# If an eigenvalue of an integer matrix lies on the unit circle, must it be a root of unity?

Let $$A$$ be a real matrix with integer entries, and suppose $$z$$ is a complex eigenvalue of $$A$$ with $$|z| = 1$$. Is it true that $$z$$ is either an odd or even root of unity? That is, must there exist an $$m$$ such that $$z^m = 1$$?

Another way of asking the same thing is (I think) whether complex units with an irrational argument are algebraic numbers.

EDIT: The accepted answer shows this is not true. However, I'm still interested under what conditions can this be claimed about $$A$$.

No. Consider the companion matrix of the polynomial $$p(x)=x^4-2x^3-2x+1$$. Its eigenvalues are the roots of $$p(x)$$, two of which are complex (non-real) numbers with absolute value $$1$$, none of which is a root of unity.
• Excellent, thank you. However, I'm still interested in the conditions under which $A$ cannot have complex units that are not roots of unity. Any pointers? – Leo Oct 13 '19 at 1:38
• $$\left( \begin{array}{cccc} 1&1&1&0 \\ 1&1&0&1 \\ 0&1&0&0 \\ 0&0&1&0 \\ \end{array} \right)$$ $$x^4 - 2 x^3 - 2x + 1$$ at his more recent question math.stackexchange.com/questions/3393377/… – Will Jagy Oct 14 '19 at 17:04