What is the order of a rotation matrix? Given $${\displaystyle R(\theta)={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{bmatrix}}}$$
What is the order of $R(2\pi/n)$? I know it depends on $n$. But I have problem finding the relation. 
 A: The order of $R(2\pi/n)$ is the smallest positive integer $k$ for which $$R(2\pi/n)^k=\begin{bmatrix}\cos (2\pi k/n) &-\sin (2\pi k/n) \\\sin (2\pi k/n) &\cos (2\pi k/n) \\\end{bmatrix}$$
is equal to the identity matrix. So you have to find the smallest positive $k$ such that $\cos (2\pi k/n)=1$ and $\sin(2\pi k/n)=0$. Can you do this?
A: With $${\displaystyle R(\theta)={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{bmatrix}}}$$
we find that $$R^2 (\theta) = \begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{bmatrix} \times{\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{bmatrix}}$$
$$ =\begin{bmatrix}\cos 2\theta &-\sin 2\theta \\\sin 2\theta &\cos 2\theta \\\end{bmatrix} =R(2\theta)$$
Similarly $$R^n(\theta ) = R(n\theta)$$
For $\theta = \frac {2\pi}{n}$ we have $$R^n(\theta ) = R(2\pi)= I $$
A: Hint: How many radians are there in a full circle? And what happens if you rotate something full circle?
A: Hint:
Prove by induction that
$$\bigl(R(\theta)\bigr)^k= R(k\theta)=\begin{bmatrix}
\cos k\theta & -\sin k\theta \\
\sin k\theta & \phantom{-}\cos k\theta 
\end{bmatrix}.$$
A: Your intuition is correct.  Of course, applying a rotation repeatedly,  say $m$ times,  is the same as rotating by $m\cdot \theta$, where $\theta$ is the angle of the original rotation.   Now if $\theta =\frac{2\pi}n$, we get that applying the rotation $n$ times gives a rotation by $n\cdot \frac{2\pi}n=2\pi$, which is the same as no rotation at all.
Also note, for $m\lt n$, $\frac mn\not\in\mathbb Z$, so we don't get the identity. 
