# Inverse of Linear Transformation - Two Non-Standard Bases

• Oct 12th 2011, 10:20 AM
divinelogos
Inverse of Linear Transformation - Two Non-Standard Bases
Let V=S2x2 (2x2 real symmetric matrices) with basis (1,1;1,0) (0,1;1,0) (1,0;0,1) and let W=P2(R) with basis {x^2,x+1,1-x}.

Let the images of the basis vectors of V be 2x-2, x^2+x+1, and 3x^2+2 respectively. Find T^-1(f).

Can someone explain this? I can do it in standard bases but this gives me trouble. Thanks!
• Oct 12th 2011, 12:14 PM
Deveno
Re: Inverse of Linear Transformation - Two Non-Standard Bases
what we need to do, is find the matrix for T with respect to these two bases, and then find the inverse of that matrix.

let's give the bases names, to keep them straight. so B =

$\left\{\begin{bmatrix}1&1\\1&0 \end{bmatrix},\begin{bmatrix}0&1\\1&0 \end{bmatrix},\begin{bmatrix}1&0\\0&1 \end{bmatrix}\right\}$

and C = $\{x^2,x+1,1-x\}$.

so to find the matrix for T, we're looking for a 3x3 matrix. now we know the columns of T will be the C-coordinate images of the B basis vectors (in B-coordinates the elements of B are just (1,0,0), (0,1,0) and (0,0,1)).

the trouble is, we aren't given the images of the B-basis vectors in C-coordinates, but instead in the standard basis of P2. now, this might seem "backwards" but what we are going to do, is find the matrix P that changes C-coordinates, into "standard coordinates", because this is very easy. i am going to use {1,x,x^2} as the standard basis, if your text uses {x^2,x,1} you'll have to reverse the order of some things.

P([1,0,0]C) = x^2 = 0(1) + 0(x) + 1(x^2) so the first column of P is (0,0,1)^T.
P([0,1,0]C) = x+1 = 1(1) + 1(x) + 0(x^2), so the 2nd column of P is (1,1,0)^T.
P([0,0,1]C) = 1-x = 1(1) + (-1)(x) + 0(x^2), so the 3rd column of P is (1,-1,0)^T.

so P (the "change of basis matrix" from C-->standard) is

$\begin{bmatrix}0&1&1\\0&1&-1\\1&0&0\end{bmatrix}$

but what we want to do is go from standard--->C, so we need to find $P^{-1}$. verify that $P^{-1}$ is indeed:

$\begin{bmatrix}0&0&1\\ \frac{1}{2}&\frac{1}{2}&0\\ \frac{1}{2}&-\frac{1}{2}&0 \end{bmatrix}$

so the C-coordinates of 2x-2, which is (-2,2,0) in the standard basis, are: [0,0,-2]C,
the C-coordinates of x^2+x+1, which is (1,1,1) in the standard basis, are: [1,1,0]C
the C-coordinates of 3x^2 which is (0,0,3) in the standard basis, are: [3,0,0]C

(you could have just figured these out directly, but i wanted to show you "how it's done in general").

so our matrix for T, relative to the bases B and C is:

$\begin{bmatrix}0&1&3\\0&1&0\\-2&0&0\end{bmatrix}$

now that we have a matrix for T, we need to find a matrix for $T^{-1}$.

verify that the matrix for $T^{-1}$ is:

$\begin{bmatrix}0&0&-\frac{1}{2}\\0&1&0\\ \frac{1}{3}&-\frac{1}{3}&0\end{bmatrix}$

so now we have a matrix for $T^{-1}$ relative to C and B. the thing is, our input will probably be in "standard form", so to compute explicitly what $T^{-1}$ is (and not just the matrix for it), we need to first apply $P^{-1}$ to p(x), or what is equivalent, to find the images of the basis elements 1,x, and x^2 under $P^{-1}$ and THEN apply the matrix for $T^{-1}$ to those.

the C-coordinates of 1 = 1 + 0x + 0x^2 = (1,0,0) are [0,1/2,1/2]C
the C-coordinates of x = 0 + 1x + 0x^2 = (0,1,0) are [0,1/2,-1/2]C
the C-coordinates of x^2 = 0 + 0x + 1x^2 = (0,0,1) are [1,0,0]C

applying our matrix for $T^{-1}$ to these C-coordinate vectors, we get:

1-->[-1/4,1/2,-1/6]B
x-->[1/4,1/2,-1/6]B
x^2-->[0,0,1/3]B

almost done....we still want to express these B-coordinates as symmetric matrices. do so, and express f(x) as a + bx + cx^2, to find T^-1(f) as a 2x2 symmetric matrix. check your answer by verifying that T^-1(2x-2) gives you the first element of B, and so on.
• Oct 12th 2011, 01:52 PM
divinelogos
Re: Inverse of Linear Transformation - Two Non-Standard Bases
I actually got a different matrix for T, so let's make sure that part is right before continuing. I setup the system:

c1x^2+c2(x+1)+c3(1-x) yields

c1x^2+(c2-c3)x+(c2+c3)= [images]

so in matrix form we have:

$\begin{bmatrix}1&0&0&-2&1&2\\0&1&-1&2&1&0\\0&1&1&0&1&3\end{bmatrix}$

and turning that into RREF we get:

$\begin{bmatrix}1&0&0&-2&1&2\\0&1&0&1&1&3/2\\0&0&1&-1&0&3/2\end{bmatrix}$

So the matrix rep. of T is (or at least I think it should be :)):

$\begin{bmatrix}-2&1&2\\1&1&3/2\\-1&0&3/2\end{bmatrix}$

Then the inverse matrix rep. of T would just be the inverse of that, correct?

