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Show that if the columns of $B$ are linearly dependent, then so are the columns of $A B .$

Columns of $A B$ are linearly dependent.

Algebra

Chapter 2

Matrix Algebra

Section 1

Matrix Operations

Introduction to Matrices

McMaster University

Harvey Mudd College

Baylor University

Lectures

01:32

In mathematics, the absolu…

01:11

06:44

Let $A=\left[\begin{array}…

02:29

Suppose $A$ is an $m \time…

06:28

Let $A$ be an $m \times n$…

00:49

Suppose columns $1,3,5,$ a…

16:05

If $\mathbf{v}_{1}$ and $\…

02:03

Prove that if $A$ is a mat…

05:58

Prove that if $y_{1}$ and …

06:16

Prove that if $\left\{\mat…

02:32

Show that the vectors$…

02:45

okay here were asked to show that if the columns of b r linearly dependent, then so are the columns of a B. And so this is ah, approved question. Have to prove this statement. Um, but this proof is really just unpacking a lot of unpacking and a lot of knowing the definitions. So we'll start with the idea of the columns would be our linearly dependent. So that means that, for example, this column here, this first column can be represented as a linear combination of all of the other columns. It's in particular. It means that this single element here is a linear combination of these elements and sing with this element in these elements and so on. And so what that means is, if you write it out, for example, be 11 is equal to some constant C two times be one too, plus some constant C three times be +13 plus and all the way up to see um and be one end. And this will be the same for B. Thio. Ah, one will be equal to again the same. It'll be the same constant, actually, because this entire column Ah, this entire first column is only a combination of all of the other entire columns. So this constant seats, who will be the same in each in each expression of of these elements be Jay. I urge a one. So be to one we'd be equal to see two times be to to plus C three times be 23 and the same all the way up to C and B to end. And this will continue in this manner. And so we go all the way to the bottom two b one b m. One week will to see to e m too, plus C three b m three and all the way up to C and B m n. Okay, um, I'm gonna go to the next page here, but we're gonna remember these equations, Okay, because now we're going to use the fact that the columns of a times B are linear combinations of the columns of a using the corresponding element or ah, using this course wanted. Call him a B as the weights in the linear combination. So, for example, the vector a B one will be equal to, um, be one one times a one plus be to one times eight to the column. Vector eight to and so on. All the way up Until be Mm M one a. M. Okay, but now we're going to sub in what we found. Where the equations of these e entries. This calling these entries in the in the first column of B. So this is gonna be ah, very long. Some. So I'm gonna break up in the lines. Eso 1st 1 will be multiplied by the vector. A one. And it will be, um, see too. Be one too. Plus C three, I mean 13 right? All the way up until C n be one end. And this will be plus the column vector a two time see too be to to plus C three be 23 see and be to end Plus and then all the way until we get to a mm time. See, too e m too. Plus C three since b m three all the way up and so see and b o. And this is yet and it read e mn. Okay, So this is all one big sum you've just subbed in our equations from before and Now what I'm gonna do is I'm going to multiply in the column vectors. I'm going to distribute the call in vectors into this product here. I'll do that on the next page again. Pardon? Because this gets very long. So again, we're still looking at the first column of The Matrix a Times B and we'll distribute in this thes Colin vectors, so that will be equal to C two. Be one too a one plus c three b +13 a one plus that dot CNN Be one n times a one plus right. That's the first line, Uh, plus C two. Me, too, too. A two this time. Let's see. Three. Beat 23 still Times Column Vector eight to all the way up to see n to end a to again. And this will continue until we get to the last one, which is seats you b m 2 a.m. Let's see. Three be mm. Three a three. All the way to see. And that's B M and a M. Okay, so now what I'm gonna do is I'm gonna factor out these. See, I cz As you can see, I kind of drew them in a nice call him here intentionally. So when a factor those out and we'll see what we get left with. So again, I'll use yet another page to do this and the result of that well, look like this. So we see too times all of be one end a one plus B hoops or be one to not be one end. You write that. Sorry. Be one too of times a one plus B 22 times a two plus and so on until B and ah, yes, or B M too. So I was AM and the next line will be C three times all of B 13 times a one plus B 23 times a two plus all we have to be m three a. M. And this will continue. So we get all the way down to CNN Times, all of B one n a one plus B two en tons, eight to all the way up until be. Mm. And since a m, I don't know why I drew parentheses there. They're not necessary. Okay, But now if we take a look at this year, this piece in the brackets is exactly a B two and this in the rackets here is exactly maybe three. And so what we found is at the first column of a B one, right, Because this whole analysis has been on the first column of Baby One is equal to C two times, maybe two plus C three times a B three all the way up until C n. So I'm just a B N. And this is exactly the formula. Or this in this show is exactly that the first column of A is a linear combination of the first column of a B one or a baby is a linear combination of the other columns of a B. And so we've shown what we intended to show and we're done.

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