In the last video, we started

with a linear transformation S, that was a mapping between

the set x, that was a subset of Rn to the set y. And then we had another

transformation that was a mapping from the set

y to the set z. And we asked ourselves,

given these two linear transformations, could

we construct a linear transformation that goes all

the way from x to z? What we did was we made

a definition. We said let’s create something

called the composition of T with S. What that is, is first you apply

S to some vector in X to get some vector in Y. And that’s your vector

right there. And then you apply T to

that, to get to z. And so we defined it that way. And our next question was, was

that a linear transformation? We show that it was. It met the two requirements

for them. And because it is a linear

transformation, I left off in the last video saying that

it should be able to be represented by some matrix

vector product. Where this will have to

be an l by n matrix. Because it’s a mapping from an n

dimensional space, which was x– it was a subset of Rn–

to an l dimensional space. Because z is a subset of Rl Now

in this video, let’s try to actually construct

this matrix. So at the beginning of the last

video, I told you that T of x could be written as some

matrix product, B times x. Let me write that and rewrite

it down here. So I told you that the linear

transformation T applied to some vector x, could be written

as the matrix vector product, B times a vector x. And since it was a mapping from

an m dimensional space to an l dimensional space, we know

this is going to be in l by m matrix. Now similarly, I told you that

the transformation S can also be written as a matrix

vector product. Where we can say A is its

matrix representation times a vector x. And since S was a mapping from

an n dimensional space to an m dimensional space, this will

be an m by n matrix. Now by definition, what was the

composition of T with S? What is this? By definition, we said that this

is equal to– you first apply the linear transformation

S to x. And I’ll arbitrarily

switch colors. So you first apply the

transformation S to x. And that essentially gets you

a vector right there. This is just a vector in Rm. Or it’s really a vector in y,

which is a subset of Rm. And then you apply the

transformation T to that vector to get you into z. Given this we can use our matrix

representations to replace this kind of

transformation representation. Although they’re really

the same thing. What is a transformation

of S applied to x? Well this right here is just

A times x, where this is an m by n matrix. So we can say that this is equal

to the transformation applied to A times x. Now, what is the T

transformation applied to any vector x? Well that’s the matrix B

times your vector x. So this thing right here is

going to be equal to B times whatever I put in there. So the matrix B times the matrix

A times the vector x right there. This is what our composition

transformation is. The composition of T with S

applied to the vector x. Which takes us from the set x

all the way to the set z is this, if we use the matrix

forms of the two transformations. Now at the end of last video I

said I wanted to find just some matrix that if I were to

multiply times this vector, that is equivalent to

this transformation. And I know that I can

find this matrix. I know that this exists because

this is a linear transformation. So how can we do that? Well, we just do what we’ve

always done in the past. We start with the identity matrix,

and we apply the transformation to every column

of the identity matrix. And then you end up with your

matrix representation of the transformation itself. So first of all, how

big is the identity matrix going to be? Well, these guys that we’re

inputting into our transformation, they are subsets

of x, or they’re members of x, which is

an n dimensional space, a subset of Rn. So all we do to figure out C

is we start off with the identity matrix. The n dimensional identity

matrix, because our domain is Rn. And of course we know what

that looks like. We have 1, 0 all the way down. It’s going to be an n by n

matrix, and then 0, 1 all the way down 0’s. These 0’s right here, and then

you have 1’s go all the way down the columns and everything

else is 0. We’ve seen this multiple times

that’s what your identity matrix looks like, just 1’s down

the column from the top left to the bottom right. Now to figure out C, the matrix

representation of our transformation, all we do is we

apply the transformation to each of these columns. So we can write that our matrix

C is equal to the transformation applied

to this first column. What is the transformation? It is the matrix B times the

matrix A times whatever you’re taking the transformation of. In this case we’re taking the

transformation of that. We’re taking the transformation

of 1, 0, 0 all the way down. There’s 1 followed by

a bunch of 0’s. That’s going to be our

first column of C. Our second column of C is going

to be B times A times the second column of our

identity matrix. And, of course, you remember

these are each the standard basis vectors for Rn. So this is going to be times

E2, which is a 0, 1, 0 all the way down. And then we’re going to keep

doing that until we do get to the last column, which is B

times A times a bunch of 0’s all the way down

until get a 1. The nth term is just

a 1 right there. Now what is this going

to be equal to? It looks fairly complicated

right now. But all you have to do is make

the realization– and we’ve seen this multiple times. If we write our vector A or we

write our matrix A as just a bunch of column vectors. So this is a column vector A1,

