TITLE: If $A^2=-I$, Prove that $\det{A}=1$
QUESTION [5 upvotes]: If $A^2=-I$ , where $A$ is a square matrix of order $n$ and which contains real entries only and $I$ is identity matrix. Then how can we prove that $\det(A)=1$?.
I could prove that $n$ should be an even integer. But could not proceed to prove that $\det(A)$ can take only $1$, finding out few matrices which satisfies such properies (of small order) also verifies the given statement that the determinant is only $1$ and not $-1$.
Can anyone help with a hint ?

REPLY [9 votes]: Let $\lambda$ be an eigenvalue of $A$, with eigenvector $v$.
Then 
$$v^HA^2v = v^H(-I)v.$$
What can you conclude about $\lambda$?
Now use the fact that the eigenvalues of a real matrix must come in complex conjugate pairs, and that the determinant is the product of the eigenvalues.