$A^n = I$ only if $n\equiv 0 \pmod{4}$

Question

Is the following proof correct? In addition could you pleases suggest a more cleaner or shorter proof.

Theorem. Given any $n\in\mathbb{Z^+}$ then $$\begin{pmatrix} 0&0&-1\\ 0&1&0\\ 1&0&0\\ \end{pmatrix}^n = \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{pmatrix} $$

only if $4 \mid n$.

Proof. Assume that $n\in\mathbb{Z^+}$ and let $A$ denote the matrix $$ \begin{pmatrix} 0&0&-1\\ 0&1&0\\ 1&0&0\\ \end{pmatrix} $$ We prove the contrapositive. Given the standard choice of basis for $\mathbb{R}^3$ we can see $A$ represents a $90^{\circ}$ clockwise rotation about the $y$-axis consequently we may write out $A^2$, $A^3$ and $A^4$ as follows:

$$ A^2 = \begin{pmatrix} -1&0&0\\ 0&1&0\\ 0&0&-1\\ \end{pmatrix} $$

$$ A^3 = \begin{pmatrix} 0&0&1\\ 0&1&0\\ -1&0&0\\ \end{pmatrix} $$

$$ A^4 = \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{pmatrix} $$

Now assume that $4\not\mid n$ then the division theorem implies that $n$ is of the form $4q+1$, $4q+2$ or $4q+3$. Considering the case where $4q+1$ we can see that $$A^{4q+1} = (A^{4})^q\cdot A = I^q\cdot A = I\cdot A\neq I.$$ By similar reasoning we can make the same conclusion in the other two cases.

$\blacksquare$

Answer 1

Yes, it is correct, but I would not use a geometric argument here, since a very simple computation is enough to obtain $A^n$ for $n\in\{2,3,4\}$.

Answer 2

You are correct and your argument handles all cases with minor change.

If natural $n>1$ and $k \in \{0,1,2,3\}$

$$A^{4n+k} = (A^4)^n \cdot A^k = I \cdot A^k = A^k$$

and so $A^{4n+k}=I$ only when $k=0$. That is $A^n=I$ only when $n$ is a multiple of $4$.