Problem With Change of Basis

1. Mar 11, 2012

TranscendArcu

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

3. 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?

2. Mar 11, 2012

vela

Staff Emeritus
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.