Given an $n \times n$ matrix $A$ with real entries such that $A^2 = -I$, prove that $\det(A) = 1$
This question is multi-part, but I happen to be stuck on this one. The previous parts showed:
$A$ is nonsingular, $n$ is even, and $A$ has no real eigenvalues.
I know that $\det(A)^2 = 1$ since $A$ has real entries and $n$ is even, but am not sure how to show that $\det(A)$, which can be either $1$ or $-1$, is not $-1$. Does anyone know how to continue from here?
from Hot Weekly Questions - Mathematics Stack Exchange
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