Linear Algebra - Change of basis question

[zeion]

Homework Statement

Let A = E4 in R4 (standard basis) and B = {x^2, x, 1} in P2 over R. If T is the linear transformation that is represented by
[tex]

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

[/tex]

relative to A and B, find the matrix that represents T with respect to A' and B' where
A' = {(1,0,0,0), (0,0,1,0), (1,-1,0,0), (0,-1,1,1)}
B' = {x^2 + 1, x, 1}

Homework Equations

The Attempt at a Solution

So by looking at this matrix T, it's clear that its a transformation from A to B, so we want the transformation matrix [tex]T_{B'A'}[/tex],
which is: [tex]T_{B'A'} = I_{B'B}T_{BA}I_{AA'}[/tex]

So I need to find [tex]I_{AA'}[/tex] and [tex] I_{B'B}[/tex].

For [tex]I_{AA'}[/tex], I write A' wrt A(which is standard basis of R4):

I get : [tex]I_{AA'} = \begin{bmatrix}1 & 0 & 1 & 0 \\0 & 0 & -1 &-1\\0 & 1 & 0 &1\\ 0&0&0&1 \end{bmatrix} [/tex]

Then for [tex] I_{B'B}[/tex], I write B wrt B', and get

[tex]I_{B'B} = \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0\\-1 & 0 & 1 \end{bmatrix} [/tex]

Now I put them together to get something with lots of zeros.. which doesn't seem right?

 

[vela]

Your answer is correct. (At least it sounds correct.) You have the representations of the A' basis in the natural basis, so try applying them to the given T and see what you get. Then look at how those results would be represented in the B' basis. You'll see why the transformed T has so many zeros.

 

[zeion]

When I put them all together I get:
[tex]
\begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0\\-1 & 0 & 1 \end{bmatrix} \begin{bmatrix}1 & 1 & 0 & 1\\0 & 0 & 1 & -1\\1 & 1 & 0 & 1 \end{bmatrix} \begin{bmatrix}1 & 0 & 1 & 0 \\0 & 0 & -1 &-1\\0 & 1 & 0 &1\\ 0&0&0&1 \end{bmatrix} = \begin{bmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 0 & 0 \end{bmatrix}
[/tex] ??