TITLE: Prove properties of $A^2 = -I$
QUESTION [8 upvotes]: Given an $n\times n$ matrix A with real entries such that $A^2=-I$, prove (a) that $n$ is even and (b) that $A$ has no real eigenvalues. How do you do this? I have no idea where to start.

REPLY [2 votes]: Let's start with (b).  Suppose $\lambda$ were a real eigenvalue of $A$.  Then there would exist a nonzero vector $v \in R^n$ such that 
$Av = \lambda v$.
Then 
$A^2v = \lambda^2 v$,
and
$(A^2 + I)v = (\lambda^2 + 1)v$;
but since $A^2 + I = 0$ this implies
$(\lambda^2 + 1)v =0$,
which since $v \ne 0$ implies
$\lambda^2 + 1 = 0$;
but no real $\lambda$ satisfies this equation.  This contradiction shows $A$ has no real eigenvalues.  
As for (a), if $n$ were odd, then the characteristic polynomial $\det(A - \lambda I)$ of $A$ would have odd degree $n$; but every polynomial with real coefficients and odd degree has at least one real root.  But we have seen that $A$ can have no real eigenvalues.  Thus $n$ must itself be even.  QED.  
Hope this helps.  Cheers.