Trace of finite-order matrix over $\Bbb C$.

Question

Suppose that $A\in M_n(\Bbb C)$ is such that $A^m=Id$ for some $m\in > \Bbb N$, is it true that if $Tr(A)\in \Bbb R$ then it must be an integer? Here, $Tr(A)$ denotes the sum of the elements in the diagonal of $A$

I've been trying to solve certain problem and if this were true, I'd be done.

My approach: I know that the eigenvalues of $A$ will be roots of unity, and that the trace is the sum of such eigenvalues. The problem then reduces to show that any sum of roots of unity (not necessarily with all sumands being different) that lies in $\Bbb R$ is an integer. I have not found any counter example to this so I'm assuming it's true, however, I can't find how to prove it. Any ideas please? Thanks!

Answer 1

No, that is not true in general. Consider any matrix of the form $\left( \begin{array}{clcr} \cos{\frac{2 \pi}{n}} & \sin{\frac{2\pi}{n}}\\-\sin{\frac{2 \pi}{n}} & \cos{\frac{2\pi}{n}} \end{array} \right)$, for $n$ a positive integer. This is a matrix of finite order $n$, which certainly has real trace. However, its trace is usually not an integer (eg it is not for $n = 5$ or $n = 7$).

It is true that if a matrix of finite order has rational trace, then that trace is an integer.

Answer 2

If one adds $\zeta_8$ to $\overline{\zeta_8}$, one doesn't get an integer. So a diagonal matrix consisting of these will have real trace, it will have order $8$, and its trace will not be an integer.