we have M rows or M
5:15:30equations therefore we also expect when
5:15:33it comes to Counting from the top that
5:15:35at the bottom we will see an M whereas
5:15:38if we count
5:15:39from this side so kind of like imagine
5:15:43it like a column then we see that it
5:15:45goes from one till
5:15:47n so those are common observations and
5:15:51reference to um number of observations
5:15:55and number of uh features that you will
5:15:58see in your data when dealing with data
5:16:01analysis or modeling data so just this
5:16:04uh just keep those things in mind this
5:16:07uh abbrevation of M and then n m
5:16:09equations and unknowns because this will
5:16:12become very handy and the same also
5:16:14holds for this indexing just to keep in
5:16:16mind that this I and this J what those
5:16:20indices are and how for instance the
5:16:23first you know the I the first index
5:16:27changes when we go from up to the bottom
5:16:30and how the second index J goes and
5:16:32changes when we go from left to the
5:16:35right when we go through the columns but
5:16:38we are going to it is also in the uh
5:16:40upcoming slides so uh we can we will
5:16:43have time to practice
5:16:45it so um this is what we are calling a
5:16:48coefficient labeling the coefficient uh
5:16:51a i j so this thing in a linear system
5:16:56they are labeled where the first index
5:16:58represents the row and the second index
5:17:01denotes the column so when we see a i j
5:17:07we know that this
5:17:09refers to the row and the J refers to
5:17:14the
5:17:15column so this is something that we use
5:17:19in order to understand where exactly in
5:17:21our metric something that we can we will
5:17:24see very soon where exactly our unit or
5:17:29our uh member that is part of our Matrix
5:17:33where exactly is that located in which
5:17:36row and in which column
5:17:39the systematic labeling is super
5:17:41important because this helps us to keep
5:17:43the structure and this helps us to
5:17:45understand uh what does this uh
5:17:48coefficient represent what what is this
5:17:50row that it belongs and what is the
5:17:52column it belongs so for which equation
5:17:55and for which unknown we have already
5:17:57solved the problem such that we can know
5:18:00what this uh coefficient
5:18:04represents so before moving on onto the
5:18:07actual linear systems and the definition
5:18:09of metrices let's quickly understand
5:18:11this distinction between homogeneous and
5:18:13non-homogeneous because this will help
5:18:15us to also get an understanding how we
5:18:17can solve a system of linear systems so
5:18:21a system is homogeneous if all the
5:18:23constant terms b i are zero otherwise
5:18:28it's non homogeneous so identifying this
5:18:31helps us to really understand the nature
5:18:34of the solution set that we need to get
5:18:37and to understand what kind of strategy
5:18:39we need to use in order to solve this
5:18:42problem now what do I mean by
5:18:44bi we is so that we had this system of M
5:18:49equations with n unknowns and we saw
5:18:52that that we have in the right hand side
5:18:55this B1 B2 up to BM which means that we
5:18:58had this m different equations with n
5:19:02different unknowns and to find a
5:19:05solution to the system it means finding
5:19:08this value
5:19:11values corresponding
5:19:17to X1 X1 here X2 X2 xn so basically
5:19:22finding the set of X1 X2 up to xn that
5:19:26solves this problem and for us to know
5:19:29how to solve this problem we need to
5:19:30know whether this B1 is equal to zero or
5:19:34not this B2 is equal to zero or not and
5:19:39then this BM is equal to zero or
5:19:42not this is very similar to this idea of
5:19:44solving any sorts of um problems that
5:19:48contain unknowns for instance if we have
5:19:52three
5:19:53x is equal
5:19:55to let's say five solving this is
5:20:00entirely different than if we know that
5:20:01the tree exal to
5:20:05Z
5:20:07so this is a simplified version of
5:20:10course but the IDE is the same knowing
5:20:12that this B1 B2 up to BM this R zero
5:20:17this gives us an idea how we can solve
5:20:20this problem and later on we will see
5:20:21this distinction between non-homogeneous
5:20:23and homogeneous system and whenever
5:20:26these BS so whenever this B1 B2 up to BM
5:20:30whenever these BS are zero then we are
5:20:33saying that the system is homogeneous
5:20:35and we need to solve a homogeneous
5:20:37system otherwi wise we are dealing with
5:20:39nonhomogeneous system so this means that
5:20:42the bis are not all zero let's now move
5:20:46on to the second module which is about
5:20:48the matrices so we are going to define
5:20:50the Matrix we are going to see the
5:20:52definition of it as well as the notation
5:20:54this idea of rows columns
5:20:57Dimensions uh some of which we have
5:20:59already touched upon but we are going to
5:21:02uh go into the depth of it we are going
5:21:04to learn properly as well as we are
5:21:06going to see many examples then we are
5:21:09going to talk about Matrix types so here
5:21:11we will talk about identity Matrix
5:21:13diagonal matrices and also special type
5:21:16of matrices like matrices containing
5:21:18only zeros and only
5:21:20ones so by definition add a matrix is a
5:21:25rectangular array of real numbers that
5:21:28are arranged in rows and in columns for
5:21:33example an M byn Matrix a can be
5:21:37represented as follows so let's look
5:21:41into this definition and this reference
5:21:44to Matrix we call this Matrix or
5:21:49Matrix
5:21:51a and every Matrix it can be described
5:21:55by this rows and columns where we always
5:22:00have this uh way of describing this
5:22:04Matrix always should be
5:22:06defined by the
5:22:09number of rows and number of
5:22:14columns so this is super
5:22:16important and let's look into this
5:22:19specific Matrix so we have a matrix a
5:22:22and all these values they are members of
5:22:25this Matrix they form the
5:22:28Matrix and we already saw this labeling
5:22:31of a i j where we said that I is
5:22:36referred to the row so you might recall
5:22:39that those were all these equations that
5:22:41we got so this horizontal lines where I
5:22:47was equal to 1 I I was equal to two I
5:22:50was equal to three up to the point of I
5:22:52was equal to M and then we had this J so
5:22:58this thing and then J was referred to
5:23:02the
5:23:04columns and we had J
5:23:09was here one and then two and then three
5:23:12up to the point of n so one 2 3 and
5:23:19N this is exactly what you can see here
5:23:22so in this Matrix we got all these
5:23:24elements a11 is a number a12 is a number
5:23:27up to the a1n is a number those are all
5:23:30real numbers and one thing that you can
5:23:33notice here is that here we got a11 so
5:23:37this is our first row and First Column
5:23:40here we got A1 two this is our first row
5:23:45and second column and then we got up to
5:23:49the point of a1n actually let me just
5:23:52write this down even at a bigger scale
5:23:55such that I can make more
5:24:00noes so let's assume we have this Matrix
5:24:06a and this Matrix a
5:24:13a if I'm
5:24:16bigger and we got all these different
5:24:19elements so we start with our first row
5:24:23and here we have A1 1 so
5:24:28here the row that I will write with
5:24:33let's say
5:24:35with blue the r is equal to 1 and then
5:24:41the column is one so this is Row one
5:24:47this is Row one row one and this is
5:24:51column
5:24:53one let me write it with red this is
5:24:58column
5:24:59one this is column
5:25:01two this is column three dot dot dot and
5:25:06this is column n
5:25:09and this is row two this is Row three
5:25:14dot dot dot and this is row M so in
5:25:17total I got M rows and N columns I will
5:25:24come to this notation that I'm putting
5:25:27here later for now let's keep track of
5:25:30the rows and the columns to get a good
5:25:32understanding what this indices were
5:25:34about that we just
5:25:36learned so every time I will also
5:25:38mention this reference to a i j to keep
5:25:42track of this and also let me write it
5:25:45with the right colors so a i this is the
5:25:51row and
5:25:53J which is the
5:25:56column so all the elements I'm just
5:26:00defining by this a because it just a way
5:26:02to reference a part that comes from a
5:26:05matrix it's a just common way to write
5:26:07the higher matrix by capital letter A
5:26:11whereas its members we will write with
5:26:13the um with the lower case
5:26:18a so this is
5:26:21Matrix
5:26:23Matrix
5:26:25a all right so here in the second row
5:26:30But First Column we got
5:26:33a two and then one because it is still
5:26:38in the First Column and then when it
5:26:40comes to this
5:26:43element we have here
5:26:46a the row is the first one because we
5:26:49are in the first
5:26:50row but then we are in the second column
5:26:53so this one should be
5:26:56two then we go on to the next element in
5:26:59our first row so a
5:27:03one and then
5:27:06three and then dot dot
5:27:10dot the last element is an a as we are
5:27:14still in the first row it will be one
5:27:17the I but then given we are in the last
5:27:20column the column index or the J will be
5:27:23equal to
5:27:25n because we got in total n
5:27:29columns so we are now ready to go into
5:27:32the second row so here given that we
5:27:36already have our first element
5:27:40a21 this is in our second row and the
5:27:43First Column so the I is equal to here
5:27:46two and G is equal to 1 let's now write
5:27:49down the element in the second draw
5:27:52second column as you might have already
5:27:54guessed I is equal to here 1 I is equal
5:27:56to here two s and then uh the J is equal
5:28:00to
5:28:022 and then we go on to the next element
5:28:05which is in the second row and the third
5:28:07column so it's a the
5:28:11row index is 2 so I is equal to 2 and
5:28:16then the column index is three dot dot
5:28:19dot and then we
5:28:22got a as we are in the second row it is
5:28:27the I is equal to two and as we are in
5:28:31the last column the J is equal to n now
5:28:34you might have already guessed when I
5:28:36was writing this down that whenever you
5:28:38are in the row and you move on to all
5:28:41the elements in the same
5:28:43Row the I so the row index it stays the
5:28:47same only you need to uh update the
5:28:50column index so here for instance you
5:28:52got one one one here also one so all the
5:28:56way down in the same row or one which
5:28:59logically makes sense because we are in
5:29:01the same row so the row index should not
5:29:04change but instead you should change the
5:29:06column index like here column one column
5:29:08two column three all the way to column n
5:29:11so those are our
5:29:16columns dot dot dot so let me make this
5:29:22distinction and
5:29:24those are our
5:29:28rows as you can
5:29:32see so this kind of mentally helps us to
5:29:35understand why we are writing all these
5:29:37indices
5:29:39over time once you practice more with
5:29:41this this will become more
5:29:48natural very quickly remove
5:29:51this so now our ride rest very quickly
5:29:55so as you might have already guessed we
5:29:57are in the third row so we have a tree
5:30:00so everywhere I will just write down
5:30:05the ace so first write down the A's and
5:30:10then the rows the row index will stay
5:30:13the same as I in the same row but then I
5:30:16will increase the columns gradually so
5:30:18we are in the column one and the column
5:30:19two column three up to the column n so
5:30:24now the remaining stuff you can actually
5:30:26write down yourself to just
5:30:29practice let's now move on on to the
5:30:32last row and last column so in the last
5:30:35row we got a a a up to
5:30:40here and in the last show the uh row
5:30:44index is M which means that here I need
5:30:47to have M M M everywhere I need to have
5:30:51M and then the column index is 1 2 3 all
5:30:57the way to
5:31:00n so this last column is very
5:31:03interesting too you can see here that we
5:31:06have the opposite of what we have here
5:31:10because in the last column we see that
5:31:14the uh column index is the same so it is
5:31:19everywhere n Only the first index the
5:31:21index of the row it changes it goes from
5:31:241 2 3 up to M which is of course logical
5:31:27because we said that in the last column
5:31:29if we are looking it from the
5:31:30perspective of column so all these
5:31:33values this A's so the all the ends they
5:31:36are logical because they we are in the
5:31:38last column we are in the same column
5:31:41but then the row changes here we are in
5:31:42the row one here we are in the row two
5:31:44Row three of two row M therefore we have
5:31:47also at the end a m
5:31:50n now let's talk about this idea of
5:31:54MN we said that our Matrix
5:31:58a
5:32:01has
5:32:03M as a number of rows
5:32:08and n as a number of
5:32:12columns which you can see by the way
5:32:14also
5:32:16here so we always refer the dimension of
5:32:21a
5:32:22matrix so the
5:32:26dimension dimension of Matrix a by these
5:32:31two
5:32:34numbers so first we always write down
5:32:37the number of rows in this case
5:32:42M then as the second element we are
5:32:46writing the number of columns in this
5:32:49case n we are always putting this small
5:32:51X in between two kind of emphasize M by
5:32:56n Matrix and we most of the time use the
5:33:00square braces to Showcase that we are
5:33:02dealing with Dimension and in this case
5:33:05we are saying the dimension of Matrix a
5:33:08is equal to M byn so we are dealing with
5:33:12M byn Matrix this is a common convention
5:33:16used in linear algebra in mathematics
5:33:18General but also used in data science uh
5:33:21in machine learning artificial
5:33:23intelligence so whenever you are dealing
5:33:25with matrices a it is a common
5:33:27convention to talk about this idea of
5:33:30dimensions and the idea of Dimensions is
5:33:33super important when it comes to the
5:33:35idea of multiplication multiplying
5:33:38Vector with Matrix Matrix with Matrix so
5:33:41this dot product Dimensions play a
5:33:44central role in here so keep this one in
5:33:47mind once we uh get to the point of that
5:33:50products this one will become very handy
5:33:52so let's now look into a specific
5:33:54example where we see simple Matrix a so
5:33:57in this case you can see that we are
5:33:59dealing with a matrix that has a 2x3
5:34:01Dimensions so like we just learned
5:34:062x3 means that we
5:34:11got two
5:34:15rows and three
5:34:19columns that's something that you can
5:34:21also see here very quickly so you have a
5:34:24small Matrix on the small matrix it's
5:34:26really easy to actually count so you can
5:34:29see that we got Row one and row two and
5:34:32we got column 1 column two and column
5:34:35three so this basically confirms this
5:34:37Dimensions therefore we are also saying
5:34:40that we have a 2
5:34:43by three Matrix and like usual we first
5:34:48write down the number of rows and then
5:34:50the number of columns you can see here
5:34:53that here we have this elements for our
5:34:56Matrix so a is equal to 1 2 3 for the
5:34:58first row and then uh 4 5 6 for the
5:35:01second row so from this actually I think
5:35:05it's a good exercise to just uh very our
5:35:08understanding of indices and from this
5:35:11um we can write down that for instance
5:35:13all these different elements uh like a 1
5:35:171 is equal to 1 A1 2 which means that we
5:35:21are in the first row and in the second
5:35:24column so we have this element is equal
5:35:29to two and then we got a and then one
5:35:34Tre so we are in the third column so
5:35:37this one 1 is equal
5:35:40to 3 and then a 21 is equal to 4 a 22 is
5:35:48equal to 5 and then a 23 is equal to 6
5:35:53so this is actually a good way to
5:35:56practice our understanding of indices
5:35:58our understanding of this Matrix
5:36:00structure and the understanding of
5:36:02dimension of the Matrix which in this
5:36:04case is 2x3 so this is yet another
5:36:07different definition of a matrix
5:36:09structure when it comes to the rows coms
5:36:11and dimensions so this is exactly what
5:36:14we just spoke about on our example and
5:36:17let's just quickly look at the formal
5:36:19definition so the rows of a matrix are
5:36:22the horizontal lines of the of the
5:36:24entries while the comms are the vertical
5:36:27lines so basically it's saying those are
5:36:32let me remove
5:36:34this so the rows are are the horizontal
5:36:39line and the columns are those vertical
5:36:42lines those are the columns this helps
5:36:45us to form these columns so column one
5:36:47column two and column three whereas this
5:36:49horizontal lines it helps us to create
5:36:51the rows so Row one and row
5:36:54two so then we have the dimensions of
5:36:56Matrix are given by the number of rows
5:36:58and columns it has so an M by n Matrix
5:37:02has M rows and N columns that's
5:37:07something that that we already
5:37:08saw so let's look into some special type
5:37:12of matrices one Matrix type is the
5:37:16identity Matrix so we saw before we had
5:37:18this Identity or unit Vector now we have
5:37:23identity Matrix so the two are quite
5:37:26similar so like before when we had our
5:37:29unit vectors we had this for instance E1
5:37:34in three dimension we had 1 0 0 then we
5:37:37had our E2 which had 0 1 0 and then we
5:37:42had our E3 which was 0 01 so you might
5:37:46recall this about our identity vectors
5:37:49or we were calling it unit factors you
5:37:52might notice very quickly that we have
5:37:55formed an identity Matrix i n which is a
5:37:59square Matrix with one on the diagonal
5:38:02and zeros elsewhere is basically a
5:38:05matrix that is built using those unit
5:38:08vectors so here we have E1 here we have
5:38:11E2 and here we have
5:38:13E3 so you can also see that this
5:38:183x3 Matrix because we got three
5:38:22rows and three
5:38:27columns so you can see that here we
5:38:31have on the diagonal so we call this
5:38:35diagonal on this diagonal we have all
5:38:38ones and in
5:38:40here outside of the diagonal they are
5:38:43all zeros and this is the definition of
5:38:46identity Matrix it is this i n Matrix
5:38:49where n is the dimension of a matrix and
5:38:54given that it's a square Matrix it means
5:38:56that the dimension of it is n by n so
5:38:59all the rows so the number of rows is
5:39:01equal to the number of columns on the
5:39:04diagonal we have all these ones and
5:39:06every where else we got
5:39:09zeros and do note that we are forming
5:39:12this identity Matrix simply by combining
5:39:15these different uh unit vectors so like
5:39:18here E1 E2 and
5:39:21E3 so let me actually uh give you yet
5:39:24another example but of much higher
5:39:28Dimension so of this identity Matrix so
5:39:31let's say we have I and then this I
5:39:38uh let us actually use this notation i
5:39:41n so let's say we got i
5:39:47n what this means is that we got
5:39:50actually this large
5:39:52matrix it's a square Matrix which means
5:39:56that it is n by n so it has n as the
5:40:01number of rows
5:40:10and n as number of
5:40:13columns so the dimension is n by n you
5:40:17got n as number of columns too because
5:40:20it's a square and let us actually write
5:40:24down that how that Matrix looks like
5:40:27it's a large Matrix the N is the size of
5:40:30that Matrix so here we got on the
5:40:34diagonal we got one here we got one here
5:40:37we got one dot dot dot up to the last
5:40:39point one and the index of this one here
5:40:45so this is the first row this the First
5:40:47Column
5:40:51basically and everything else is simply
5:40:55zero so here we got z0 0 dot dot dot
5:41:00zero here we got 0 0 all the way down to
5:41:04zero here also zero all the way down to
5:41:06Z
5:41:08and then here also zero so everywhere
5:41:11here and here we all got zeros only on
5:41:15this
5:41:16diagonal we actually got
5:41:19once so basically by using our common
5:41:24notation we can say that in the D in the
5:41:28identity Matrix we got a 1 1 = to a 22 =
5:41:34to a 33 equal to all the way to a NN
5:41:40equal to 1 and then when it comes down
5:41:45to the rest of
5:41:48the cases so all the other
5:41:52observations let's
5:41:55say a
5:41:5921
5:42:02a 31 or a 41
5:42:08anything so anything that is not um a11
5:42:12or a22 anything that is not on the
5:42:14diagonal it is simply equal to zero we
5:42:17also say in those cases that a i j is
5:42:21equal to
5:42:241 if I is equal to J because then it
5:42:29means that we are talking about item
5:42:31that is on diagonal because both the row
5:42:33index is equal to the column index
5:42:37otherwise
5:42:40the a i
5:42:43j is equal to Zer if I is not equal to
5:42:50J so this is in the nutshell how a large
5:42:54identity Matrix in general can be
5:42:58defined so let's now move on to another
5:43:01type of Matrix which is the diagonal
5:43:03matrix so by definition a diagonal
5:43:06matrix is a matrix where all of diagonal
5:43:09elements are zero so what does this mean
5:43:14we is saw um example of a diagonal
5:43:18matrix which was our identity Matrix
5:43:21because identity Matrix is an example of
5:43:25a diagonal matrix and what do I mean by
5:43:27that in our just seen example we saw
5:43:31that only on the diagonal we had all
5:43:33these nonzero elements but the rest were
5:43:37all zeros so all the off diagonal
5:43:39elements were
5:43:41zeros like in here and in here exactly
5:43:44the same holdes for the diagonal
5:43:46matrices only unlike in the identity
5:43:49Matrix we no longer need to have this
5:43:52diagonal elements equal to one those can
5:43:55be any other numbers so as long as we
5:43:58have this um elements D1 D2 D3 that are
5:44:03not zeros but then of the diagonal
5:44:06numbers so all these elements they are
5:44:08zero then we are dealing with the
5:44:10diagonal matrix so in this case we got a
5:44:143X3 diagonal matrix because we have uh
5:44:17three rows and three columns and here we
5:44:21can see that the um the first so the a11
5:44:25the first element from the first draw
5:44:27and First Column is equal to D1 so a 22
5:44:32is equal to D2 and then a33 is equal to
5:44:37D3 so D1 D2 and D3 those are all so D1
5:44:43D2 and D3 those are all real
5:44:48numbers now when it comes to the uh this
5:44:51numbers for example it can be that D is
5:44:55let's say 2 five 6 on diagonal then we
5:45:00have those zeros this is a diagonal
5:45:04matrix it can also be
5:45:08that D is equal to minus 3 and then 0 0
5:45:15and then 5 8 and then here we have zeros
5:45:20so again we have on the diagonal all
5:45:23these elements and the off diagonal
5:45:26elements so if all the off diagonal
5:45:28elements are zero then we are dealing
5:45:30with diagonal matrix and if you
5:45:33wondering well what happens if on the
5:45:35diagonal we got zero do we still have a
5:45:38diagonal matrix it's actually a great
5:45:40question but yes indeed we are dealing
5:45:43with the diagonal matrix as long as all
5:45:46the off diagonal elements are zero so
5:45:49for instance if we got D is
5:45:53equal here we have zero here we have 0 0
5:45:570 and then 7 and then 0o and then 8 and
5:46:02then 0 0 so we got this of diagonal
5:46:06elements so here are the diagonal
5:46:09elements and all the of diagonal
5:46:11elements are those given that all the of
5:46:15diagonal elements are zero which is the
5:46:18definition of the diagonal matrix then
5:46:20we can say that our D Matrix in here is
5:46:24indeed a diagonal
5:46:29matrix let's now look into yet another
5:46:31type of Matrix which is a special type
5:46:33of Matrix and it's called one's Matrix
5:46:37so by definition one's Matrix is denoted
5:46:40by 1 M1 so you can see here and here it
5:46:44mens the dimension of it so the number
5:46:46of rows and number of columns and it's a
5:46:49matrix in which all the elements are
5:46:52one so this is a very unique Matrix we
5:46:56often use it during the programming so
5:46:58in data science data analytics but also
5:47:01in um uh when creating like data
5:47:03structures when designing algorithms
5:47:06this becomes very very handy and this
5:47:09idea of one's Matrix is that all the
5:47:12elements are just one it means that if
5:47:15we want to create a placeholder in such
5:47:17way that we can then multiply any number
5:47:20in here with some other number and get
5:47:22that number then it can be done very
5:47:24easily because we know that when we
5:47:27multiply a number with one then we get
5:47:29that number so a * 1 is = to a x * 1 is
5:47:34= to X now this is a exactly this
5:47:37property exactly is what motivates us to
5:47:40create and to have this type of ones
5:47:43matrices it means that we are defining
5:47:47matrix by its Dimension so it is M by n
5:47:51and here the m is equal to two and then
5:47:54n is equal to three because we got two
5:47:57rows and three columns but you can see
5:48:00that all the elements are the same and
5:48:02they are equal to one so a11 is equal to
5:48:05A1 2 is equal to a13 is equal to a uh 21
5:48:11and is equal to a 22 and a 23 and they
5:48:15are all equal to one and this is the
5:48:18definition of one's Matrix you can have
5:48:21um on Matrix of the size 4 by 10 On's
5:48:27Matrix of the size
5:48:30th let say 10,000
5:48:34by 100 Etc so any number any real number
5:48:39so M and then n are real numbers you can
5:48:42use in order to create this large M by
5:48:46n1's
5:48:49matrix let's now look into our final
5:48:52special type of Matrix before moving on
5:48:54onto the next module which is about zero
5:48:57matrices so similar to this one Matrix a
5:49:01zero Matrix denoted by 0 m by N is a
5:49:06matrix in which all the elements are the
5:49:08same with the one difference that this
5:49:11time all the elements are equal to zero
5:49:13so in the on Matrix all the elements
5:49:15were ones but in the zero Matrix all the
5:49:17elements are zero this type of matrices
5:49:21become very handy also during the
5:49:23programming creating um different
5:49:25algorithms during design encoding um but
5:49:29for slightly different purposes usually
5:49:32we create the zero matrices as a
5:49:34placeholder such that in the beginning
5:49:36we can have this uh tups or we can have
5:49:38this um uh arrays or nested Loops um
5:49:42that we want to perform and then
5:49:45gradually add these values to the
5:49:47existing Mt array so if we create this
5:49:51Zer Matrix and um this is a placeholder
5:49:55then next time we can always add on this
5:49:58this new data that we get and then we
5:50:01know that zero plus a number is always
5:50:04equal to number which means that once we
5:50:07have this updated information of a we
5:50:09can add this to the zero and we will
5:50:11then have this new updated information
5:50:13in our system therefore the zero Matrix
5:50:16is often used as a way to uh have this
5:50:20placeholder with the provided Dimension
5:50:22where we can always add new information
5:50:25and the information can be
5:50:27updated so in this specific case we got
5:50:30um a zero Matrix that has two rows and
5:50:33three columns so you can see two rows
5:50:37and three
5:50:40columns so m is equal to 2 and then n is
5:50:44equal to three three perfect so we are
5:50:48done with module 2 and now we are ready
5:50:50to go on to our next module which is the
5:50:54core Matrix operations so when it comes
5:50:57to matrices we often perform Matrix
5:51:01additions Matrix subtraction but also
5:51:04Matrix um scalar multiplication of this
5:51:06Matrix so multiplying Matrix with a
5:51:08scaler and then Matrix um multiplication
5:51:12just in general so taking two matrices
5:51:15and multiplying them we are going to
5:51:17look into this concept in detail we are
5:51:19going to see many examples like before
5:51:22we are going to dive deeper into this
5:51:24such that we lay the ground on uh to the
5:51:27next module which is solving a system of
5:51:30M equations with an unknown so solving
5:51:33this General um linear system
5:51:37so for the beginning uh we will be
5:51:40looking into this Matrix operations
5:51:42where we are adding or subtracting
5:51:44matrices so by definition the sum of two
5:51:47matrices A and B of the same dimensions
5:51:50is obtained by adding their
5:51:51corresponding elements so by taking the
5:51:55element i j from both matrices and
5:51:59adding them to each other so in this
5:52:01case you can see that Matrix A and B are
5:52:04here and uh the uh definition says we
5:52:09just simply need to take the
5:52:10corresponding elements corresponding
5:52:12elements from the row I and the column J
5:52:16take them add them and this will become
5:52:19an element in our final um Matrix
5:52:24because when we are adding two matrices
5:52:27of the same size the result is yet
5:52:29another Matrix so we will use the Matrix
5:52:32a to add to Matrix B and this will give
5:52:36us a matrix A + B and this i j simply
5:52:42refers to the indices corresponding to
5:52:44the row and the
5:52:46column we will look into an example in a
5:52:48bit and this will make much more sense
5:52:51and the same holds also for the
5:52:53difference so by definition the
5:52:55difference of the two matrices A and B
5:52:57of the same dimensions is obtained by
5:52:59subtracting their corresponding Elements
5:53:02which means that in order to obtain this
5:53:05Matrix a minus B this is a new Matrix we
5:53:10simply need to look for each element so
5:53:13we are going to index them for a row I
5:53:16and J we are going to do this pairwise
5:53:19element wise subtractions we are going
5:53:22to see what is that element
5:53:24corresponding to the row I and column G
5:53:26in The Matrix a which we say is a i j we
5:53:31are going to subtract from this the
5:53:33element in the row I
5:53:36and column G that comes from Matrix B
5:53:39and this will give us our new Matrix
5:53:41which is a minus
5:53:44B so let's now look into an example in
5:53:47this Matrix Matrix um uh a and Matrix B
5:53:51are used and Matrix a is of the size 3x3
5:53:54Matrix 3 Matrix B is of the
5:53:57size 3x 3 in order to obtain a plus b
5:54:03what we are doing is that we are
