Matrix representation with non-standard bases. In chaprer 2.2 of Fiedberg's Linear Algebra is wroten about matrix representation. But all examples are only with standard ordered bases. I made a task to understand it. Please, could you show me matrix representation on this task or your own example?

Let $$ T\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x+y\\x-y\end{pmatrix}$$ is transformation from vector space V with a basis $$ v=   \{ \begin{pmatrix}1\\2\end{pmatrix}, \begin{pmatrix}3\\4\end{pmatrix}\} $$ to vector space W with a basis $$ w=   \{ \begin{pmatrix}1\\4\end{pmatrix}, \begin{pmatrix}2\\3\end{pmatrix}\} $$. 
How to find the matrix representation $$ [T]_v^w $$?
 A: Problem: Let 
$$ T\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x+y\\x-y\end{pmatrix}$$ 
is transformation from vector space V with a basis 
$$ v=   \{ \begin{pmatrix}1\\2\end{pmatrix}, \begin{pmatrix}3\\4\end{pmatrix}\} $$ to vector space W with a basis 
$$ w=   \{ \begin{pmatrix}1\\4\end{pmatrix}, \begin{pmatrix}2\\3\end{pmatrix}\} $$. Find $[T]_v^w$ for $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ as given.
Solution: there is a simple way to calculate for given examples. Basically, for each domain basis element (in $v$) we need to express it as a linear combination of the basis of the codomain ($w$). Let's begin:
$$ T(v_1)=T(1,2) = (1+2,1-2) = (3,-1) = A_{11}w_1+A_{21}w_2 \qquad \star$$
where $w_1 = (1,4)$ and $w_2 = (2,3)$. Likewise,
$$ T(v_2)=T(3,4) = (3+4,3-4) = (7,-1) = A_{12}w_1+A_{22}w_2 \qquad \star^2$$
So we must solve $\star$ and $\star^2$ for $A_{ij}$. That is,
$$ (3,-1)=A_{11}(1,4)+A_{21}(2,3) \ \ \& \ \ (7,-1)=A_{21}(1,4)+A_{22}(2,3)$$
These can be solved in many ways, for example:
$$ \left[ \begin{array}{c} 3 \\ -1 \end{array}\right]  = \left[ \begin{array}{cc} 1 & 2 \\ 4 & 3 \end{array}\right]\left[ \begin{array}{c} A_{11} \\ A_{21} \end{array}\right] \ \ \& \ \ 
\left[ \begin{array}{c} 7 \\ -1 \end{array}\right]  = \left[ \begin{array}{cc} 1 & 2 \\ 4 & 3 \end{array}\right]\left[ \begin{array}{c} A_{21} \\ A_{22} \end{array}\right]$$
reduce to a single matrix equation
$$ \left[ \begin{array}{cc} 3 & 7 \\ -1 & -1 \end{array}\right]  = \left[ \begin{array}{cc} 1 & 2 \\ 4 & 3 \end{array}\right]\left[ \begin{array}{cc} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}\right]$$
which I solve by multiplication by inverse
$$ \left[ \begin{array}{cc} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}\right] = 
\frac{1}{-5}\left[ \begin{array}{cc} 3 & -2 \\ -4 & 1 \end{array}\right]\left[ \begin{array}{cc} 3 & 7 \\ -1 & -1 \end{array}\right]  = \frac{1}{-5}\left[ \begin{array}{cc} 11 & 23 \\ -13 & -29 \end{array}\right] = [T]_v^w.$$
Let $\Phi_v: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the coordinate map defined by linearly extending $\Phi_v(v_i)=e_i$ and likewise $\Phi_w (w_i)=e_i$ for the standard basis $e_1=(1,0)$ and $e_2 = (0,1)$. Then I also denote $\Phi_v(x) = [x]_v$ and $\Phi_w(x) = [x]_w$; the coordinate charts with respect to bases $v$ and $w$ respective. Given all this notation the meaning of $[T]_v^w$ is simply that it is the matrix which acts as $T$ when we swap vectors in the domain and codomain for their corresponding coordinate vectors. That is:
$$ [T]_v^w[x]_v = [T(x)]_w $$
To actually calculate the coordinate charts (and hence the matrix of $T$) it generally requires we solve some systems of equations because it it not immediately obvious how to express $T(x)$ in the $w$-basis. I have examples and more discussion in a notation pretty close to the one I use here at:my linear algebra lectures on You Tube.
A: I suggest you to procede in the following way: let $V$ your vector space, and $\mathcal{B}$, $\mathcal{D}$ two different basis. You can calculate first the matrix $A$ associate to $T$ respect the canonical base $\mathcal{E}$. Then you can easily compute the matrix $M$ associate to the change of coordinates from $\mathcal{B}$ to $\mathcal{E}$ and finally the matrix $N$ associate to change of coordinates from $\mathcal{E}$ to $\mathcal{D}$.
Then the matrix associate to $T$ respect $\mathcal{B}$ and $\mathcal{D}$ is the matrix obtain by the following product:
$$[T]_{\mathcal{B}}^{\mathcal{D}}=MAN.$$
PS: I'm sorry, I change your notation.
