Find T(V) by using the standard matrix and the matrix relative to B and B'
$T: R^3 -> R^2, T(x,y,z) = (x-y + 0, 0 + y-z), v = (2,4,6)$
$B = {(1,1,1), (1,1,0), (0,1,1)}$
$B' = {(1,1),(2,1)}$
Standard Matrix Way
A = $\begin{bmatrix}1&-1&0 \\ 0&1&-1\end{bmatrix}$
$T(v)$ = $Av$ = $\begin{bmatrix}1&-1&0 \\ 0&1&-1\end{bmatrix} \begin{bmatrix}2 \\ 4 \\ 6\end{bmatrix}$
= $\begin{bmatrix}-2 \\ -2 \end{bmatrix}$
Matrix relative to B and B' Matrix
$T(v_1) = T(1,1,1) = (1, -1, 0) + (0, 1, -1) = (1,0, -1)$
$T(v_2) = T(1,1,0) = (1, -1, 0) + (0, 1, 0) = (1, 0, 0)$
$T(v_3) = T(0,1,1) = (0, -1, 0) + (0, 1, -1) = (0, 0, -1)$
$T(v)$ = $\begin{bmatrix}1&1&0 \\ 0&0&0 \\ -1&0&-1\end{bmatrix}$
$[v]_B$ = $2(1,1,1) + 4(1,1,0) + 6(0,1,1)$ = $(2,2,2) + (4,4,0) + (0,6,6)$
$[v]_B$ = $(6,14,8)$
$[T(v)]_B$ = $\begin{bmatrix}1&1&0 \\ 0&0&0 \\ -1&0&-1\end{bmatrix} \begin{bmatrix}6 \\ 14 \\ 8\end{bmatrix}$ = $\begin{bmatrix}20 \\ 0 \\-14\end{bmatrix}$
Obviously this is not the right answer, both should come out to be (-2,-2). What am I doing wrong?
Best Answer
You shouldn't get the same answer: you should get $\begin{pmatrix}-2\\-2\end{pmatrix}$expressed in terms of $\beta '$.
So $\begin{cases}x+2y=-2\\x+y=-2\end{cases}$. So $x=-2,y=0$, that is, $\begin{pmatrix}-2\\0\end{pmatrix}$.
Note $[T]_B^{B'}$ should be $2×3$.
Since $T(1,1,1)=(0,0)=0(1,1)+0(2,1), T(1,1,0)=(0,1)=2(1,1)-1(2,1)$ and $T(0,1,1)=(-1,0)=1(1,1)-1(2,1)$, we get $[T]_B^{B'}=\begin{pmatrix}0&2&1\\0&-1&-1\end{pmatrix}$.
So, $(2,4,6)=4(1,1,1) -2(1,1,0)+2(0,1,1)$. Hence $\begin{pmatrix}2\\4\\6\end{pmatrix}_B=\begin{pmatrix}4\\-2\\2\end{pmatrix}$.
Finally check that $[T]_B^{B'}[v]_B=\begin {pmatrix}-2\\0\end{pmatrix}=\begin{pmatrix}-2\\2\end{pmatrix}_{B'}$.