I read the following exercise and I am quite stuck:
Let $A \in GL_n(\mathbb{R})$, whose coefficients are in $\mathbb{Z}$. We assume that $A$ has $n$ distinct complex eigenvalues, and that the moduli of these eigenvalues are $\leq 1$. Prove that the eigenvalues are roots of unity.
My try : of course $A$ is diagonalizable. I can see that the fact that the moduli of the eigenvalues are $\leq 1$ implies that $(A^k)_{k \in \mathbb{N}}$ is bounded, and I think we can deduce from the fact that $A$ has integer coefficients that the eigenvalues have modulus $=1$. But I am not really sure how, and how to deduce that the eigenvalues are roots of unity ? Can we maybe prove that $A^k=I_n$ for some $k$ ?
Thanks in advance !