Is this the correct matrix rep. of T?
• Oct 12th 2011, 02:29 PM
Deveno
Re: Inverse of Linear Transformation - Two Non-Standard Bases
as i pointed out in my previous post, if you use {x^2,x,1} as a basis instead of {1,x,x^2} you will wind up with a different "change of basis matrix" (which is P, by the way, not T).

"your P" is my P "upside-down" (at least for the first 3 columns). however, what you are doing with the last 3 columns is a mystery to me, what are those?
• Oct 12th 2011, 03:53 PM
divinelogos
Re: Inverse of Linear Transformation - Two Non-Standard Bases
Quote:

Originally Posted by Deveno
as i pointed out in my previous post, if you use {x^2,x,1} as a basis instead of {1,x,x^2} you will wind up with a different "change of basis matrix" (which is P, by the way, not T).

"your P" is my P "upside-down" (at least for the first 3 columns). however, what you are doing with the last 3 columns is a mystery to me, what are those?

I think once I explain this it will show you where I am misunderstanding the problem, so here goes my explanation of those "mystery" columns :)

In general, to get the matrix representation of T under 2 different non-standard basis, you do the following:

1. Calculate the images of the first basis by passing them through the given transformation. In our case, this was given in the problem statement (ie, T((1,1;1,0))=2x-2 T((0,1;1,0))=x^2+x+1, and T((1,0;0,1))=3x^2+2)

2. Now, we need to write the images (ie, 2x-2,x^2+x+1, and 3x^2+2) with respect to the given basis.

3. To do this (and this might be where I need help), we need to write the images as a linear combination of the given basis vectors for that space, ie:

c1x^2+c2(x+1)+c3(1-x), which, after distributing and re-organizing, gives:

c1x^2+(c2-c3)x+(c2+c3).

Setting above equation equal to the first image gives:

c1x^2+(c2-c3)x+(c2+c3)=2x-2

which can be written as the system (here, i also added the other 2 images as columns, so we can solve all of them at the same time)

1 0 0 -2 1 2
0 1 -1 2 1 0
0 1 1 0 1 3

Now, reducing this to RREF gives a 6x6 matrix, the left 3x3 part is the identity matrix, and the right 3x3 part is, well, what I thought was the final matrix representation of T, but apparently this is not the case.

Can you tell me where I'm going wrong?
• Oct 12th 2011, 05:35 PM
Deveno
Re: Inverse of Linear Transformation - Two Non-Standard Bases
Quote:

Originally Posted by divinelogos
I think once I explain this it will show you where I am misunderstanding the problem, so here goes my explanation of those "mystery" columns :)

In general, to get the matrix representation of T under 2 different non-standard basis, you do the following:

1. Calculate the images of the first basis by passing them through the given transformation. In our case, this was given in the problem statement (ie, T((1,1;1,0))=2x-2 T((0,1;1,0))=x^2+x+1, and T((1,0;0,1))=3x^2+2)

so far, so good

Quote:

2. Now, we need to write the images (ie, 2x-2,x^2+x+1, and 3x^2+2) with respect to the given basis.
i agree here, as well. i called the "given" basis, C.

Quote:

3. To do this (and this might be where I need help), we need to write the images as a linear combination of the given basis vectors for that space, ie:

c1x^2+c2(x+1)+c3(1-x), which, after distributing and re-organizing, gives:

c1x^2+(c2-c3)x+(c2+c3).

Setting above equation equal to the first image gives:

c1x^2+(c2-c3)x+(c2+c3)=2x-2
if you are solving this for the c's, this is:

1c1 + 0c2 + 0c3 = 0 <---x^2 term
0c1 + 1c2 - 1c3 = 2 <----x term
0c1 + 1c2 + 1c3 = -2 <--- constant term

Quote:

which can be written as the system (here, i also added the other 2 images as columns, so we can solve all of them at the same time)

1 0 0 -2 1 2
0 1 -1 2 1 0
0 1 1 0 1 3

Now, reducing this to RREF gives a 6x6 matrix, the left 3x3 part is the identity matrix, and the right 3x3 part is, well, what I thought was the final matrix representation of T, but apparently this is not the case.

