Eigenvalues of a unimodular matrix Let $U$ be a unimodular matrix, i.e. $U \in \mathbb{Z}^{n \times n}$, and $\text{det}(U) = \pm 1$.
Do the real (or complex for that matter) eigenvalues of $U$ admit a special structure?
Edit:
It is not hard to show that the integral eigenvalues must necessarily be $\pm 1$, but this is not the case for all eigenvalues.
In all of the examples I can think of the eigenvalues are of the form $a \pm \sqrt{a^2 \pm 1}$, and they always come in conjugated pairs. Is this the case in general?
 A: I think that the best you can say is that the characteristic polynomial has integer coefficients.  That is, the characteristic polynomial has the form
$$
x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0
$$
with $a_i \in \Bbb Z$. In the case of unimodular matrices, we of course have
$$
a_0 = \prod_{i=1}^n \lambda_i = \det(U) = 1
$$
similarly, each coefficient can be expressed in terms of the eigenvalues, which are the roots of this equation.  For example, we have
$$
a_{n-1} = \sum_{i=1}^n \lambda _i = \text{trace}(U)
$$
Perhaps you could find something useful in Vieta's formulas.
A: You speak about $GL_n(\mathbb{Z})$. Let $E=\{x\in \mathbb{C};\text{ there is a monic irreducible polynomial }P\in \mathbb{Z}[u]\text{ s.t. }P(0)=\pm 1,P(x)=0\}$. $E$ is a subset of the ring of algebraic integers. Moreover $(x,y)\rightarrow xy$ is stable for $E$ ; yet $(x,y)\rightarrow x+y$ is not.
We show: let $a$ be  a complex number. Then $a\in E$ iff there are $n$ and $A\in GL_n(\mathbb{Z})$ s.t. $a\in spectrum(A)$.
Proof:($\Leftarrow$) is clear. ($\Rightarrow$):  let $P$ be the minimal polynomial of $a\in E$ and $degree(P)=n$. Then the companion matrix of $P$ is in $GL_n(\mathbb{Z})$ and works.
