Matrix for a linear transformation for a non-standard basis I wanted to make sure I understand how to find the matrix for a linear transformation, but for a non-standard basis. Lets say for example you have the linear operator in $\mathbb{F}^2$ defined in the standard coordinates by T$\begin{pmatrix}x\\y\end{pmatrix}$ = $\begin{pmatrix}x+y\\y-x\end{pmatrix}$ for the basis $(2,3)^T$, and $(1,2)^T$. Would I just the matrix as (T$\begin{pmatrix}2\\3\end{pmatrix}$ | T$\begin{pmatrix}1\\2\end{pmatrix}$)? Which comes out to be $\begin{pmatrix}5&3\\1&1\end{pmatrix}$
 A: If that were true, one would expect the matrix of $T$ in the non-standard basis to take the coordinates of $v$ in that basis to $Tv$ in that basis. However:
$\begin{bmatrix}5&3\\1&1 \end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}5\\1\end{bmatrix}$
So your matrix takes $1(2,3)^T + 0(1,2)^T \to 5(2,3)^T + 1(1,2)^T = (11,17)^T$.
However: $T((2,3)^T) = (5,1)^T$, so this cannot be right.
The matrix for $T$ in the standard basis is:
$A = \begin{bmatrix}1&1\\-1&1\end{bmatrix}$.
Let us denote the coordinates of $(x,y)^T$ in the non-standard basis by:
$[u,v]^T = u(2,3)^T + v(2,1)^T$.
The matrix which sends $[u,v]^T$ to its "standard basis coordinates" is:
$P = \begin{bmatrix}2&1\\3&2\end{bmatrix}$, for example, this sends:
$[1,0]^T = 1(2,3)^T + 0(1,2)^T \to (2,3)^T$
$[0,1]^T = 0(2,3)^T + 1(1,2)^T \to (1,2)^T$, as expected.
Thus $AP([u,v]^T)$ returns the standard-basis coordinates of $T$'s image on non-standard basis coordinate vectors. To get our final answer in non-standard basis coordinates, we must apply:
$P^{-1} = \begin{bmatrix}2&-1\\-3&2\end{bmatrix}$, so that the desired matrix is:
$P^{-1}AP = \begin{bmatrix}9&5\\=13&-7\end{bmatrix}$
and this matrix takes $(2,3)^T = [1,0]^T \to [9,-13]^T = (5,1)^T$ and:
$(1,2)^T = [0,1]^T \to [5,-7]^T = (3,1)^T$, as desired.
