Consider the group of matrices $G =\operatorname{GL}(n,\mathbb{Z})$ with integer entries and determinant $\pm 1$. For each matrix $D \in G$, the product of the eigenvalues of $D$ is equal to $\det D =\pm 1$, and so the spectral radius $\rho(D)$, which is the size of the largest eigenvalue, is at least one. Moreover, if $\rho(D) =1$, then all the eigenvalue have size 1 and can be proven to be roots of unity.


My question is this: is there a lower bound $B>1$ such that if $\rho(D) \neq 1$ then $\rho(D) \geq B$? If so, does this $B$ depend on $n$?



Lower bound for spectral radius on $\operatorname{GL}(n,\mathbb{Z})$
Every monic integer polynomial $f(x) \in \mathbb{Z}[x]$ of degree $n$ is the characteristic polynomial of an $n \times n$ matrix, namely its [companion matrix](http://en.wikipedia.org/wiki/Companion_matrix). The companion matrix is invertible iff the constant term of $f(x)$ is $\pm 1$. Conversely, every characteristic polynomial of an element of $GL\_n(\mathbb{Z})$ has this form. So the question reduces to a question about largest eigenvalues of monic integer polynomials of degree $n$ with constant term $\pm 1$. Let me use "spectral radius" to mean the absolute value of the largest (in absolute value) root of a polynomial to save space. 


Now it's clear that a bound of the desired form must exist (depending on $n$). The reason is that if any coefficient $e\_k$ of the characteristic polynomial gets large, then at least one of the eigenvalues must be large. More formally, if $e\_k$ has absolute value at least ${n \choose k} R^k$, then the spectral radius is at least $R$. Hence, for any $R$, the space of possible coefficients $e\_k$ with the appropriate bounds is bounded, and so the set of possible characteristic polynomials with spectral radius less than $R$ is finite. But this argument is very inefficient: it only tells you that the number of such polynomials is at most


$$2 \prod\_{i=1}^{n-1} \left( 2 {n \choose k} R^k + 1 \right).$$


For example, when $n = 2$ we are looking at characteristic polynomials of the form $x^2 + kx \pm 1, k \in \mathbb{Z}$. The polynomial $x^2 - x - 1$ has largest eigenvalue the golden ratio


$$\phi = \frac{1 + \sqrt{5}}{2} = 1.618 \dots$$


and the only polynomials that can have smaller largest eigenvalue than this (not equal to $1$ in absolute value) must satisfy $|k| < 2 \phi = 3.236 \dots$. There aren't very many of these and we can just verify by hand that they don't. 


The bound must depend on $n$; to see this consider the sequence of polynomials $f\_n(x) = x^n - x - 1$. If a root $x\_0$ of $f\_n(x)$ has absolute value $R = 1 + r \ge 1$ then


$$(1 + r)^n \ge 1 + nr$$


but on the other hand, since $x\_0^n = x\_0 + 1$ we must have $(1 + r)^n \le 2 + r$. It follows that $2 + r \ge 1 + nr$, so $1 \ge (n - 1) r$, or $r \le \frac{1}{n - 1}$, so


$$R \le 1 + \frac{1}{n - 1}.$$


