# Linear Algebra - Change of basis question

1. Feb 1, 2010

### zeion

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

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

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}

2. Relevant equations

3. 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 $$T_{B'A'}$$,
which is: $$T_{B'A'} = I_{B'B}T_{BA}I_{AA'}$$

So I need to find $$I_{AA'}$$ and $$I_{B'B}$$.

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

I get : $$I_{AA'} = \begin{bmatrix}1 & 0 & 1 & 0 \\0 & 0 & -1 &-1\\0 & 1 & 0 &1\\ 0&0&0&1 \end{bmatrix}$$

Then for $$I_{B'B}$$, I write B wrt B', and get

$$I_{B'B} = \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0\\-1 & 0 & 1 \end{bmatrix}$$

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

2. Feb 1, 2010

### vela

Staff Emeritus
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.

3. Feb 2, 2010

### zeion

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