Let E be the standard basis and S another basis for $\mathbb{R}^2$. Assume the matrix M is understood to be in terms of E. Find $[M]_S.$ The question: Let $E = \left\{\begin{pmatrix} 1\\ 0\end{pmatrix}, \begin{pmatrix} 0\\ 1\end{pmatrix} \right\}$ be the standard basis for $\mathbb{R}^2$. The set $S = \left\{\begin{pmatrix} 1\\ 2\end{pmatrix}, \begin{pmatrix} -3\\ 2\end{pmatrix} \right\}$ is another basis for $\mathbb{R}^2$, Assume the matrix $M =\begin{bmatrix} 3 & 5\\ -3 & 2 \end{bmatrix}$ is understood to be in terms of $E$. Find $[M]_S.$
I am really confused on what this is asking of me and how I should approach this. I think that it is saying that I have $[M]_E$, so then to find $M$. I have tried
$
\left[
\begin{array}{cc|cc}
3 & 5 & 1 & 0 \\
-3 & 2 & 0 & 1 \\
\end{array}
\right]
\rightarrow
\left[
\begin{array}{cc|cc}
1 & 0 & \frac{2}{21} & \frac{-5}{21}\\
0 & 1 & \frac{1}{7}& \frac{1}{7}\\
\end{array}
\right]
$, so that I am left with $M = \frac{1}{21}\begin{bmatrix}
2 & -5\\
3& 3\\
\end{bmatrix}$. I then used this new $M$ to solve for $[M]_S$ by doing $MS = \frac{1}{21}\begin{bmatrix}
2 & -5\\
3& 3\\
\end{bmatrix}
\begin{bmatrix}
1 & -3\\
2& 2\\
\end{bmatrix}=
\frac{1}{21}
\begin{bmatrix}
-8 & -16\\
9& -3\\
\end{bmatrix}
$.
But this seems extremely wrong. My professor didn't do a great job going over bases, so any advice would be appreciated.
 A: $$[M]_E =\begin{bmatrix} 3 & 5\\ -3 & 2 \end{bmatrix}$$
means that the linear map $M$ acts on the basis $E$ as
$$M\begin{bmatrix}1 \\ 0 \end{bmatrix} = 3\begin{bmatrix}1 \\ 0 \end{bmatrix}-3\begin{bmatrix}0 \\ 1 \end{bmatrix}, \quad M\begin{bmatrix}0 \\ 1 \end{bmatrix} = 5\begin{bmatrix}1 \\ 0 \end{bmatrix}+2\begin{bmatrix}0 \\ 1 \end{bmatrix}$$
or in general
$$M\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 & 5\\ -3 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix}3x+5y \\ -3x+2y \end{bmatrix}.$$
You have to find a matrix $$[M]_S = \begin{bmatrix} \alpha & \gamma \\ \beta & \delta\end{bmatrix}$$
such that
$$M\begin{bmatrix} 1 \\ 2\end{bmatrix} = \alpha \begin{bmatrix} 1 \\ 2\end{bmatrix} + \beta \begin{bmatrix} -3 \\ 2\end{bmatrix}, \quad M\begin{bmatrix} -3 \\ 2\end{bmatrix} = \gamma \begin{bmatrix} 1 \\ 2\end{bmatrix} + \delta \begin{bmatrix} -3 \\ 2\end{bmatrix}.$$
A: It is known that the components of a vector $v$ in a new base $S$ are given by $S^{-1}v$.
So for the transformation $M$ having $v\mapsto Mv$, we do
$S^{-1}Mv$ to get the new components for $Mv$ and observe that
$$S^{-1}v\mapsto S^{-1}Mv=(S^{-1}MS)S^{-1}v,$$
then we see how the matrix $S^{-1}MS$ relates the new versions $S^{-1}v$ and $S^{-1}Mv$.
