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MOOC_LinearAlgebra_Lesson05

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Hello, and welcome to Lesson Five Of this Introduction to Linear Algebra with Wolfram U. The topic of this lesson is matrix equations. So, let's begin with a brief overview of the lesson. So, let's say you have a matrix, like the one shown over here. It's got two rows and three columns. And you have a vector <i>x</i>, a column vector, {2, 5, 1}. Then, in this lesson You learn how to find the product of <i>A</i> and <i>x</i>. And we're just going to call it <i>A x</i> Although you might want to put a dot in it if you like. This product would be used to define a matrix equation Such as <i>A x</i> = <i>b</i>, where <i>b</i> is a vector. And what we've shown is that Such a matrix equation is equivalent to First of all, a vector equation, and secondly, a linear system. And of course, matrices are much easier to work with So we will always work with matrix equations wherever possible. Let's begin by trying to define the product of a matrix and a vector. So, here you have a matrix, capital <i>A</i>, with <i>n</i> columns <i>a</i>_1, <i>a</i>_2, up to <i>a</i>_<i>n</i> And your vector with entries <i>x</i>_1 up to <i>x</i>_<i>n</i>. Then, the product of <i>A</i> and <i>x</i> is defined by taking <i>x</i>_1 • <i>a</i>_1 and then <i>x</i>_2 • <i>a</i>_2 and then <i>x</i>_<i>n</i> • <i>a</i>_<i>n</i> So that's like multiplying a scalar by a vector in each case. Now, of course, to do that You must make sure that the number of columns of <i>A</i> Is equal to the number of entries in <i>x</i>. You need the same <i>n</i> over here and over there. Then, of course, you'll notice that the right-hand side over here Is a linear combination of the columns of <i>A</i> With weights from <i>x</i>. So, that's a linear combination. Of course, one last thing is that this matrix equation <i>A x</i> = <i>b</i> is then the same as the vector equation <i>x</i>_1 • <i>a</i>_1, up to <i>x</i>_<i>n</i> • <i>a</i>_<i>n</i>, = <i>b</i> OK, so as a simple example, here is a matrix-vector product. So you have a matrix over here, your vector over there. So what you do is you do 4 times the first column 3 times the second column, and 7 times the third column And you get back 3 and 6. Let's check that with actual calculations. So here is the answer from actual calculations. We have 4 • {1, 0}, 3 • {2, -5}, and 7 • {-1, 3}. Now, the point is that we have A very simple but very powerful function in the Wolfram language Namely Dot, which will do exactly that for you. So, here's the matrix, here's the vector and then if you do <i>A</i> . <i>x</i> (That dot in the middle is multiplication) Then you get back just the same answer. Dot is by far one of the most useful functions In linear algebra. If we look back over here, you see that the first entry, 3 Really came from doing 4 • 1, and then 3 • 2, and then 7 • -1. Although, you're really doing a row times the vector. And that motivates the other approach to multiplication Which is the row-vector rule. The idea is that if you want an entry in <i>A</i> . <i>x</i> You simply take the <i>i</i>th row of <i>A</i> And then you multiply it by the vector <i>x</i>. So again, here's an example. You have a matrix, you have a vector And if you want the first entry of <i>A x</i> What you do is take the first row of <i>A</i> The {2, 3, -1} And then kind of lift it and place it on the first column. So you get 2 • 4, 3 • 1 and -1 • 3, that's 8 And for the second entry You do 3 • 4 and then 5 • 1 and then 2 • 3. That's the other entry. You can check that with the Dot function And you get back just the same result, {8, 23}. So the row-vector rule Is a very nice way of doing the multiplication in this case. OK, so, a few properties of these products. If you have a matrix <i>A</i> which is <i>m</i> by <i>n</i> And you have two vectors <i>u</i> and <i>v</i> From <b>R</b>^<i>n</i> Then for every scalar <i>c</i>, for every number <i>c</i> <i>A</i> . (<i>u</i> + <i>v</i>) is (<i>A</i> . <i>u</i>) + (<i>A</i> . <i>v</i>) Just like in ordinary arithmetic algebra And then <i>A</i> . (<i>c</i> <i>u</i>) is <i>c</i> (<i>A</i> . <i>u</i>). That says that the dot product kind of commutes with The addition and scalar multiplication Of vectors. What it means is that if Over here you do the plus first, and then the dot And over there you do the dot first And then you do the plus, etc. So that’s the meaning of 'commutes' You have two operations And they can be more or less interchanged. The other thing is that, in future We'll see that a matrix is a linear transformation Because of these very properties. Let's verify these properties in an example. So here is a matrix <i>A</i> And a pair of vectors <i>u</i> and <i>v</i>. We have got <i>A</i>, <i>u</i> and <i>v</i>. Let's check that if you do <i>A</i> . (<i>u</i> + <i>v</i>) And you do (<i>A</i> . <i>u</i>) + (<i>A</i> . <i>v</i>) You get the same answer, and you can check that With the double equals sign you get back True. Let's check the second property. Suppose <i>c</i> is 5, and you do <i>A</i> . (<i>c u</i>) You get an answer. You do <i>c</i> (<i>A</i> . <i>u</i>); you get another