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
<br /> <br /> \begin{bmatrix}1 &amp; 1 &amp; 0 &amp; 1\\0 &amp; 0 &amp; 1 &amp; -1\\1 &amp; 1 &amp; 0 &amp; 1 \end{bmatrix}<br /> <br />

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 T_{B&#039;A&#039;},
which is: T_{B&#039;A&#039;} = I_{B&#039;B}T_{BA}I_{AA&#039;}

So I need to find I_{AA&#039;} and I_{B&#039;B}.

For I_{AA&#039;}, I write A' wrt A(which is standard basis of R4):

I get : I_{AA&#039;} = \begin{bmatrix}1 &amp; 0 &amp; 1 &amp; 0 \\0 &amp; 0 &amp; -1 &amp;-1\\0 &amp; 1 &amp; 0 &amp;1\\ 0&amp;0&amp;0&amp;1 \end{bmatrix}

Then for I_{B&#039;B}, I write B wrt B', and get

I_{B&#039;B} = \begin{bmatrix}1 &amp; 0 &amp; 0 \\0 &amp; 1 &amp; 0\\-1 &amp; 0 &amp; 1 \end{bmatrix}

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:
<br /> \begin{bmatrix}1 &amp; 0 &amp; 0 \\0 &amp; 1 &amp; 0\\-1 &amp; 0 &amp; 1 \end{bmatrix} \begin{bmatrix}1 &amp; 1 &amp; 0 &amp; 1\\0 &amp; 0 &amp; 1 &amp; -1\\1 &amp; 1 &amp; 0 &amp; 1 \end{bmatrix} \begin{bmatrix}1 &amp; 0 &amp; 1 &amp; 0 \\0 &amp; 0 &amp; -1 &amp;-1\\0 &amp; 1 &amp; 0 &amp;1\\ 0&amp;0&amp;0&amp;1 \end{bmatrix} = \begin{bmatrix}1 &amp; 0 &amp; 0 &amp; 0\\0 &amp; 1 &amp; 0 &amp; 0\\0 &amp; 0 &amp; 0 &amp; 0 \end{bmatrix}<br /> ??