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}<br /> 1&amp;0&amp;1 \\<br /> 0&amp;1&amp;0 \\<br /> 1&amp;0&amp;-1 \end{array} \right|

I have the transformation relative to the basis:

\left| \begin{array}{ccc}<br /> 3&amp;0&amp;0 \\<br /> 0&amp;-1&amp;0 \\<br /> 0&amp;0&amp;2 \end{array} \right|

and the inverse of the basis matrix,

\left| \begin{array}{ccc}<br /> 1/2&amp;0&amp;1/2 \\<br /> 0&amp;1&amp;0 \\<br /> 1/2&amp;0&amp;-1/2 \end{array} \right|

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

\left| \begin{array}{ccc}<br /> 1/2&amp;0&amp;1/2 \\<br /> 0&amp;1&amp;0 \\<br /> 1/2&amp;0&amp;-1/2 \end{array} \right|\left| \begin{array}{ccc}<br /> 3&amp;0&amp;0 \\<br /> 0&amp;-1&amp;0 \\<br /> 0&amp;0&amp;2 \end{array} \right|\left| \begin{array}{ccc}<br /> 1&amp;0&amp;1 \\<br /> 0&amp;1&amp;0 \\<br /> 1&amp;0&amp;-1 \end{array} \right| = \left| \begin{array}{ccc}<br /> 5/2&amp;0&amp;1/2 \\<br /> 0&amp;-1&amp;0 \\<br /> 1/2&amp;0&amp;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.