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!

Given a linear transformation $T : V \to V$ and an ordered basis $\mathcal{B} = \{b_1, \dots, b_n\}$ of $V$, the standard matrix of $T$, with respect to $\mathcal{B}$, is given by $$[T]_{\mathcal{B}} = [T(b_1)\ \cdots\ T(b_n)]$$ where $T(b_i)$ is expressed in the basis $\mathcal{B}$. That is, $T(b_i)$, expressed in the basis $\mathcal{B}$, is column $i$ of $[T]_{\mathcal{B}}$. 
For your particular linear transformation, note that $(1, 0)^t = (\cos 0, \sin 0)^t$, so $R_{\psi}((1, 0)^t) = (\cos\psi, \sin\psi)^t$, the first column of $[R_{\psi}]_E$. Now note that $(0, 1)^t = (\cos\frac{\pi}{2}, \sin\frac{\pi}{2})^t$, so $R_{\psi}((0, 1)^t) = (\cos(\psi + \frac{\pi}{2}), \sin(\psi + \frac{\pi}{2}))^t = (-\sin\psi, \cos\psi)^t$, the second column of $[R_{\psi}]_E$.
