Question about negative matrix involution

Question

If we have matrix $A\in Mat_{n}(\mathbb{R})$ with property $A^{2}=-I$. How to show n must be even?

Answer 1

Just take the determinant on both sides. You get $\det (A)^2 = (-1)^n$.

Answer 2

Let $\lambda$ be an eigenvalue of $A$ with associated eigenvector $v$:

$Av = \lambda v; \tag 1$

then

$A^2v = A(\lambda v) = \lambda Av = \lambda (\lambda v) = \lambda^2v; \tag 2$

thus,

$(\lambda^2 + 1)v = \lambda^2v + v = (A^2 + I)v = 0v = 0, \tag 3$

from which we infer, since

$v \ne 0, \tag 4$

that

$\lambda^2 + 1 = 0, \tag 5$

and hence that $\lambda$ is purely imaginary; in fact,

$\lambda \in \{i, -i \}; \tag 6$

now if $n$ were odd, the degree of the characteristic polynomial $\det(A - \lambda I)$ of $A$ would also be odd, and hence have at least one real root, in contradiction to (6); therefore $n$ must be even. .