Shifting basis for a transformation matrix "In the vectorspace $\mathbb{R}^2$ two vectors are given (Which also form the basis a): $a_1=(8,-3)$ and $a_2=(5,-2)$
A linear mapping is determined by: $f(a_1) = 2a_1-4a_2$ and $f(a_2)=-a_1+2a_2$
"
How can one determine the transformation matrix of f with respect to the standard e-basis?
I determined that that transformation matrix of f with respect to basis a is:
$$ \left( \begin{matrix} 2 & -1 \\ -4 & 2 \end{matrix} \right)$$ 
Then I tried taking the inverse matrix of the basis change matrix from basis e to a which I think is $$_aM_e= \left( \begin{matrix} 8 & 5 \\ -3 & -2\end{matrix} \right) \text{ and } _eM_a=(_aM_e)^{-1}= \left( \begin{matrix} 2 & 5 \\ -3 & -8\end{matrix} \right)$$
Can I just change basis of the transformation matrix directly by: $$_eF_a=_eM_a \cdot aFa$$
This gives me the matrix $\left( \begin{matrix} -16 & 8 \\ 26 & -13 \end{matrix} \right)$
I don't know how to tell whether this is the correct approach?
 A: *

*Given a linear map $T: V \to W$ suppose  $A=\{a_1,...,a_n\}$ is a basis for $V$ and $B=\{b_1,...,b_m\}$ is a basis for $W$ i.e they are finite dimensional vector spaces. We will write $T = \begin{pmatrix} T(a_1) & \cdots & T(a_n) \end{pmatrix}$.

*Given a vector $\vec{v}$, you would like to find $[T\vec{v}]_B$. First we have $\vec{v} = \sum_k v^k a_k$ which we identify with $(v^1,...,v^n)^T:=[v]_A$ i.e its coordinates representation. And now, by the definition of matrix multiplication,
\begin{align*} T\vec{v} =  \sum _{j=1}^nT(a_j)\  v^j = \begin{pmatrix} T(a_1) & \cdots & T(a_n) \end{pmatrix} \begin{pmatrix} v^1 \\ \vdots \\ v^n \end{pmatrix} \Rightarrow [T\vec{v}]_B &= \sum_{j=1}^n[T(a_j)]_B\ v^j\\ \\ & = \underbrace{\begin{pmatrix} [T(a_1)]_B & \cdots & [T(a_n)]_B \end{pmatrix}}_{\color{blue}{:= T_{BA}}} [v]_A\end{align*}


*

*Therefore, given the linear map $f: (\mathbb{R}^2,\{a_1,a_2\}:=A) \to (\mathbb{R}^2,\{(1,0)^T, (0,1)^T\}:=B)$, we have that the coordinate transformation matrix is given by,


$$ f_{BA} = \begin{pmatrix} [f(a_1)]_B & [f(a_2)]_B \end{pmatrix}$$
i.e we have to write $f(a_1) = 2a_1-4a_2$ and $f(a_2)= -a_1 + 2a_2$ in terms of the unit basis vectors $(1,0)^T$ and $(0,1)^T$. Given $a_1 = (8,-3)^T,a_2 = (5,-2)^T$ then $f(a_1) =(-4,2)^T $ and $f(a_2) = (2,-1)^T$. It follows that the matrix is,
$$f_{BA}=\begin{pmatrix} \begin{pmatrix} -4 \\ 2\end{pmatrix} & \begin{pmatrix} 2 \\ -1\end{pmatrix}\end{pmatrix}$$


*

*$\textbf{What's the connection}$? If you want the transformation matrix with respect to basis $A$ then you are trying to compute $f_{AA}$. It follows that $\color{red}{f_{AA} = I_{AB} \circ f_{BA}}$ where, 


$$I_{AB} = \begin{pmatrix} 2 & 5 \\ -3 & -8 \end{pmatrix} $$
i.e we have,
$$ f_{AA} = \begin{pmatrix} 2 & -1 \\ -4 & 2\end{pmatrix}$$
which is precisely what you wrote.