5:54:05performing element wise additions now
5:54:08let's verify
5:54:10this so what we are doing here is that
5:54:13we are saying a plus
5:54:16b let me actually get a larger area
5:54:21here so let's say we have the two
5:54:23matrices I want to add the two in such
5:54:26way that we do everything one by one
5:54:28such that this idea of a plus b and
5:54:31addition of the matrices will make
5:54:33sense so we want to find out a plus b
5:54:37for that what we are going to do is that
5:54:40we are going to make use of this
5:54:42definition that A + B and then I J is
5:54:46equal to a i j+ b i j which is a fancy
5:54:53way or mathematical way or describing
5:54:55that for each element we need to go and
5:54:57look for the row I and column J and take
5:55:01that element from the um column from
5:55:04that uh Matrix a and from the Matrix
5:55:07B so this means
5:55:10that for a + b this is going to be a
5:55:16matrix that will have the same number of
5:55:18rows and the same number of columns as
5:55:20two matrices because both A and B are
5:55:233x3 which means also their sum is going
5:55:26to be 3x3 so this going to be 3x3 and
5:55:30here we are going to do so we are going
5:55:32to take for the first row in the First
5:55:36Column so for
5:55:40A+
5:55:42b 1 1 so first row and First Column we
5:55:46need to go to the first row and First
5:55:48Column of Matrix a and the first row and
5:55:51First Column of Matrix B and we need to
5:55:53add these two elements so we need to do
5:55:571 + 1 and then we need to go on to the
5:56:01second column so the first row and the
5:56:03second column which means that we need
5:56:05to be here
5:56:08in both
5:56:09matrices so here we have 0 + 2 and then
5:56:14we got 2 + 3 and then we got 0 + 0 so
5:56:18you can see it in
5:56:20here and then we have 1 + 0 and then we
5:56:23have 3 + 1 0 + 1 and then 0 + 2 and then
5:56:291 + three which gives us
5:56:37so 1 + 1 is = to 2 0 + 2 is = 2 and then
5:56:432 + 3 is = to 5 0 + 0 is equal to 0 0 +
5:56:481 is = to 1 1 + 0 is = to 1 0 + 2 is =
5:56:52to 2 and then 3 + 1 is = 4 1 + 3 is = to
5:56:574 which means that our A + B is equal to
5:57:01this Matrix that we got in here so you
5:57:05can see that we are getting exactly what
5:57:07we uh what we have here only we have
5:57:10done it manually one by one so the same
5:57:13idea holds exactly when we have a minus
5:57:16B only instead of adding you will have
5:57:19to do here minuses so minus minus so
5:57:23everywhere minus so 1 - 1 0 - 2 2 - 3
5:57:30Etc so let's look into another addition
5:57:33so in this case by definition it is
5:57:36defined as this element wise uh of the
5:57:39adding of these two matrices here the
5:57:42only difference in this definition is
5:57:44that it's saying it's calling this a
5:57:46plus b as C so this new Matrix that we
5:57:51are getting as a result of adding a to B
5:57:54it's calling C so basically it's the
5:57:56same as calling this Matrix as C you
5:57:59will see also this type of definitions
5:58:01so in this case The Matrix C is equal to
5:58:04a plus b which basically means that for
5:58:06each row with index I and with each
5:58:10column with index J go and look for row
5:58:14I and index J take the corresponding
5:58:16elements from Matrix a and Matrix B add
5:58:19them in order to get that corresponding
5:58:21element in our new Matrix C and you can
5:58:25see that in this example that's exactly
5:58:27what we are doing we have a we have B we
5:58:29are taking this element and this one so
5:58:321 + 1 we are getting here two and then 0
5:58:35+ 2 we are getting two here 2 + 3 is 5
5:58:39and then 0 + 0 is = 0 1 + 0 is = to 1
5:58:43and then 3 + 1 is equal to
5:58:464 so now we already go to the next topic
5:58:49which is about scalar multiplication of
5:58:51a matrix so by definition scalar
5:58:54multiplication of a matrix a by scalar
5:58:57Alpha results in new Matrix where each
5:59:00entry of a is multiplied by Alpha the
5:59:04idea of scalar multiplication matrices
5:59:06is actually quite similar to this idea
5:59:09of scaled multiplication in vectors so
5:59:12uh we have already seen in the lecture
5:59:15of the vector multiplication that when
5:59:17we were having this scaler C and we had
5:59:22this Vector
5:59:24a then uh when we are multiplying C
5:59:28which is a real number with Vector a
5:59:30then we simply need to take all the
5:59:33elements of vector a so A1 A2 all the
5:59:37way down to a n and we need to multiply
5:59:40them by this same scaler so
5:59:44see this is what we were doing with
5:59:47vectors and that's exactly the idea
5:59:49behind matrices and when uh doing the
5:59:52scalar multiplication of matrices only
5:59:55instead of multiplying only just one
5:59:58vector with this scaler C now we need to
6:00:02apply this to all the rows and all the
6:00:04columns so here we got this one column
6:00:06and Matrix is simply a combination of
6:00:09multiple vectors which means that we
6:00:11need to multiply all these elements of
6:00:14all the vectors of all the columns in
6:00:17this Matrix so let's actually look into
6:00:20a specific
6:00:22example so in this case we have a matrix
6:00:25a and this Matrix a is this thing and we
6:00:30have a scaler which is three so in here
6:00:32our Alpha is equal to three or you can
6:00:34call it C or anything so you can see
6:00:39that when we are scaling The Matrix with
6:00:43a scaler in this case Tre with this
6:00:46Matrix what we are doing is that we are
6:00:48simply taking each of these elements and
6:00:50multiplying it with this scum so 1 by 3
6:00:53is 3 2x 3 is 6 3x 3 is 9 and 4x 3 is
6:01:0012 this is the idea behind this entire
6:01:04scal multiplication ofation Matrix in
6:01:07more general terms if we for instance
6:01:10have a matrix a so let's actually look
6:01:13into a high level General example where
6:01:16we have a DA Matrix M by n so we got M
6:01:21rows and N columns and we want to get a
6:01:24scal multiplication of this Matrix and
6:01:28um scaler that we have here as in our
6:01:31definition it is defined by this alpha
6:01:34alpha is just a number you can qu C you
6:01:36can qu B anything so in this case our
6:01:40scaler
6:01:42alpha alpha is coming from R so it's a
6:01:46real number so Alpha time a is then
6:01:52simply equal to to this new
6:01:56Matrix where all of these elements are
6:02:01simply multiplied by this scal so I will
6:02:04just take over all these values H1 up to
6:02:09a M1 and then A1 2 a22 all the way down
6:02:15to a M2 and then let me also add the
6:02:19last column just for fun here a 2 N and
6:02:25then here a m
6:02:30n so here this new scaled M multiplies
6:02:35so so scaled uh Matrix a so Alpha * a is
6:02:40simply equal to Alpha time all these
6:02:42elements are simply multiplied by the
6:02:46scale it is as simple as
6:02:55that so that's the simple idea behind um
6:02:59Matrix as scaling so when you are doing
6:03:02scalar multiplication of this Matrix you
6:03:05simp take all the values and you
6:03:08multiply them element by element per row
6:03:11and per column by that single scalar
6:03:14Alpha do note that you are multiplying
6:03:17them all without exclusion with exactly
6:03:20the same number which is that
6:03:24Alpha so let's now look into the
6:03:27definition of matrix
6:03:29multiplication so here we are no longer
6:03:32multiplying a matrix with a scalar but
6:03:34we are multiplying Matrix with Matrix so
6:03:37the product of an M by n Matrix a and an
6:03:40N by P Matrix B results in an M by P
6:03:44Matrix C where each entry cig is
6:03:47computed as the dotproduct of the e Road
6:03:50of a and the J column of B now what does
6:03:54this mean firstly let's look and unpack
6:03:58this part of the definition so we got
6:04:02Matrix
6:04:04a that is
6:04:06M by n and then we got Matrix B which is
6:04:11n by P what this means is that in this
6:04:16case Matrix a has M
6:04:19rows and N
6:04:22columns and Matrix B has n
6:04:26rows and P
6:04:30cups so this is then simply the
6:04:34dimension dimension of the two
6:04:39matrices so then it's saying that by
6:04:43definition the product of these two
6:04:45matrices so the product of A and B the
6:04:50product of the
6:04:57two
6:05:02B is equal to to this Matrix
6:05:07C and each entry cig
6:05:11J so c i j is computed as the dotproduct
6:05:18of the each row of a and the Jade column
6:05:23of B now this part might seem bit
6:05:27difficult but once we look into the
6:05:29actual example and we illustrate this on
6:05:32our common high level General
6:05:34expressions of Matrix am and their
6:05:36multiplication this will make much more
6:05:39sense for now before coming to this one
6:05:43I just wanted to refresh our memory on
6:05:45one thing I said before when discussing
6:05:48also this idea of improving uh this uh
6:05:51different properties of vectors that
6:05:54when we want to
6:05:55multiply a vector with a matrix or
6:05:58Matrix with Matrix or vector with a
6:06:00vector we need to ensure that from the
6:06:04first element the number number of comms
6:06:06is equal to the number of rows of the
6:06:07second element this is also very
6:06:10important for this specific case and
6:06:12just in general for matrix
6:06:14multiplication so you can notice here
6:06:17that the number of coms here is equal to
6:06:20the number of rows in here and the order
6:06:24is very important so in case of matrix
6:06:27multiplication the order is really
6:06:29important which means that if you have a
6:06:32matrix a and you want to multiply with
6:06:36the Matrix B then
6:06:38the number of
6:06:43columns of
6:06:46a should be equal to the number of
6:06:52rows of
6:06:54B otherwise you cannot multiply those
6:06:59two matrices with each other so in case
6:07:01you got a matrix a that doesn't have the
6:07:04same number of columns as the rows of
6:07:06number of the Matrix B then there are
6:07:09some alternative things that you can do
6:07:11including this idea of the transpose
6:07:13that we saw also doing when Computing
6:07:15the dot product between this Vector a
6:07:17and Vector B that's something that we
6:07:19also do in programming when we are
6:07:21dealing with this Matrix and we want to
6:07:23compute this relationship between two
6:07:26matrices but the number of columns of
6:07:29one of the first one is not equal to the
6:07:31number of rows of the second one we are
6:07:33simply uh manipul ating this matrices or
6:07:36removing some data if that's not hurting
6:07:39our problem maybe uh flipping so
6:07:42transposing our Matrix or applying any
6:07:45other source of operation to it to
6:07:48ensure that the two matrices that we are
6:07:50multiplying with each other the first
6:07:53one's number of columns is equal to the
6:07:55second one's number of rows that's just
6:07:58the low and that's something that you
6:08:00should follow if you want to multiply
6:08:02these two
6:08:03matrices all right so now let's move on
6:08:06onto this idea of multiplying and Dot
6:08:08product let's look into a specific
6:08:11example and this will uh help us to
6:08:14understand this process
6:08:16better so before doing that I just want
6:08:20to quickly show you this general idea so
6:08:23if we have a matrix a that is M by n
6:08:28which means that it looks something like
6:08:30this like
6:08:31A1 1 a 2 one up to the point of a
6:08:37M1 and then here we got let's say a 1 2
6:08:42a 22 up to the point of a M2 and then at
6:08:48the end we got a MN and here we got A1
6:08:58n so let me also add this one 2 N and we
6:09:03got a matrix B this Matrix B is n by P
6:09:08so it has n rows and P columns so we are
6:09:13fine in terms of Dimension
6:09:15here and we got here
6:09:18b11 B21 up to the point of b m sorry b n
6:09:25in this case let's not confuse the
6:09:28letters so b n 1 B1 2 B 22 up to the
6:09:35point of b n 2 because n now is the
6:09:40number of rows for Matrix B unlike for
6:09:43the Matrix a up to
6:09:46B1 p and here b 2 p and here after the
6:09:52point of B and then n p this is the last
6:09:56element in order to perform um
6:09:59multiplication between these two
6:10:00matrices so to obtain a matrix C which
6:10:05is a equal to a *
6:10:08B what we need to do is we simply need
6:10:12to take pair case or pair Row for the
6:10:15row I for
6:10:18instance we need to take this element so
6:10:21this row and we need to multiply it with
6:10:24this so we need to find the dot product
6:10:26between this row and this column then we
6:10:30need to move on on to the next one and
6:10:33then for the second element we will then
6:10:36take this row and we will multiply it
6:10:40with this
6:10:43one so this is then something that we
6:10:46need to do in order to obtain these
6:10:49elements and you might have already
6:10:50noticed that we got this m by n and n by
6:10:54P so you might have already guessed what
6:10:57will be the dimension of the C if we got
6:11:00that the dimension of a is equal to M by
6:11:04n
6:11:06and the dimension of B is equal
6:11:112 N by P then the
6:11:15results Matrix after M multiplying the
6:11:18two so Matrix c will be will be having a
6:11:22number of rows equal to this and the
6:11:25number of columns equal to this so This
6:11:27middle part basically disappears and the
6:11:30number of rows of the first Matrix will
6:11:33be then the number of rows of this
6:11:34result Matrix C and the number of
6:11:37columns or the second Matrix so Matrix B
6:11:40will then be our final number of columns
6:11:42so we will then have a matrix C that
6:11:44will have a
6:11:46dimension so
6:11:50Dimension so dimension of C will then be
6:11:54equal
6:11:57to M
6:12:00by P so we will have M rows and P
6:12:07columns so how we are going to compute
6:12:11this so for c i j which means row
6:12:18I and column J let's look into the
6:12:23definition of it it's saying c i j is
6:12:26computed as a dotproduct of the each row
6:12:30and the Jade column so each row from a
6:12:32and Jade column of B what where is the
6:12:35each Road of a the each Road of a is
6:12:39somewhere here so each Road of
6:12:42a it is uh the A and then
6:12:50I
6:12:53one then
6:12:56a and then I 2 and then a and then I Tre
6:13:01dot dot dot and then
6:13:03a i and then then we got in total n
6:13:07columns
6:13:09n and we always do the transpose right
6:13:12when Computing this um dot product so we
6:13:15then take the transpose so we take this
6:13:19row row I and we multiply it so we do
6:13:23the dot product between this one this is
6:13:25the a
6:13:28i and the
6:13:32B J
6:13:35this is column J it is somewhere
6:13:39here so it is B and then we got the
6:13:44first element which is one and then J
6:13:48and then b 2 J
6:13:51B 3j dot dot dot up to B and then in
6:13:56total we got n rows in B so n and then
6:14:02the J is the
6:14:05column so it stays the
6:14:08same so this is then the dot product
6:14:13between
6:14:14e row that comes from Matrix a and the J
6:14:19column that comes from Matrix B so it's
6:14:22always like that actually so we always
6:14:24take row by row so we
6:14:29take this different so every time we
6:14:32take just a row
6:14:35and we multiply with the corresponding
6:14:38column and then we get the dot product
6:14:41between this row that comes from the
6:14:42first Matrix and then the column that
6:14:44comes from the second Matrix in that
6:14:46specific order in order to get our DOT
6:14:48product and that specific valum and what
6:14:51is this amount actually
6:14:54so when we calculate this do product you
6:14:57can quickly see that we have a
6:15:02i1 multiplied by B 1 J plus a I2
6:15:09multiplied by b 2 J and then dot dot dot
6:15:14a i n multiplied by b n
6:15:19g and this new Matrix
6:15:23c will then have all these elements so
6:15:26C11 C 21 and then c31 dot dot dot and
6:15:31then
6:15:32C the last
6:15:35row as the number of rows of C is m c m
6:15:41see here M so C and then here it will be
6:15:45one 2 C 22 C3 and then 2 up to the point
6:15:51of
6:15:53cm and then two and then here the last
6:15:56col will be C1 and then p is the number
6:16:00of coms in C so C1 p and then C2 p and
6:16:04then
6:16:05c3p dot dot dot and then
6:16:10c m and then
6:16:15P
6:16:17okay so this is what we get this is our
6:16:20final Matrix C when multiplying Matrix a
6:16:25and Matrix B so let me clean this
6:16:32up C is to now if you want to find out
6:16:37what is C11 you can easily fill in this
6:16:40general formula that uh that we just
6:16:42calculated the I is equal to 1 and then
6:16:45J is equal to 1 and this will give you
6:16:47C11 by using this formula if you want to
6:16:51get the C and P then just fill in the I
6:16:53is equal to M and then J is equal to P
6:16:56in order to get this value C and P so
6:16:59you can already see the amount of
6:17:01calculations you need to do in order to
6:17:03get all these elements from this l large
6:17:05matrices A and B let's actually look
6:17:07into a simple example to clarify this so
6:17:10we have a matrix a here and Matrix B
6:17:13here and we want to do a multiplication
6:17:15of the two and we have just learned how
6:17:17to do it let's actually do it one by one
6:17:19so we got a matrix a which is equal
6:17:24to 1 2 3 4 with Dimensions 2 by 2 then
6:17:31we got a matrix B which has values two Z
6:17:36and then one two so it is 2 by two and I
6:17:41want to find what is c that is equal to
6:17:44a * B and I know already by looking at
6:17:47these Dimensions that c is going to be
6:17:50equal to 2 by 2 so you might recall that
6:17:54I said that when looking at this final
6:17:56result the number of rows or the final
6:17:59um Matrix will be this so the number of
6:18:03rows of the initial Matrix a and then
6:18:06the number of columns of this final
6:18:07Vector c will be the number of vectors
6:18:10number of columns of this second Matrix
6:18:13B so two therefore I know already before
6:18:17even doing calculations that the uh
6:18:20product Matrix c equal to a * B is going
6:18:23to have a dimension 2x two let's
6:18:25actually do a calculation to check this
6:18:28so C is then equal to a * B and it's
6:18:31equal
6:18:32to 1 2 3 4 4 multiplied by 2 0 1
6:18:392 okay
6:18:47so I expect to have four different
6:18:49elements here here here and here so to
6:18:54obtain the C11 so it is C11 in here what
6:18:59I need to do is that I need to look at
6:19:02the first row and in the first column in
6:19:06here so first row from a and the First
6:19:09Column of B and I'm doing the dot
6:19:11product which means 1 * 2 + 2 * 1 1 * 2
6:19:16is 2 2 * 1 is 1 so here I'm getting 1 *
6:19:212 + 2 * 1 which basically gives me 2 + 2
6:19:28and that's equal to 4
6:19:37so here I'm just writing
6:19:39[Music]
6:19:43down 1 * 2 + 2 * 1 now when I want to
6:19:50get this value which is
6:19:52C12 this means that I want to get the
6:19:55first row and the second column and
6:19:57that's exactly what I'm doing so I'm
6:19:59going back and I'm saying let's look at
6:20:03the first row but this time will look at
6:20:05the second column coming from the uh
6:20:07from The Matrix B so 1 *
6:20:120 0 + 2 * 2 and then I do the same only
6:20:17this time for the second row which means
6:20:19I'm picking this row and then this
6:20:23column
6:20:25so it
6:20:27is three * 2 + 4 * 1 and for the final
6:20:33element c22
6:20:38I'm taking the second row and the second
6:20:45column which gives me three * 0 plus 4 *
6:20:522 now what does this gives
6:20:58me this gives me this 4x4 Matrix where 1
6:21:03* 2 + 2 * 1 is 4 1 * 0 + 2 * 2 is 4 3 *
6:21:092 + 4 is = to 6 + 4 which is
6:21:1410 and then 3 * 0 + 4 * 2 is = to
6:21:188 so let's check 4
6:21:224108 that's exactly what we have here so
6:21:26as you could see here the idea is that
6:21:28every time to follow what element I'm
6:21:31looking for for the CI J and then I just
6:21:34go to the E rows from the first Matrix
6:21:38and the J column from the second Matrix
6:21:41and I do the dot product of the A and
6:21:45then
6:21:46I and then K let's say so I'm going to
6:21:51the E Row from the first Matrix and I'm
6:21:55taking all the elements which means I
6:21:57don't even need to mention this index it
6:22:00just means the entire each row coming
6:22:03from the Matrix a and then I'm doing the
6:22:05dot product between this row and the
6:22:11column that comes from the Matrix B
6:22:14which means B and then
6:22:17J which then will give me the cig so I'm
6:22:22looking at this and taking this
6:22:23multiplying this dot product and this
6:22:25gives me the first element then the
6:22:27first row and then the second column
6:22:29which gives me the uh second element in
6:22:32the first row in my Matrix so this one
6:22:34and so on so hope this makes sense uh if
6:22:38it doesn't make sure to reach out
6:22:40because it's a very important concept
6:22:43and uh let's also look into another
6:22:45example to make sure that we got this
6:22:47right so in this case as you can see we
6:22:49have another matrices so set of A and B
6:22:52matrices again 2 by two a simple
6:22:55one and we want to know what is a so
6:22:59let's say we call this C we already know
6:23:01C should be 2 by 2 and what we are doing
6:23:05is basically for
6:23:07C11 we are saying let's look at the
6:23:11first row so first row and the First
6:23:15Column coming from the second Matrix B
6:23:19and let's do the dot product so 2 * 1 2
6:23:23* 1 2 * 1 4 * 5 4 * 5 we get this and
6:23:29then when we want to find what is C oh
6:23:34what is C and then one two so in the
6:23:37first row but in the second element in
6:23:39our final Matrix so I is equal to one
6:23:42and J is equal to 2 it means we need to
6:23:45look at the first row from The Matrix a
6:23:50but this time
6:23:52the second column from The Matrix B so
6:23:56it is 2 by 3 2 by 3 4 * 7 4 * 7 and this
6:24:03gives us a number 13 four even if you
6:24:05calculate you can see that 2 * 1 is
6:24:07equal to 2 4 * 5 is 5 so 2 4 * 5 is 20
6:24:11so 2 + 20 is 22 in here and then you can
6:24:15do the rest of calculations and this
6:24:17will be a good practice to see how we
6:24:20can do a basic matrix multiplication the
6:24:23idea is actually quite straightforward
6:24:25when it comes to multiplying it it just
6:24:27it comes with a practice when we see all
6:24:30these uh much bigger matrices
6:24:34so um this is another example I will
6:24:37leave this one to you to complete it
6:24:40just uh to keep in mind we always do uh
6:24:44so we always look at the dimension first
6:24:46in here 2 * 2 and 2 * 2 which gives me
6:24:49an impression already what I can expect
6:24:51the result will be 2 by two and when it
6:24:55comes to the uh cross elements just
6:24:58ensure to always look to the E row and
6:25:02the J column
6:25:05this comes from Matrix a and this comes
6:25:07from Matrix B take them compute the dot
6:25:09product and then you will find your C uh
6:25:13your final result let's call it
6:25:17um kig because in this case we have a
6:25:20matrix C already welcome to the module 4
6:25:23of this course when we are talking about
6:25:26matrices and linear systems so in this
6:25:28module we are going to dive deeper into
6:25:30this uh idea of linear systems with
6:25:33matrices and solve linear systems using
6:25:35different techniques and specifically we
6:25:37are going to learn the uh concept behind
6:25:41solving linear systems using matrices
6:25:43named gausian elimination and gaussian
6:25:46reduction welcome to the module one in
6:25:48this unit so in this uh case we are
6:25:51going to talk about algebraic lows for
6:25:53matrices we are going to discuss four
6:25:55different properties for matrices uh and
6:25:59the first one is the communative laow
6:26:00for Matrix addition the associative law
6:26:03for matrices the distributive laow for
6:26:05matrices both the left and the right one
6:26:07and then finally we're going to talk
6:26:09about the scalar multiplication laow for
6:26:12matrices so the algebraic lows or
6:26:14matrices they are like in case of real
6:26:17numbers like in case of vectors they
6:26:19help us to do different operations on
6:26:22these entities they are very similar to
6:26:24the real numbers and the vector cases
6:26:27where we for instance um so that if for
6:26:30instance A + B uh is equal to B+ C C or
6:26:34a * um b + C is equal to a b + a c those
6:26:40are all sorts of lows that we uh learn
6:26:43as part of high school prealgebra and we
6:26:46have applied it to real numbers we know
6:26:48how helpful those can be and similar
6:26:51type of lows we have also for the
6:26:54matrices and we got in this case four
6:26:58different laws that we will be
6:27:00discussing the first one is what we are
6:27:01referring as associative law the second
6:27:04one is the distributive low the SEC the
6:27:07third one is the scalar multiplication
6:27:09low and the fourth one is the
6:27:10communative low for addition so these
6:27:12laws help us to do different metrix
6:27:14operations they help us to manipulate
6:27:17algebraically this metrices and then uh
6:27:20this can help us to solve different
6:27:22sorts of problems including solving a
6:27:24system of linear
6:27:26equations so let's start with the
6:27:28commutative law for Matrix addition so
6:27:31the Matrix addition is cumulative um
6:27:34which means that A+ B is equal to B plus
6:27:37a so unlike the matrix multiplication
6:27:40that we have seen in the previous
6:27:43lessons uh where the order did matter
6:27:46and we said that we um had to uh ensure
6:27:49that the number of columns of the first
6:27:51Matrix is equal to the number of rows of
6:27:54the second Matrix in case of addition
6:27:57that's this is not the case so we should
6:27:59not care about the order whenever we
6:28:01want to add two matrices the other thing
6:28:04that we need to keep in mind though is
6:28:06that the two matrices needs to have the
6:28:08same size so I mean that both Matrix a
6:28:13and Matrix B need to have a dimension so
6:28:17dimension of Matrix a should be equal to
6:28:19dimension of Matrix B and let's say
6:28:21should be equal
6:28:23to M
6:28:27by n but for the rest we don't really
6:28:30need to care uh which one we will put
6:28:33first
6:28:34will we put first a and then add the b
6:28:37or we will do the other way around so we
6:28:40will then First Take B and then we will
6:28:42add
6:28:43a so this is the idea behind communative
6:28:46low for Matrix
6:28:48addition so first let's look into all
6:28:50this uh lows and then we will also look
6:28:53into the corresponding
6:28:55examples so for this specific case it
6:28:58might actually also be helpful to write
6:29:00down the general formula which will um
6:29:03make sense out of this um low for the uh
6:29:08uh which is a communative low for the
6:29:10Matrix additions so let's say we got a
6:29:14matrix
6:29:15a which is M by
6:29:18n and this Matrix can be represented as
6:29:22a11 dot dot dot a M1 so this is
6:29:26something that we saw time and time
6:29:28again so I'll just quickly write it down
6:29:30the common notation for this and then
6:29:34here we have the last column which is a
6:29:37MN and this is the Matrix a then we got
6:29:42Matrix B which is again M by n and can
6:29:47be represented as
6:29:48b11 and then dot dot dot and then B M1
6:29:53dot dot dot b1n dot dot dot b
6:29:59MN
6:30:01so the communative lows says that A +
6:30:09B should be equal
6:30:12to B + a let's check that whether this
6:30:16is the case let's first compute this
6:30:19part and then we will do this
6:30:21one
6:30:23well the first
6:30:25one means that we get so
6:30:31A+ B is and we remember remember how we
6:30:34add matrices right so we know that we
6:30:37just need to pick their corresponding
6:30:39elements and add them to each other so
6:30:41we get
6:30:44a11 plus
6:30:47b11 then dot dot dot and then a
6:30:52M1 plus b M1 this is why also D is
6:30:57really important that they got um the
6:30:59same Dimension which means that they got
6:31:01exactly the same amount of elements the
6:31:04same uh amount of columns and the same
6:31:06amount of rows um in terms of the uh
6:31:10Matrix size so then here we have a 1 n
6:31:17and then
6:31:19plus B1
6:31:21n then dot dot dot and then a
6:31:27M1 and then plus b m here I need to put
6:31:34n we are in the last element of the
6:31:36Matrix so
6:31:38BMN and that is it this is our Matrix A
6:31:42+ B let's now look into the Matrix b + a
6:31:48so what that amount
6:31:53is so the
6:31:56Matrix b + a will then be equal to
6:32:07b11 plus
6:32:09a11 dot dot dot and then B
6:32:15M1 plus a
6:32:18M1 then dot dot dot and then b 1
6:32:23n plus a 1 n then dot dot dot and then
6:32:29the last element will be B
6:32:31MN and then Plus
6:32:34a m
6:32:41n
6:32:42so in here if we remember from the real
6:32:47numbers we know that a
6:32:51+ b is equal to B + a for instance if a
6:32:59is equal to 2 and then B is equal to 1
6:33:02then a + b is = to 2 + 1 which is equal
6:33:06to 3 and then b + 1 is = to 1 + 2 and
6:33:10it's equal to 3 so we know that indeed
6:33:12for the real numbers a plus b is equal
6:33:15to B + a and making use of that property
6:33:18we can already state that b11 + a11 is
6:33:24equal to A1 1 + B1 1 and then the
6:33:29general case is that a
6:33:33i
6:33:36j plus b i
6:33:40j is equal
6:33:42to b i
6:33:46j plus a i j where I is the index of the
6:33:52rows and then J is the index of the
6:33:55columns from the coefficient
6:33:57labeling so using this property from the
6:34:01real numbers given that all these values
6:34:03in the m Matrix are real numbers we can
6:34:05quickly see that the Matrix B+ a that we
6:34:10just got in
6:34:11here is equal to this Matrix a plus b
6:34:16which means
6:34:18that 1 is equal to B and this proves
6:34:23that A + B is equal to B + a this is the
6:34:27communative property of the Matrix
6:34:31additions so the next law is the
6:34:34associative law for matrices which says
6:34:37that the in case of Matrix addition a +
6:34:40B+ C isal to A + B +
6:34:45C so
6:34:48basically this
6:34:50time we go from here to adding one more
6:34:54element which is the third Matrix Matrix
6:34:57C so we are saying A + B + C is equal to
6:35:02a +
