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Given a matrix $A\in \operatorname{SL}_d(\mathbb{Z})$ (the special linear group) satisfying the two conditions: (1) no eigenvalue of $A$ is a root of unity, (2) the characteristic polynomial of $A$ is irreducible over $\mathbb{Q}$.

QUESTION. Does it follow that at least one eigenvalue $\lambda$ of $A$ fulfills $\vert\lambda\vert>1$? Assuming true, it seems that there must be a theorem of a sort here, but I couldn't recall. It would also be nice if one can relax the conditions to gain the same conclusion, if possible.

UPDATE. After exploring papers by Kronecker (thanks Terry Tao) and others, I realized that we don't quite need "irreducibility" but "monic" is enough.

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  • $\begingroup$ Can't we simply say that the product of eigenvalues equals 1 (if we are considering modulus)? $\endgroup$ Commented Feb 10, 2022 at 18:18
  • $\begingroup$ @richrow the point is to discard eigenvalues in the unit circle such as $(3+4i)/5$ (which is not a root of unity). In particular the conclusion is false if $\mathrm{SL}_d(\mathbf{Z})$ is replaced with $\mathrm{SL}_d(\mathbf{Q})$. $\endgroup$ Commented Feb 10, 2022 at 18:20
  • $\begingroup$ @YCor Indeed, now it makes sense. $\endgroup$ Commented Feb 10, 2022 at 18:22
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    $\begingroup$ This is a result of Kronecker. mathoverflow.net/questions/10911/… $\endgroup$ Commented Feb 10, 2022 at 19:28
  • $\begingroup$ @TerryTao: that is what I thought! It can't be new, although I appreciate the work below. $\endgroup$ Commented Feb 10, 2022 at 19:35

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Since the characteristic polynomial is irreducible, eigenvalues are simple and hence the matrix is $\mathbf{C}$-diagonalizable. If all were on the unit cercle, it would follow that $\{ A^n:n\in\mathbf{Z}\}$ is bounded. But since it is contained in the set of integral points, this would force it to be finite, and hence all eigenvalues would be roots of unity, contradiction.

Hence one is not on the unit circle. Since the product of eigenvalues is $\pm 1$ (the determinant), it follows that some eigenvalue has modulus $>1$.

PS: Here's the straightforward way to extend this the general case (without irreducibility). Let $A\in\mathrm{M}_d(\mathbf{Z})$ be a matrix whose eigenvalues are not only among $0$ and roots of unity. As any integral matrix, we can block-triangulate it (over $\mathbf{Z}$) so that all diagonal blocks are $\mathbf{Q}$-irreducible. Hence at least one diagonal block, with characteristic polynomial $P=P(t)$, also satisfies the assumption. So $P$ is $\mathbf{Q}$-irreducible, and not equal to $t$. If $|P(0)|\ge 2$ then clearly some root has modulus $>1$. Otherwise, $P(0)\in\{\pm 1\}$ and we are in the previous (main) case.

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  • $\begingroup$ @T.Amdeberhan this is a subset of the $d^2$-dimensional space $M_d(\mathbf{C})$ of complex matrices. $\endgroup$ Commented Feb 10, 2022 at 19:27
  • $\begingroup$ "bounded" in a finite-dimensional vector space is a perfectly well-defined notion. $\endgroup$ Commented Feb 10, 2022 at 19:38
  • $\begingroup$ @T.Amdeberhan this is really ultra standard. Yes, it means that all matrix entries are bounded. $\endgroup$ Commented Feb 10, 2022 at 19:38
  • $\begingroup$ @YCor: I know you are clear about your points, I just want readers to follow as well. Thanks for the input, too. $\endgroup$ Commented Feb 10, 2022 at 19:44
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    $\begingroup$ By the way, for context, there exists matrices in $\mathrm{SL}_4(\mathbf{Z})$ whose characteristic polynomial is irreducible over $\mathbf{Q}$ but has roots of modulus 1. Indeed $(X^2+(1+\sqrt{2})X+1)(X^2+(1-\sqrt{2})X+1)$ is such a polynomial. $\endgroup$ Commented Feb 10, 2022 at 23:07

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