Is Change of Basis the Solution to My Linear Algebra Problem?

[TranscendArcu]

Homework Statement

The Attempt at a Solution

So first I thought to myself that the proper way of doing this problem was to construct each of the standard basis vectors as a linear combination of the basis given us. I have,

$T(1,0,0) = \frac{1}{2} T(1,0,1) + \frac{1}{2} T(1,0,-1) = (\frac{3}{2},0,1)$
$T(0,1,0) = (0,-1,0)$
$T(0,0,1) = \frac{1}{2}T(1,0,1) - \frac{1}{2} T(1,0,-1) = (\frac{3}{2},0,-1)$

Thus I have the matrix with columns that are the images of the standard basis vectors.

However, it seems conceivable that I should also be able to do this problem via change of basis. I have, taking the bases in matrix form,

$\left| \begin{array}{ccc} 1&0&1 \\ 0&1&0 \\ 1&0&-1 \end{array} \right|$

I have the transformation relative to the basis:

$\left| \begin{array}{ccc} 3&0&0 \\ 0&-1&0 \\ 0&0&2 \end{array} \right|$

and the inverse of the basis matrix,

$\left| \begin{array}{ccc} 1/2&0&1/2 \\ 0&1&0 \\ 1/2&0&-1/2 \end{array} \right|$

If I compose the matrices to find the matrix composition, I have

$\left| \begin{array}{ccc} 1/2&0&1/2 \\ 0&1&0 \\ 1/2&0&-1/2 \end{array} \right|$$\left| \begin{array}{ccc} 3&0&0 \\ 0&-1&0 \\ 0&0&2 \end{array} \right|$$\left| \begin{array}{ccc} 1&0&1 \\ 0&1&0 \\ 1&0&-1 \end{array} \right|$ = $\left| \begin{array}{ccc} 5/2&0&1/2 \\ 0&-1&0 \\ 1/2&0&5/2 \end{array} \right|$

But clearly, my results for the two matrices are not identical, so what have I done wrong?

 

[vela]

Your results for T(1,0,0) and the rest of the basis vectors are still expressed in terms of the given basis. In other words, the coordinate vector (3/2, 0, 1) corresponds to 3/2(1,0,1) + 1(1, 0,-1) = (5/2, 0, 1/2) with respect to the canonical basis.