Matrix Representation and Linear Transformation 
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*Let $E_1$, $E_2$ denote the standard basis and $L:\mathbb{R}^{2}\rightarrow\mathbb{R^2}$ defined by


$L(E_1)=8E_1-4E_2$
$L(E_2)=9E_1-4E_2$.
a) Find the matrix representation of $L$ with respect to the standart basis.
b) Find the matrix representation of $L$ with respect to the basis {$V_1, V_2$}
where $V_1 = 3E_1 - 2E_2$ and $V_2 = -E_1 + E_2$.
My solutiıon-trying for a):  $L(E_1)=8E_1-4E_2=8\begin{pmatrix} 1 \\ 0 \end{pmatrix}-4\begin{pmatrix} 0 \\ 1 \end{pmatrix}=\begin{pmatrix} 8 \\ -4 \end{pmatrix}$, similarly we get $L(E_2)=9E_1-4E_2=\begin{pmatrix} 9 \\ -4 \end{pmatrix}$. Therefore, Matrix rep. of $L$ is $\begin{pmatrix} 8 & 9 \\ -4 & -4 \end{pmatrix}$.
Can you check my answer? If it's false, then can you help, can you give a hint? And can you give a hint for b)
 A: Your answer for part a) is in fact correct since the linear transformation is determined by its action over the canonical basis.
This means that $A=\begin{pmatrix} 8&9 \\ -4 &-4\end{pmatrix}$ is the matrix representation of $T$ with respect to the standard coordinates.
Now given $v_1=3e_1-2e_2=\begin{pmatrix} 3 \\ -2\end{pmatrix}$ and $v_2=-e_1+e_2=\begin{pmatrix} -1 \\ 1\end{pmatrix}$ we know that there is a change of basis matrix from the canonical basis to $\{v_1,v_2\}$. Let $B$ be the change of basis matrix.

 This matrix is obtained by taking $v_1, v_2$ as column vectors and arranging them into a matrix. This would be: 
 $$B=\begin{pmatrix} 3&-1 \\ -2 &1\end{pmatrix}$$

Then our transformation $T(x,y)=A\begin{pmatrix} x \\ y\end{pmatrix}$ is represented in the canonical basis. Since $B$ takes us from the canonical basis to $\{v_1,v_2\}$, its inverse will take us on the other direction.

 We know $B$ has an inverse since $\{v_1,v_2\}$ is, by hypothesis, a basis. Also this can be checked by row-reducing $B$ or computing its determinant.
 Finally the matrix representation on the transformation with respect to the new basis is obtained by taking the product of $B^{-1}$ and the matrix of $T$ in the canonical basis.
$$B^{-1}A=\begin{pmatrix} 4&5 \\ 4 &6\end{pmatrix}$$