6:35:04B+ C so it doesn't matter whether we
6:35:07will First Take The Matrix a and then B
6:35:10and then add them up and then we add
6:35:12Matrix C or if we first take the Matrix
6:35:16B and C add them up and then we add a to
6:35:19this sum it doesn't
6:35:21matter we will see an example of this in
6:35:24a
6:35:25bit and then the uh second part of this
6:35:28associative law for matrices says that
6:35:30for matrix multiplication a * B * C is
6:35:35equal to a * B * C so again in terms of
6:35:40the
6:35:41um order when it comes to this specific
6:35:45multiplication so it doesn't matter
6:35:47whether we will first multiply a by B
6:35:50and then by C or we will first multiply
6:35:52B by C and then we add the a at the end
6:35:56we will end up with the same amount so a
6:35:59* B and then * C is equal to B * C C and
6:36:04then in the left hand side we add the a
6:36:06so a * B * C so these properties help us
6:36:10to add or multiply matrices without
6:36:13really worrying about this idea of
6:36:15grouping of the terms so we can always
6:36:18group them and perform all sorts of
6:36:21operations so this is this first
6:36:23property that we see in
6:36:25here let's say we have this uh Matrix a
6:36:29matrix B and Matrix C so let's prove
6:36:33that in
6:36:34the order doesn't matter and this
6:36:36associative property holdes so let's
6:36:39prove
6:36:41that so for that the first thing we need
6:36:44to do is to
6:36:46compute a plus b so this part so A + B +
6:36:53C what is that first I need to compute
6:36:56this
6:36:57part and then I will add C which is the
6:37:02second part
6:37:04so A + B is equal to my a is equal
6:37:10to
6:37:121 2 3 4 plus and my B is equal
6:37:18to 5
6:37:216 7
6:37:248 this is then equal to so 1 + 5 is = to
6:37:296 2 + 6 is = 8 3 + 7 is = to 10 and then
6:37:334 + 8 is equal to 12 this is my A + B
6:37:38this is the first part now the second
6:37:41part is then to add to this A +
6:37:46B this C this I can by the way also call
6:37:50some Matrix D so I can say that this is
6:37:55equal
6:37:56to D+
6:37:58C so let's find out what is this amount
6:38:03so a plus b or what we're referring as D
6:38:07is 6 8 10 12 we just calculated it in
6:38:11here I'm also adding now my Matrix C
6:38:15which is 91 10 12 so 9 10 11 12 what is
6:38:21this amount it is 6 + 9 is 15 8 + 10 is
6:38:2618 10 + 11 is 21 12 + 12 is
6:38:3124 this is my final Matrix
6:38:35so I have checked
6:38:38that D A + B + C is equal to 15 18
6:38:4821 and
6:38:5024 this is the first part let's now go
6:38:53ahead and check whether this is equal to
6:38:56the second part which is this part so
6:38:59this is one this is two so
6:39:04this is then A + B + C as you can see it
6:39:09in here let's now calculate that amount
6:39:12and like previously we will do it in an
6:39:14order so first we need to calculate this
6:39:17part and then the entire
6:39:23thing so B+ C is then equal
6:39:29to the B was 5 6 7 8
6:39:355 6 7 8 plus and the C
6:39:41was 9 10 11
6:39:4412 9
6:39:4710 11
6:39:5012 what is the much 5 + 9 is 14 6 + 10
6:39:55is 16 7 + 11 is 18 and 8 + 12 is
6:40:0220 this is the first amount let's refer
6:40:06refer this as a
6:40:08d or we can even call it by some other
6:40:11letter let's say k this is Matrix K so B
6:40:14plus C is
6:40:17k then the second part is to take this
6:40:20B+ C so B+
6:40:25C which we have referred as
6:40:28K say
6:40:30K and then we are adding to this
6:40:33the A and specifically just to ensure
6:40:37that we stay with the same order I'm
6:40:38saying I will add from the left side the
6:40:42a to this Matrix
6:40:45K and this obviously
6:40:53means this is equal to so A+ b + C this
6:40:59is what I'm referring by just uh in a
6:41:02more simpler note a I'm just using K in
6:41:05here so this is my B plus C or what I'm
6:41:09referring also as a
6:41:11k and this amount is equal
6:41:15to what is my a my a is 1 2 3
6:41:204 1 2 3 4 plus and what is B plus C we
6:41:26just calculated that that's the K so 14
6:41:3016 18 20
6:41:34T So 1 + 14 is 15 2 + 16 is 18 3 + 18 is
6:41:4121 4 + 20 is
6:41:4524 so we have learned that the A + B + C
6:41:51is this
6:41:54Vector now is this Vector equal to the a
6:41:59plus b and then plus C well here we got
6:42:02this 15 18 21
6:42:0624 15 18 21 24 so we have just proved
6:42:13that the first part is equal to second
6:42:16part which means that we have proved
6:42:18that
6:42:19indeed the order doesn't matter and A +
6:42:23B + C is equal to a + B+
6:42:27C so this calculation confirms that the
6:42:30both sides of this equations they are in
6:42:33indeed equal and this confirms the
6:42:35associative low for the Matrix
6:42:42addition so let's now look into the
6:42:45distributive law for matrices which says
6:42:48that Matrix addition and multiplication
6:42:51they satisfy the distributive property
6:42:54which means that if we have a left
6:42:56distribution a * b + C is equal to a + a
6:43:02c
6:43:03and then in the right distribution we
6:43:05basically have the Matrix multiplying
6:43:07from the right from hence the name right
6:43:10distribution A + B * C is equal to a C +
6:43:15BC you might very quickly see and
6:43:19recognize from here that we have very
6:43:22similar actually exactly uh the same
6:43:26rule only for real numbers we know that
6:43:28a * b + C is equal to and then we open
6:43:32the parentheses with say this is equal
6:43:33to this times this so AB plus this times
6:43:37this a c you can see that we have
6:43:41exactly the same here only in the
6:43:43capital letters so in the real numbers
6:43:45we have exactly the same low so the same
6:43:48we have also for our left
6:43:51distribution when it comes to Matrix
6:43:53additional multiplication and the same
6:43:56we have only with a different order here
6:43:58you can see the C so this one is
6:44:01basically uh with the different order
6:44:04instead of having the Matrix multiplied
6:44:05in the left here we have from the right
6:44:08and this is similar to the property that
6:44:11A + B * C is equal to C * a which is a c
6:44:17plus c * B which is BC an example uh
6:44:22where we will prove that the
6:44:23distributive law for matrices um is
6:44:25indeed true and I have skipped
6:44:30deliberately the uh example for this one
6:44:34because uh this a b * C is equal to a *
6:44:38b c so the associative law for matrix
6:44:41multiplication it's something that you
6:44:43can calculate for yourself using the
6:44:46same a b and c matrices only this
6:44:49includes multiplication of these two
6:44:51matrices and it's something that we are
6:44:53going to do as part of this example so
6:44:56instead of doing and redoing this
6:44:59multiplication I thought that it's great
6:45:01to leave that for you as a practice and
6:45:03instead focus on bit more complex
6:45:06problem like this one that one way or
6:45:09the other includes the same matrix
6:45:12multiplication so I need to calculate
6:45:14the a * B in this case which means that
6:45:18by providing you this example I'm also
6:45:20including what is needed to do the
6:45:23previous example only it would be a
6:45:24great way to practice the material for
6:45:26yourself so let's now move into proving
6:45:29the distributive low for matrices so we
6:45:32got this mat matrices a b and c and here
6:45:36I'm going to apply matrix multiplication
6:45:39the same as that is needed for the
6:45:40previous uh case and here what we need
6:45:44to prove is that a * b + C is equal to a
6:45:49* AC so this is the first part this is
6:45:52the second part so let's go and
6:45:55calculate them
6:45:57separately so for the first one we need
6:46:00to calculate
6:46:05a * b +
6:46:08C which is
6:46:12then something that we can calculate by
6:46:15first doing the addition so we will
6:46:18first do the addition of matrices B and
6:46:21C and then once we are done with that we
6:46:24will then do a * b +
6:46:28C so that's the second part
6:46:39so let's go ahead and do that
6:46:41calculation so first we
6:46:43got b + C what is B plus C B is 5 67 8 5
6:46:506 7
6:46:538 plus and the C is minus one 0 0 minus
6:46:57one so on the diagonal we got min-1 and
6:47:00minus one and then of diagonal lower and
6:47:02upper part we got zero and what is this
6:47:05Matrix this is equal to 5 - 1 so 5 + -1
6:47:09is equal to 4 6 + 0 is = 6 7 + 0 is = 7
6:47:15and then 8 - 1 is = 7 this is our B plus
6:47:19C which we can refer also as Matrix D so
6:47:23let's call this D which means that now
6:47:26we are interested in a
6:47:29* B
6:47:33so for this second
6:47:36part we need to take this Matrix
6:47:40a so a
6:47:44* D is then equals to we need to take
6:47:49the Matrix a which is
6:47:521 2 3 4 and we need to multiply it with
6:47:57this Matrix that we just got because
6:47:59this is the B plus C or the D that we
6:48:00were referring 46
6:48:0477 okay so let me remove this
6:48:09part cuz we are going to need some space
6:48:12for
6:48:13this and let's do this calculation this
6:48:16is 2x two and this is 2x
6:48:25two I will do the calculations in here
6:48:29so we need to end up with the Matrix
6:48:31that is also 2 by 2 because we know 2x 2
6:48:36Matrix times 2x two we will pick this
6:48:39part so the number of rows and the
6:48:41number of columns of the second one this
6:48:43will be our resulting Matrix which is 2
6:48:45by
6:48:47two all right so for the matrix
6:48:51multiplication we know that for this
6:48:53element in the place of so one a or
6:48:58let's call this Matrix we don't even
6:49:00actually need to call this anything we
6:49:02we can keep it simple so let's say that
6:49:04we are in the first draw in the First
6:49:06Column so this is the first draw in the
6:49:08First Column for this what we need to do
6:49:11is we need to take the first row from
6:49:14the first Matrix so Matrix a and then
6:49:16the First Column of the Matrix D so this
6:49:19one and we need to do the dot product
6:49:22which means that we do
6:49:26basically 1 * 4 1 * 4 plus 2 *
6:49:387 so for this element which is in the
6:49:42second row and the First Column we need
6:49:46to take the second row and First
6:49:51Column in here so we end up with three
6:49:56times 4 so 3 * 4 and then 4 * 7 so Plus
6:50:024 * 7 The Dot product between this one
6:50:05and then this
6:50:07one so for this element which is in the
6:50:10first row and then the second column of
6:50:12the final Matrix so first row and second
6:50:15column we need to pick the first row and
6:50:19second column and do a DOT product which
6:50:21means
6:50:24one 1 * 6 + 2 * 7 2 *
6:50:317 and then in here in this element we
6:50:35got the second column and second row so
6:50:38second row second
6:50:39column which means that we need to pick
6:50:42the second
6:50:45row and the second column the dot
6:50:48product of the second row of Matrix a
6:50:51and the second column of Matrix D which
6:50:56is 3 *
6:51:006 Plus 4 * 7 4 *
6:51:077 so let's quickly calculate what this
6:51:10amount
6:51:13is so this is the a * B+ C basically and
6:51:21this Matrix is 1 + 4 is 4 2 * 7 is 14 4
6:51:28+ 14 is
6:51:3018 1 * 6 X is 6 2 * 7 is 14 and 6 + 14
6:51:38is
6:51:3920 3 * 4 is 12 4 * 7 is
6:51:4728 which means that we got here
6:51:5140 3 * 6 is 18 4 * 7 is 28 which means
6:51:58we got here 36 and 46
6:52:06this is our final a * b +
6:52:11C let's now go ahead and calculate the
6:52:15second part so the second
6:52:18part says that we
6:52:22got a * b + a *
6:52:26C so a * b + a * C
6:52:32which means that first we need to do
6:52:34this calculation and then this
6:52:37one and then we need to add them to each
6:52:41other so let's quickly then calculate
6:52:44what is a * B and then a * C and then
6:52:48add them to each
6:52:51other let me clean up some space in
6:52:55here we're going to KN
6:53:00that when we write this one in a smaller
6:53:03format so this is equal to
6:53:0718 20 40 and
6:53:1646 and let me take over the second
6:53:19element which we still need to calculate
6:53:22which is AB plus a
6:53:28c first we will do this and then this
6:53:32and then we will add them to each
6:53:35other so a * B is equal
6:53:41to 1 2 3 4 multiplied
6:53:46by 5 6 7 8 5 6 7
6:53:518 now following the
6:53:54same
6:53:55approach from the previous example when
6:53:58we calculate this Matrix I will then
6:54:00quickly calculate what is is a * B so in
6:54:04here we got first row and First Column
6:54:07so 1 * 5 + 2 * 7 so the dot product
6:54:11between the first row and the First
6:54:13Column from
6:54:14here now for this element here we got
6:54:18the second row and the First Column we
6:54:21need to take the second row in the First
6:54:23Column from here and we do the dot
6:54:26product which means 3 * 5 + 4 * 7 in
6:54:32here here we got the first draw and
6:54:33second column so the first draw and
6:54:35second column which
6:54:38means that we need to have 1 * 6 + 2 *
6:54:468 then here we got the second row and
6:54:49then second column which means 3 * 6 + 4
6:54:53*
6:54:558 and then this is equal
6:54:59to 1 * 5 is 5 5 2 * 7 is 14 so this is
6:55:0819 1 * 6 is 6 2 * 8 is 16 6 + 16 is
6:55:1822 in here 3 * 5 is 15 4 * 7 is 28 so
6:55:24this is then 33 and then 43 so
6:55:3043 and in here we got 3 * 6 6 is 18 4 *
6:55:368 is 32 so we end up with
6:55:4650 so hope I haven't made any mistakes
6:55:49in the
6:55:51calculations so this is the a *
6:56:00B so a * B is then equal
6:56:06to 19 22 43
6:56:1150 let's clean this pce and let's move
6:56:14ahead to the second part of the
6:56:16calculation which is a *
6:56:21C what is a * C well a * C
6:56:29is 1 2 3 4 1 2 3 3 4 multiplied by
6:56:36-1 0 0 -1 so here we are then
6:56:42getting -1 +
6:56:450 here we are
6:56:47getting -3 + 0 here we
6:56:53have -1 +
6:56:560
6:56:58so no so the first row and second column
6:57:04which is 0us
6:57:062 and then in here we got the second row
6:57:10and the second column which is 0 -
6:57:144 which means that we end up with this
6:57:17Matrix and it's equal to
6:57:20-1
6:57:22-3 then -2 and then
6:57:29-4 which means that we are getting
6:57:33as a final step AB plus a which
6:57:37means 19 202 43 and then 50 then
6:57:44plus -1 - 2 - 3 - 4 and what is this 19
6:57:51- 1 is 18 43 - 3 is 40 22 - 2 is 20 50 -
6:58:004 is 46
6:58:04okay so we got
6:58:08that this amount ab+ a c is equal to 18
6:58:1420 40
6:58:1646 and as you can see already
6:58:20here this Matrix that we got in the
6:58:23previous calculation from one is equal
6:58:25to this Matrix that we got as part of
6:58:27second calculation which means that now
6:58:30we have proved that for this specific
6:58:32example indeed 1 is equal to 2 which
6:58:35means that a * b + C is equal
6:58:40to AB
6:58:43plus
6:58:46AC there we go so let's now look into
6:58:50another law which is the scalar
6:58:53multiplication law for
6:58:55matrices so the scalar multiplication
6:58:58law for matrices says
6:59:01that if we got a scalar R and a matrix A
6:59:05and
6:59:06B then R * a * B is equal to R * a * B
6:59:13and is equal to a * R *
6:59:16B so here the r is just a
6:59:21scalar so it's a real
6:59:23number and then A and B are
6:59:28matrices and what this low basically
6:59:31says is is that it doesn't matter what
6:59:34in which stage you will do your matrix
6:59:37multiplication with the scaler if you
6:59:39have this external scaler you can first
6:59:42take the two matrices multiply them with
6:59:45each other so the A and then B and then
6:59:49multiply it with
6:59:51r
6:59:53or you can
6:59:55take the scalar R multiply with your
6:59:58first Matrix and then multiply with B
7:00:02or you can take your second Matrix
7:00:06multiply with the scaler and then
7:00:08multiply with a it doesn't matter they
7:00:11will all result in the same Matrix so
7:00:15let us actually prove this by making use
7:00:18of our skills from matrix multiplication
7:00:20and scalar multiplication here I've
7:00:23picked up bit more uh Advanced example
7:00:26where uh a is
7:00:292x3 and B is 3x3 in this way we will
7:00:32train our multiplication skills for
7:00:35matrices and at the same time we will
7:00:37also prove that the scalar
7:00:39multiplication law of matrices holds so
7:00:42let's go ahead and do the
7:00:46multiplications so first we have a
7:00:48matrix
7:00:49a what is that Matrix Matrix a
7:00:56is 1 - one 2 so 1 - one and then
7:01:03two then we got 0
7:01:082 and then
7:01:121 which is 2 by 3 and then we got B
7:01:19which is equal 2 it is 3x
7:01:253 with elements 1
7:01:28Z 1 1
7:01:332
7:01:35Z one one and then 3
7:01:411 0 2 so it
7:01:46is
7:01:483x 4 so it's 3x 4 not 3x 3 but 3x 4
7:01:55Matrix now the final part that I need
7:02:00here is this which is R is equal to 2
7:02:03the scalar value so R is equal to
7:02:072 so the first thing that I'm going to
7:02:09do is to calculate this amount which is
7:02:12R * a * B for that what I need to do is
7:02:16to First calculate this a * B so let me
7:02:20quickly go and calculate this for us
7:02:55so given that the a has Dimension 2x3
7:02:57and then B has a dimension 3x4 I can see
7:03:00that quickly that my Dimension criteria
7:03:03is satisfied the number of columns of a
7:03:06is equal to number of rows of B so
7:03:08that's fine and then I know also know
7:03:11that the final dimension of a * B is
7:03:13going to be 2x4 so it's going to be a
7:03:182x4 Matrix and how do I know that well
7:03:21because I know that from our um all the
7:03:26problems that we have solved we have
7:03:28already seen that we always need to pick
7:03:30the number of rows of the First Column
7:03:33and the number of columns of the second
7:03:36uh Matrix in order to get the final
7:03:38Dimension which is
7:03:402x4
7:03:42so let me then go ahead and do the
7:03:45calculation so we are going to have a
7:03:482x4 Matrix let me write it even
7:03:52bigger so it's going to be a
7:03:572x4 Matrix
7:04:12so for the first
7:04:15row and First
7:04:18Column I need to look in here the first
7:04:22row and the First Column which means I
7:04:25need to
7:04:27take
7:04:30one so it's equal to
7:04:331 * 1 so + 1 * 1 is 1 - 1 * 2 is -
7:04:412 2 * 3 is 6 + 6 this is my first value
7:04:48and what is this amount it is equal to 1
7:04:50- 2 is - 1 and 6 - 1 is equal to 5 so
7:04:56this amount is five
7:05:03five so what is this
7:05:08amount well this is my second row in the
7:05:11First Column so I need to make
7:05:15use
7:05:18of second row and First Column which is
7:05:22equal to 0 * 1 is 0 2 * 2 is 4 and 1 * 3
7:05:28is 3 4 + 3 is 7 so this value is 7 seven
7:05:33we are ready to go on to the next column
7:05:36so column number two so then this
7:05:43time I need to look at the first row and
7:05:48second column so we are going to use
7:05:50this one so first we will use this first
7:05:53Row 1 * 0 is 0 - 1 * 0 is 0 0 + 0 is 0
7:05:58and then 2 * 1 is the only nonzero
7:06:00element 2 * one is two so I already know
7:06:03that for my second column I got here
7:06:10two and what is this element well for
7:06:13this I need to look at the second row
7:06:18and second column so this thing so 0 * 0
7:06:21is 0 2 * 0 is 0 1 * 1 is one which means
7:06:27that here I get a
7:06:30one let's not move on to on uh towards
7:06:33the third column so in here first I need
7:06:37to look at the first row so 1 - one and
7:06:39two and then this
7:06:43time remove
7:06:48this I need to look at the third
7:06:52column because I'm here in the third
7:06:55column
7:06:58so 1 * 1 is 1 -1 * 1 is 1 so here I got
7:07:051 - 1 and then 2 * 0 is 0+ 0 1 - 1 + 0
7:07:13is 0 because those two cancel
7:07:16out this means that here in this element
7:07:19I got a zero and what about this element
7:07:22where I need to look here in the second
7:07:24row and here I need to look at the third
7:07:26column 0 * 1 is 0 2 * 1 is 2 1 * 0 is 0
7:07:32so 0 + 2 + 0 is equal to 2 so this is
7:07:372 and now we are left with the fourth
7:07:41column so for that I need to
7:07:44look in
7:07:46here so for the first row which means
7:07:50first row in here and then the fourth
7:07:51column in here so first row in here and
7:07:53fourth column here 1 * 1 is 1 - 1 * 1 is
7:07:581 and then 2 * 2 is 4 which means that I
7:08:01end up with 1 - one and then + 4 and
7:08:05what is this this two cancel out I end
7:08:08up with four which means that here I
7:08:10need to fill
7:08:11in
7:08:13four and what is this final element it
7:08:16is the second row in the fourth column
7:08:17so the second row in the fourth column 0
7:08:20* 1 is 0 2 * 1 is 2 1 * 2 is 2 0 + 2 + 2
7:08:26is =
7:08:284 so now we obtained that a * B is this
7:08:342 * 4 Matrix as we have
7:08:39expected so this is then equal to 5 2 04
7:08:45and then 7 1 2 4
7:09:03so then the next step would be to take
7:09:05the scaler R and multiply it with a *
7:09:11B let me actually keep the colors
7:09:14consistent so a * B this is a * B so the
7:09:20only thing that I need to do is to take
7:09:24that in here and multiply this two with
7:09:29each of those elements so I will end up
7:09:32with the same size Matrix so 2x 4 only
7:09:37all these elements need to be multiplied
7:09:39with the scaler which means that I will
7:09:41get 5 * 2 is 10 2 * 2 is 4 0 * 2 is 0 4
7:09:47* 2 is 8 and then 7 * 2 is 14 1 * 2 is 2
7:09:532 * 2 is 4 and then 4 * 2 is 8 so this
7:09:58is the result of the multiplication
7:10:04so this is the first part this is what
7:10:06we are referring as
7:10:08one so we have then
7:10:11checked in
7:10:14here that the
7:10:18r times actually we have already in here
7:10:22so I won't be writing again so as part
7:10:24of the first
7:10:26section we have already seen that R * a
7:10:29* B is this Matrix
7:10:31let's now move on to the next one which
7:10:34is
7:10:36calculating the second part so this is
7:10:39the first part this is the second and
7:10:41this is the third we have this already
7:10:44let's now move on and calculate this one
7:10:47so for this second case so the second
7:10:50case what we want to calculate is R * a
7:10:54* B so it is R * a and then * B this is
7:11:01what we need to calculate so the first
7:11:03thing that we will do is to calculate
7:11:05this
7:11:06part and then to calculate the entire
7:11:09thing the second
7:11:11point so let's go ahead and do
7:11:14that first we will take the A and then
7:11:18we will multiply all its elements by
7:11:20scaler two to get the r and then a this
7:11:24amount is equal
7:11:272 1 * 2 is = 2 - 1 1 * 2 is - 2 2 * 2 is
7:11:33= 4 0 * 2 is = 0 2 * 2 is = 4 2 * 1 is
7:11:39equal to 2 this is R * a now in The Next
7:11:44Step so this was one the next step we
7:11:47need to take this amount this
7:11:53Matrix to minus 2 4
7:11:58042 and multiply it with
7:12:041 2 3 0 0 1 and then 1 1 Zer and then 1
7:12:13one 2 so basically the Matrix
7:12:16B let's now move and work our way out
7:12:21with that one actually let me remove
7:12:23this from here and keep the space bit
7:12:26more clean R time a and I will
7:12:31multiplying this with the Matrix 1 2 3
7:12:35and then 0 0 1 and then 1 1 1 1 and then
7:12:410
7:12:422 well I know that this one is 2x3 and
7:12:46this one is 3x 4 which means that the
7:12:48result will be 2x 4 let's now go ahead
7:12:52and calculate that Matrix which is equal
7:12:56with a dimension of 2x 4
7:13:04well for the first row and First Column
7:13:07let me actually go and quickly do those
7:13:11calculations let's now go ahead and do
7:13:13those calculations so we are going to
7:13:16have four columns as previously the
7:13:19dimension is going to be
7:13:212x4 so let's do it column by column in
7:13:24here it means that we are in the row one
7:13:27and then column 1 so 2 * 1 is equal to 2
7:13:31- 2 * 2 is - 4 and then here we got 4 so
7:13:354 * 3 is 12 so we got 2 - 4 and then +
7:13:4212 and what is this amount 2 - 4 is - 2
7:13:47+ 12 is
7:13:5010 so here we got 10 let me remove
7:13:57this
7:13:5910 this is the second row and the First
7:14:02Column which means we got 0 * 1 is 0 4 *
7:14:062 is 8 and 2 * 3 is 6 so 8 + 6 is equal
7:14:14to 14 so here we got
7:14:2014 this is the first row and second
7:14:23column which means that we are looking
7:14:24at this row and second column this
7:14:29time so 2 * 0 is 0 - 2 * 0 is 0 the only
7:14:33thing that we care about is this one and
7:14:36this element which is 4 * 1 so this
7:14:38should be four let's now do the same for
7:14:40the second row 0 * 0 is 0 4 * 0 is 0 0 +
7:14:440 is 0 which means we are left with 2 *
7:14:461 so here it comes
7:14:49two let's now do the third column so for
7:14:54the third column we got First Row 2 * 2
7:14:582 * 1 is 2 - 2 * 1 is - 2 and then 4 * 0
7:15:02is 0 which means that here we get 0
7:15:06because 2 - 2 + 0 is 0 then we got the
7:15:12second row and third column which is
7:15:15this row and then third Comm so 0 * 1 is
7:15:200 4 * 1 is 4 2 * 0 is 0 0 + 4 + 0 is
7:15:25four so this is four and then for the
7:15:28first row and then fourth coln so it
7:15:31means that we need to look at this
7:15:32specific column the first row is 2 * 1
7:15:38it is 2 - 2 * 1 is - 2 and then 4 * 2 is
7:15:438 so 2 - 2 + 8 is
7:15:488 and then finally for the second row
7:15:51and the fourth column 0 * 1 is 0 4 * 1
7:15:55is 4 2 * 2 is 4 4 + 4 is 8
7:16:01this is the final Matrix which means
7:16:04that this entire amount that we just
7:16:06calculated step by step this is equal to
7:16:10this Matrix in
7:16:13here okay so this is the second element
7:16:16let's check whether the first element is
7:16:18equal to the first one so we see here 10
7:16:2148 142
7:16:23248 as you can see we are dealing with
7:16:26exactly the same Matrix which proves
7:16:30that indeed
7:16:32R * a * B is equal to R * a * B so this
7:16:36part we have already proven because we
7:16:38have seen that 1 is equal to
7:16:412 perfect so the only thing that is
7:16:43remaining is to calculate this third
7:16:45part and to see whether this is equal to
7:16:49this matrices because we have seen that
7:16:52the two of those are equal so the
7:16:55remaining thing that is left to prove
7:16:57this theorem is to calculate this third
7:16:58part let's go ahead and do that
7:17:06that so the third element says let's
7:17:09first calculate the r * B and then
7:17:11multiply it by a so we need to
7:17:14calculate
7:17:16b r *
7:17:19B * a this is what we need to calculate
7:17:22which means first we need to calculate
7:17:24this and then we need to can calculate
7:17:26the entire thing all right so let's go
7:17:29ahead and do that
7:17:33so R * B is equal
7:17:43to so we need to multiply each of the
7:17:47elements of B by two so we end up with
7:17:50this Matrix
7:17:532 0 2 2 and then 2 * 2 is 4 2 * 0 is 0
7:18:01and then 2 2 and then 2 * 3 is 6 2 * 1
7:18:06is 2 2 * 0 is 0 2 * 2 is four this is
7:18:11that first Matrix let's now go ahead and
7:18:15calculate the second part which is a *
7:18:19Matrix a so it
7:18:22is 1 - 1 2 and then 0 2 and then 1
7:18:32multiplied by this Matrix which is 2 4 6
7:18:370 0 2 and then 2 two 0 and then 2 2
7:18:444 okay
7:18:46perfect so this is then what we need to
7:18:50calculate well this is 3 * 4 this is 2 *
7:18:543 which means the result should be 2 * 4
7:18:58let's go ahead and do those calcul
7:19:09ations this first amount will be the
7:19:13first row and the First Column
7:19:15dotproduct of those which means 1 * 2 is
7:19:182 -1 * 4 is - 4 and then 2 * 6 is 12 so
7:19:25here we got 2 - 4 + 12 and what is this
7:19:31amount well 2 - 4 is - 2 12 - 2 is = to
7:19:4010 so this one this element is
7:19:4410 then for the second show we need to
7:19:46look in here so 0 2 one and the
7:19:49dotproduct of the one with the First
7:19:51Column so this
7:19:54thing and that is 0 * 2 is 0 2 * 4 is 8
7:19:59and then 1 * 6 is 6 so what is 8 + 6
7:20:06that is
7:20:0814 and then for the first row and then
7:20:11the second column we need to look to
7:20:16the first row in here and then the
7:20:19second column in here and the dotproduct
7:20:21of the two well 1 * 0 is 0 Min - 1 * 0
7:20:25is 0 and then 2 * 2 is four so that's
7:20:28what we are left with four
7:20:31and then for the second row and then the
7:20:35second column so this element we
7:20:38got 0 * 0 is 0 2 * 0 is 0 1 * 2 is
7:20:462 for the first row and the third column
7:20:51so we need to look in
7:20:55here 1 * 2 is 2 - 1 * 2 is - 2 and then
7:21:012 * 0 is 0 so we are left with
7:21:06zero 0 and then once we do the
7:21:10calculation for the second row we will
7:21:11see that we end up with 0 * 2 is 0 2 * 2
7:21:15is 4 and then 1 * 0 is 0 so we end up
7:21:20with four and then here for the final
7:21:26column 1 * 2 is 2 - 1 * 2 is - 2 and
7:21:32then 2 * 4 is 8 the first two cancel out
7:21:36and we end up with 8 and then for the
7:21:39second row and the fourth column we look
7:21:41into here again this time the second row
7:21:440 * 2 is 0 2 * 2 is 4 and 1 * 4 is 4 and
7:21:50then 4 + 4 is
7:21:518 so if we look in here this is our
7:21:55third amount we will quickly see that
7:21:59again we have the same Matrix with
7:22:02exactly the same elements so now we have
7:22:05also proved this third part and we have
7:22:08seen that in all cases the r * a * B is
7:22:12equal to R * a * B is equal to a * R * B
7:22:17now we are ready to move on towards the
7:22:19second module in this unit which is
7:22:21about the determinants and their
7:22:23properties we are going to look into the
7:22:26uh determinants at high level we are
7:22:28going to Define them and going to
7:22:30understand what why they matter and why
7:22:31they are important then we are going to
7:22:34see how we can calculate the
7:22:35determinants we are going to see the
7:22:38calculation for 2x two Matrix then 3x3
7:22:41Matrix and then just in general how we
7:22:43can do it and then we are going to see
7:22:46the properties of determinants one by
7:22:48one and then finally we are going to see
7:22:50the determinants interpretation from the
7:22:52geometric perspective so when we
7:22:54visualize it using
7:22:57python so by definition the determinant
7:23:00is a scalar value that can be computed
7:23:03from the elements of a square
7:23:06Matrix so this important Square Matrix
7:23:09and encodes certain properties of the
7:23:12Matrix so the determinant provides a
