Powers of Rotation matrix

Question

Consider the rotation matrix $A= \begin{bmatrix} \cos \ x & \sin \ x \\ -\sin \ x & \cos \ x \end{bmatrix} $.

Then $A^2,A^4 A^6$ etc. gives identity. Why?
Do this have any connection with de moivre's theorem.
Thanks in advance.

Answer 1

$A^2$ is not the identity, $A^2$ is

$$A^2=\begin{bmatrix}\cos^2 x-\sin^2 x & 2\sin x \cos x\\ -2\sin x \cos x & cos^2 x-\sin^2x\end{bmatrix}$$ This simplifies to $$A^2=\begin{bmatrix}\cos(2x)&\sin(2x)\\-\sin(2x)&\cos(2x)\end{bmatrix}$$

which is the matrix of rotation by $2x$ (because rotating twice by an angle of $x$ is the same as rotating once by an angle of $2x$.

Answer 2

Hint: $A^2$ gives rotation by $2x$. If $nx$ is a multiple of $2\pi$ you’ll get identity for $A^n$, not otherwise.

Answer 3

If you have a $2\times2$ irrep of $SO(2)$, $R(\varphi)$, you can show $$R^{n}(\varphi)=R(n\varphi)$$ If $\varphi\in2\pi\mathbb{Z}$, then you get identity