Given an $n\times n$ matrix A with real entries such that $A^2=-I$, prove (a) that $n$ is even and (b) that $A$ has no real eigenvalues. How do you do this? I have no idea where to start.
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Since the matrix has real entries, $\det A$ is real, so $\det A^2 = (\det A)^2$ is positive. But $\det -I = (-1)^n$, because we can just multiply down the main diagonal, so we must have that $n$ is even.
Suppose $A$ has a real eigenvalue $\lambda$ with corresponding eigenvector $v$. Then $$-v=-Iv=A^2v=A(Av)=A(\lambda v)=\lambda^2v.$$
This products a contradiction, because no real number squares to $-1$.
Take a look at Cayley Hamilton Theorem. An overkill but the two people above me already answered it! In particular, look at the minimal polynomial.
Let's start with (b). Suppose $\lambda$ were a real eigenvalue of $A$. Then there would exist a nonzero vector $v \in R^n$ such that
$Av = \lambda v$.
$A^2v = \lambda^2 v$,
$(A^2 + I)v = (\lambda^2 + 1)v$;
but since $A^2 + I = 0$ this implies
$(\lambda^2 + 1)v =0$,
which since $v \ne 0$ implies
$\lambda^2 + 1 = 0$;
but no real $\lambda$ satisfies this equation. This contradiction shows $A$ has no real eigenvalues.
As for (a), if $n$ were odd, then the characteristic polynomial $\det(A - \lambda I)$ of $A$ would have odd degree $n$; but every polynomial with real coefficients and odd degree has at least one real root. But we have seen that $A$ can have no real eigenvalues. Thus $n$ must itself be even. QED.
Hope this helps. Cheers.
To prove (a):
Consider the matrix $I_n$, the $n*n$ identity matrix. Clearly, $\det I_n = (-1)^n$. Now, consider $\det A$. As $A$ has only real entries, $\det A$ is real. Thus, $(\det A)^2$ is positive. Since $A^2 = I_n$, it is also the case that $(\det A)^2 = \det I_n = (-1)^n$, and as $(\det A)^2 > 0$, it must be the case that $\det I_n > 0$, so $n$ must be even.