7:23:15critical information about The Matrix
7:23:17such as whether it's
7:23:19invertible and the volume scaling factor
7:23:24for the linear transformation it
7:23:26represents so we see that the uh concept
7:23:29of theer detent is highly related to
7:23:33many other concept that we have seen
7:23:34before so first here it's talking about
7:23:36the square Matrix then it's talking
7:23:38about encoding certain properties so
7:23:41having the determinant it contains
7:23:43certain information that um is related
7:23:46to the properties of the system that
7:23:50that Matrix is representing and then it
7:23:53provides critical information about the
7:23:55underlying metrix because the
7:23:57determinant is calculated from Matrix we
7:24:01say the determinant of a matrix so it
7:24:03contains a critical information about
7:24:05that Matrix such as whether it's
7:24:07invertible or not and this goes back to
7:24:09the concept of inverse we will see this
7:24:12once we learn the concept of determinant
7:24:14because the inverse calculation is
7:24:16dependent on the
7:24:18determinant but keep this thing in mind
7:24:21that the determinant contains
7:24:23information whether we can get um
7:24:26inverse from a matrix or not we will see
7:24:29this concept over also in detail in the
7:24:32next section but for now we can remember
7:24:35that the determinant contains important
7:24:37information related to the invertibility
7:24:40of the Matrix so having an inverse or
7:24:42not and then it also contains
7:24:44information about the volume scaling
7:24:46factor for the linear transformation it
7:24:50represents so here we then go back to
7:24:52this concept of a x is equal to B and
7:24:55then knowing the determinant we can then
7:24:58comment on this volume scaling factor
7:25:02for this linear transformation that it
7:25:04represents so let's go uh on to the next
7:25:08slide to find out bit more about the
7:25:11determinants and specifically how we can
7:25:13calculate the determinant in the
7:25:15mathematical terms when it comes to the
7:25:172x two Matrix because the uh determinant
7:25:22of a 2X two Matrix is quite
7:25:24straightforward so for 2x2 matrix a with
7:25:28this elements where a b c and d they are
7:25:32all real
7:25:35numbers the
7:25:37determinant which we Define by this de a
7:25:41so that is a short way of saying
7:25:43determinant and then in here we always
7:25:45write the Matrix of which we are
7:25:47Computing the determinant is then equal
7:25:50to and then we are taking this a * D so
7:25:54we are taking this diagonal elements a *
7:25:59d so they are on the diagonal and then
7:26:02we are subtracting from this this other
7:26:05two the remaining two
7:26:08elements of the diagonal so B * C and
7:26:14this gives us the determinant of 2x2
7:26:17matrix this is just a formula that you
7:26:20need to uh remember whenever you want to
7:26:22calculate the determinant of a matric by
7:26:25hand
7:26:26manually so the calculation for larger
7:26:29matrices it involves bit more uh
7:26:32difficult uh calculation we will see
7:26:35also in a bit the uh determinant of a
7:26:383X3 Matrix It relies on the determinant
7:26:40of a 2X two Matrix and the idea is that
7:26:44every time we uh increase the dimension
7:26:46of our problem so let's say we are in R4
7:26:49then we will go back to the R3 and then
7:26:51given that R3 relies on the determinant
7:26:54of the underlying 2x two matrices
7:26:56anytime we increase the dimension we
7:26:58again go back to this IDE of using 2 by
7:27:01two matrices that form the entire Matrix
7:27:03in order to compute the determinant only
7:27:06when it is R4 R5 Etc so it becomes much
7:27:10more difficult to describe and to do it
7:27:12manually therefore there are other
7:27:14algorithms which we will see at the end
7:27:16of this course like uh the composition
7:27:19algorithms and factorization algorithms
7:27:21that can be used in order to calculate
7:27:23the determent of a matrix that has
7:27:25higher Dimension higher than the tree
7:27:27for instance but in this specific unit
7:27:30we are going to discuss both the
7:27:32calculation of the 2x two matrices
7:27:33determinant and the determinant of a 3X3
7:27:36matrices and we will also see detailed
7:27:38examples of
7:27:40them so without further Ado let's then
7:27:42go ahead and calculate the determinant
7:27:45of this 2x2 matrix so let's now look
7:27:47into this specific example where we are
7:27:49calculating the determinant of this 2x2
7:27:51matrix so this is the A and let's keep
7:27:55in mind that this is the um uh a this is
7:27:59the
7:28:00uh B in this not in this uh way of
7:28:05writing the Matrix a so the uh letters
7:28:08corresponding of the uh elements of this
7:28:11Matrix a so this is the a this is the
7:28:13B and then this is the C this is the D
7:28:17and we said that the
7:28:19determinant
7:28:21that of a is equal to the diagonal
7:28:26elements so 1 * 4
7:28:31minus the of diagonal Elements which is
7:28:362x3 because we said that the definition
7:28:41of
7:28:43this determinant is that is equal to a *
7:28:47D and then minus B * C which is exactly
7:28:51what we are doing in here so if we
7:28:54calculate 1 * 4 is = to 4 and then 2 * 3
7:28:57is = 6 4 - 6 is = to - 2 therefore we
7:29:01say that the
7:29:03determinant of Matrix a is equal to Min
7:29:07- 2 let's now go ahead and uh practice
7:29:11with calculation of determinants on two
7:29:14other matrices so in this case we are
7:29:16still in the two dimensional space so we
7:29:19have 2 by two matrices we'll first
7:29:22calculate the the determinant for Matrix
7:29:24a so we see that we got this element
7:29:285061 and we know that by definition the
7:29:32determinant of the 2x two Matrix so that
7:29:36of
7:29:38Matrix is equal
7:29:41to a * d - B * C where the
7:29:50Matrix has the following form so we got
7:29:53a and then D in here and then B and the
7:29:56C in here so we see that this is basic
7:29:59Bally our a this is our D this is our B
7:30:04and this our C the way you can also said
7:30:08is that those are the diagonal elements
7:30:12and those are the of diagonal
7:30:16elements so therefore it means that we
7:30:19can calculate the
7:30:23determinant of Matrix
7:30:26a by taking the five multiplying with
7:30:31one so it is 5 * 1 minus the off
7:30:37diagonal element which is 6 *
7:30:400 and this amount is equal to 5 - 0 and
7:30:45is equal to 5 let's go ahead and also
7:30:48calculate the determinant of Matrix B we
7:30:52see here that on the diagonal we have
7:30:54this two elements one one and of the
7:30:57diagonal elements are both zero
7:31:00those two therefore we can calculate the
7:31:04determinant of this 2x2 matrix which is
7:31:07also sometimes referred as I2 so it is
7:31:10the identity Matrix because we got here
7:31:13the E1 and then E2 in the two
7:31:16dimensional
7:31:18space and the determinant of the Matrix
7:31:21B using this definition is then equal to
7:31:241 * 1 - 0 * 0 so 1 * 1 - 0 *
7:31:350 and this is equal to 1 and this is
7:31:39actually a special case of determinant
7:31:43and later on we will see why it is so
7:31:46important to uh have this relationship
7:31:49of identity Matrix having a determinant
7:31:52and having it equal to one um and this
7:31:56relationship between determinant
7:31:57identity Matrix is something that we see
7:31:59so uh in the upcoming lesson so keep
7:32:02this one in mind so now when we are
7:32:04clear on how we can calculate the
7:32:06determinant for 2 by2 Matrix so this is
7:32:09quite simple and straightforward
7:32:10calculation by taking the diagonal
7:32:13elements a and then D and then
7:32:16subtracting from that from that product
7:32:18a * C we are subtracting the off
7:32:20diagonal elements products B * C we can
7:32:23then get our determinant and now when we
7:32:26are clear on that we are ready to go on
7:32:28to bit more Advanced calculations which
7:32:31is calculating the determinant this time
7:32:34for the 3X3 Matrix so now we increase
7:32:38the um the dimension size and we go from
7:32:42R2 to
7:32:45R3 because now we have a 3X3 Matrix and
7:32:49by definition given a 3X3 Matrix a which
7:32:53has the following elements so a11 a21
7:32:57a31 and then a12 a32 a 32 so we have
7:33:01already seen this coefficient labeling
7:33:03this should look very familiar this is
7:33:053x3 Matrix 2 and the determinant of a
7:33:09matrix a denoted as that a is calculated
7:33:13using the formula and here we see the
7:33:17formula we are basically using the 2x
7:33:22two matrices that form this Matrix a in
7:33:27order to calculate the determinant of
7:33:29the 3X3 Matrix and how we are doing that
7:33:33well we are using this element and then
7:33:37this element and this element and every
7:33:40time we are
7:33:44hiding part of the Matrix so when we
7:33:48have for instance this a11 so for this
7:33:51first part we are saying well let's hide
7:33:57the row and the column
7:34:00corresponding to this
7:34:02element which means that we need to hide
7:34:06this this row and this column and what
7:34:11is left is this 2x two
7:34:13Matrix we will calculate the determinant
7:34:16of this 2x2 matrix and we will multiply
7:34:19this with this element that we use in
7:34:21order to remove the corresponding row
7:34:24and
7:34:26column this will form the first element
7:34:29in here
7:34:30so you can see a11 which is a simple
7:34:34value so this is the um uh entry volume
7:34:39which is in the first draw and First
7:34:41Column a11 multiplied by the determinant
7:34:45of this Matrix so this
7:34:50Matrix so once we have that and we
7:34:53already know how we can calculate a
7:34:56determinant of a 2X two Matrix because
7:34:59this 2X two so taking the diagonal
7:35:01elements and then multiplying them
7:35:03together subtracting from that the of
7:35:05diagonal elements product now we are
7:35:07ready to go on to the next part of the
7:35:11calculation which is this time adding a
7:35:13minus here so you can see here this here
7:35:16is plus and then here is
7:35:19minus so we do here
7:35:23minus and for this second step what we
7:35:27need to do is kind of similar only this
7:35:29time
7:35:30the element that we will be using to
7:35:33understand how we can remove the row and
7:35:35the column so we will then dark it out
7:35:37it is this
7:35:39one
7:35:41a12 so then we will need to remove this
7:35:44column and this row and then the
7:35:47remaining Matrix which
7:35:50is this one this 2x two and here I mean
7:35:54a
7:35:5621 a 31 and then a a
7:36:0023 and then
7:36:03a33 this is the Matrix that you can see
7:36:06in here remaining which means remove
7:36:08this
7:36:09one and then this one and then the
7:36:12remaining 2 by two Matrix is what you
7:36:15need to use in order to do your
7:36:18calculations so you can see that I got
7:36:21exactly the same in here and once again
7:36:24we are Computing the determinant of this
7:36:27Matrix we are multiplying this with this
7:36:29a want to element so this
7:36:31element and now we have also the second
7:36:34element in our
7:36:36calculation and then we go on to the
7:36:39next step which is a plus sign here let
7:36:42me use the same colors plus sign here
7:36:45and then we are using this time our
7:36:47final third
7:36:50element to understand which row and
7:36:53which car we need to dark out which
7:36:56is this
7:36:57element so we then remove the first row
7:37:01and the last column and this is then the
7:37:04Matrix the 2x two Matrix that we use in
7:37:07order to do our calculation so deter the
7:37:09determinant of this Matrix multiplied by
7:37:11the
7:37:13a13 so we could also use in the same
7:37:17manner this row or this row it really
7:37:20depends on the kind of values the the
7:37:23tip that um I will provide to you or the
7:37:25trick is that to always look for these z
7:37:28z values wherever I see Z zos or I see
7:37:31one one I'm thinking that hey those s u
7:37:35values that um give me the more
7:37:38straightforward and easy calculations
7:37:41because if I have zeros in my Cal in my
7:37:46entry so if I got a zero here for
7:37:48instance 0 times any determinant is zero
7:37:52I don't even then need to calculate the
7:37:54determinant right because then I know
7:37:56that I'm multiplying that determinant
7:37:58with zero therefore if I know that that
7:38:01entry for instance this row contains the
7:38:03majority uh of zero so it is 0 01 then
7:38:07of course it's a great uh row to pick to
7:38:10use these Target
7:38:12elements so in that way I will then know
7:38:15that this is the row that I need to
7:38:16Target but if it is like that that for
7:38:19instance I got a matrix 10 3 4 and here
7:38:23I got 0 1 Zer and here I have 100 three
7:38:27and four of of course the easiest thing
7:38:30would be to not use this row but instead
7:38:34use this one so in that case I will then
7:38:37have this zero and zero as my target
7:38:40values which means that I will only need
7:38:42to calculate the determinant of a 2x2
7:38:45matrix this uh for this one for the two
7:38:49cases I don't need to do it because I
7:38:50know that the corresponding Target
7:38:53values the target elements from my
7:38:55Matrix will be zero so let me show you
7:38:58what I mean by that so if for instance I
7:39:02go for this second row and not the first
7:39:06one what I need to do is that I can
7:39:09calculate the determinant of a by taking
7:39:13the
7:39:15a21 this then will be my target I will
7:39:18then need to remove this uh column and
7:39:22this row then I will need to do the
7:39:25determinant of A1 2 A1 3 a 3 2
7:39:33a33 this is what then I need to do then
7:39:37the next thing I need to do is of course
7:39:40here I have a plus here I need to do
7:39:42minus because we always need to
7:39:44Interchange the values so here is a plus
7:39:46here is a minus here is a plus so I do
7:39:49PL a minus in here then I
7:39:52do the next element in my row which is
7:39:55this
7:39:56one let me use red color so a22 so I'm
7:40:01then doing a 22 multiply the determinant
7:40:07of so I'm re removing this row and this
7:40:10column A1 1
7:40:14a13 and then
7:40:16a31
7:40:20a33 and then the final part is of course
7:40:23as you might have already guessed is to
7:40:24look into this element so it is
7:40:31plus
7:40:33a23
7:40:35multiplied the determinant of let me
7:40:38actually write it down in here the
7:40:41determinant of
7:40:43a11 a12 and then a31 and then
7:40:50a32 so in this way basically independent
7:40:54of what row I
7:40:57will take as my leading row that I will
7:41:01do my calculations and I will just need
7:41:03to pick one row I can always get the
7:41:06same value for determinant of a but
7:41:09choosing intelligently which row to
7:41:12pick it will save you a lot of time and
7:41:15headache in terms of calculations
7:41:17because if you are dealing with a row
7:41:20that contains many zeros for instance
7:41:22you have 0 0 one or one 0 0 or even
7:41:27better 00 0 then you know know
7:41:29automatically that you will need to
7:41:31calculate your DET the determinant once
7:41:34here also once and here you don't even
7:41:36need to calculate it you know that you
7:41:38got zero here Z here zero here so it's
7:41:40automatically equal to zero so I hope
7:41:43this makes sense because this is a trick
7:41:45that usually you will not come across
7:41:47but this just helps you to save a lot of
7:41:49time uh when it comes to calculation of
7:41:51your determinants in a tree by3
7:41:54settings so in this case uh we have this
7:41:58um we now we have this definition and we
7:42:01know the tricks that we can use but I
7:42:03think it's really uh helpful to go ahead
7:42:06and to solve a problem so basically this
7:42:10is the higher level summary of the steps
7:42:12that we just discussed um so the
7:42:14determinant of a 3X3 Matrix it simply
7:42:16involves multiplying the a11 by the
7:42:21determinant of the 2x2 matrix that that
7:42:24remains after excluding the row and
7:42:26column of a11 so what we did in here by
7:42:30doing this and subtracting the product
7:42:33of A1 2 and the determinant of its
7:42:36respective 2 two Matrix so this
7:42:40part and then adding the product of a13
7:42:43and the determinant of its respective 2x
7:42:46two Matrix so this part and the signs
7:42:50alternate so it means first you always
7:42:53got the
7:42:57plus then you always is get the minus
7:43:00and then the
7:43:04plus so they interchange you start with
7:43:07plus then you do the minus and then the
7:43:17plus so let's go ahead and calculate the
7:43:20determinant of this
7:43:22Matrix so before even looking at the
7:43:25answer let's actually go ahead and do
7:43:28that on this page paper so we got a
7:43:32matrix a which is equal to 1 2 3 4 so
7:43:37basically from 1 till 9 1 2 3 4 6
7:43:434 5 6 and then 7 8 9 and for this 3x3
7:43:51Matrix we need to calculate the
7:43:53determinant so the determinant of a the
7:43:57first thing that I'm seeing is that
7:43:58there are no no rows with zeros or
7:44:00columns which means that I cannot use my
7:44:02uh trick and instead I will just need to
7:44:05go with let's say the first dra and it's
7:44:08also convenient given that I got as
7:44:10scaler this values this much smaller
7:44:12values relatively to the other
7:44:15ones all right so first things first
7:44:20let's go ahead and write down that
7:44:21formula so the determinant of a is equal
7:44:24to first we are going to take this one
7:44:28so our one
7:44:29one times and then we
7:44:32got determinant of and then we
7:44:37have this Matrix which is 5 6 8 and
7:44:449 this is our remaining Matrix then the
7:44:49next thing we need to do is to
7:44:51Interchange the size uh the the sign
7:44:54which is minus and then we got
7:45:06so the remaining Matrix is
7:45:14then determinant
7:45:19of 4 6
7:45:2579 and then
7:45:27finally Plus plus three
7:45:31times and then determinant
7:45:36of what do we
7:45:40have well this is the
7:45:42target so it is
7:45:454 5 7
7:45:508 right so let's go and do those
7:45:53calculations
7:45:54quickly this is equal to 1 * the
7:45:57determinant of this is the diagonal
7:45:59element so 5
7:46:03* it is 5 * 9 - 8 * 6 - 2 * 4 * 6 - 7 *
7:46:146 Sorry 4 *
7:46:189 so the diagonal elements 4 * 9 - 7 * 6
7:46:24and plus three * and then 4 *
7:46:318 - 5 *
7:46:347 this is equal
7:46:36to so 9 * 5 is = 45 8 * 6 is =
7:46:4548 then - 2 * 4 * 9 is
7:46:4936 7 * 6 is =
7:46:5449 7 * 6 is = to
7:46:5742 + 3 * 4 * 8 is = 32 - 5 * 7 is = 35
7:47:05so this is equal to 1 * - 3 - 2 * and
7:47:10then here we got 36 - 42 so that's - 6 +
7:47:173 * -
7:47:213 which is that = to - 3 + 12 - 9 which
7:47:28which is equal to
7:47:31zero so let's check it indeed we got the
7:47:36right answer perfect so now when we are
7:47:39clear on how we can do this calculation
7:47:41let's now go ahead and calculate yet
7:47:43another determinant of a 3X3 Matrix and
7:47:46this time I want to show you this uh
7:47:48simplified version by you making use of
7:47:51this trick that I uh specified so
7:47:54instead of using this first dra as an
7:47:56indicator I will be using the uh second
7:47:59row as my indicator one thing to keep in
7:48:03mind when making use of this trick is
7:48:05that when you start from the second row
7:48:10so from the even
7:48:13rows even
7:48:15rows second fourth or sixth then in
7:48:19those cases you need to flip the signs
7:48:22that you will be using so while in here
7:48:26you had Des Sign Plus
7:48:29in the beginning then you got a minus
7:48:31and then a plus when doing all these
7:48:33calculations so you remember here we got
7:48:36plus minus plus when you start from the
7:48:40second row instead of first one you need
7:48:42to flip the order of this so you need to
7:48:45start with minus you have minus you got
7:48:47plus and then
7:48:49minus so knowing this trick it also
7:48:52means that you go One Step Beyond and
7:48:55you know how you need to intelligent
7:48:58ently uh reduce the time that you are
7:49:01spending on calculation calculation of
7:49:04the determinant but it also means that
7:49:06you need to be careful on knowing what
7:49:09kind of signs you need to use because if
7:49:11you start from the first row you start
7:49:14with plus and then you do minus plus
7:49:16minus plus so knowing how to start you
7:49:19already know how you can go on but when
7:49:21it comes to the second row so the even
7:49:24rows you need to start with a minus so
7:49:26you need to do minus plus minus plus dot
7:49:29dot
7:49:30dot all right so let's now go ahead and
7:49:33use that technique in
7:49:36here so here I see that my first row
7:49:39doesn't contain zeros but my second row
7:49:41does so this gives me indication that I
7:49:44can reduce the time that I spent on
7:49:46calculating the determinant at least one
7:49:48time because I didn't no longer need to
7:49:51calculate that
7:49:52determinant so the determinant of B is
7:49:56then equals you I will then start with
7:49:59minus given that I'm going to use this
7:50:01rope and then I have zero times
7:50:05so because this is my element the
7:50:10determinant the determinant of
7:50:152306 and then
7:50:20plus then this time the second element
7:50:24Target element is this one so it's four
7:50:31times and then we
7:50:33got determinant of 1 3 1 6 and then
7:50:42minus the five
7:50:48times so this
7:50:52five determinant of 1 2 1 0
7:50:59and what is this amount it is equal to
7:51:02this I don't need to calculate because I
7:51:05got a zero in here this this trick is
7:51:07all about this to not calculate the
7:51:09determinant too often and
7:51:13then this equals you four
7:51:18times four times determinant of this is
7:51:226 - 3 so 1 * 6 - 1 * 3 which is equal to
7:51:283
7:51:32and then
7:51:37minus 5
7:51:39* determinant of 1 2 1 0 which is 0 - 2
7:51:49so this equal to 4 * 32 - 5 * - 2 which
7:51:55is equal to 12 + 10 and this is equal to
7:51:5922 let's go ahead and check this and
7:52:02this is the more detailed and formal
7:52:05derivation so one uh interesting thing
7:52:08is that I calculated with my second row
7:52:11and in here in this slides you can see
7:52:13calculation with the first dra this is
7:52:15just a nice way of seeing the difference
7:52:17that you can do and here uh in this
7:52:20solution what we have is that we have
7:52:22manually calculated this first
7:52:24determinant too so in total three
7:52:26determinants but we again in end up with
7:52:29the same determinant so independ what
7:52:31kind of row you will use in order to
7:52:33calculate your uh determinant of Matrix
7:52:36B you will all always end up with the um
7:52:39with the same similar volume unless you
7:52:42have made a mistake in your calculations
7:52:45so you just need to keep track of the uh
7:52:47rows that contain many zeros and you
7:52:50need to um be careful in terms of the
7:52:53signs that you need to use and the sign
7:52:56that you will need to start if you start
7:52:58with the first row then start with plus
7:53:00if you start with the second row then it
7:53:02is minus and then plus Etc so as you can
7:53:05see here it's a plus and then minus and
7:53:08then Plus in my case I did with my
7:53:10second row therefore I started with
7:53:14minus all right so let's now move on to
7:53:17the properties of determinants so the
7:53:20determinant of an identity Matrix is one
7:53:23that's something that we have also seen
7:53:25when doing our calculations because we
7:53:28are so that in one specific case when we
7:53:30had this example so this Matrix B and
7:53:34the Matrix B was the identity 2 in the
7:53:36two dimensional space we have calculated
7:53:39its determinant and we saw that it's
7:53:41equal to one and this was not a
7:53:42coincidence because the determinant of
7:53:45identity matrices is always equal to
7:53:49one then the second property is that
7:53:52swapping two rows or Columns of a matrix
7:53:55changes the sign of its
7:53:57determinant so if you swap rows or
7:54:00columns in your Matrix so if you end up
7:54:04with Matrix A and B they are exactly the
7:54:06same only one swaps the two columns or
7:54:10two rows then you are changing the
7:54:12determinant of that Matrix uh the sign
7:54:16of that determinant but not devalue
7:54:18itself it means that if you got a and
7:54:20you got B and your a is equal to let's
7:54:25say uh A1 and then uh A2 and then A3 so
7:54:31it contains these
7:54:33columns and then Matrix
7:54:37B is equal to um let's say
7:54:43A2 and then
7:54:45A1
7:54:47A3 then the
7:54:51determinant determinant of a will be
7:54:55equal to minus of the determinant of B
7:55:00you can also say determinant of B will
7:55:03then be equal to the minus determinant
7:55:07of a this is basically the idea of this
7:55:11property let's now move on to the third
7:55:14property which says that if a matrix has
7:55:17a row or a column of zeros its
7:55:19determinant is zero
7:55:23so if you got a matrix a that contains
7:55:27this different values a11 H1 dot dot dot
7:55:31a and uh M1 and then here you got
7:55:35suddenly um column that contains all
7:55:38zeros and then the rest are nonzero even
7:55:41so in that
7:55:44case you know that your
7:55:47determinant is equal to
7:55:50zero so for a specific
7:55:54example if you got for instance Matrix 1
7:55:592 0 0 0 3 13
7:56:04then the determinant of this Matrix is
7:56:08equal to
7:56:09zero and
7:56:12otherwise if you got a matrix B that has
7:56:17a column of zeros so column that is
7:56:20entirely of
7:56:22zeros so let's say here we have 1 one
7:56:25one and we have a zero Vector here so we
7:56:29got in here 0 0 0 and then 3 4 five then
7:56:34given that we have here this zero Vector
7:56:38then the determinant of Matrix
7:56:43B is equal to zero and this actually
7:56:47straightforward to be seen from this
7:56:50calculations that we saw because if you
7:56:52do the uh if you pick this specific row
7:56:56and then you do zero times the ter DET
7:56:58minant of the remaining Matrix 0 times
7:57:00determinant of the other Matrix and then
7:57:03plus so PL and then so minus and then
7:57:06plus and then minus 0 * determinant of
7:57:09the third
7:57:10Matrix it is obvious that 0 * a
7:57:13determinant is 0 0 * determinant is z 0
7:57:15* determinant is zero which means that
7:57:17you got a whole bunch of zeros to be
7:57:20added to each other or subtracted from
7:57:23each other this means that if you have a
7:57:25row or a Col with zeros this already
7:57:28gives you an idea that your determinant
7:57:30is equal to zero you don't even need to
7:57:32do
7:57:34calculations so the final property of
7:57:38determinants is that if a determinant of
7:57:40a product of matrices equals the product
7:57:43of their determinants so the
7:57:47determinant determinant of a b is equal
7:57:51to determinant of a multip by
7:57:55determinant of B this is basically what
7:57:57this property is
7:58:00about so let's quickly go through
7:58:02examples to ensure that we are at the
7:58:03same page with all these properties and
7:58:05we can prove them so let's say we have
7:58:09an identity
7:58:12Matrix n by n which is we are dealing
7:58:15with I in now according to this first
7:58:21property when we
7:58:24calculate the determinant of this Matrix
7:58:27so determinant of i n is equal to
7:58:361 Let's actually look at a specific
7:58:38example so here we got um identity um
7:58:42Matrix in the two dimensional space in
7:58:45the
7:58:46R2 and we can quickly calculate the
7:58:50determinant of this I2 and we can see
7:58:54that it is equal to this diagonal
7:58:56element so 1 by one - 0 * U it's
7:58:59actually something that we did as part
7:59:01of my previous examples so this equal to
7:59:051 - Z and is equal to 1 one thing that I
7:59:10wanted to show you before moving on to
7:59:12the next example about the swapping rows
7:59:15is that when we are swapping some of the
7:59:18rows or some of the comms or two rows or
7:59:20two cars of Matrix a we are referring to
7:59:23this matrix by this notation so we add
7:59:26this nod in here and we say that that
7:59:28this is basically the manipulated
7:59:31version of Matrix a so if we have for
7:59:33instance Matrix a equals u a b
7:59:39and and c and those are
7:59:43vectors and then we are
7:59:46swapping two of The Columns let's say we
7:59:50are swapping this two we get B and then
7:59:53a and then
7:59:55C then this Matrix will are referring as
7:59:58a not this is just an a matter of
8:00:01notation and we just learned that as
8:00:04part of the properties that the
8:00:06determinant of this new
8:00:10Matrix is equal to minus the determinant
8:00:14of
8:00:18a so if a matrix a has a row or column
8:00:22of zeros then the determinant of it is
8:00:24zero so let's actually quickly look at
8:00:27this specific example example in here we
8:00:29got a which is uh having a column of
8:00:32zeros and another column of B and D
8:00:35where B and D are real
8:00:39numbers so let's prove that this
8:00:41determinant is actually equal to zero so
8:00:44the determinant of a 2X two Matrix we
8:00:46have already seen is equal to the
8:00:48diagonal elements so 0 * D minus the of
8:00:54diagonal Elements which is 0 * B 0 * B
8:01:01and what is number * 0 is equal to 0 0 -
8:01:050 * B is also Z it's equal to Z
8:01:08therefore the determinant of a is equal
8:01:11to
8:01:13Z so when it comes to the uh determinant
8:01:16of a product of a matrices let's prove
8:01:19that the determinant of a * B is equal
8:01:22to the determinant of a times the
8:01:25determinant of B so therefore the first
8:01:28thing we need to do is to calculate this
8:01:30a * B let's quickly go ahead and do that
8:01:35so let me add here this um blank
8:01:42file so a is equal
8:01:46to 1 2 3 and 4 B is equal
8:01:51to 5
8:01:546
8:01:5678 and I I want to prove that the
8:01:59determinant of a b is equal to
8:02:02determinant of a Time determinant of B
8:02:07first I will be calculating this and
8:02:09then I will be calculating
8:02:11this so for the first one what I need to
8:02:17do is that first I need to
8:02:19calculate d a * B which is equal to 1 2
8:02:253 4
8:02:28times 5 6 7
