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 $\begin{pmatrix}5&3\\1&1\end{pmatrix}$
[Math] Matrix for a linear transformation for a non-standard basis
linear algebra
Related Solutions
The matrix of a transformation is the matrix that turns the vector of coordinates of the input into the vector of coordinates of the output in certain bases. Coordinate vectors are always column vectors of some $\mathbb{R}^n$.
For example: In the standard basis $\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin{bmatrix}0&1\\0&0\end{bmatrix},\begin{bmatrix}0&0\\1&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix}\}$, of the space of matrices. The coordinate vector of the matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$, is the column $\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$.
In your case the transformation is $$T(\begin{bmatrix}a&b\\c&d\end{bmatrix})=\begin{bmatrix}2ia&b+ci\\c+bi&2id\end{bmatrix}$$
When you compute the matrix of this transformation in the standard basis you get $$A_T:=\begin{bmatrix}2i&0&0&0\\0&1&i&0\\0&i&1&0\\0&0&0&2i\end{bmatrix}$$
What this means is that when you multiply the vector of coordinates $$\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$$ of some vector $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ of your space, by $A_T$ you get the vector of coordinates $$\begin{bmatrix}2ia\\b+ci\\c+bi\\2di\end{bmatrix}=A_t\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$$
of the output of $T$. This is, the vector of coordinates of the vector $\begin{bmatrix}2ai&b+ci\\c+bi&2di\end{bmatrix}$ in the standard basis.
You need to check again what a matrix $M:V_1\to V_2$ represents. In particular for an endomorphism like in your case (meaning both vector spaces are the same $V_1=V_2$), each column $j$ of the matrix $M$ gives the image of the basis vector $b_i$ by that matrix, i.e., $Mb_i$.
Thus, in the standard basis $$M=\begin{pmatrix}1 & -2& 0\\0 & 1& 1\\-1 & 1& 0\end{pmatrix}$$
Let's call $v_1=(1,1,1)^T,\,v_2=(1,1,0)^T,\,v_3=(1,0,0)^T$. We have $$Mv_1=(-1,2,0)^T=2v_2-3v_3\\Mv_2=(-1,1,0)=v_2-2v_3\\Mv_3=(1,0,-1)=-v_1+v_2+v_3$$ Therefore, in the new basis $$M'=\begin{pmatrix}0 & 0& -1\\2 & 1& 1\\-3 & -2& 1\end{pmatrix}$$ Which is the answer you are seeking.
Notice, if we call $C$ the matrix providing the change of basis from the new $\{v_i\}$ to the old one $\{e_i\}$, it is $$C=\begin{pmatrix}1 & 1& 1\\1 & 1& 0\\1 & 0& 0\end{pmatrix}$$ and it holds that $$M'=C^{-1}MC$$
You could have also proceeded by determining $C,\,C^{-1}$ and doing the above matrix multiplication. For a "small" matrix ($3\times 3$) like in the present case, however, it is simpler and more elegant to do as we did.
Best Answer
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.