answer Which is the same as the earlier one And hence, the two are exactly the same. So, with that background on matrices and their products We can now talk about the relationship Between matrix equations and linear systems. The point is that Every linear system can be written as a matrix equation. Now, the matrix <i>A</i> over here is the coefficient matrix For the left-hand sides of the equations Whereas the vector <i>b</i> Is a vector of constants on the right-hand side. Let's take a particular linear system But I first need to clear the variable <i>x</i> I used it early on. You know, the equation, there are Three equations and three unknowns. So, the matrix <i>A</i> now is {2, 3, 7}, from the first equation left-hand side Then {5, -1, 11} from the second equation And {4, -5, 1} from the third equation. Whereas, the vector <i>b</i> is {29, 36, -3} From that, 29 and 36 and -3. So, once you have <i>A</i> and <i>b</i> You can then solve the equations using LinearSolve. And now, that's a solution from LinearSolve You can check it By plugging it back in the equations. I've used the thread function Which basically sets <i>x</i> to 1 and <i>y</i> to 2 and <i>z</i> to 3 And you see that you actually get back True in all the cases. So, you have a relationship Between a linear system and a matrix equation over here. OK, once you understand that, then we can go a bit deeper And talk about when do solutions for a mixed equation exist. So, let's say you fix <i>A</i> and you fix <i>b</i>. Then this system, or this equation Has a solution if and only if <i>b</i> is a linear combination of the columns of <i>A</i>. So let's check that. Here is the matrix, here is the vector <i>b</i>. Write in the columns of <i>A</i>, a1, a2, a3. The first column is 4 and -2 and 2 The second column is 5, 1 and 4 The third column is 2, 0, and 5 Just like you can see over here. Now, let's try a particular linear combination. I'll do 2 times the first column 3 times the second one and 7 times the third one And I get back an answer, which is {37, -1, 51} Which is exactly the same as <i>b</i>. Now, the question is how to define it I'll tell you in just a minute, but the main point Is that this says that you actually have A solution {2, 3, 7} for the equation. Let's check the answer's correct using LinearSolve. You do get back 2, 3 and 7. But the point is that every time you have a solution You really have a linear combination Which lets you express <i>b</i> In terms of the columns of <i>A</i>. On the other hand Suppose you fix your <i>A</i>, and then you vary <i>b</i>. Question, what happens then? Well, the point is that you have three statements over here. First of all, suppose you know that for all <i>b</i> This equation has a solution. Then, that's the same thing as saying that <i>b</i> is a linear combination of the columns of <i>A</i> Or that the columns of <i>A</i> span the space from which <i>b</i> comes. So that's a relationship between <i>A</i> and <i>b</i> But <i>A</i> is fixed and <i>b</i> is varying. So here's the matrix <i>A</i>, for example Here's an arbitrary vector in two dimensions. Now, if I solve the system <i>A x</i> = <i>b</i> using LinearSolve I get back a solution. And let's check it's actually correct. So, you're trying to plug back. You do the first entry times the first column of <i>A</i> The second entry times the second column of <i>A</i> Equal to <i>b</i> And, oh, that doesn't look right. Well, I didn't simplify far enough So if I apply the Simplify function Then I actually get back True. The point is it looked like we didn't get the answer But in fact, just a bit of simplification Gives you exactly the required result. The main point is, in this case we have had a solution found for all <i>b</i> not just for a particular <i>b</i>. And the statements over here Simply mean that the columns of <i>A</i> Span the plane <b>R</b>^2 Because every vector <i>b</i> can be written as A linear combination of the columns of <i>A</i>. So that brings me to the end of this lesson And the main point is that One can compute the product of a vector and a matrix Either by using linear combinations Or by using row-vector multiplication And overall, the row-vector multiplication Is a bit easier to understand. Second, we have a very nice, powerful function Called Dot in the Wolfram Language Which can be used to compute such matrix-vector products And that's the function we'll often encounter in the course So it'd be nice if you can get familiar with it. Now, the dot product has got nice properties Commutes with the addition and scalar multiplication Of vectors, that's good to know. Finally, every linear system can be written as a matrix equation And vice versa. So, in the next lesson, we'll talk about The important concept of linear independence But before that, do review this lesson. It's an important and basic lesson. And be ready for the discussion of linear independence. So stop here, thank you very much.

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Duration: 12 minutes and 11 seconds
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Posted by: wolfram on Sep 30, 2020

MOOC_LinearAlgebra_Lesson05

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