# Matrix Representation of a Linear Map

1. Feb 18, 2007

### playboy

1. The problem statement, all variables and given/known data

Consider the linear map A : R3 ----> R3 given by
A(x1, x2, x3) = (x1 − x2,−x1 + x2, x3).
(a) Find the adjoint map $A^*$.

(b) Obtain the matrix representations of A and A* with respect to the canonical basis $f_1$ = [1, 2, 1], $f_2$ = [1, 3, 2], $f_3$ = [0, 1, 2].

2. Relevant equations

None!

3. The attempt at a solution

I first want to find the matrix representation of the linear map A.

so we have A(x1, x2, x3) = (x1 − x2,−x1 + x2, x3).

[1 -1 0]
[-1 1 0] = A
[ 0 0 1]

Is that correct?

If that is correct, I then find the adjoint by using the formula, and I get,

[1 1 0]
[1 1 0] = A*
[0 0 1]

Last edited by a moderator: Feb 18, 2007
2. Feb 18, 2007

### matt grime

Why do you need us to confirm the first answer? Is it beyond you to multiply A into a vector? I don't think it is - you're very much capable of doing that. What you've written doesn't look like A referred to the given basis anyway, but to the standard basis (what makes the f_i in anyway canonical?)

3. Feb 18, 2007

### playboy

I feel so stupid matt grime, yeah, that was very simple.

I just really confused myself and made something straight forward complicated.

I am now focusing on part (b)

I have no idea how $f_1$ = [1, 2, 1], $f_2$ = [1, 3, 2], $f_3$ = [0, 1, 2] are connonical...

the cannonical basis for R3 is { [1, 0, 0] [0, 1, 0] [0, 0, 1] }

Anyways, here is how I went about finding the matrix representation for A:

It consists of 2 parts

Part 1:

A(1 0 0) = [1 -1 0]
A(0 1 0) = [-1 1 0]
A(0 0 1) = [0 0 1]

Hence, we have a matrix:

[1 -1 0]
[-1 1 0]
[0 0 1]

which is the same matrix as I have in post 1 :shy:

now,

we find the matrix from the f_i basis to the standard basis above.

[1 1 0]
[2 3 1]
[1 2 2]

and now our matrix representation is:

[1 -1 0] [1 1 0]
[-1 1 0]x[2 3 1]
[ 0 0 1] [1 2 2]

[-1 -2 -1]
[1 2 1]
[1 2 2]

is the matirx representation of the map A.

Is this correct? Could somebody please check this for me?
I am allways unsure of my work..

4. Feb 18, 2007

### AKG

What is the definition of adjoint of A? What is the definition of matrix representation of A with respect to basis {f1, f2, f3}?

5. Feb 20, 2007

### playboy

The adjoint of a linear map A:V--->W is the unique map A*:W--->V satisying <A*w, v> = <w, Av>

in my text book, it says;

"If V=R^n, then the adjoint of a linear map V = R^n to W = R^m with respect to the dot product is obtained by taking the transpose of its corresponding matrix, as in the identity"

So in my case above, the corresponding matrix is:

[1 -1 0]
[-1 1 0] = A
[0 0 1]

And A transpose is A,

I am completely lost on matrix representation of A with respect to basis {f1, f2, f3}

6. Feb 20, 2007

### AKG

{f1,f2,f3} is a basis. Any vector in your vector space, v, can be expressed in a unique way as a linear combination of these vectors, i.e.:

v = a1f1 + a2f2 + a3f3

where a1, a2, and a3 are scalars. So the (column) matrix representation of v with respect to basis {f1,f2,f3} is (a1 a2 a3)T. Let's call this column matrix v'. Now if you apply the transformation A to the vector v, you get another vector Av. Since this too is a vector in your vec. space, it can be represented in a unique way as a linear combination of the fi, i.e. Av = b1f1 + b2f2 + b3f3 for some unique scalars bi. So the (column) matrix representation of Av with respect to this basis is (b1 b2 b3)T. Let's call this column matrix (Av)'. The matrix representation of transformation A, then, with respect to this basis is whatever matrix A' performs the action:

A'v' = (Av)'

for any v in V. So now you have the definition of the matrix rep. of A w.r.t. a basis, but you need to know how to find this matrix representation. Think about it for yourself for a bit, and if you still have trouble, then ask.

7. Feb 21, 2007

### playboy

Thank you for your help AKG and explaing to me the definition.

It is hard for me to answer the question working straight from the definition - i need a concrete example.

I looked in books and online for a question like this - I could not find one, not even in my text book.

Would you (or anyboy) care to help me with the alogritm or calculation?

8. Feb 21, 2007

### AKG

A(f1) = A(1,2,1) = (1-2,-1+2,1) = (-1,1,1)

Now express (-1,1,1) as a linear combination of f1, f2, and f3. You want to find c1, c2, c3 scalars such that:

c1f1 + c2f2 + c3f3 = (-1,1,1)

Let F denote the 3x3 matrix that has it's first row equal to f1, second row f2, and third row f3. So the first row will be 1 2 1, the second is 1 3 2, the third is 0 1 2. Then

(c1 c2 c3)F = (-1 1 1)
(c1 c2 c3) = (-1 1 1)F-1

So you need to find F-1. So now you'll have c1, c2, c3 that correspond to the representation of A(f1) as a linear comb. of f1, f2, f3. Similarly, find d1, d2, d3 that corrsepond to the rep. of A(f2) as a lin. comb. of the fi, and find e1, e2, e3 that correspond to the rep. of A(f3) as a lin. comb. of the fi.

Then the matrix which I originally called A' will be (c1 c2 c3)T as its first column, (d1 d2 d3)T as its second, and (e1 e2 e3)T as its third. I'll explain more details later if you need it. Try this process out, and see if you understand why it's right too.

9. Feb 22, 2007

### playboy

Thank you AKG for showing me a worked example, I really appreciate that.

I checked back to see how you used the definition to come up with this, and I do see how it is linked to the definition.

I had another way of calulating the constands c1 c2 c3, d1 d2 d3, e1 e2 e3, it was similar to your way. both gave the same answers.

the matrix turns out to be:

[-5 -10 -4]
[4 8 3]
[1 2 0]

Now I have this matrix, but i am uncertain on what it actually does? I mean, it does represent the linear transformation with respect to the fi bases... i am missing the entire idea of what it does.

Skipping ahead, A(x1, x2, x3) = (x1 − x2,−x1 + x2, x3) is a self adjoint linear map.

Therefore, A=A*

Thus,

A*(x1, x2, x3) = (x1 − x2,−x1 + x2, x3)

So wouldn't

[-5 -10 -4]
[4 8 3]
[1 2 0]

be the matrix representation of A* with respect to the fi basis? common sense tells me it does.