8:02:328 and then this is equal 2 should be 2
8:02:38by two so first I take this 1 * 5 is 5 2
8:02:43* 7 is 14 14 + 5 is
8:02:4719 then for this one I need to pick this
8:02:51row so 3 * 5 is
8:02:5415 and then 4 * 7 is 28
8:02:59so 15 + 28 is so there we have 33
8:03:0543 so I got here
8:03:1643 then I'm going on to the next column
8:03:19which is in this
8:03:23case 1 * 6 is 6 6 + 6 16 is
8:03:2922 and now the second column 3 * 6 is
8:03:3418 4 * 8 is 32 and this gives me
8:03:4550 all right so now I have the a * B
8:03:49then as the next step what I need to do
8:03:52is to
8:03:53calculate the
8:03:55determinant of this a * B which is equal
8:04:00to the determinant of this Matrix 19 22
8:04:054350 that I just
8:04:08calculated and what is this amount the
8:04:11diagonal elements 19 * 50 - 43 * 22 19 *
8:04:1950 is then equal to
8:04:2195 and 43 * 22 is 94 6 which means that
8:04:29we end up with four this means that the
8:04:32determinant of the a * B is equal to 4
8:04:36let's quickly check what are the parts
8:04:39of the second amount so for that I need
8:04:44to calculate determinant of a which is
8:04:46equal to 1 * 4 - 2 * 3 1 * 4 is 4 3 * 2
8:04:53is 6 so 4 - 6 is = - 2 determinant of B
8:05:00is equal
8:05:01to 5 * 8 which is equal to 4T and then 7
8:05:06* 6 is equal to 42 and this is equal
8:05:11to - 2 and determinant of a *
8:05:17determinant of B is equal to - 2 * -2
8:05:21which is equal to 4 so we can see that
8:05:25now we just provve that the determinant
8:05:27of a * B is equal to 4 so we have seen
8:05:32that determinant of a is equal to 4 and
8:05:37we see that that's exactly the same as
8:05:40determinant a * determinant of B which
8:05:42is equal to 4 so we have just proven
8:05:45that the this equation indeed
8:05:51holds so the determinants they are not
8:05:54just um some calculations or some
8:05:58amounts but they are actually uh
8:06:00important concept and their
8:06:02interpretation um is highly relevant
8:06:05from geometric perspective so the
8:06:08terminant have a geometric
8:06:10interpretation and the for example the
8:06:13terent of a 2X two uh Matrix or 3x3
8:06:16Matrix they represent the area in case
8:06:20of 2x two or the volume in case of 3x3
8:06:24Matrix uh of the parallelogram
8:06:28that they are
8:06:29forming so uh this is often referred as
8:06:33a parallel uh piped um I hope I'm
8:06:37pronouncing this correctly and it's
8:06:39formed by the con vectors of the Matrix
8:06:43so if we have for instance this uh
8:06:46Matrix a and then we have a b and then C
8:06:48and D we have this A and C which is the
8:06:52first vector and then B and D which is
8:06:54the second vector and the uh the shoe
8:06:58vectors they actually form a
8:07:01parallelogram um when it comes to the uh
8:07:06two dimensional
8:07:07space and the area that this uh
8:07:11parallelogram um is
8:07:13forming that is equal to the determinant
8:07:17of this
8:07:18Matrix so the determinant ofer this
8:07:21scalar value that summarizes this linear
8:07:25transformation that we describe by this
8:07:28Matrix because we saw that we had this a
8:07:31x is equal to
8:07:32B linear system that we were describing
8:07:35using this coefficient Matrix and this
8:07:37was our unknowns this was our variable
8:07:42and then this B was the um amount that
8:07:46we were uh putting this as equal to if B
8:07:48was equal to zero then we were solving
8:07:50the homogeneous system otherwise we had
8:07:52this non-homogeneous system and in the
8:07:54geometric terms the determinant of this
8:07:57Matrix
8:07:57a so the determinant of a um in case of
8:08:022x two space so in
8:08:06R2 um when we got two vectors
8:08:09basically in our Matrix a this is equal
8:08:13to the area that is spent by these
8:08:16vectors in the two dimensional space in
8:08:18a bit I will also show you specific
8:08:20example such that um we will be on the
8:08:23same page when it comes to this concept
8:08:25of parallelogram the deter determinant
8:08:28and those vectors that form the column
8:08:31um uh space of the uh Matrix a uh when
8:08:36it comes to the three dimensional space
8:08:37when we have R3 so we got 3x3 Matrix of
8:08:43a then the determinant of this Matrix a
8:08:48is the volume that is um formed by these
8:08:52uh threedimensional vectors because
8:08:55unlike the 2D
8:09:00in R3 we got the three vectors that form
8:09:03the a let's say this one this one and
8:09:07then this one and then here we can
8:09:09create this
8:09:11area covered by this Tre vectors and the
8:09:14area that is formed by the tree vectors
8:09:17from a it is equal to the determinant of
8:09:21that Matrix a so in terms of the 3D it's
8:09:25bit harder to uh visualize it but in uh
8:09:29case of the two-dimensional space I
8:09:31think this will help uh to improve our
8:09:34understanding of the determinants and
8:09:35make this interpretation uh from
8:09:38geometry uh from geometrical perspective
8:09:41so given the two vectors A and B in the
8:09:44two dimensional space the determinant of
8:09:46this Matrix uh is then equal to the um
8:09:50diagonal elements we already know minus
8:09:53the of diagonal elements right so we are
8:09:56also saying
8:09:59we have seen this notation already very
8:10:01often you will see this volume this is
8:10:04the absolute we already know this from
8:10:06high school this is the absolute volume
8:10:09because the determinant can also be a
8:10:11negative number we have seen minus 20 or
8:10:13minus 2 and we know that the area cannot
8:10:17be a negative number therefore we are
8:10:19adding this absolute term here so
8:10:24knowing for example that we have this m
8:10:27matx a which consists of the elements 3
8:10:312 and then 1 14 we know that the
8:10:34determinant of this a is equal to 3 * 4
8:10:4012 - 1 * 2 it is 10 and the absolute
8:10:44value of it so absolute value of 10 is
8:10:47equal to 10 given that is positive and
8:10:49this is exactly what we have here and
8:10:52this is referred as the area of the
8:10:56parallelogram that the two vectors are
8:10:58forming and how does that look like in
8:11:02uh the uh coordinate space so this is
8:11:04the parallelogram that we were referring
8:11:07by and
8:11:09this area that is formed by this
8:11:13parallelogram is equal to the
8:11:16determinant of the
8:11:19a The Matrix
8:11:25a so so one thing that we need to keep
8:11:28in mind is the definition of
8:11:30parallelogram which means that those two
8:11:31are parallel and they are the same so
8:11:34this and this lines those two are the
8:11:37same and then of course the same holes
8:11:39for those two they are parallel and they
8:11:41have the same um length therefore this
8:11:46figure in here this is what we are
8:11:49referring as
8:11:51parallelogram and those two
8:11:54vectors that we can see in here
8:11:58this one and this one they form this
8:12:01parallelogram and they are the two
8:12:04vectors that are part of the Matrix
8:12:08a hence if we got two vectors that the
8:12:12uh that come from The Matrix a so Matrix
8:12:15a and we got here this two vectors in a
8:12:192X
8:12:21two Matrix then the determinant of this
8:12:26Matrix is then describing the area that
8:12:30these
8:12:30two vectors are using or are spanning
8:12:35when creating this
8:12:39parallelogram so the determinants they
8:12:41play an important Ro in understanding
8:12:44the geometric properties of the spaces
8:12:47that uh spent uh by these vectors they
8:12:51provide valuable insights when it comes
8:12:53to the scaling effect effect of linear
8:12:55transformation the or orientation and
8:12:58the um the locations of them in the
8:13:02cordan system as well as the Practical
8:13:04applications in calculating areas in
8:13:06calculating volumes welcome to another
8:13:09unit in our fundamentals to linear
8:13:11algebra course where we are going to
8:13:13talk about Advanced linear algebra
8:13:15Concepts so uh in the first module we
8:13:18are going to talk about Vector spaces
8:13:20and the projections we are going to
8:13:23define the bases in a couple of examples
8:13:26of them we have already touched upon
8:13:28this concept briefly when we are
8:13:30calculating the basis of a no space and
8:13:33the basis of a comp space we are going
8:13:35to do a similar example in this case and
8:13:38then we are going to uh look into this
8:13:40concept of the uh standard bases for uh
8:13:44different spaces including the R2 we're
8:13:47going to introduce the concept of
8:13:49projections what is the definition of
8:13:51projections what is a Formula how we can
8:13:54calculate it we are going to look into
8:13:55detailed examples of that
8:13:58then we are going to talk about the
8:13:59concept of uton normal basis in this
8:14:01module we are going to introduce this
8:14:03concept and we are going to understand
8:14:04the orog gonality normalization we are
8:14:07going to then discuss a very important
8:14:10topic in linear algebra which is a
8:14:12gramme process we're going to Define it
8:14:15we are going to see the overview the
8:14:17step-by-step process of applying grme uh
8:14:21algorithm then we are going to see an
8:14:23example of it and the calculations step
8:14:26by step
8:14:27and then we are going to talk about
8:14:29applications of auton normal bases the
8:14:32application of gram Smiths process and
8:14:34the importance of this auton normal
8:14:37basis this is the module one of this
8:14:41part so let's first Define the basis a
8:14:45basis of a vector space is a set of of
8:14:48linearly independent vectors that spend
8:14:50the entire Vector space every Vector in
8:14:54the space can be expressed as a unique
8:14:56linear combination of the basis
8:15:00vectors so there are a couple of parts
8:15:02in this definition they are really
8:15:05important and first thing that we need
8:15:08to uh mention here is this Vector space
8:15:13that says it is a set of linearly
8:15:16independent vectors that spend the
8:15:18entire Vector space this is very
8:15:20important because um here we are with
8:15:25the basis is simply this Vector space
8:15:30that is a set of linearly independent
8:15:33vectors which means that one of these
8:15:36vectors cannot be Rewritten as a linear
8:15:39combination of the other one so we have
8:15:42a linearly independent vectors and they
8:15:45span the entire Vector
8:15:48space so for instance if we are in
8:15:52R2 then the basis of a vector space is
8:15:56then a set of linearly independent
8:15:58vectors that span this entire
8:16:01R2 so we do we then need to
8:16:05have for vectors forming a basis so
8:16:09let's say we have a basis of vector
8:16:15space for us to say that this is the
8:16:17basis of this Vector space let's say in
8:16:22R2 we need to First
8:16:25say we need to First prove that these
8:16:30vectors this
8:16:34vectors are linearly
8:16:39independent and
8:16:42two they
8:16:45span the entire R2 which means that span
8:16:49of this
8:16:51vectors is equal to
8:16:54R2 we can actually be even more spefic
8:16:59specific in a
8:17:01example of let's say having a vectors A
8:17:06and
8:17:07B we can say that this set that we have
8:17:11here consisting of vectors A and B in
8:17:17R2 form the
8:17:20bases of a vector space
8:17:28if the first criteria
8:17:33is that
8:17:35a and b are
8:17:41linearly
8:17:44independent and the second criteria is
8:17:48that those two vectors together they
8:17:52spend the entire Vector space of R2
8:17:56which means that
8:18:00span of a and b Vector space is equal to
8:18:07R2 on more specific
8:18:11example and then the second part of this
8:18:13definition says that every Vector in the
8:18:18space can be expressed as a unique
8:18:20linear combination of the basis factors
8:18:23which means in our specific example when
8:18:26we had this A and B forming the bases of
8:18:28a vector space this means that if we
8:18:31prove that this is indeed the
8:18:34basis of this Vector
8:18:40space then any
8:18:42combination every
8:18:45Vector let's say a vector
8:18:49C that consists of this C1 and C2
8:18:55elements that this Vector this random
8:18:58Vector from
8:19:01R2
8:19:03C can be represented as a linear
8:19:07combination of these vectors A and B so
8:19:12let's say we have a coefficient
8:19:15K1 * a plus K2
8:19:20*
8:19:22B then here we are representing this
8:19:26random Vector c as a linear combination
8:19:30of these vectors A and B which form the
8:19:33bases of a vector space of this Vector
8:19:38space so we have previously spoken about
8:19:41the no space and Comm space so let's now
8:19:46go ahead and do one more example when we
8:19:49are calculating the new space and the
8:19:51Comm space and then we are again
8:19:54calculating this concept of basis of
8:19:56Vector space and a b basis of the com
8:19:59space and then we will be uh finding the
8:20:02basis of a vector space uh with um R2
8:20:06example so given that we have already
8:20:10looked into this concept the basis of
8:20:12Comm space and based of no space I will
8:20:14try to uh go through this example bit
8:20:16more quickly to save time on more
8:20:18complex
8:20:22Concepts so let's say we have an example
8:20:26Le of a matrix and that Matrix is a is
8:20:31equal to 1 2
8:20:3536 this is our 2x two Matrix
8:20:39a and the first thing that I want to do
8:20:43is to understand look into my Matrix and
8:20:46understand whether I'm dealing with
8:20:48unique vectors or not and by unique I
8:20:51mean whether I'm dealing with two
8:20:53vectors that are linearly dependent or
8:20:55linearly independent this kind of
8:20:58inspection always helps us to save time
8:21:01when we are doing our calculation for
8:21:03the no space and for the Comm space and
8:21:05for the basis of no space and base of
8:21:07Comm space now here we can see that this
8:21:10is our A1 the first Vector the first com
8:21:14Vector forming the Matrix a and this
8:21:18Vector is the
8:21:20A2 another thing that uh we can notice
8:21:23here is that we can easily take the
8:21:27First Column A1 multiply it by two and
8:21:31get the A2 because 1 * 2 is 2 3 * 2 is 6
8:21:37that is that 2
8:21:42*
8:21:44A1 is equal to
8:21:50A2 which means that we can say that A1
8:21:57and
8:21:57H2
8:21:59are
8:22:02linearly
8:22:06dependent okay so seeing this and
8:22:09knowing this this can help us to quickly
8:22:11go through our calculations of the bases
8:22:14of the co space and the bases of a no
8:22:16space so let's go ahead and first
8:22:19calculate what is the
8:22:22basis of no space
8:22:28of
8:22:30a so we have already learned that the um
8:22:34basis of a no space can be calculated
8:22:38when looking into the first no space so
8:22:41we C we need to calculate the no space
8:22:44and then we need to calculate the basis
8:22:46of that no
8:22:47space
8:22:49so this means that we need to get the
8:22:53na and we have learned that in order to
8:22:56get
8:22:58DNA we for that need to solve the a x is
8:23:02equal to zero
8:23:04problem and this
8:23:10x will give
8:23:13us the no space of a we have also
8:23:17learned that the no space of a is equal
8:23:20to the no space of r r EF of a which
8:23:25means that using gausian reduction or
8:23:28gausian elimination we can quickly find
8:23:31the solution to this problem of a x is
8:23:35equal to Z and find this x this is
8:23:39simply solving a similar problem only in
8:23:41this case the B so this is equal to zero
8:23:46because we are dealing with the
8:23:49homogeneous
8:23:53case I want do the calculation for this
8:23:56we have done a ton of examples when we
8:23:58were doing this step by-step calculation
8:24:01getting the uh argumented Matrix of a
8:24:04and then uh doing all these different
8:24:06draw operations normalizations and then
8:24:09eliminations in order to uh get this uh
8:24:13complex Matrix a to the point of uh
8:24:16basic representation from which either
8:24:18we can visibly see the solution to the
8:24:21problem or we can at least simplify it
8:24:23and describe it as a linear combination
8:24:25of vectors
8:24:27in this case if you go ahead and solve
8:24:29this problem you will find that the
8:24:33x that
8:24:35solves the a x is equal to Z
8:24:39problem is unique and this x is equal to
8:24:44minus
8:24:460.894 as the first element and then
8:24:500.447 as a second element this can be a
8:24:53good practice also to refresh um the
8:24:55memory when it comes to the gaion
8:24:57elimination and reduction the example
8:24:59itself is quite simple the a is just a
8:25:022x2 matrix um and um by performing
8:25:06couple of operations uh in terms of
8:25:08normalization and elimination you can
8:25:10find this
8:25:12X for your a is equal to
8:25:18zero given that now we know what is the
8:25:22solution to a is equal to Z problem now
8:25:25we know what no space is because in this
8:25:28case
8:25:29the all this help us to understand that
8:25:34the no space of a is then equal to the
8:25:38set the vector set where as part of this
8:25:42we got just single column which is -
8:25:470.894 and
8:25:510.447 this is the no space
8:25:56this is the first part I will say it
8:26:011.1 and then 1.2 will
8:26:06be to get the
8:26:11basis
8:26:13basis of this
8:26:16Na and we have just seen what is the
8:26:19definition of the bases so the basis of
8:26:22vector space is a set of linearly
8:26:24independent vectors that spend the
8:26:25entire Vector
8:26:29space therefore given that we got just
8:26:32this single Vector as a solution to our
8:26:36problem we can see then very quickly
8:26:40that the new space of a is based on this
8:26:43and then the basis of the no space is
8:26:46simply this entire
8:26:48set so knowing what the solution is to
8:26:51our homogeneous problem a is equal to
8:26:55zero so let me also write down in here
8:26:58then we know that the no space the N A
8:27:03is then equal
8:27:072D
8:27:09Vector minus
8:27:120.894 and
8:27:140.447 this is my Vector X that solve
8:27:16this ax isal to zero
8:27:20problem and this is simply the no space
8:27:23of a and given that we
8:27:27have calculated and we have got this
8:27:30unique solution to our problem we can
8:27:34say that any Vector in R2 can be
8:27:37represented as a linear combination of
8:27:40this
8:27:42Vector so
8:27:451.2 any
8:27:49Vector in
8:27:51R2 can be represented
8:27:58as linear
8:28:00combination
8:28:07combination
8:28:09of this
8:28:13Vector
8:28:20X therefore we are saying that the
8:28:24bases of of no
8:28:29space of
8:28:33a is this entire set consisting of the
8:28:37single
8:28:45Vector so this is about the basis of a
8:28:49no space
8:28:56let's Now quickly look into the concept
8:28:58of the basis of a calm
8:29:01space so the first thing we need to then
8:29:04uh get is the column
8:29:07space
8:29:09so to get the
8:29:13basis of Comm
8:29:19space we need to get the ca first which
8:29:22is the Comm space of a and what is the
8:29:25Comm c space of a the Comm space of a is
8:29:29the uh setle and the space of the
8:29:32vectors that we can see in
8:29:35here in this A1 and A2 is it's quite
8:29:40straightforward so this two vectors they
8:29:43form the Comm space of this Matrix
8:29:47a so then the ca is
8:29:52simply the set of one three
8:29:56and then two six vectors this is
8:30:00A1 this is
8:30:03A2 now we have just seen in the
8:30:06beginning before even starting our
8:30:08calculations that A1 and A2 are linearly
8:30:11dependent because A2 can be right
8:30:14written as 2 * A1 so one of these
8:30:17vectors can be written as a linear
8:30:19combination of the other one this means
8:30:21that we got just a single linearly
8:30:24independent vectors and why is this
8:30:26important because we have seen in the
8:30:29definition of the basis that for us to
8:30:32have a basis we need to have a linearly
8:30:36independent vectors so the basis of
8:30:40vector space in this case the Comm space
8:30:42is a set of linearly independent vectors
8:30:44that need to spend the entire Vector
8:30:46space in this case
8:30:49R2 so
8:30:51therefore we need to look into the ca
8:30:54that we got in here and select one of
8:30:58these two
8:31:00vectors that can be considered as
8:31:02linearly independent let's say we pick
8:31:05one
8:31:07three now we know that we can then write
8:31:12any Vector in R2 as a linear combination
8:31:16of this Vector 1 3 so we can scale this
8:31:20Vector one Tre and get a new Vector in
8:31:24R2 therefore or we are saying that the
8:31:28basis of Comm space
8:31:31basis of Comm
8:31:35space
8:31:37space of a is then the set of one
8:31:44Tre because one Tre so
8:31:48A1 is then
8:31:50linearly
8:31:52independent and the span of
8:31:56A1 is
8:31:59R2 now when it comes to the uh basis of
8:32:03the entire R2 one thing that we can
8:32:06notice is that
8:32:08this A1 so one
8:32:12three it's not forming it's not spanning
8:32:14the entire
8:32:15R2 because because we cannot uh write
8:32:20any random Vector in r two as a linear
8:32:23combination of this two therefore we are
8:32:25saying that this is the basis of Comm
8:32:27space but we are not saying that this is
8:32:29the basis of R2 and the final element in
8:32:32this definition that I want you to uh
8:32:34focus on is that every Vector in the
8:32:37space can be expressed as a unique
8:32:38linear combination of the basis
8:32:41vectors so in here we have looked into
8:32:45this idea of bases of a new space and
8:32:47the base of Comm space and we saw that
8:32:51we are talking about specifically the
8:32:53new space and comp space but when it
8:32:55comes to the entire space for instance
8:32:58the basis for R2 then the basis of Comm
8:33:04space for instance is no longer um
8:33:07helping us because the basis of Comm
8:33:10space it consists of this Vector one
8:33:12tree and this one tree alone is not
8:33:15satisfying the second criteria that says
8:33:18that this Vector needs to spend the
8:33:20entire Vector space because this one Tre
8:33:25vector it's a single vector and this
8:33:29Vector it is not forming the entire R2
8:33:34it's not um the basis for R2 it's not
8:33:38spinning the entire uh R2 so
8:33:43given that the one tree is not
8:33:50spinning the entire
8:33:54R2 because of that we know that the one
8:34:00tree is not the set of one Tre is not
8:34:05the
8:34:08basis of
8:34:10R2 so this distinguishing of the basis
8:34:15of R2 basis of Comm space basis of no
8:34:18space is really important because basis
8:34:21for R2 it means that we need to find set
8:34:24of linearly independent vectors that
8:34:27they together form the entire R2 they
8:34:30span the R2 which means any random
8:34:33Vector that we can see in R2 we can
8:34:35represent as a linear combination of the
8:34:38vectors in this space so in here let me
8:34:43also prove that this one tree alone is
8:34:46actually not forming the R2 it's not
8:34:49spinning the R2 which then uh concludes
8:34:52that they are not the it is not the
8:34:54basis of R2 CU and after this I will
8:34:58then provide you an example where we
8:35:01have a set of vectors that span R2 and
8:35:05are linearly independent which means
8:35:06that they are the bases of the entire R2
8:35:10so first I want to show you why this
8:35:12single Vector one Tre is not the basis
8:35:17of
8:35:18R2 so being the base of R2 we have the
8:35:22criteria that the vectors need to be
8:35:25linearly
8:35:31dependent so let me actually clear up
8:35:34some space
8:35:38here so I want to see and find the basis
8:35:41of
8:35:45R2 first I want to
8:35:47prove that this
8:35:50set which is the base of Comm
8:35:54space I want to
8:35:56prove that this is not the
8:36:01basis of
8:36:04R2 then I will also as part of the
8:36:08second part of this proof look in look
8:36:10into the case when we do have vectors
8:36:13and the set of vectors it forms the base
8:36:16of
8:36:17R2 so the first thing the first criteria
8:36:21of the basis of R2
8:36:23says that quote 1.1 the first criteria
8:36:28says that this Vector in this Vector
8:36:31space it need to be they need to be
8:36:34linearly independent well that criteria
8:36:37is valid given that one3 is
8:36:43linearly
8:36:47independent this means that criteria one
8:36:50is satisfied
8:36:59so whenever you got just one vector this
8:37:01criteria is automatically
8:37:03satisfied so then you have the
8:37:061.2 which says that we need to have this
8:37:10spin of these vectors equal to
8:37:14R2
8:37:15so is the
8:37:17span of
8:37:21one3 the R2
8:37:29well no and how we can prove that
8:37:32because the idea is that any Vector
8:37:35including an example where I have for
8:37:38instance uh let's say four and five this
8:37:42Vector that I need to be able to find a
8:37:47scalar that will help me to create a
8:37:50linear combination let's say
8:37:53C linear combination using this Vector
8:37:5713 which will then set this amount this
8:38:01to be equal to this which means that I
8:38:03need to be able to write my random
8:38:07Vector 45 as a linear combination of
8:38:09this Vector that forms my uh Vector
8:38:14space so let's see whether that is even
8:38:17possible well here I got four and
8:38:21five if I do this multiplication in the
8:38:24right hand side I
8:38:26get C and here I got 3
8:38:31C because C * 1 is C and 3 * C is
8:38:37C and this means that I have an
8:38:41equation 4 is equal
8:38:44to
8:38:46C and 5 is equal
8:38:51to 3 * C
8:38:57from this I get that the C is equal to 4
8:39:01and C is equal to 5 / to
8:39:053 but that is impossible because 4 is
8:39:09not equal to 5 / to 3 which means that
8:39:12I'm proving in here and I got to prove
8:39:15that the uh any random chosen Vector 45
8:39:20cannot be written as a linear
8:39:23combination of this Vector that 4 forms
8:39:25this uh space
8:39:28therefore as
8:39:32random
8:39:35Vector from
8:39:39R2
8:39:41can't be
8:39:45written as
8:39:48linear
8:39:50combination
8:39:52of one three
8:40:02criteria two is not
8:40:11satisfied because for that we had to say
8:40:15that this pen of One Tree is equal to R2
8:40:19which we saw that it's not the case
8:40:21because then we would have been able to
8:40:22represent this four five as a linear
8:40:24combination of the one Tre Vector okay
8:40:27so now we have proven that the one Tre
8:40:30is not forming the bases of R2 let's now
8:40:33look into what then does form the basis
8:40:36of R2 an example of
8:40:38it so we are familiar with the unit
8:40:42vectors E1 and E2 into
8:40:46R2 which form the identity Matrix
8:40:51I and this is 1 0 and this is 0 1
8:40:55also 1 0 0 1 in the form of a
8:41:04matrix so in this example we have a set
8:41:10consisting of E1 and
8:41:14E2 where this is this E1 this is the
8:41:20E2 and the set corresponding to this
8:41:23Vector space is then
8:41:261 Z and then
8:41:3001 and now I will be proving that this
8:41:34space this Vector space does
8:41:38indeed equal to the bases of
8:41:42R2 so this
8:41:45is the
8:41:47basis of
8:41:50R2 so the first
8:41:52criteria of the bases is that these two
8:41:56vectors should be linearly
8:42:00independent now we can quickly uh
8:42:02remember from our previous theory that
8:42:06the two unit vectors one z01 are
8:42:12actually linearly independent that's
8:42:13something that we have proven and you
8:42:15can easily see it also from here there
8:42:18is no way that you can find um scalar
8:42:21C that you can multiply this Vector we
8:42:25and get a vector 0 one because for that
8:42:29for this one to become a zero you need
8:42:31to multiply this with zero but then 0 *
8:42:340 is not equal to 1 which means that
8:42:37there is no way that you can find a
8:42:39scaler C to multiply this E1 to get the
8:42:44E2 so let me write this down
8:42:55E1 and E2 are
8:43:00linearly
8:43:06independent
8:43:09because
8:43:11there is
8:43:14no scaler
8:43:17C which is a real
8:43:20number such that
8:43:25such
8:43:26that c
8:43:29*
8:43:31E1 is equal to
8:43:34Ich
8:43:35so this
8:43:40means you
8:43:43can't
8:43:46write hu as linear
8:43:51combination of A1
8:43:57or vice
8:44:01versa this means that E1 and E2 are
8:44:06linearly
8:44:08independent and this
8:44:11satisfies our first
8:44:13criteria so
8:44:15criteria one is satisfied
8:44:27what we have also learned is that any
8:44:30Vector in R2 can be actually written as
8:44:33a linear combination of a unit vectors
8:44:37that form that um
8:44:39R2 in this case 1 0 and
8:44:4201 so let's assume that this random
8:44:47Vector is C1 C2 so this is C vector
8:44:55and what we want to prove is that we can
8:44:58always write this C in terms of linear
8:45:00combination of these two vectors and how
8:45:03can we do that
8:45:06well let's say here we got a
8:45:13K1 K1 which is a real
8:45:17number and we multiply this by one
8:45:21Z and then we add
8:45:29K2
8:45:31K2 and then here
8:45:3401 so this is our E1 this is our E2 can
8:45:38we do this well what is this this is
8:45:41equal to K1 0
8:45:48plus 0
8:45:53K2 and and what does this give
8:45:57us
8:45:59well this means this
8:46:03amount let me write it
8:46:06over K1 * 1 which is the E1 plus K2 * 01
8:46:14which are which is our second Vector E2
8:46:17this is equal to K1
8:46:200+ 0 K2 and this is equal to K1
8:46:27K2 so I got on one hand this Vector C1
8:46:34C2 which I want to write as a linear
8:46:39combination of K1 E1 plus K2
8:46:46E2 if I take
8:46:50D
8:46:51K1 equal to C K1 and
8:46:57K2 K2 equal
8:47:03to
8:47:05C2 well then in that case I can prove so
8:47:10this is basically equal to C1 and C2
8:47:14which means if I take this K1 and K2
8:47:16equal to C1 and C2 respectively and
8:47:19those numbers are given then I can
8:47:22represent this vector
8:47:26c as a linear
8:47:29combination of
8:47:30E1 and
8:47:33E2 which is what I had to prove in order
8:47:36to say that
8:47:39the
8:47:41Spen
8:47:44of one Z which is the
8:47:47E1 and 01 which is
8:47:51E2 is equal to R2
8:47:55because any random Vector that will be
8:47:58provided to me with an element C1 and C2
8:48:01and those are just real numbers can be
8:48:03written as a linear combination of these
8:48:06two vectors this means that the spend of
8:48:09these two vectors is equal to
8:48:11R2 and this is basically the second
8:48:15criteria so
8:48:18criteria
8:48:20to
