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!

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. 