A2, all the way to An. We already learned that this

was and n buy m matrix. Then what is the vector A times,

for example, x1, x2 to all the way down to xn. We’ve seen this multiple

times. This is the equivalent to x1

times A1 plus x2 times A2, all the way to plus xn times An. We’ve seen this multiple

times. It’s a linear combination of

these column vectors where the waiting factors are the terms

in our vector that we’re taking the product of. So given that, what is this

guy going to reduce to? This is going to be A1 times

this first entry right here, times x1, plus A2 times a second

entry, plus A3 times a third entry. But all these other

entries are 0. The x2’s all the way

to the xn are 0’s. So you’re only going to end

up with 1 times the first column here in A. So this will reduce to–

let me write this. So the first column is going

to be B times– now A times this E1 vector, I guess we could

call it, right there is just going to be 1 times the

first column in A plus 0 times the second column in A plus

0 times the third column. So it’s just 1 times the

first column in A. So it’s just A1. That simple. Now what is this one going

to be equal to? It’s going to be 0 times the

first column in A, plus 1 times a second column in A, plus

0 times a third column in A, and the rest are

going to be 0’s. So it’s just going to be 1 times

the second column in A. So the second column in our

transformation matrix is just going to be B times A2. And I think you get

the idea here. The next one is going to be B

times A3 and all the way until you get B times An. So that’s how you would

solve for your transformation matrix. Remember what we were

trying to do. We were trying to find some–

let me write down and summarize everything that

we’ve done so far. We had a mapping S, that was

a mapping from x to y. But x was a subset of Rn. Y was a subset of Rm. And so we said that this linear

transformation could be represented as some matrix A

where A is an m by n matrix times a vector x. Then I showed you another

transformation, we already called it T, which was a

mapping from y to z. z is a subset of Rl. And of course, the

transformation T applied to some vector in y, can be

represented as some matrix B times that vector. I shouldn’t have drawn

parentheses there, but you get the idea. And this, since it’s a mapping

from a subset of our Rm to Rl, this will be an l by m matrix. And then we said, look, if

we actually just take the composition of T with S,

of some vector in x, this reduced to B. So first we applied the

S transformation. We multiplied the matrix

A times x. And then we applied the T

transformation to this. So we just multiplied

B times that. Now we know this is a linear

transformation, which means it can be represented as a

matrix vector product. And we just figured out what the

matrix vector product is. So this thing is going to be

equal to C times x, which is equal to this thing right there,

which is equal to that thing right there. Which is equal to– let me write

it this way– B, A1, where A1 is the first column

vector in our matrix A. And then the second column

here is going to be B. And then we have A2, where

this is the second column vector in A. And you can keep going all the

way until you have B times An times x, of course. Now this is fair enough. We can always do this if you

give me some matrix. Remember this is an

l by m matrix. And you give me another matrix

right here that is an m by n matrix, I can always do this. And how do I know I can

I always do that? Because each of these

A’s are going to have m entries, right? They’re going to be Ai. All of them are going to

be members of our Rm. So this is well-defined. This has m columns. This has m entries. So each of these matrix vector

products are well-defined. Now, this is an interesting

thing, because we were able to figure out the actual matrix

representation of this composition transformation. Let’s extend it a little

bit further. Wouldn’t it be nice if this

were the same thing as the matrices B times A. All of that times x. Wouldn’t it be nice if these

were the same thing? Because then we could say that

the composition of T with S of x is equal to the matrix

representation of B times a matrix representation of S. And you take the product

of those two. And that will create a new

matrix representation which you can call C. That you can then multiply

times x. So you won’t have to do it

individually every time, or do it this way. And I guess the truth of the

matter is there is nothing to stop us from defining this

to be equal to B times A. We have not defined what a

matrix times a matrix is yet. So we might as well. This is a good enough motivation

for us to define it in this way. So let’s throw in

this definition. So if we have some matrix B. B is an l by m matrix. And then we have some other

matrix A– and I’ll actually show what A looks like, where

these are its column vectors. A1, A2, all the way to An. We’re going to define

the product. So this is a definition. We’re going to define the

product BA as being equal to the matrix B times each of

the column vectors of A. So it’s B times A1. That’s going to be

its first column. This is going to

be B times A2. All the way to B times An. And you’ve seen this before in

algebra two, but the reason why I went through almost two

videos to get to here, is to show you the motivation for

why matrix products are defined this way. Because it makes the notion

of compositions of transformations kind

of natural. If you take the composition of

one linear transformation with another, the resulting

transformation matrix is just the product, as we’ve just

defined it, of their two transformation matrices. For those of you who might not

have a lot of experience taking products of matrices, and

who think this is fairly abstract to look at, in the next

video I’ll actually do a bunch of examples and show you

that this definition is actually fairly straightforward.

makes me want to kill myself aaaaaaaaaaaaaa

Thanks so much Sal!!! This video directly correlates to what my assignments are on. Tytyty!

Thank you for all of the linear Algebra videos!!!!!

Thanks a lot for your effort Sal. Greetings from Sweden!

Where did you find this??? Wow 🙂

excellent

there is small mistake at 7:46 A dimension should be (m x n )

it confused me at the beginning thanks a lot Sal .

7 comments with around 50k views. Seems like someone's been deleting comments.

thanks for excellent tutorials. 13:28 i think it should be "Rm" not "Rn"

I love you man