Prove properties of $A^2 = -I$

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.

Solution 1:

(a)

Since the matrix has real entries, $\det A$ is real, so $\det A^2 = (\det A)^2$ is positive. But $\det -I = (-1)^n$, because we can just multiply down the main diagonal, so we must have that $n$ is even.

(b)

Suppose $A$ has a real eigenvalue $\lambda$ with corresponding eigenvector $v$. Then $$-v=-Iv=A^2v=A(Av)=A(\lambda v)=\lambda^2v.$$

This products a contradiction, because no real number squares to $-1$.

Solution 2:

HINT: $\det(-I)=\det(A^2)$. If $Ax=\lambda x$, then $-x=A^2x=\ldots\;$?