TITLE: Determinant of a finite order matrix
QUESTION [5 upvotes]: Let $M$ be a $5\times 5$ with real entries. Suppose $M$ has finite
  order and $\det(M-I_5)\neq 0$. Find $\det(M)$.

I am trying to do this old algebra qual problem. So far I know that since $M$ is of finite order, say $|M|=n$, then $1=\det(I_5)=\det(M^n)=[\det(M)]^n$ and so $\det(M)$ is a root of unity. $M$ has real entries and so the determinant is real, but the only real roots of unity are $1$ and $-1$. I don't know how to use the fact that $M$ is $5\times 5$ or that $1$ is not an eigenvalue. Can we perhaps say something about the value of $n$?

REPLY [3 votes]: Since $M$ has real entries, its non-real eigenvalues must come in complex conjugate pairs.  So the eigenvalues of $M$ are some number of copies of $-1$, and then some number of non-real roots of unity coming in conjugate pairs.  Since the conjugate of a root of unity is its inverse, the non-real eigenvalues will cancel out in $\det(M)$.  Furthermore, there must be an odd number of $-1$s, since $5$ is odd.  So $\det(M)$ must be $-1$.