1. ## Linear transformations

Let $T$ be the rotation in the plane $\mathbb{R}^2$ through angle $\frac{\pi}{3}$ in the anti-clockwise direction. Find the matrix which represents the linear transformation $T$.

Not too sure how to start this question off. The only thing I can think of doing for this is to use complex numbers.

If someone could give me a clue on how to start off this question and I'll try and do it myself.

2. Look here.

3. Ok thanks, so is that just a general result that should be remembered

This is what I thought of doing.

Consider a complex number $a+ib$

Multiply it by $\cos{\frac{\pi}{3}}+i\sin{\frac{\pi}{3}}$

$\Rightarrow (a+ib)(\frac{1}{2}+\frac{\sqrt{3}}{{2}}i)$

$=\frac{1}{2}a-\frac{\sqrt{3}}{2}b+i(\frac{\sqrt{3}}{2}a+\frac{1} {2}b)$

$T\begin{pmatrix}a\\ b\end{pmatrix}=\begin{pmatrix}\frac{1}{2}a-\frac{\sqrt{3}}{2}b\\ \frac{\sqrt{3}}{2}a+\frac{1}{2}b\end{pmatrix}$

$=\begin{pmatrix}\frac{1}{2}-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}+\frac{1}{2}\end{pmatrix}\begin{ pmatrix}a\\b\end{pmatrix}$

Is this correct?

4. As to why rotation through angle $\theta$ is given by
$\begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{bmatrix}$

look at what it does to the "basis vectors"
$\begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix}= \begin{bmatrix}cos(\theta) & sin(\theta)\end{bmatrix}$
That is, (1, 0) is mapped to the point $(cos(\theta), sin(\theta)$. The distance from (0, 0) to that point is $\sqrt{cos^2(\theta)+ sin^2(\theta)}= 1$ and that point lies along a line with slope $\frac{sin(\theta)}{cos(\theta)}= tan(\theta)$. Since the slope of a line is the tangent of the angle the line makes with the x-axis, that says that (1, 0) has been mapped to a point on the line that makes angle $\theta$ with the x-axis. That is, it has been rotated through angle $\theta$.

$\begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{bmatrix}\begin{bmatrix}0 \\ 1\end{bmatrix}= \begin{bmatrix}-sin(\theta) & cos(\theta)\end{bmatrix}$ tells us the same thing.

5. acevipa: Yep, it all looks good.