Can you tell me where I'm going wrong?
i think i see what you are doing, but you're not matching up the terms correctly.

so the matrix you should be bringing to RREF is:

$\begin{bmatrix}1&0&0&0&1&3\\0&1&-1&2&1&0\\0&1&1&-2&1&0\end{bmatrix}$

now, here, i have a question: your original post lists the image of the 3rd 2x2 matrix basis vector as 3x^2, and yet in your more recent posts it is 3x^2 + 2. which is it?

for the matrix listed above (using 3x^2), the RREF is:

$\begin{bmatrix}1&0&0&0&1&3\\0&1&0&0&1&0\\0&0&1&-2&0&0\end{bmatrix}$

(you can check that HERE)

which leads to the same matrix i derived earlier (the 3x3 part on the right).

so...check on what the 3rd basis image is, and then do another RREF with the proper 3x6 matrix (if necessary), and get back to me.
• Oct 13th 2011, 10:30 AM
divinelogos
Re: Inverse of Linear Transformation - Two Non-Standard Bases
The third image should be 3x^2+2 (I fixed it in the original post).

I'm at work right now but I can get back to you on my lunch break :)

Thanks for the help man.
• Oct 13th 2011, 12:17 PM
divinelogos
Re: Inverse of Linear Transformation - Two Non-Standard Bases
Here's what I have after the corrections:

New matrix representation of T:

0 1 3
0 1 1
-2 0 1

New inverse of T (ie, T^-1):

1/4 -1/4 -1/2
-1/2 3/2 0
1/2 -1/2 0

So, we have

T^-1(ax^2+bx+c)= [T]^-1 * [a b c] (< this is a column vector)

Now, I know it isn't [a b c], and this is where I think I'm getting stuck. Is it correct that we have to re-write [a b c] as a linear combo of the given basis for that space? For example, we'd solve:

c1x^2+(c2-c3)x+c2+c3)=ax^2+bx+c for c1,c2, and c3? So we'd have:

c1=a
c2=(b+c)/2
c3=(c-b)/2

then,

T^-1(ax^2+bx+c)= [T]^-1 * [a (b+c)/2 (c-b)/2] (this is a column vector)

Now, I won't write it out, but this yields a 3-component column vector.

Do I use these three components to create the symmetric matrix? If so, how do I know which component is the repeated one on the diagonal?

Thanks again deveno for the help! I have a feeling we almost have this one defeated :)
• Oct 14th 2011, 06:07 AM
Deveno
Re: Inverse of Linear Transformation - Two Non-Standard Bases
sorry for the late reply...this particular problem is quite calculation-intensive. i have checked your rref, and the inverse matrix of T, and they look good.

remember, a matrix is dependent on our choice of basis. so the "input" for our matrix is C-coordinate polynomials,

and the "output" is B-coordinate vectors (which yield 2x2 matrices). so we can just read off the B-coordinates of the C-basis vectors,

they are just the columns of our matrix for T^-1. that is:

T^-1(x^2) = [1/4,-1/2,1/2]B
T^-1(x+1) = [-1/4,3/2,-1/2]B
T^-1(1-x) = [-1/2,0,0]B

but we can't just take as "input" the polynomial ax^2 + bx + c as (a,b,c), because those are not C-coordinates. what we need to do is find a',b',c', where:

ax^2 + bx + c = a'x^2 + b'(x+1) + c'(1-x). equating terms, we have:

ax^2 = a'x^2 <---x^2 term (so a' = a)
bx = (b'-c')x <---x term
c = b'+c' <---constant term.

it's not hard to see that b' = (b+c)/2, and c' = (c-b)/2

so the C-coordinates of ax^2+bx+c are [a,(b+c)/2,(c-b)/2]C.

now, if we've done everything right up to this point, T^-1(2x-2) should be:

[1 1]
[1 0] = [1,0,0]B

first: convert 2x-2 = (0,2,-2) to C-coordinates: (0,2,-2) = [0,0,-2]C.

apply our matrix for T^-1 to the column vector [0,0,-2]:

$\begin{bmatrix}\frac{1}{4}&-\frac{1}{4}&-\frac{1}{2}\\-\frac{1}{2}&\frac{3}{2}&0\\\frac{1}{2}&-\frac{1}{2}&0\end{bmatrix}\begin{bmatrix}0\\0\\-2\end{bmatrix}= \begin{bmatrix}1\\0\\0\end{bmatrix}$

and [1,0,0]B is indeed our first basis matrix in B. you can verify that we get [0,1,0]B as the image of x^2+x+1 in C-coordinates, and [0,0,1]B as the image of 3x^2+2 in C-coordinates.

now, ax^2+bx+c in C-coordinates is [a,(b+c)/2,(c-b)/2]C, so multiplying the (column) vector (a,(b+c)/2,(c-b)/2) by our matrix for T^-1, i get (and you should verify this, it's messy):

T^-1(ax^2+bx+c) = [(2a+b-3c)/8, (-2a+3b+3c)/4, (2a-b-c)/4]C

$=\begin{bmatrix}\dfrac{6a-b-5c}{8}&\dfrac{-2a+7b+3c}{8}\\ \dfrac{-2a+7b+3c}{8}&\dfrac{2a-b-c}{4}\end{bmatrix}$

and that (i believe) officially makes this the matrix problem from hell :)