Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

2 Answers 2

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 $[R_{\psi}]_E$.

share|improve this answer
    
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
add comment

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

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.