3
$\begingroup$

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 $A$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ 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? $\endgroup$ – Leo Oct 13 '19 at 1:38
  • $\begingroup$ I would love to help here, but I have no idea. $\endgroup$ – José Carlos Santos Oct 13 '19 at 9:02
  • $\begingroup$ $$ \left( \begin{array}{cccc} 1&1&1&0 \\ 1&1&0&1 \\ 0&1&0&0 \\ 0&0&1&0 \\ \end{array} \right) $$ $$ x^4 - 2 x^3 - 2x + 1 $$ at his more recent question math.stackexchange.com/questions/3393377/… $\endgroup$ – Will Jagy Oct 14 '19 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.