Is it true that if $A^4 = I$, then $A^2 = \pm I$? So for linear algebra, I either need to prove that, for all square matrices, if $A^4 = I$, then $A^2 = \pm I$, or find a counterexample of this statement. Can anyone help please?
Thanks!
 A: Check
$$A=\begin{bmatrix} i & 0 \\0 & 1 \end{bmatrix}$$
A: Let consider
$$A^2=\begin{bmatrix} 0 & 1 \\1 & 0 \end{bmatrix}$$
A: Or
$$A=\begin{bmatrix} 0 & -1 & 0 \\1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
A: This is not true. Take, for instance, the matrix 
$\begin{bmatrix} 1 & 0 & 0 & 0 \\
0 & -1 & 0 &0 \\
0& 0 & i & 0 \\ 0& 0& 0& -i
 \end{bmatrix}$
which is maybe a little too much, but I liked it.
A: Completing the previous answers, it's enough to look for matrices $n \times n$ that satisfy $$(A^{2}) = (A^{2})^{-1}$$
where $A^{2} \neq I$. Check this in the previous answers.
A: This answer expands and expounds upon Botund's answer above; if $S= \{ \pm 1, \pm i \}$, then $x^4 = 1$ for every $x \in S$.
Consequently, if $D$ is any $n$-by-$n$ diagonal matrix with diagonal entries from $S$, then $D^4 = I_n$; however, the diagonal entries can be chosen such that $D^2 \ne \pm I_n$. For example, select any diagonal matrix containing at least one element of $\{\pm i\}$ and at least one element of $\{ \pm 1 \}$ in the diagonal. 