8:48:21satisfied and if the criteria one and
8:48:24criteria 2 are both satisfied it means
8:48:29that this Vector
8:48:33space of 1 Z and
8:48:3901 this is the
8:48:42basis of the entire
8:48:45R2 so let's now talk about the concept
8:48:48of projections by definition a
8:48:51projection of a vector a onto another
8:48:53Vector B is the orthogonal projection of
8:48:57a along B it's denoted by approach and
8:49:01then B underneath here we see the index
8:49:03and then a so projection of a onto B so
8:49:07here is the A and here is the B and
8:49:10represents the component of a in the
8:49:12direction of
8:49:14B so component of a in the direction of
8:49:21B all right so in order to properly
8:49:24understand this concept the intuition of
8:49:26it let's actually make use of the R2
8:49:30space so let's first start by picturing
8:49:34in our flat world the R2 coordinate so
8:49:37the Cartesian coordinate system so let's
8:49:40say here we got our y AIS here we got
8:49:44our
8:49:45x-axis so this is the X this is the
8:49:49Y and uh here we of course we need to
8:49:54keep in mind this is just an example
8:49:56when it comes to projections we can
8:49:58always go beyond R2 but for keep it
8:50:00simple and truly understand this
8:50:02Concepts and this intuition behind the
8:50:05projection I want to simplify this and
8:50:07do the example in
8:50:09R2 so here uh imagine that we got this
8:50:15line
8:50:17and this is
8:50:20our a line that goes through the
8:50:23center that let's call this
8:50:29line
8:50:31B so B is
8:50:35line in
8:50:37R2 let's say this is that
8:50:43line and now that imagine that we have
8:50:47this
8:50:48Vector which is part of this
8:50:52line let's say this is this line
8:50:58and this line is the representing by uh
8:51:02on this line we got this Vector B and
8:51:06this Vector is basically part of that
8:51:09line as you can
8:51:14see this is the vector B on this line B
8:51:20so we know from this concept of the line
8:51:24spanning the R2 and then vectors we know
8:51:27that in this case independent what is
8:51:30the magnitude of this Vector what is the
8:51:32direction of this Vector we can
8:51:34represent this line
8:51:37B by this linear
8:51:40combination based on this Vector so
8:51:42linear combination of this Vector which
8:51:44is in this
8:51:46case
8:51:49D
8:51:52set set then here we got some
8:51:58C where C is a real
8:52:05number multiplied by this
8:52:08Vector
8:52:13B knowing that this C is just a real
8:52:21number so let's make it actually green
8:52:28so we can basically say that this entire
8:52:31line B can be represented as this set of
8:52:35this linear combinations of these
8:52:37vectors so for instance if this
8:52:41is one and we do the C is equal to two
8:52:46then we can get this
8:52:48part of so we can get this Vector
8:52:51otherwise this is equal to three we can
8:52:53get this vector or C is equal to for
8:52:55this vector and then and so on which
8:52:58means that we can always come up with a
8:53:00linear combination forming a part of
8:53:03this line therefore we are seeing that
8:53:05this line can be represented as all
8:53:07these linear
8:53:08combinations uh of this Vector B which
8:53:11is part of this
8:53:12line and here the C is just a scaler so
8:53:17a number which is a real number so this
8:53:20C * Vector B represents this uh entire
8:53:24line
8:53:26so we will knit this in a bit but for
8:53:28now imagine this line and part of this
8:53:30line which is this Vector
8:53:33B so imagine then that we got yet
8:53:37another
8:53:39Vector which is let's say in
8:53:44here again going from the center but
8:53:47this time in this different
8:53:50direction
8:53:52so in here
8:53:55this
8:54:00is
8:54:02Vector
8:54:04a we call this Vector an
8:54:08A so you can see that this Vector a is
8:54:11actually much longer than the vector B
8:54:14and we see that Vector a is not lying on
8:54:17the same line as B so B is lying on the
8:54:19line b and a is not lying on the line B
8:54:25now let's say we want to
8:54:29project this Vector a onto this Vector B
8:54:34which means that we want
8:54:36to project this a in this
8:54:41direction so we want
8:54:44to bring this Vector a onto this
8:54:50line let me actually use a different
8:54:55color and the word of the projection
8:54:58actually does make sense in here as you
8:55:00might notice because we're trying
8:55:03to cast the shadow of a onto this line
8:55:07of B and how can we do that we can only
8:55:12do that if we connect
8:55:16this Vector
8:55:18a like this with this orthogonal line
8:55:23let's Say by using a different color
8:55:28of
8:55:30this so with this perpendicular
8:55:34line we then will be connecting the
8:55:37vector a to the line B because we want
8:55:39to project our Vector a onto this
8:55:48direction so this perpendicular line
8:55:51that you see in
8:55:52here that
8:55:54goes from Vector a to the line B where
8:55:59on line B we have the vector B so here
8:56:03is the line a line B and this
8:56:07perpendicular line it goes from
8:56:11a to line B and on line V we have the
8:56:15vector
8:56:17B that is represented like this then the
8:56:21projection of a onto Line B
8:56:24is this Shadow Vector that you see in
8:56:28here and the word projection or the name
8:56:32projection actually does make sense
8:56:35because we are projecting this Vector a
8:56:38onto this line and it creates this
8:56:40Shadow so we are casting this Shadow on
8:56:43here and this Vector is what we are
8:56:47referring as
8:56:49projection of vector a onto l line B
8:56:55notice that we don't say projection of B
8:56:58on Vector B but instead we are saying
8:57:00projection of a on the line B then
8:57:03another thing we can notice is that we
8:57:05are getting this
8:57:07projection of a on B so this
8:57:11vector by taking the vector
8:57:17a so Vector a and subtracting from that
8:57:26projection of a on
8:57:31B that is the
8:57:33formula for this
8:57:36Vector that we refer as a
8:57:40perpendicular that
8:57:42goes from a to line B so when drawing
8:57:47this perpendicular line from a to line B
8:57:50we are referring this as a minus
8:57:52projection of a b because you can see
8:57:55that this Vector is simply this Vector
8:57:57minus this Vector that is the um
8:58:00mathematical expression for this
8:58:02perpendicular
8:58:05line so how we can then find out what is
8:58:10this C that we got in here because we
8:58:14understand that to get this exact
8:58:17formula for
8:58:19the
8:58:21projection of a
8:58:24on the line B we need to understand what
8:58:28is the scaler specifically what value
8:58:31are we using to multiply this Vector B
8:58:33to get to this
8:58:36point so what is that
8:58:42c what is C what is
8:58:47C such
8:58:50that c times
8:58:54a is then equal
8:58:58to
8:59:01projection of
8:59:04a on the line B because we can have
8:59:08different sorts of a linear combination
8:59:11of vector
8:59:12B on this line uh B and in fact B this
8:59:17line B is the set of all linear
8:59:20combinations of this Vector B and I want
8:59:23to know
8:59:25specifically what is the vector that we
8:59:29see in here what is the shadow Vector
8:59:32because this is the projection of a on
8:59:34the line
8:59:35B what we see in
8:59:37here now how can we do
8:59:41that well let's first formally Define on
8:59:45this specific case what is the
8:59:47projection of a on this line
8:59:51B so projection
8:59:54of
8:59:55a on line
8:59:58B is some
9:00:02Vector that is
9:00:05also
9:00:07on line
9:00:09B
9:00:17where
9:00:18a
9:00:22minus projection
9:00:25of
9:00:26a on
9:00:30B
9:00:32is per
9:00:38pendicular or
9:00:43ortogonal to this is basically the
9:00:46definition of the projection of a on
9:00:49line B under this specific example
9:00:54so in this case the way we can find this
9:01:00projection is by looking into this C so
9:01:04this is what we are interested this
9:01:07specific
9:01:09specific C
9:01:13* B
9:01:17vector and knowing C and knowing B we
9:01:20already know what is B what B is knowing
9:01:23C
9:01:24we can then describe this specific
9:01:29projection so one thing that we can know
9:01:32is the condition under which we say two
9:01:35vectors are autal that's something that
9:01:37we already have learned as part of the
9:01:39previous lessons so let's go ahead and
9:01:42find that amount so now what we need to
9:01:44do is to calculate this value of C
9:01:46because value of C calculation will then
9:01:48lead us to the exact uh Vector that we
9:01:52are interested in which is this
9:01:54projection so our end goal is to find
9:01:57out what is this projection of a on B
9:02:02this is what we want and for that we
9:02:04need to calculate this C because we
9:02:05already know the vector B so let me
9:02:09quickly remove this
9:02:16part cuz here we will then do our
9:02:21calculation so one thing that we need to
9:02:24make use of is this part when it says
9:02:26orthogonal because we know that if two
9:02:29vectors are orthogonal then they dot
9:02:32product is equal to zero so we know that
9:02:36this Vector is orthogonal to this target
9:02:39Vector which means that we can say that
9:02:43the
9:02:45vector
9:02:46a and
9:02:48then minus
9:02:55projection of
9:02:57a on
9:03:02B
9:03:06multiplied with Vector
9:03:10B that this is equal to
9:03:14zero this is something that we know by
9:03:17definition of orthogonality two vectors
9:03:19are orthogonal it means that their
9:03:20dotproduct is then equal to zero
9:03:24now let's make use of that part
9:03:27so this means that we need to describe
9:03:32this projection of A and B we need to
9:03:35make use of the fact that we know that
9:03:37this projection of a onto B is actually
9:03:43some linear combination of vector
9:03:47B so let me actually go ahead and remove
9:03:51this part we already know the definition
9:03:55so let us go ahead and calculate that c
9:03:58that we need in order to find out what
9:04:00is this entire projection so few things
9:04:03that we need to clear out is those
9:04:05formulas because then we can make use of
9:04:07them to find the C so we know that by
9:04:09definition the projection of a on the
9:04:11line B it is this
9:04:15Vector that we get where we draw this
9:04:18perpendicular line from Vector a onto
9:04:21Line B and we said that this line is
9:04:23equal to this amount this is simply the
9:04:26vector a minus this Vector the shadow
9:04:29Vector which we said it's defined by
9:04:31projection of A and B this thing so we
9:04:34can make use of that because we also see
9:04:36in here that this we are saying isogonal
9:04:40to this
9:04:43Vector so given that this uh Vector a
9:04:48minus projection a b is orthogonal to
9:04:51line B that is also orthogonal
9:04:54on this specific Vector which is the
9:04:56projection itself so from this we can
9:05:01make use of the fact that two vectors
9:05:03when they are autal their dotproduct is
9:05:06equal to zero in order to find this uh
9:05:09value of C so firstly we just set that
9:05:14the a minus projection of a on the line
9:05:21B that this
9:05:25multiplied by this Vector B is equal to
9:05:29zero because those two lines they should
9:05:31be
9:05:36perpendicular but at the same time we
9:05:39know that this is simply the linear
9:05:42combination of this Vector because this
9:05:47line is perpendicular to this one and
9:05:50this line is some linear combin a of
9:05:55this Vector B because if I have here a
9:05:59vector and then I have the longer
9:06:03version of that Vector on the same line
9:06:05which is then a linear combination of
9:06:08this original Vector let's say this is
9:06:09my Vector B then this second Vector that
9:06:14I have in here is then equal to some C *
9:06:19Vector
9:06:20B this is also exactly what we said in
9:06:23here here we said any Vector on line B
9:06:26can be represented as a linear
9:06:28combination of vector B and this is
9:06:31exactly what we are seeing in here so
9:06:34this
9:06:35projection is simply that c
9:06:41times Vector
9:06:44B this is something that we have already
9:06:47said so we are just making use of that
9:06:49to fill in that volum so this then
9:06:52results
9:06:54in a
9:06:56minus this C
9:06:59*
9:07:01B multiplied by this Vector B is equal
9:07:04to zero
9:07:07formula so here we are simply making use
9:07:10of the fact that the
9:07:14projection of a onto B is the shadow
9:07:17Vector which is then equal to some
9:07:20linear combination of this original
9:07:23Vector B which is on this
9:07:27line
9:07:32B then I can easily find the scaler C
9:07:36from here because we know how we can
9:07:38easily calculate this dot product so let
9:07:42us actually go ahead and do that let's
9:07:44first multiply
9:07:47this a
9:07:50by
9:07:51B and then my minus so I'm simply
9:07:55opening the parenthesis C * then I got B
9:08:00by B and this equal to
9:08:04zero so C
9:08:07* B time B is then equal
9:08:12to a and b which means that
9:08:18c is equal to a * B / to B *
9:08:35B now when we have the C we can easily
9:08:39derive the formula for the projection of
9:08:43a on the line
9:08:48B so this is the first part this is the
9:08:53second
9:08:54part so then the
9:08:57projection of a on
9:09:04B so
9:09:06projection of a on B is equal
9:09:10to this
9:09:13c
9:09:17c times the B and we just found out that
9:09:23this is equal to the C was equal
9:09:26to a * B / to B *
9:09:33B and now we need to take this C and
9:09:37then
9:09:39multiply by Vector
9:09:41B this is then the projection of a on
9:09:49B this Vector so projection
9:09:53of
9:09:55a on line
9:10:02B so you will notice that this is the
9:10:05same that we just got so whether you
9:10:09compute the projection of a on the
9:10:11entire line b or projection of a on the
9:10:13specific Vector b as we are using the
9:10:16vector b as a source for drawing our
9:10:20line this is the same as the project
9:10:23rection of vector
9:10:27a on Vector
9:10:32B and this is the same formula as we SE
9:10:35in here so this is the projection
9:10:37formula that we have just uh found out
9:10:40so projection of a onto B is given by
9:10:43this
9:10:44formula a * B so the dot product of the
9:10:48vector A and B divided to the dot
9:10:50product of the bay withd itself and
9:10:52multiply with the vector
9:10:54B and this is the in here this is
9:10:57something that we have calculated time
9:10:59and time again in our examples so if we
9:11:02go back to our
9:11:04example then here we can see that this
9:11:09is our Vector B this is our Vector a and
9:11:15we are saying if we take the vector a
9:11:18and we project it onto this Vector B
9:11:21then we can calculate this Pro
9:11:23projection which is in here the formula
9:11:26for this entire
9:11:29Vector which we are calling projection
9:11:31of a on b or projection of
9:11:35a on
9:11:38B this can be find out so the the length
9:11:41of that Vector we can find by using this
9:11:44formula so the dot product of vector A
9:11:47and B divided to the dotproduct of B
9:11:49with itself and then multiplied with
9:11:51Vector B so again a DOT product
9:11:54produ and this is of course something
9:11:57that we get as a vector so this is a
9:12:00vector something that is equal to this
9:12:03entire Vector in
9:12:06here this
9:12:08Vector so uh I know that this uh might
9:12:12look bit messy because it contains many
9:12:14moving Parts but I wanted to provide
9:12:17this detailed explanation and the step
9:12:19by-step process even if it is bit
9:12:21confusing and bit messy um in the
9:12:25beginning because this help us to
9:12:27understand what this uh formula is about
9:12:30and what is the intuition behind it
9:12:32because what we are doing is that we are
9:12:34making use of the fact that the line can
9:12:38be represented as a linear combination
9:12:41of all the
9:12:44um vectors that we use in here so this
9:12:47is Vector B and this entire line B is a
9:12:51linear combination of this vector B and
9:12:54we can make use of that in order to find
9:12:56that scalar that we are multiplying to
9:12:58create this single linear combination
9:13:01that will end up giving us this Vector
9:13:04that we see in here which is the
9:13:06projection the projection that we are
9:13:09interested which is this line This is
9:13:11the projection that we are defining by
9:13:14this projection a on to
9:13:16B and we can get that by making use of
9:13:20the fact that this this perpendicular
9:13:23line that we are creating in here which
9:13:25is simply the vector a minus this
9:13:29projection this is this Vector this
9:13:31projection Vector that this is
9:13:34perpendicular to this line
9:13:38B and if the vector B is part of this
9:13:42line B this means also that this line a
9:13:45minus projection a is also perpendicular
9:13:47to that vector vector B making use of
9:13:50that formula we can then uh make use use
9:13:53of the product of the two we know that
9:13:55the dot product of two perpendicular
9:13:57vectors is equal to zero making use of
9:13:59that we can then obtain this specific
9:14:01scaler C that we
9:14:03need in order to get the final formula
9:14:08for our projection we are interested in
9:14:11this C because knowing C we can then
9:14:14multiply with this Vector B to get our
9:14:16final projection and we have found that
9:14:19that projection a on B is is defined as
9:14:24the dotproduct of the A and B divided to
9:14:27the dotproduct of the B with the B and
9:14:28multiply with the vector B and this is
9:14:31again a
9:14:34vector now let's look into a couple of
9:14:36numeric examples to clarify this topic
9:14:39and practice with it so given vectors A
9:14:42and vectors B find the projection of a
9:14:44on to B so without looking into answer I
9:14:49will quickly go onto that example itself
9:14:52so Vector a
9:14:53is this Vector 3 4 can also represent
9:14:59this by our more common notation which
9:15:02is three and four and then Vector
9:15:07B
9:15:10is one and zero so let's quickly draw
9:15:15our coordinate
9:15:17system this our xaxis this our y axis
9:15:21and then what is the a
9:15:25the a is three and
9:15:30four three and
9:15:39four so this is our
9:15:43a and what is the B the B is one and
9:15:51zero which which means that our line
9:15:56B is
9:16:00then C times the vector B given that the
9:16:05C is a real number and one thing that
9:16:08you can notice is that the line B is
9:16:12actually our x-axis it is this line this
9:16:15is our line
9:16:20B this is our line l
9:16:24b
9:16:26so the projection is then this line this
9:16:30is our projection because we can know
9:16:32that by drawing a perpendicular
9:16:35line in here from a to the line B we can
9:16:41get then the connection between
9:16:44our Vector a and Vector B and create our
9:16:48projection so this is then the A minus
9:16:54projection of a on line
9:16:59B and this part is
9:17:04then this is then this
9:17:08projection
9:17:10a on
9:17:13B and how we can get this projection
9:17:17well we just learned that the projection
9:17:24of
9:17:28a on
9:17:31B is equal
9:17:34to dotproduct
9:17:36of
9:17:38a with B divide it to dotproduct of B
9:17:43with B
9:17:44itself and multiply it by
9:17:48B this is the formula that we can use
9:17:50and even if you don't remember the
9:17:52formula by heart you can make use of
9:17:54this visualization to figure out what
9:17:55that formula is because we know that if
9:17:58this line is perpendicular to this one
9:18:00then a minus projection of a on B
9:18:03multiplied by this projection a on B
9:18:06should be equal to zero and this
9:18:09projection of a on B is equal to some
9:18:12scalar C multiplied by Vector B that's
9:18:16something that we see in
9:18:20here the first thing we need to do to
9:18:23compute the dot product between a and
9:18:26b a * B is equal
9:18:33to
9:18:3634 multiplied by 1 0 this is the dot
9:18:43product which is then equal to 3 * 1 + 0
9:18:47* 4 and this is equal to
9:18:503 the next thing we need to to do is to
9:18:53compute the dotproduct between B itself
9:18:56so B * B and what's that that is 1 0
9:19:02with 1 Z multiplied this is equal to 1 1
9:19:07* 1 + 0 plus 0 * 0 is equal to
9:19:111 then the third thing that we can do
9:19:15then is to obtain the final value which
9:19:20is
9:19:25projection of
9:19:28a on
9:19:32B is then equal
9:19:36to three / 2 1 multiplied by the vector
9:19:43B which is 1 0 which is equal to 3
9:19:500 and this actually makes sense visually
9:19:53too as you can see in here this is the
9:19:57tree for the xaxis and here we have the
9:20:00center Z so this projection is then the
9:20:03vector 3 0 so even without calculation
9:20:06we could see just from plotting the uh
9:20:09on the coordinate system the vectors A
9:20:11and B that the projection of a on B will
9:20:13be this Vector 3 but we have followed
9:20:15the formula in order to do calculation
9:20:17step by
9:20:18step which is something that you can see
9:20:21in this answer too
9:20:23so the projection of this Vector a onto
9:20:25B is then this Vector of a length tree
9:20:28in the direction of B so you can see
9:20:31that it is of the length of
9:20:36three so this is the tree on the
9:20:39direction of B so on the line
9:20:43B let's now move ahead and look into a
9:20:46different example but this time we will
9:20:47do the calculation in a quicker way so
9:20:50we got two vectors 4 three and B is
9:20:52equal to
9:20:5320 and we need to find this projection
9:20:56of a on to B so the first thing we need
9:21:00to do is to calculate the a * B which is
9:21:05equal to 43 multili 2 0 and that's equal
9:21:12to 4 * 2 + 3 * 0 and it's equal to 8 the
9:21:16second thing we need to calculate is the
9:21:18B do product with B which is equal to 2
9:21:220
9:21:232 0 this is then equal to four and the
9:21:27final part is to take and uh from this
9:21:31one and two this values and then bring
9:21:33them all together so then
9:21:36the
9:21:38projection
9:21:40of a on B is equal
9:21:51to H
9:21:53ided to
9:21:544 multiplied by the vector
9:21:58to0 and this is equal 2 8 / 2 4 is 2 2 *
9:22:042 is 2 2 * 0 is 0 so we are getting this
9:22:10two Vector so projection of a on B is
9:22:14then this Vector 20 which is actually on
9:22:16this xaxis similar to what we had before
9:22:19only with the length of t uh towards the
9:22:21direction of
9:22:23which is then equal to 4
9:22:26and0 and this is again similar to what
9:22:28we had before uh where we got the
9:22:31projection of a on B on that end up on
9:22:34the x axis but now with the length of
9:22:36four so now our
9:22:42projection has the following Vector so
9:22:45the uh following magnitude and
9:22:48Direction so this is the step by-step
9:22:50process that I just followed if if you
9:22:52want to do it bit slowly and this is the
9:22:55final
9:22:56result so uh the interpretation of this
9:23:00projection is that this projection a
9:23:02onto B is simply this 4 zero this means
9:23:06that the A's component in the direction
9:23:09of B it spends uh four units along this
9:23:12x-axis that we saw in
9:23:18here because this is the value X this is
9:23:21the value of y
9:23:25so this projection shows us that A's
9:23:28influence in the direction of B is
9:23:30completely horizontal with this
9:23:32magnitude of four because we saw that we
9:23:35end up with the projection on the x-axis
9:23:39again so this was four this was our
9:23:42projection vector and if you plot this
9:23:45entire Vector a and Vector B on this
9:23:49x-axis and y axis then you can clearly
9:23:51see that the uh horizontal line that we
9:23:55end up with the uh
9:23:58projection of a n b is very similar to
9:24:03what we had
9:24:05before in
9:24:08here let's now talk about a concept of
9:24:10auton normal bases so let's now Define
9:24:14what the auton normal bases are so by
9:24:16definition auton normal basis for a
9:24:19vector space is a basis where all vector
9:24:22vors are orthogonal or perpendicular to
9:24:25each other and each Vector is of unit
9:24:28length so as you can notice here here we
9:24:32have a special type of basis it's called
9:24:35auton normal basis because in the
9:24:38beginning of this section of this module
9:24:40we defined formally this concept of
9:24:42bases we talked about the concept of
9:24:45colal uh space and then the uh basis of
9:24:49a comp space the no space the basis of a
9:24:52n space and then we talked about the
9:24:55concept of the bases of the entire space
9:24:59for instance the R2 and now we are
9:25:02defining a special type of bases which
9:25:05we are referring as auton normal bases
9:25:08and this auton normal basis as you can
9:25:10see from this definition it contains two
9:25:12criteria for it to be auton normal so an
9:25:16auton basis for a vectory space is a
9:25:18basis where a all vector are orthogonal
9:25:23or perpendicular to each other and B
9:25:27each Vector is of unit length we already
9:25:30have learned that when we have vectors
9:25:34let's say Vector A and B
9:25:36perpendicular it means that A and B
9:25:40their dot product is equal to zero
9:25:42that's the first criteria that we need
9:25:44for calling our basis an auton normal
9:25:48basis then the second criteria is that
9:25:53each of these vectors they need to have
9:25:56a
9:25:57length of
9:26:03one if we have this condition satisfied
9:26:06then we are saying that our vectors they
9:26:09help us to form this auton normal basis
9:26:13if we got three vectors forming this
9:26:15Vector space it means that we need to
9:26:18have the a * B = to
9:26:220 a * C = 0 and then B * C = 0 this is
9:26:30if we are in in case we are using three
9:26:34different vectors that Define our Vector
9:26:39space in this
9:26:41case let me make this part smaller so
9:26:46let's put the length of B in here in
9:26:51this case
9:26:53the second
9:26:54criteria becomes that the length of a is
9:26:59equal to the length of B and then is
9:27:02equal to the length of c and is equal to
9:27:06one so depending on the number of
9:27:09vectors that you use to form your vector
9:27:11space the proof that you are dealing
9:27:14with auton normal bases will be
9:27:15different here we got just two vectors
9:27:18here we got three vectors but in both
9:27:20case we first need to prove that we are
9:27:22dealing with uh vectors Each of which
9:27:25are set of orthogonal perpendicular
9:27:28vectors and all of them pairwise they
9:27:32need to be perpendicular and at the same
9:27:34time the second criteria says that they
9:27:36all need to have a unit length so their
9:27:40length should be equal to
9:27:42one we need this auton normal bases in
9:27:45order to simplify our calculations
9:27:47including the calculations of
9:27:49projections and Transformations that we
9:27:51just so before when we were discussing
9:27:54this concept of projecting a vector onto
9:27:57a line or projecting a vector onto not a
9:27:59vector because we were in this basic
9:28:01case when we had just two vectors in
9:28:05R2 and calculating projection in R2 is
9:28:09very easy because we can make use of
9:28:10this formula um a and then B uh the dot
9:28:15product of them and then divided two
9:28:17product of the B and then times the B
9:28:19this was quite straightforward right but
9:28:22when it
9:28:23came so this is the projection of a on B
9:28:30but when it comes to projection in
9:28:32higher dimensional space let's say you
9:28:34have R5 or you have R 100 or R th000
9:28:37then it becomes much more difficult to
9:28:40do those projections and to calculate
9:28:42the projections and for those cases we
9:28:45can make use of this concept of auton
9:28:47normal basis to simplify our
9:28:49calculations and we will see that in a
9:28:51bit
9:28:53so let's first understand this
9:28:55orthogonality and the normalization part
9:28:57so orthogonality refers then to the part
9:28:59of uh when we are saying that the
9:29:01vectors should be orthogonal to each
9:29:03other and the normalization refers to
9:29:06the fact uh to the fact that the length
9:29:09should be one this is basically the set
9:29:11of two criteria that I just discussed
9:29:14this is uh the summary slide that will
9:29:17give you an indication what is meant by
9:29:19that so if we have two vectors V and W
9:29:23then we say that the first criteria is
9:29:25that those two vectors are orthogonal
9:29:27which means their dot product is equal
9:29:29to zero and we are saying that their
9:29:31length is equal to one which we are
9:29:33referring as a normalized vector so if
9:29:36the
9:29:37vector has a length of one then we are
9:29:41calling a vector
9:29:42v
9:29:49normalized so if both of this criteria
9:29:52of normalization and
9:29:53orthogonality is satisfied that we are
9:29:56saying that we are dealing with an uton
9:29:58normal basis so now where we have
9:30:00learned this idea of projections also
9:30:02this idea of autog colonization and the
9:30:06uh concept of auton normal basis we are
9:30:08ready to discuss the concept of the
9:30:10grade process so the grade process is
9:30:14this method for orthogonalizing a set of
9:30:16vectors in an inner product space and
9:30:19turning them into an auton normal
9:30:24set so let's say we have a set of
9:30:27vectors we want to uh bring and
9:30:30transform all these vectors onto this
9:30:33auton normal set of vectors which means
9:30:36that we want them to be aized so we want
9:30:40them to be perpendicular and we want
9:30:43them to be normalized because we know
9:30:45that the two criteria were specified
9:30:47right so the first criteria was that we
9:30:49need to have vectors
9:30:55ortogonal hence we are doing
9:31:02orthogonalization and the second
9:31:04criteria was that they need to be
9:31:07normalized because we want the
9:31:12vectors to
9:31:15have
9:31:17length one so we are doing normalization
9:31:24this process of turning this set of
9:31:29vectors onto this uton normal
9:31:32set by using this method of
9:31:35orthogonalization which is something
9:31:37that we are referring as a grme
9:31:39process this is something that we can
9:31:42use in order to simplify later this
9:31:45different sorts of Transformations which
9:31:47we need in order to perform bit more
