Given an $n \times n$ matrix $A$ with real entries such that $A^2=-I$. Prove that A has no real eigenvalues.
We can easily prove the following additional statements about $A$ by taking determinants of both sides of the given equality –
(a) $A$ is nonsingular;
(b)$n$ is even;
(c)$det(A)=1$.
Best Answer
Well, what does it mean for something to be a real eigenvalue of A? It means there's some vector we can multiply into A on the right and get back a scalar multiple of that vector.
But for any eigenvector, if we do this twice: $-v = (-I)v = A^2 v = A(A v) = A(\lambda v) = \lambda^2 v$.
You should be able to take it from here.