Basic homework question about rotations in linear algebra Math.SE! I'd like some help understanding the premise of the following question:


A rotation of $\mathbb{R}^2$ about the origin is a linear mapping $R_\psi$ given by 
  $R_\psi$ $\begin{pmatrix}
        r\cos\phi  \\
        r\sin\phi  \\
        \end{pmatrix}$ = $\begin{pmatrix}
        r\cos(\phi+\psi)  \\
        r\sin(\phi+\psi)  \\
        \end{pmatrix}$  
  for $0\leq\psi<2\pi$ and where any vector $v\in \mathbb{R}^2$ can be written as $\begin{pmatrix}
        r\cos\phi  \\
        r\sin\phi  \\
        \end{pmatrix}$ where $r$ is the length of $v$ and $\phi$ is the angle between $v$ and the positive $x$-axis. Verify that $R_\psi = T_A$ where $A=[R_\psi]_E=\begin{pmatrix}\cos \ \psi&-\sin \ \psi\\ \sin \ \psi&\cos\ \psi\\ \end{pmatrix}$ and $T_A(v)=Av$ for $v \in V$.

It wasn't difficult to actually verify this result - my real question is this: how can I obtain the fact $[R_\psi]_E=\begin{pmatrix}\cos \ \psi&-\sin \ \psi\\ \sin \ \psi&\cos\ \psi\\ \end{pmatrix}$ (where $E$ is the standard basis), and, more generally, how do I determine what $[R_\psi]_B$ is for any arbitary basis $B$ of $\mathbb{R}^2$?
Thanks in advance for any help!

${\large\mbox{First question}:}$
$$
\overbrace{\vec{r}'\,\cdot\vec{r}' = \vec{r}\,\cdot\vec{r}}
^{\mbox{Rotation definition}}\
\Longrightarrow\
x'\,\hat{x'} + y'\,\hat{y'}
=
x\,\hat{x} + y\,\hat{y}
\quad\Longrightarrow\quad
\left\vert%
\begin{array}{rcl}
x' & = & x\ \hat{x}\cdot\hat{x'} + y\ \hat{y}\cdot\hat{x'}
\\ & = & x\cos\left(\psi\right) - y\sin\left(\psi\right)
\\[3mm]
y' & = & x\ \hat{x}\cdot\hat{y'} + y\ \hat{y}\cdot\hat{y'}
\\ & = & x\sin\left(\psi\right) + y\cos\left(\psi\right)
\end{array}\right.
$$
${\large\mbox{Second question:}}$
Given a new base $\left\lbrace \vec{v}_{i}\right\rbrace$, a matrix ${\bf M}$ can be written as:
$$
{\bf M}
=
\sum_{ij}M_{ij}\,\vec{v}_{i}\,\vec{v}_{j}^{\rm T}
\quad\mbox{where}\quad
M_{ij} \equiv \vec{v}_{i}^{\rm T}\,{\bf M}\,\vec{v}_{j}
$$