9:31:49advanced uh Transformations like Matrix
9:31:52uh factorization different decomposition
9:31:55techniques so given this set of linearly
9:31:58independent vectors this process which
9:32:01we are referring as grme process
9:32:03produces this auton normal set that is
9:32:06spinning the same
9:32:08Subspace so we have the same
9:32:11Subspace it's just that we are turning
9:32:13the set of vectors into an auton normal
9:32:16set of vectors that is spanning the same
9:32:20Subspace so the gr Street process step
9:32:23by step looks like something like this
9:32:26so given the vectors A1 A2 up to a n the
9:32:30first thing we need to do is to start
9:32:33with the vector V1 which is equal to our
9:32:36first Vector A1 and first we need to
9:32:39normalize this vector and how we can
9:32:42normalize this Vector well we need to
9:32:45take this vector and we need to divide
9:32:47it to its length so the grme process
9:32:50step by step will look like like
9:32:52something like this so in the first step
9:32:54what we need to do when starting with
9:32:56these vectors of A1 A2 up to a n so in
9:32:59RN we need to First Take the first
9:33:02vector and we need to normalize it and
9:33:04how we can normalize the vector and
9:33:06ensure that its length is equal to this
9:33:09length of P1 well we need to take that
9:33:13vector and we need to divide it to this
9:33:15length because
9:33:18when we take the
9:33:20vector the 1 and we divide it to its
9:33:24length of V1 then we will ensure that
9:33:27the length of that Vector is equal to
9:33:32one we can actually prove that very
9:33:34easily but I won't do it in here uh feel
9:33:37free to go through the process assuming
9:33:40that
9:33:41the
9:33:43length of the vector what what you want
9:33:46to achieve at the end is that the length
9:33:48of a vector v is equal to one
9:33:51this is something that we want to
9:33:53achieve and this normalization process
9:33:56can be
9:33:57done if we find a way to ensure that we
9:34:02uh get this E1 because E1 means that we
9:34:06end up with this Vector y 000000 0 this
9:34:10will be for first Vector so
9:34:13V1 this is E1 so the one is really
9:34:17important here so we want to normalize
9:34:20this Vector View 1 by uh ensuring that
9:34:25we get the E1 so we go from V1 to E1 and
9:34:30the way we do that is that we take the
9:34:32V1 and we divide it to the length of
9:34:35V1 and in this way we get the E1 so the
9:34:40normalized version of P1 is
9:34:45E1 so then for each subsequent Vector a
9:34:49k which means A2 A3 A4 up to a n we need
9:34:54to subtract its projection on all the
9:34:57previously computed orthogonal
9:35:04vectors in this way by using this tab
9:35:07two we are ensuring that all these
9:35:11different each pair wise set of A1 A2
9:35:15and then A2 A3 Etc they are all
9:35:17orthogonal to each
9:35:19other and we know that this projection
9:35:24is something that we got when we had
9:35:26this two perpendicular lines so we had
9:35:29this Vector we're projecting onto this
9:35:31Vector we got that by finding this
9:35:36perpendicular line and making use of
9:35:38that using this property we are then
9:35:41making use of that in order to see how
9:35:43we can ensure that the subsequent Vector
9:35:46that we have is always perpendicular to
9:35:49this
9:35:50one so let me actually write down what
9:35:54is in this
9:35:55formula so here VK is equal to a minus
9:36:01the sum of all the projections so then
9:36:04we need to normalize the VK to get the
9:36:07EK and then we need to repeat this step
9:36:10two and three for all vectors which
9:36:13means that first here we apply this
9:36:16normalization on the vector A1 so V1 is
9:36:21to A1 and then we get the normalization
9:36:25by getting this E1 so E1 is normalized
9:36:29version and then we need to apply a bit
9:36:33different tactique for our V2 V3 up to
9:36:38VN and then let me actually write down
9:36:41this for this General
9:36:44case so what this processed this the GR
9:36:52let me ensure that I'm not making a typo
9:36:58Schmid
9:37:01process step by step means step number
9:37:08one for
9:37:12vectors A1
9:37:15A2 A3 dot dot dot a n so we are in the
9:37:19RN
9:37:25then step number one basically says take
9:37:30the
9:37:32V1 and set it equal to this first
9:37:35element
9:37:38V1 this is
9:37:46A1 then what we need to do is to
9:37:51normalize it to get the
9:37:55E1 so normalize
9:37:59normalized
9:38:01V1 to get E1 which is equal to 1 0 0 0
9:38:09and then dot dot dot zero and the size
9:38:11of this n by one and how we can do
9:38:16that by taking this Vector V1 and
9:38:21divided it to the length of V1 which
9:38:25basically means in this specific case A1
9:38:29divided to the length of A1 this will
9:38:33then give us our
9:38:35A1 this Vector this is basically what
9:38:39the step one entails then in the step
9:38:42number
9:38:44two we have for each
9:38:48subsequent a where K is just an index
9:38:52referring to whether we are dealing with
9:38:55K is equal to 2 so uh A2 A3 and then dot
9:39:00dot dot a
9:39:02n this is what basically the K is used
9:39:05for to refer to which Vector we are
9:39:07dealing
9:39:09with we need to subtract its projection
9:39:13on all previously computed orthogonal
9:39:18vectors by using this formula so let's
9:39:22actually do a couple of those case to
9:39:24see what is going on for instance for K
9:39:27is equal to
9:39:322 so K is equal to 2 and here is the
9:39:37formula by the way
9:39:40so
9:39:42VK VK is equal to
9:39:47AK
9:39:50minus some
9:39:51K is equal to K starts with
9:39:54one and
9:39:56then let me use a different
9:40:00index so I is = to 1 till K minus
9:40:08one and then
9:40:11projection
9:40:13of a
9:40:15k a k
9:40:24on E1 or
9:40:29eii so the EI that we have just computed
9:40:32because every time you are then
9:40:34normalizing and normalizing every time
9:40:37your vectors and then you are uh finding
9:40:40out what is the projection of your
9:40:42vector a onto that
9:40:49EI and then you are substract in that
9:40:51from your vector so what this means in
9:40:54Practical terms when for instance your K
9:40:56is equal to 2 it means that V 2 is equal
9:41:04to a 2 minus sum of I is = 1 and then K
9:41:12is = 2 K - 1 this means this is
9:41:16one
9:41:18projection of a a and then
9:41:242 on
9:41:27A1 given that this is one this is simply
9:41:30equal to
9:41:32A2
9:41:33minus
9:41:36projection of
9:41:38A1 that's normalized version of E1 and
9:41:43then A2 so projection of A2 on E1
9:41:53and then in the step number three we
9:41:56need to do we need to go from VK to get
9:41:58EK so basically we are ensuring with the
9:42:01step number two the orthogonally uh
9:42:03orthogonality condition and with step
9:42:06number
9:42:07three I
9:42:10me add some space in here so in the step
9:42:14number three step number three we then
9:42:16saying let's
9:42:19normalize normalized
9:42:22this VK that we have just
9:42:26computed in
9:42:29here because we remember that the second
9:42:32criteria after tonality is normalization
9:42:34that the unit or the length of the
9:42:37vector should be equal to
9:42:39one so then VK in this case for K is
9:42:44equal to 2 for K is equal to 2 means
9:42:47that we need to go from V2 to E2
9:42:52and the way we can do that is by taking
9:42:55the
9:42:57V2 by V2 and then divide it to the
9:43:02length of
9:43:04V2 this will then give us the E2 this is
9:43:08the normalization
9:43:10part and the step number four basically
9:43:14means
9:43:16repeat
9:43:18repat step two
9:43:22entry for all
9:43:25case which means that if we go back so
9:43:29we are done with V2 so we have obtained
9:43:32V2 and then we have obtained normaliz
9:43:34normalized version of V2 by getting this
9:43:38E2 we are ready to come back and do the
9:43:41same for
9:43:44K is equal to three and for K is equal
9:43:47to three in Step number two we got
9:43:51V3 is equal to
9:43:55A3 minus making use of this
9:43:59formula sum overall I is = to 1 K - 1 is
9:44:04= to 2 and then
9:44:07projection of this time a Tre see three
9:44:12k is equal to three and then on E2
9:44:21actually it says EI let me remove this
9:44:25this otherwise we would have made a
9:44:30mistake this should be I because an I
9:44:34will change per K this is the entire
9:44:37idea we need to re um subtract all the
9:44:40um
9:44:41projections what this basically means is
9:44:44that we need to take A3 and this time
9:44:47given that here we have two instead of
9:44:49one in here we need you have an extra
9:44:52step which means A3
9:44:56minus and then what this formula
9:44:58basically says this is the sum of the
9:45:01projections of A3 on e i where I goes
9:45:04from one till two so
9:45:07projection of a Tre on
9:45:14a one when K so when I this is the I is
9:45:19equal to one case plus
9:45:23projection of a Tre on a 2 this is the I
9:45:30equal to 2 case this is
9:45:32basically what this
9:45:35summation says this is this element and
9:45:38we have seen this as part of the high
9:45:40school but also the pre-algebra
9:45:43course okay so now when we are clear on
9:45:49how we can calculate the V3 in the step
9:45:52number two for K is equal to 3 we are
9:45:55ready to go onto the step number three
9:45:57and what was step number
9:46:00three the step number three
9:46:05for K is equal to 3 was saying let's
9:46:09take the V3 and normalize it to go from
9:46:13V3 to E3 and how we can do that by
9:46:17taking the V3 and dividing it to the
9:46:21length of V
9:46:23tree to get on to E
9:46:28tree and this cycle goes on and on until
9:46:32we cover all the case so all the vectors
9:46:37so the idea is that we first for our
9:46:41initial step we set the V1 equal to
9:46:44A1 we normalize it then starting from
9:46:49the K is equal to two we don't go first
9:46:53on and on WE autog it by formula in here
9:46:58by using this we can ensure that each of
9:47:02these vectors is then orthogonal to all
9:47:06the other vectors so for K is equal to 2
9:47:08we ensure that this uh Vector that we
9:47:12get is orthogonal to all the other ones
9:47:16and the case equal to treat that the
9:47:18third Vector isogonal to all the other
9:47:20ones and we are doing that in Step
9:47:21number
9:47:23two so for each case for each K we
9:47:26basically are ensuring that in this case
9:47:30we have an a vector that is orthogonal
9:47:33to all the other vectors in this
9:47:35set and for each Vector we are also
9:47:38normalizing it to satisfy the second
9:47:40criteria because we had these two
9:47:42criterias to create this auton normal
9:47:45set so we are doing this in subsequent
9:47:49uh way so first for K isal to 1 so
9:47:53basically for A1 and then we are doing
9:47:55this for K is equal to 2 so A2 and then
9:48:01until K is equal to n so a n what we are
9:48:05doing every time is that we are
9:48:06obtaining this V1 and then we go from V1
9:48:10to E1 to normalize it and then here we
9:48:13are getting the V2 here to go to E2 by
9:48:18normalizing it so this basically
9:48:21the step two and step three and then we
9:48:24do this every time up until to the point
9:48:28of obtaining VN and then from VN we go
9:48:32to en n to normalize
9:48:35it so this is the idea of this entire
9:48:40process step by step to start with V1 as
9:48:43part of the step number one and then as
9:48:45part of Step number two for each
9:48:47subsequent Vector a k so K is equal to
9:48:50two obtain the VK and then normalize it
9:48:54for K is equal to 3 obtain the V3 and
9:48:56then normalize it to get E3 up to the
9:48:58point of the last Vector which is a n
9:49:01the vector a n we compute the
9:49:05VN and then we normalize it to get
9:49:09the and this is what this part is which
9:49:13is the St number two that says repeat
9:49:15steps 2 and three for all vectors it
9:49:17means that every time when you increase
9:49:19your K when you go to the next Vector we
9:49:22first compute the V so VK and then you
9:49:25normalize it you get the EK and then you
9:49:27go back to the step number two at three
9:49:29because you then again need to calculate
9:49:31the VK and then EK and then for the next
9:49:34case so this is something that you will
9:49:37see also a lot when you are writing the
9:49:40code for
9:49:42your uh algorithms because in many cases
9:49:45you need to do this reputation of the
9:49:48steps so uh you for one Vector you do
9:49:51something or for one iteration you do
9:49:53process and then you uh go back and do
9:49:56for the next one and for the next one
9:49:58this process is what we are referring by
9:50:00repeat step number two and three for all
9:50:06vectors so let's now look into an
9:50:08example let's apply this grumme process
9:50:11to vectors A1 and A2 where A1 is 1 1 0
9:50:16and A2 is 101
9:50:22so let's go ahead and do that so A1 is
9:50:25equal
9:50:28to 1 1
9:50:320
9:50:34A2 is equal
9:50:37to 1 0
9:50:401 we want to apply this gret process to
9:50:43create this Aon normal basis for the
9:50:46Subspace that is Pinn by A1 and A2 so
9:50:49now we have this
9:50:51set 1 1
9:50:540 and one
9:50:561 and what we want is to create an auton
9:51:00noral
9:51:02basis so
9:51:05creating
9:51:06creating or to normal normal
9:51:13basis
9:51:16which
9:51:18CR meet
9:51:27process so here we got only two vectors
9:51:31so obviously it's this and it's a very
9:51:33simplified version of it what was the
9:51:36first step in our case uh in our
9:51:38algorithm it was to set the V1 equal to
9:51:44A1 what we need to do step number one we
9:51:49need to set the V1 equal to
9:51:54A1 and we need to
9:51:56normalize
9:51:59normalize the
9:52:02V1 to get E1 that's what our goal is so
9:52:07let's go ahead and do
9:52:09that V1 is equal to A1 and is equal
9:52:15to 1 1 0er that's our A1 so 1 1
9:52:22Z and in order to normalize
9:52:28V1 and get the
9:52:30E1 we know that this is equal to V1 / to
9:52:34the length of
9:52:37V1 which is then equal to take the V1 so
9:52:42that is 1 1 0 and then divide it to the
9:52:47length of V1
9:52:51and you can very quickly see
9:52:55that
9:52:57given
9:53:00V1 is equal to V1 * V1 that's something
9:53:04that we learned in the very beginning of
9:53:06our fundamentals to linear algebra
9:53:07course that the length of V1 is simply
9:53:10the dotproduct between uh V1 and V1 and
9:53:13it's equal
9:53:14to 1 1 0 * 1 1 0
9:53:21which is equal to 1 + 1 so two so this
9:53:27is then equal to 1 1 0 / 2 which is
9:53:32equal to 1 / 2 1 / to 2 and then
9:53:360 this is our E1 so we are done with our
9:53:40step number one because now we have
9:53:42V1 and we got
9:53:46E1 so what was the step number two in
9:53:49the step number
9:53:51two we need to set the k equal to
9:53:562 this is the next K so for
9:54:03A2 what we need to do is we want to
9:54:07get
9:54:09V2
9:54:11and
9:54:14normalize V2 by getting E2 and how can
9:54:19we get that
9:54:21well first let's find what is the V2
9:54:25well V2 was and using that formula that
9:54:27we saw before which was this formula so
9:54:31it's equal to a k minus and the sum I is
9:54:35equal to 1 K minus one and then
9:54:37projection of a onto EI
9:54:40I so let's take this formula
9:54:43over this is equal to a 2 because K is
9:54:49equal to 2 a a k minus Su and then I is
9:54:53equal to one till K -1 and then K -1
9:54:58which is equal to basically 1 given that
9:55:02K is equal to
9:55:032 and then projection of A2 onto e
9:55:13i and this is equal to
9:55:16A2 minus given that we got k - Y is
9:55:20equal to 1 so the limit for our
9:55:22summation is equal to
9:55:241 so this one this means that like
9:55:29before we got just one part as part of
9:55:31our summation so minus and then
9:55:37projection of let me actually keep the
9:55:40same color I want it to be consistent so
9:55:44projection of
9:55:47A2 on d e
9:55:521 so you see here the i i is equal to 1
9:55:57and the limit of the I is K minus one
9:56:00which is equal to 1 so we got here just
9:56:06E1 so we got the V2 formula we can then
9:56:10now calculate it because we know
9:56:14A2 and the A2 is this so one one
9:56:211 0 1 but now we got a problem we don't
9:56:26know what this is so let's
9:56:29quickly go and calculate this
9:56:34part so
9:56:36projection of
9:56:39A2 on
9:56:42a
9:56:431 and we learned from the projection
9:56:47formula that this is equal to H 2 * E1 /
9:56:532 E1 * E1 so the dot product multip by
9:56:58E1 and what is this this is equal
9:57:01to 1 1 multiplied by and what is the E1
9:57:07E1 we just calculate in here so it is 1
9:57:11/ to 2 1 / to 2 and then zero here / two
9:57:16and then 1 / 2 1 / 2 and then 0 0
9:57:21multiplied by 1 / 2 1 / to 2 and then Z
9:57:25here multiplied by the same Vector so
9:57:33E1 so this two cancel out this two also
9:57:37cancel out and as you can see we are
9:57:40getting that
9:57:42the projection of A2 on E1 is equal to
9:57:46this vector
9:57:52we can also manually check that actually
9:57:55so let's let's do that so let's see we
9:57:57are not canceling out these
9:58:01vectors and instead we are manually
9:58:04calculating
9:58:10this so here we got 101 ultip by 0.5 and
9:58:170.50 this is equal to 1 *
9:58:201 / 2 is 1 / 2 0 * 1 / 2 is 0 1 * 0 is 1
9:58:27so + 1 /
9:58:312 this amount is
9:58:351/4 +
9:58:371/4 this multiplied by the vector 1 / 2
9:58:421 / 2 and then 0 in here this is equal
9:58:46to 1 / 2 1 + 1 / 2 is = to 3 / to 2 and
9:58:54then 1 1/4 + 1/4 is equal to 1 / 2 2
9:59:01multiplied by 1 2 and then 1 2 and then
9:59:07zero what is this amount well those two
9:59:10cancel out so we end up with three * and
9:59:15then 1 / 2 2 and then
9:59:201 / 2 and then
9:59:22zero this is then the projection 3 / 2 3
9:59:26/ to two and then
9:59:32zero so let me remove all these
9:59:35calculations
9:59:53and then we can take over the projection
9:59:56value which
9:59:57is 3 / to 2 3 / to 2 and then zero to
10:00:04get our Vector
10:00:07V2 which is equal to 1 - 3 / to 2 0 - 3
10:00:15/ to 2 and then 1 - 0 and this is equal
10:00:20to here it is 1 here it is - 3 / 2 and
10:00:24here is minus and then 1 / 2
10:00:292 because 3 / 2 is minus uh it is 1.5
10:00:34and then 1 - 1.5 is simply minus
10:00:370.5 so this is then the vector
10:00:40V2 then what we need to do is to
10:00:43normalize this Vector to get D
10:00:46E2 which is then equal to V2 ided to V2
10:00:53length which is simply equal to
10:00:57V2 / to V2 * V2 so the dot product and
10:01:03this is equal to let's take the V2 which
10:01:06is -1 / 2 and then - 3 / 2 and then 1
10:01:11and then divided 2 and this amount let's
10:01:14quickly calculate that it is equal to so
10:01:19the length of V2 is equal
10:01:23to - 1 / 2^
10:01:272 + - 3 / 2^ 2 + 1 this is equal to 1/4
10:01:34+ 9/ to 4 + 1 which is 4 / to 4 and then
10:01:41this is equal to 1 + 9 is 10 10 + 4 is
10:01:4514 so 14 / 2 4 this is the length of it
10:01:57so 14 / 2
10:02:044 so then this is equal to this Vector
10:02:08to this threedimensional
10:02:11Vector min-1 * 14 - 1 / 2 * 14 / 4 is
10:02:18equal to this is 7 so
10:02:23minus 7 / 2
10:02:294 and then - 3 /
10:02:342 think I just made a mistake here
10:02:38actually so Min - 1 / to 2 so the first
10:02:43element and then ided 24 / 4 is actually
10:02:47actually equal to this multipli by four
10:02:50ided to 14 so you take this element then
10:02:55divide it to this one and we know that a
10:02:58/ to B * C / 2 D is equal to a *
10:03:05D and then B * C so we are basically
10:03:10flipping this
10:03:12side this is
10:03:14from
10:03:15pre-algebra and then here
10:03:19this is equal
10:03:25to 2 and then is equal to -1 /
10:03:3827 then let's do the second one
10:03:41twoo so we got minus 3 / to 2 / to 14 /
10:03:47to 4 is actually equal to - 3 / to 2 * 4
10:03:51/ to
10:03:5314 and then if we remove this this is
10:03:58then 2 this is 7 this cancel out this
10:04:02equal to - 3 / to 7 - 3 / to 7 and then
10:04:08finally we got 1 /
10:04:122
10:04:1514 / 2 4 which is equal to 4 / 2 14 this
10:04:19equal to 2 / to 7 so 2 / to 7 and this
10:04:23is our A2 and given that we got just two
10:04:27vectors so we have already reached the
10:04:29end of our solution so now when we have
10:04:34already the V1 and the V2 the E1 and the
10:04:37E2 We have basically completed the
10:04:40process of this grummet uh procedure
10:04:43because we have already uh only two
10:04:47vectors that means that we need to have
10:04:49V1 and V2 and then uh E1 and E2 and this
10:04:53is all that you need in case you got two
10:04:55vectors if you have three vectors of
10:04:57course the process will include um the
10:05:00same process of Step number two and
10:05:02three so the V2 and the normalization of
10:05:05it two times for your K is equal to 2
10:05:08and K is equal to 3 and then if you have
10:05:12more vectors than every time you will
10:05:14have more of the steps but at the end
10:05:17what we want to have is the set of
10:05:19vectors that are orthogonal and at the
10:05:21same time they are normalized in this
10:05:23case we say that this vectors form this
10:05:26oron normal bases now why is this
10:05:29important the applications of
10:05:31orthonormal bases well firstly it
10:05:34simplifies a complex Vector
10:05:37operations and uh this is the basis of
10:05:40many uh more difficult mathematical
10:05:43Concepts uh like foror series or quantum
10:05:46mechanics it's used also um when when it
10:05:50comes to this auton normal basis uh also
10:05:53signal processing and it's a critical uh
10:05:56process in numerical methods especially
10:05:59in machine learning algorithms and in
10:06:01data compression so we will see this
10:06:03process to be used also as part of uh
10:06:05decomposition techniques which is really
10:06:08important when it comes to different
10:06:10algorithms uh whether it's optimization
10:06:12algorithms but also um algorithms that
10:06:16are used for recommender systems for
10:06:18example and those uh Concepts they all
10:06:22come together and we will see later on
10:06:24when we will be discussing the concepts
10:06:26of the compositions and metrics
10:06:29factorization so this Aon normal basis
10:06:31and this grum process they are really
10:06:33foundationally in linear algebra they
10:06:36provide tools for simplifying and also
10:06:39solving these higher dimensional
10:06:40problems efficiently their application
10:06:43include different fields of science
10:06:45engineering demonstrating their
10:06:47versatility and utility
10:06:52let's now talk about the special
10:06:54matrices and their properties so we are
10:06:56going to talk about special matrices
10:06:59like symmetric matrices and their
10:07:01example diagonal matrices and their
10:07:03corresponding example but also the
10:07:05ortogonal matrices with the
10:07:06corresponding
10:07:08example so when it comes to the special
10:07:11matrices special matrices have unique
10:07:14properties such as being symmetric or
10:07:16all nonzero elements on the diagonal
10:07:19like diagonal matrices or
10:07:21orthogonality uh in matrices which means
10:07:24that we have orthogonal
10:07:26matrices so when it comes to the
10:07:28symmetric Matrix it means that uh the a
10:07:33The Matrix a is equal to its transpose
10:07:36the a so a is equal to a in this case we
10:07:40can confirm and say that the Matrix a is
10:07:43symmetric so in this case we have Matrix
10:07:46a and we know that the way we need to to
10:07:50transpose this Matrix is to taking this
10:07:53rows and making them The Columns of our
10:07:56transpose Matrix so a is then equal
10:08:01to 2 - 1 and then zero then the second
10:08:05row which is min -1 and then 2 and then
10:08:09Min -1 and then the third row which is 0
10:08:13- one and two so the third row then
10:08:15becomes my third column so as you can
10:08:17see those two are the same so I'm using
10:08:20then the definition of the transpose of
10:08:23the um Matrix and then here then we end
10:08:28up with two matrices they are actually
10:08:31the same so we can see that the A and
10:08:34the a in both the First Column they got
10:08:372 minus one and then zero the second
10:08:40column minus one 2 and minus one the
10:08:42third column 0 - 1 and two so their
10:08:45columns and their rows they are the same
10:08:48which means that we are dealing with a
10:08:50symmetric Matrix so whenever we want to
10:08:52check whether the Matrix is symmetric we
10:08:54just need to take the transpose of it
10:08:56and see where the the Matrix is equal to
10:08:59its transpose in that case we are
10:09:00dealing with symmetric
10:09:02Matrix do also note therefore for uh
10:09:05Matrix to be symmetric it needs to be a
10:09:08square Matrix so it needs to be 2x two
10:09:11into two dimensional space or 3 by3 in
10:09:14the three dimensional space or n by N in
10:09:16N dimensional space which means that the
10:09:19number of rows should be equal to number
10:09:23of
10:09:27columns because otherwise when you flip
10:09:30your number of rows with number of
10:09:32columns on in case there is no um uh
10:09:37Square version of that Matrix so m is
10:09:40not equal to n in that case a will have
10:09:45a dimension of M by n and then then a t
10:09:51so a
10:09:56t will have a dimension of n
10:10:00by m which means that there is no way
10:10:03that a can be equal to a this is not
10:10:07then possible therefore we need to have
10:10:10a square Matrix for them to be
10:10:15symmetric let's now talk about diagonal
10:10:17matrix so a diagonal Matrix has a
10:10:20nonzero element only on its diagonal
10:10:24which means that in this case we have
10:10:27this nonzero elements on the diagonal so
10:10:30let's call it
10:10:31d11 d22 and then d33 this equal to three
10:10:36this equal to 5 this equal to 7 and all
10:10:39the other elements as you can see in
10:10:41here they are
10:10:44zeros so the concept of diagonal
10:10:47matrices is very uh simple therefore we
10:10:50will then go through the next example
10:10:52which is about orthogonal Matrix now
10:10:55this is a concept that we haven't yet
10:10:57seen and we spoken about so let's Cod
10:11:00read through this bit slowly so an
10:11:02orthogonal Matrix is a square Matrix
10:11:05whose columns and rows are orthogonal
10:11:09unit vectors so oron normal vectors and
10:11:13its transpose equals its
10:11:17inverse so there are two two part of
10:11:19this elements so firstly it
10:11:22says that for the Matrix to be octogonal
10:11:24Matrix it should be a square Matrix so
10:11:29Square
10:11:33Matrix and then its
10:11:36columns and rows are orthogonal unit
10:11:41vectors so
10:11:44columns and
10:11:46rows
10:11:48are
10:11:52orthogonal unit
10:11:55vectors which means they need to be
10:11:57normalized so
10:12:01normalized so um we have seen when
10:12:05forming this orthonormal basis that we
10:12:07had this process of uh this condition of
10:12:11orthogonality the vectors had to be AAL
10:12:14and they had to have a length of one
10:12:16which means that they they had to be
10:12:18normal I we can see exactly the same in
10:12:21here so hence the name oron normal
10:12:25vectors so they are utal and they are
10:12:28unit vectors which means they are
10:12:31normalized so then the final condition
10:12:35is added in here which actually is not
10:12:38so much a condition but rather than a
10:12:40property something that we can prove
10:12:42that once we have all this we can also
10:12:46say that if we are dealing with toal
10:12:49Matrix then
10:12:51Q T * Q so the dotproduct of the
10:12:58transpose with that Matrix Q is equal to
10:13:04the Q * QT is equal to I why because the
10:13:11QT is equal to the Q minus one because
10:13:16the transpose of that Matrix Q is
10:13:19actually equal to its
10:13:21inverse and given that we learned that
10:13:23the Q minus one so the inverse time Q is
10:13:27= to Q inverse * Q is equal to
10:13:31I and given that here we are learning
10:13:34that Q T is = to Q minus one we are then
10:13:40making use of this to claim
10:13:44this so instead of minus On's that we
10:13:48are used to when we are dealing with
10:13:51inverses here we have t the
10:14:02transpose so in this case this
10:14:04orthogonal Matrix that we have just
10:14:07learned about this is the square Matrix
10:14:09whose columns and rows are orthogonal
10:14:11and they are also normalized meaning
10:14:14that we are dealing with
10:14:16QT QT sh Q minus one it will look like
10:14:21this so this q1 you can see that here we
10:14:24got the first row here we got the second
10:14:27row and if we
10:14:29calculate
10:14:31the dot product between this row and
10:14:34this row we can quickly see that we are
10:14:36getting a value of zero so we can prove
10:14:39that they are actually autogo those two
10:14:42rows let's go ahead and actually prove
10:14:44that so let's call this R1 let's call
10:14:47this R2 this is Row one and row two and
10:14:50I will leave the uh column version so q1
10:14:54* Q2 that's do product on you to prove
10:14:57that the columns are perpendicular I
10:15:00will work with the rows so R1 * R2 for
10:15:06me to prove that they are orthogonal I
10:15:08need to prove that this equal to
10:15:10zero can we do that well let's try so 1
10:15:15/ 2 < of 2 1 / 2 of 2 ultied by 1 / 2 <
10:15:24of
10:15:252 needs some bigger space in
10:15:29here so 1 / 2 of 2 and then
10:15:34minus 1 / 2 of 2 that's how I can
10:15:38calculate the dot product between R1 and
10:15:43R2 R2 and
10:15:46R1 you can see that the elements in here
10:15:49are the same and here the elements are
10:15:51also the
10:15:54same so then this is equal
10:15:58to 1 / 2 < of 2 * 1 / 2 of 2 - 1 / to of
10:16:072
10:16:11* 1 / 2 of 2 I'm simply taking this
10:16:15minus and given that the dotproduct is
10:16:18basically Plus
10:16:19and then this amount I'm just taking
10:16:21this and bringing up in here to a avoid
10:16:24one more step uh given the space is
10:16:27quite limited now what do we see in
10:16:31here this value is the same as this
10:16:34value which means that this is equal to
10:16:38zero and we know that the two vectors to
10:16:42be or token they need to have a DOT
10:16:45product equal to zero so here we have
10:16:47proven that dotproduct
