Given an n x n matrix A with real entries such that A^2 = -I, prove the following about A:
1) n is even
2) A has no real eigenvalues
3) det A = 1
Hints :
1) $\displaystyle (\det A)^2=(-1)^n$
2) If $\displaystyle \lambda \in\mathbb{R}$ is an eigenvalue of $\displaystyle A$ , then $\displaystyle Ax=\lambda x$ for some $\displaystyle 0\neq x\in \mathbb{R}^n$ .
Prove that $\displaystyle (\lambda^2+1)x=0$ (contradiction).
3) No hint, try it.
Fernando Revilla