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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!

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up vote 0 down vote accepted

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 $[R_{\psi}]_E$.

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No you should say: column $i$ of the matrix is $T(b_i)$ expressed in the basis $\cal B$. The basis plays a double role here: at departure it furnishes the argument vectors $b_i$ that $T$ acts on, and at arrival it furnishes the coordinate functions to convert the resulting vector into a column of numbers. – Marc van Leeuwen Sep 22 '12 at 16:13
@MarcvanLeeuwen: My apologies, I did not make that clear. Of course, you are entirely correct. I will edit my post accordingly. – Michael Albanese Sep 22 '12 at 16:19
Thank you, Michael. This makes more sense to me now. I would upvote if I had the requisite reputation! – Bachmaninoff Sep 23 '12 at 0:31
No problem. I'm glad I could be of assistance. – Michael Albanese Sep 23 '12 at 2:48

${\large\mbox{First question}:}$ $$ \overbrace{\vec{r}'\,\cdot\vec{r}' = \vec{r}\,\cdot\vec{r}} ^{\mbox{Rotation definition}}\ \Longrightarrow\ x'\,\hat{x'} + y'\,\hat{y'} = x\,\hat{x} + y\,\hat{y} \quad\Longrightarrow\quad \left\vert% \begin{array}{rcl} x' & = & x\ \hat{x}\cdot\hat{x'} + y\ \hat{y}\cdot\hat{x'} \\ & = & x\cos\left(\psi\right) - y\sin\left(\psi\right) \\[3mm] y' & = & x\ \hat{x}\cdot\hat{y'} + y\ \hat{y}\cdot\hat{y'} \\ & = & x\sin\left(\psi\right) + y\cos\left(\psi\right) \end{array}\right. $$

${\large\mbox{Second question:}}$

Given a new base $\left\lbrace \vec{v}_{i}\right\rbrace$, a matrix ${\bf M}$ can be written as: $$ {\bf M} = \sum_{ij}M_{ij}\,\vec{v}_{i}\,\vec{v}_{j}^{\rm T} \quad\mbox{where}\quad M_{ij} \equiv \vec{v}_{i}^{\rm T}\,{\bf M}\,\vec{v}_{j} $$

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