10:16:52of R1 and R2 is equal to
10:16:59zero so this proves that R1 and R2 so
10:17:04the two
10:17:08rows of this Matrix so R1 and
10:17:12R2 are orogo
10:17:19we can also prove that the second
10:17:21criteria of auton normal vectors is also
10:17:24satisfied in here we can prove that when
10:17:27we look at the
10:17:30length of this vector and of this one
10:17:35then they are of the unit one so let's
10:17:38actually go ahead and do for one of them
10:17:41so let's prove that
10:17:45for 1 / 2 of 2 and then 1 / 2 Ro of 2
10:17:52this is a
10:17:56vector that the length of
10:17:59it this is let's say
10:18:01our first row so this is
10:18:05R1 then the R1 length is equal
10:18:10to 1 / 2 of 2 2 + 1 / 2 of 2 2 this is
10:18:21equal to 1 / 2 + 1 / 2 and what is 1 / 2
10:18:28+ 1 / 2 it's equal to 1 so we have
10:18:32proven that the length of R1 is equal to
10:18:371 you can quickly and easily also
10:18:40compute that for the second row and you
10:18:42will then also prove that the R2 the
10:18:45length of it is also one which is then
10:18:47the second criter area which said that
10:18:51for the vectors to form this auton
10:18:53normal bases so to be auton normal
10:18:55vectors they uh also had to have a
10:18:58length of one so they had to be a unit
10:19:01vectors in this case then we can make
10:19:04use of the property that the
10:19:07Q
10:19:102 transpose is equal to Q
10:19:15inverse and this then result in Q2
10:19:20transpose * Q2 Q2 which is equal to Q2 *
10:19:24Q2 transpose which is equal to the
10:19:26identity Matrix and specifically I2
10:19:30because we are in the
10:19:34R2 so both this Q2 and the previous
10:19:37example those are orthogonal
10:19:40matrices and in here we have proven that
10:19:44the rows are indeed oroginal and we have
10:19:47also seen that the length of them are
10:19:49unit vectors meaning that we have
10:19:52automatically got
10:19:54this I will leave this one to you to do
10:19:58those proofs so to uh ensure that the
10:20:01row one and row two are orthogonal so
10:20:04they are perpendicular which means that
10:20:06the product of their uh the dot product
10:20:09of these two vectors is equal to zero
10:20:11and also that they are normalized which
10:20:13means the length of them is equal to one
10:20:16and this means that then this holds you
10:20:20can actually even go ahead and uh
10:20:24practice the material that we are uh
10:20:26learned as part of the previous units by
10:20:28calculating the inverse of this Matrix
10:20:32and checking that the inverse of this
10:20:34Matrix is indeed equal to the transpose
10:20:36of the Matrix so that QT is equal to Q
10:20:40minus 1 because we learned how we can
10:20:42compute the inverse of a matrix because
10:20:45the inverse of a matrix was equal to 1 /
10:20:48the determinant of this Matrix
10:20:52times and then the manipulated version
10:20:54of it which was in this case 0 0 and
10:20:58then we need to have here 1 so -1 *
10:21:021 and then 1 * - one so we have to
10:21:06multiply this and this by minus so one
10:21:09and then here minus
10:21:11one so in this way you can also prove
10:21:15that this inverse is actually equal to
10:21:18the Q to
10:21:20transpose because then you can prove
10:21:24that indeed and you can see for yourself
10:21:26that this formula is in equal to the Q2
10:21:30transpose because then you can prove
10:21:34that indeed and you can see for yourself
10:21:37that this formula is indeed true in this
10:21:40module we are going to talk about Matrix
10:21:42factorization we are going to discuss
10:21:44the significance of Matrix factorization
10:21:47we are going to Define Matrix
10:21:49factorization we are going also to
10:21:51discuss the common applications of
10:21:52Matrix factorization across different
10:21:54fields and then we are going to see
10:21:56detailed examples of metrix
10:21:59factorization so let's talk about why
10:22:02Matrix factorization matters so metrix
10:22:05factorization techniques they are
10:22:07essential for various reasons they are
10:22:09used for simplifying metrix operations
10:22:11like solving linear systems or when we
10:22:14have this um many uh matrices but we
10:22:18want to to um simplify these operations
10:22:21that we apply to these matrices and we
10:22:22want to solve the problem then we can
10:22:25make this uh complex Matrix operations
10:22:28more manageable and make these uh
10:22:31calculations more manageable by using
10:22:33Matrix factorization techniques we can
10:22:36also use Matrix factorization directly
10:22:38to solve system all linear equations
10:22:41efficiently we can also use Matrix
10:22:43factorization to perform igon value de
10:22:45composition singular value de de
10:22:47composition or called SVD and other
10:22:51operations which are crucial in machine
10:22:53learning and data analysis so ion values
10:22:57and igon vectors you might have heard
10:22:59already they are part of also PCA which
10:23:02is the principal component analysis and
10:23:04this comes from uh fundamentals of
10:23:06statistics and the uh PCA is used as a
10:23:10dimensionality technique and in fact
10:23:12it's one of the most popular damage s
10:23:14techniques that you will find in the
10:23:16industry used in the data science using
10:23:19data analytics machine learning even in
10:23:22the Deep learning so Matrix
10:23:25factorization can also be used to reduce
10:23:28the computational complexity by making
10:23:30use of this factorization we can then
10:23:33simplify the process and also make it
10:23:35more efficient for computation and it's
10:23:38especially important when we are dealing
10:23:40with this High dimensional data when we
10:23:42have many features or we have a very
10:23:45large model and complex model then this
10:23:48uh meing factorization technique can
10:23:51make a huge difference in our data
10:23:53processing process so this techniques
10:23:56underpin many algorithms in numeric
10:23:59analysis in optimizations and Beyond so
10:24:03whenever it comes to machine learning or
10:24:06data science or many other fields you
10:24:08will see this uh process and this term
10:24:11metrix authorization appearing a lot
10:24:14even um in the example of a streaming
10:24:16company Netflix which I'm sure that you
10:24:18are aware of netrix is using uh metrix
10:24:22uh factorization to build a recommender
10:24:24system and uh metrix authorization usage
10:24:28in building recommender algorithm for
10:24:30personalized recommendations is actually
10:24:32one of the most popular applications of
10:24:34metric factorization therefore I wanted
10:24:37to specifically discuss this topic as
10:24:39part of our Advanced linear algebra
10:24:41course and some of the concepts might
10:24:44seem bit more complex than the ones that
10:24:46we have discussed as part of the
10:24:47previous units but once we go through
10:24:50them step by step and I will give you
10:24:53all the details in all these examples
10:24:55this entire process of this different
10:24:58metrix factorization techniques should
10:25:00become much more clear and
10:25:02straightforward so we will be discussing
10:25:04not just one but multiple fundamental
10:25:07metrix factorization techniques beside
10:25:10of talking the high level where they are
10:25:12used and how you can choose for what
10:25:15type of applications so we are going to
10:25:17demes this entire concept of metrix
10:25:20factorization and we are going to uh
10:25:23start from high level then we are going
10:25:25to go into the deepest details let's now
10:25:28formally Define the metrix factorization
10:25:31so metrix factorization refers to
10:25:34decomposing a matrix into product of two
10:25:37or more matrices revealing its structure
10:25:41and simplifying further
10:25:43analysis so what is this idea behind
10:25:46metric factorization the idea is that if
10:25:48we have a matrix
10:25:50a and we want to simplify our process of
10:25:54calculation or multiplication anything
10:25:57that's related to this a but this a in
10:26:00itself it contains this weird numbers or
10:26:03it is just too complex you know it
10:26:04contains this ton of different numbers
10:26:07you don't recognize whether the columns
10:26:09are linearly independent it's not very
10:26:11readable from the first View and you
10:26:13just want to make your life easier when
10:26:15performing this calculations well for
10:26:17that you you can make use of this Matrix
10:26:19factorization to write this a in terms
10:26:23of some other matrices let's say um and
10:26:26I'm calling here randomly Q or t so it's
10:26:30equal to for instance the dotproduct of
10:26:31these two matrices Q * T where Q is much
10:26:35simpler and the t is also much simpler
10:26:38so those may contain vectors that are um
10:26:42for instance this can be a diagonal
10:26:44matrix or it can be a matrix uh with
10:26:47specific properties when using those you
10:26:50will feel much more comfortable so it
10:26:52will be easier for you to use them in
10:26:54order to multiply uh with other matri
10:26:57matrices it can be easier for you to
10:27:00solve this problem but of course if you
10:27:02are in the two-dimensional space let's
10:27:04say you are in R2 or in R3 then most
10:27:07likely it will be quite straightforward
10:27:09for you to use the a itself but if you
10:27:12are in the r 100 or R 1,000 then of
10:27:17course this uh entire computations they
10:27:20become super complex it will be
10:27:23difficult to understand and compute this
10:27:26linear combinations find out whether you
10:27:28are dealing with a linearly independent
10:27:30columns find out um the um new space the
10:27:35calm space the basis of the new space
10:27:38and calm space and all these they might
10:27:41seem uh much more difficult when you are
10:27:43in high dimensional space for in those
10:27:45cases we can then make use of metric
10:27:47factorization
10:27:48to make the entire process much more
10:27:52simplified and also more efficient this
10:27:55entire calculation process so common
10:27:58types of Matrix factorization include
10:28:00lower upper uh Matrix factorization or
10:28:03in short
10:28:04L QR factorization an Infamous type of
10:28:08factorization which is called orthogonal
10:28:11triangular
10:28:13factorization and then we have SVD
10:28:16singular value de compos ition yet
10:28:19another in famous metrix
10:28:21factorization and then finally the igon
10:28:24de composition also another Super
10:28:26popular metrix factorization
10:28:28technique so uh the QR SVD and igod
10:28:33composition are in fact highly popular
10:28:37the composition and Metric factorization
10:28:40techniques that you will see appearing
10:28:43in the 90% of all the statistics related
10:28:46and machine learning related ated books
10:28:49so this just comes to prove how
10:28:51important these concepts are when it
10:28:54comes to properly learning and mastering
10:28:57these more applied uh science related
10:29:00fields like machine learn so if you want
10:29:03to go beyond the level of knowing
10:29:05algorithms but rather than to also be
10:29:08able to edit the algorithms tweak them
10:29:11adjust them be able to understand
10:29:14machine learning algorithms deep
10:29:15learning algorithms data science and at
10:29:18its core and in order to become a
10:29:20professional well-rounded professional
10:29:23then this I composition the singular
10:29:25valid composition and the QR metrix
10:29:28authorization techniques are techniques
10:29:30that you want to know and you want to
10:29:33understand at least higher level such
10:29:35that you can easier grasp more advanced
10:29:39concepts that come from the applied
10:29:41sciences like machine learning and
10:29:44AI so let's first discuss at high level
10:29:47what this C q r DEC composition is so
10:29:50what the QR DEC composition does is that
10:29:52it decomposes a matrix into an
10:29:55orthogonal Matrix which we are referring
10:29:58by q and then an upper triangular Matrix
10:30:03R so in here you can see that we have
10:30:07this two different matrices so we are
10:30:10basically saying a is equal to this
10:30:12product of this Matrix q and r R where
10:30:19the first one this Matrix
10:30:22Q this one should be orthogonal Matrix
10:30:26so this part is really important and we
10:30:29have learned as part of the previous
10:30:31module the definition of orthogonal
10:30:34Matrix we learned that the rows or
10:30:37columns they had to be orthogonal to
10:30:39each other and we also learned that they
10:30:42need to have a length of one they need
10:30:44to um be normalized
10:30:48and we learned that this means that the
10:30:51transpose of those matrices is equal to
10:30:53the inverse of the matrices so this was
10:30:56just the last part of the previous
10:30:58module and this is exactly what this
10:31:01Matrix Q is about so we are saying that
10:31:04we will decompose I into this two
10:31:06matrices as a product of these two
10:31:08matrices Q andr one of which this Matrix
10:31:11Q should uh be uh an autal Matrix which
10:31:14means the rows and the columns they
10:31:16should be or toal to each other so their
10:31:20uh dot product each of them should be
10:31:22equal to zero and they need to be
10:31:25normalized so the length of them should
10:31:27be one for each of those rows and
10:31:29vectors and then the second part of this
10:31:33U the composition is this Matrix R which
10:31:36says that the Matrix R should be an
10:31:39upper triangular Matrix and what is the
10:31:42definition of upper
10:31:43triangular well in this case you can
10:31:47think of this r as this
10:31:49Matrix where we have here all
10:31:54zeros and then here on the diagonal you
10:31:57have numbers nonzero numbers and then
10:32:00here let's say 1 2 3 4 5 and then here
10:32:04in the upper part you will also have
10:32:06numbers so unlike in the lower part of
10:32:08this Matrix R where you will have zeros
10:32:12in
10:32:13here you will have also noner numbers
10:32:16numbers let's say seven
10:32:18uh 10 uh
10:32:208 and I'm just writing these numbers
10:32:22randomly so of course in a real case
10:32:25when we have this Matrix a and we go
10:32:27through this process of QR de
10:32:28composition of course we will have an
10:32:30appropriate q and appropriate R where
10:32:33these numbers will be different and they
10:32:35will be specific numbers that we will be
10:32:39calculating but the idea is that we need
10:32:41to get this upper triangular Matrix R
10:32:45for this calculation to make sense
10:32:48so we will then be using this qard
10:32:50composition for solving linear um linear
10:32:54Le squares problems for instance which
10:32:56is part of the linear regression too
10:33:00because linear regression from machine
10:33:03learning and from statistics uh it is
10:33:06based on the least Square technique the
10:33:10estimation technique that we are using
10:33:12for linear regression in machine
10:33:14learning um to solve this linear
10:33:17regression problem is called Ordinary
10:33:19leas squares so the algorithm is based
10:33:21on this idea of Le squares which is
10:33:23trying to minimize a squared uh
10:33:26residuales of the morel and that can be
10:33:29done by using this idea of QR
10:33:34decomposition so it helps us to provide
10:33:37numerically stable solutions for this
10:33:39type of problems too and C de
10:33:41composition is used extensively in
10:33:44Signal processing and statistical
10:33:46analysis
10:33:49let's now briefly talk about the L
10:33:51decomposition so L decomposition
10:33:53decomposes a matrix into lower
10:33:56triangular Matrix so this is the
10:33:58opposite of what we had before we can
10:34:00have an upper
10:34:03triangular
10:34:05triangular
10:34:07Matrix like we had in the QR the
10:34:11composition we can also have a lower
10:34:13triangular Matrix so you might have
10:34:15already guessed how it will look like I
10:34:17want go into that very soon in the QR
10:34:20composition example you will see the
10:34:22idea of the upper triangular I will also
10:34:25show the idea of a lower
10:34:27triang so the composition in case of Lu
10:34:32uh is done by decomposing a matrix into
10:34:36lower triangular Matrix l and an upper
10:34:39triangular Matrix U so basically the
10:34:43difference between the QR de composition
10:34:46and L de composition is that in the QR
10:34:49DEC composition we are decomposing a
10:34:51matrix into orthogonal Matrix and an
10:34:53upper triangular Matrix while in case of
10:34:56L DEC composition we are decomposing a
10:34:59matrix into lower triangular Matrix and
10:35:02an upper triangular Matrix so here you
10:35:05can see that we no longer have this idea
10:35:06of orthogonal matrix but instead of that
10:35:09we are talking about lower triangle
10:35:13Matrix so in that aspect uh L U the
10:35:17composition is different from QR DEC
10:35:19composition so what the L de composition
10:35:22does is that it facilitates the solving
10:35:25of linear equations and Matrix
10:35:27inversions it is common in engineering
10:35:30and in physical sciences for systems of
10:35:33this linear equation to be solved by
10:35:35using lud de composition and in fact if
10:35:38you are learning Quantum uh mechanics
10:35:41that this L decomposition can definitely
10:35:45help you to better understand many
10:35:46Concepts but if your target fields are
10:35:49machine learning deep learning or
10:35:51artificial intelligence then for those
10:35:54using QR composition will be uh much
10:35:57more often a case than using this lud de
10:36:03composition let's now talk about the
10:36:05singular value decomposition so what the
10:36:08SVD does is that it decomposes a matrix
10:36:11into three matrices so first one is the
10:36:16orthogonal Matrix C
10:36:19the second one is a diagonal
10:36:22matrix and then the third one is this V
10:36:26which is the conjugate transpose of an
10:36:28orthogonal
10:36:30Matrix so for now this might s bit
10:36:34complex and you can see that unlike the
10:36:36QR or Lu de composition where we got
10:36:39just uh two uh matrices as a result of
10:36:42our decomposition in case of SD we got
10:36:45three matrices like the name suggests
10:36:47two by the way so three
10:36:54parts and this might seem bit complex
10:36:57but we are going to go through this
10:37:00process step by step and I'm going to
10:37:02provide you detailed example such that
10:37:04this will all make sense but for now
10:37:07let's focus at the high level usage of
10:37:09SVD so singular value decomposition is
10:37:13one of the most popular decomposition
10:37:15techniques and it is also Direct Al used
10:37:18as part of machine learning algorithms
10:37:20to form uh those machine learning
10:37:22algorithms it is also used in the data
10:37:25compression in the noise reduction so
10:37:27when we are trying to clean our data and
10:37:30remove the Noise by using SVD because
10:37:32SVD can help us to identify those
10:37:34outliers and then remove them from the
10:37:37data by performing noise
10:37:40reduction and it is also used in the
10:37:42principal component analysis the
10:37:45PCA the uh same dimensionality reduction
10:37:48technique that I just uh mentioned
10:37:51related to the ion de composition
10:37:53because SVD and the igon de composition
10:37:55are highly related to each other so this
10:37:58SVD is used as part of this PCA
10:38:00algorithm and PCA is the most popular
10:38:04the infamous dimensionality reduction
10:38:07technique that is used both in the
10:38:09advanced statistical studies in the
10:38:11statistics in general also in finance
10:38:14and is also used as part of many machine
10:38:16learning and deep learning applications
10:38:19so knowing PCA is a must if you want to
10:38:22get into uh data analytics or data
10:38:24science machine learning and AI but
10:38:29also uh it helped you it will help you
10:38:31also to uh understand uh many other
10:38:35Concepts when it comes to this Fields so
10:38:38the SVD provides insight into the
10:38:40structure but also the rank of the
10:38:42Matrix so we are going to see this as
10:38:44part of our example too
10:38:47let's now also briefly talk about the
10:38:49igen dec composition so igen DEC
10:38:52composition which is highly related to
10:38:54this concept of igen values and ion
10:38:56vectors it decomposes a matrix into this
10:38:59ion values and ion vectors which then
10:39:02shows the matrix's fundamental
10:39:05properties which are related to this
10:39:06idea of correlation what kind of
10:39:08information does this uh Matrix contain
10:39:11what is the variation in what direction
10:39:13is the variation the largest and this
10:39:16icon that composite which is then
10:39:18related to also this idea of SVD and in
10:39:21general this dimensions and correlations
10:39:24is critical for understanding linear
10:39:26Transformations the stability analysis
10:39:29and systems of differential
10:39:32equations but beside this uh
10:39:35mathematical side of uh Concepts and
10:39:38understanding these mathematical topics
10:39:41the ion de composition is also the basis
10:39:43for many algorithms in numerical deor
10:39:46algebra
10:39:48uh but also many applied linear algebra
10:39:50topics like in the data science in
10:39:53machine learning and it's used heavily
10:39:55in artificial intelligence for feature
10:39:58extraction for dimensionality reduction
10:40:01related again to the concept of PCA
10:40:03because PCA is based entirely on this
10:40:06concept of IG de composition PCA is the
10:40:10direct result of computing the igon
10:40:14values and igon vectors
10:40:21so without knowing what are Dion values
10:40:24and ion vectors you cannot perform PCA
10:40:26because the first step of the PCA is the
10:40:30computation of the icon values and icon
10:40:32vectors and then using different rules
10:40:34which we are referring uh as the elbow
10:40:37rule or Kaiser rule we can then use this
10:40:40icon values and icon vectors to
10:40:43understand what are the features in our
10:40:45data that contain the most
10:40:48variation so the most information and
10:40:52then we can use that in order to
10:40:54understand what are the most important
10:40:56features in our data and reduce the
10:40:59dimension of our model by selecting this
10:41:02most important features because what PCA
10:41:05basically does is that it it uses these
10:41:08icon values and icon vectors to
10:41:10understand how we can uh create a linear
10:41:13combination out of our features and
10:41:16understand the the amount of those
10:41:19linear combinations that contain the
10:41:21most variation and then select those and
10:41:24uh select the largest amount of
10:41:27information in the data and then Skip
10:41:30and drop those uninformative being your
10:41:33combinations while still keeping the
10:41:35most information and this definition of
10:41:38the most will then be decided by this
10:41:40differentials this is just higher level
10:41:42Insight background information on what
10:41:44you can expect when you are talking
10:41:46about applying this highly technical
10:41:50linear algebra concept of Icon de
10:41:52composition into an applied science
10:41:55Fields like data science or machine
10:41:57learning or AI but we will see this
10:41:59later and I'll also make comments
10:42:01regarding this and though PCA won't be
10:42:05discussed as part of this course because
10:42:07here we are talking about linear algebra
10:42:09but PCA is part of the fundamentals
10:42:12statistics course and in there we are no
10:42:15longer providing all these different
10:42:17details on how you can uh perform this I
10:42:20composition therefore knowing how to
10:42:23perform I composition will then set you
10:42:27for success to actually understand the
10:42:30mathematics behind the statistical
10:42:32Concepts like PCA and also later on
10:42:37understand how you can use that PCA in a
10:42:39machine learning Concepts and in AI
10:42:42Concepts like outo encoders and how you
10:42:44can relate your um out to encoders to
10:42:48this concept of PCA how they are related
10:42:50what are their commonalities and what
10:42:52are their
10:42:55differences so everything is about the
10:42:58choice and choosing the right tool for
10:43:01your problem when it comes to the
10:43:03decomposition tools metrix authorization
10:43:05tools we have seen that there are many
10:43:07options and the question is which one
10:43:09should we pick in what cases so choosing
10:43:11the right tool is really important when
10:43:13it comes to this different metrix
10:43:15factorization techniques because they
10:43:16are many choices and each of them they
10:43:19can be used for different sorts of
10:43:21problems so therefore in order to
10:43:24understand which one you need to pick in
10:43:27what kind of cases what kind of
10:43:28requirements you have and what kind of
10:43:30solve uh problem you are trying to solve
10:43:33that in those cases you will need to
10:43:36have this knowledge that you will learn
10:43:38as part of this course in order to make
10:43:40that right choice of the
10:43:43twool so the choice among Q are the
10:43:46composition
10:43:47the L de composition the SVD and IG de
10:43:50composition it really depends on your
10:43:53specific problems requirements and the
10:43:55data characteristics so are you dealing
10:43:57with a complex data are you dealing with
10:44:00a simple data with low Dimensions what
10:44:03is the goal that you uh want to uh
10:44:06achieve what is the problem that you are
10:44:07trying to solve is it to uh reduce the
10:44:11dimension of your feature space is it to
10:44:15solve a problem with linear
10:44:17equations is it to solve a quantum
10:44:20mechanics problem or is it to um
10:44:24incorporate this as part of your machine
10:44:27learning algorithm so QR and LU DEC
10:44:30compositions are usually preferred for
10:44:32solving linear systems while SVD and I
10:44:35Anda compositions they help us for
10:44:37deeper insights when it comes to the
10:44:39data and what kind of information it
10:44:41contains how we can reduce the dimension
10:44:44of the data or how we can use it as part
10:44:47of machine learning algorithm for uh
10:44:49noise reduction identifying outliers Etc
10:44:53so um this type of algorithms like SVD
10:44:57and ion de composition it helps us to
10:44:59also uh intuitively using geometry and
10:45:03our knowledge of geometry to um
10:45:06visualize the data for instance the PCA
10:45:08helps us to visualize this High
10:45:10dimensional data using just couple of
10:45:13principal components let's say we have
10:45:1510 features in our
10:45:17model so we have a dimension of 10 we
10:45:21are in r10 but we want to visualize our
10:45:24data by using PCA we can then reduce the
10:45:26dimension and come up with uh three
10:45:29principal components which are a linear
10:45:31combination of our original 10 vectors
10:45:34and then we can use the three principal
10:45:36components to visualize our data in
10:45:393D and this basically helps us to
10:45:42geometrically visualize our data and
10:45:45then make presentation s make much more
10:45:48sense of our story so to do uh
10:45:50storytelling for our data and uh much
10:45:53more and these two uh models and tools
10:45:58they are invaluable when it comes to uh
10:46:01applications in machine learning in deep
10:46:03learning in data science and artificial
10:46:06intelligence
10:46:08so meing factorization techniques they
10:46:10are super important when it comes to
10:46:12computational mathematics they are also
10:46:15directly affecting the data science Ai
10:46:17and many other algorithms so they are
10:46:19not only important in terms of the
10:46:21problem that they are trying to solve
10:46:23but also in order to make the
10:46:24computation process so when coding in
10:46:27python or in other programming languages
10:46:30to make that process much more efficient
10:46:33they also help us to uh make these
10:46:35computations efficient and provide
10:46:37insights into different properties that
10:46:40we have in our
10:46:42data as part of this course we are not
10:46:45only going to discuss one but actually
10:46:48three of these four the composition
10:46:51techniques and this metrix factorization
10:46:53techniques in detail we are going to
10:46:55talk about the qard de composition we
10:46:58are going to not just discuss it but uh
10:47:01also to learn it step by step and we are
10:47:04going to do a detail example with all
10:47:07the steps involved such that you will
10:47:09feel confident doing a QR decomposition
10:47:12all by yourself then we are also going
10:47:15to do an SVD de comp position as well as
10:47:18ion the composition and then we are
10:47:21again going to discuss them in terms of
10:47:23their mathematical formulation the
10:47:25definition but also the application
10:47:27step-by-step process and a detailed
10:47:30example such that you can conduct each
10:47:32of those metric factorization techniques
10:47:34and these decomposition techniques by
10:47:36yourself manually doing all these
10:47:39calculations this understanding and this
10:47:42examples and this Concepts will help you
10:47:44to not just be able to formulate what
10:47:47these techniques are about but really
10:47:49and truly understand and then use them
10:47:53later on whether when doing your own
10:47:55research writing scientific papers or
10:47:58tweaking the algorithm all by yourself
10:48:00when inventing new
10:48:03algorithms I won't be discussing this L
10:48:06de composition technique because we
10:48:08already know uh that the QR and LU they
10:48:11are both used for similar type of
10:48:13problems therefore to save us time I
10:48:16have selected carefully the uh most
10:48:20important the composition techniques and
10:48:22Metric factorization techniques that you
10:48:25will most likely be dealing with will be
10:48:27dealing with in your future career in
10:48:29applied sciences

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