I have encountered the following linear algebra/number theory question in my work (low-dimensional topology), so I thought I should ask the experts.


Let $A \in SL(n,\mathbb{Z})$ be a matrix , $n \geq 3$. One can easily show that there is some $j \in \mathbb{N}$ such that $Tr(A^j) \geq 3$ (proof given at the end). My question is that how long does it take for this to happen?


**Question 1** Given $n \geq 3$, is there a number $k \in \mathbb{N}$ (depending only on n) such that for all $A \in SL(n,\mathbb{Z})$ there is some $1 \leq i \leq k$ such that $Tr(A^i) \geq 3$?


**Question 2**: If the answer to the above question is yes, then how $k$ depends on $n$? 


Proof of the claimed fact: Assume the contrary that $Tr(A^i) \leq 2$ for all natural numbers $i$. Let $\lambda\_1 , ... , \lambda\_n$ be the eigenvalues of $A$, therefore $T\_i := Tr(A^i) = \lambda\_1 ^i + ... + \lambda\_n ^i$. Since $T\_i$ are bounded above and $\lambda\_1 ... \lambda\_n =1$ then we should have $|\lambda\_i| = 1$ for each $1 \leq i \leq n$. This implies that all $\lambda\_1 , ..., \lambda\_n$ are roots of unity (because all of their Galois conjugates are on the unit circle) and therefore for some $j \in \mathbb{N}$ we have $\lambda\_1 ^j = ... = \lambda\_n ^j = 1$. This implies $T\_j \geq n \geq 3$. Contradiction.


PS: Replace the assumption $A \in SL(n,\mathbb{Z})$ by $A \in Sp(2n,\mathbb{Z})$, if it makes things easier.



Traces of powers of integral marices
Yes: for every $d\ge 3$ and all $A\in\mathrm{GL}\_d(\mathbf{Z})$, there exists $n\in\{1,\dots,12^d\}$ such that the trace of $A^n$ is $\ge 3$. (This is probably far from sharp, and I don't know if we can do better than exponential.)


First part: I claim that if $t=(t\_1,\dots,t\_d)$ is a $d$-tuple in the unit circle, then there exists $n\in\{1,\dots,12^d\}$ such that all $t\_1^n\dots t\_d^n$ have argument in $[-\pi/6,\pi/6]$. 


Indeed otherwise, denote $P\_\theta$ as the set of $d$-tuples in the unit circle of argument in $\mathopen]-\theta,\theta\mathclose[$. Then for all $m\neq n$ with $m,n\in\{-12^d/2,\dots,12^d/2\}$, we have $t^{m-n}$ not in $P\_{\pi/6}$. Define $V\_n=t^nP\_{\pi/12}$. Then the $V\_i$, for $i\in\{-12^d/2,\dots,12^d/2\}$, are pairwise disjoint. If we endow the torus with the normalized Haar measure, the measure of $P\_{\pi\theta}$, for $0\le\theta\le 1$ is equal to $\theta^d$. Hence the measure of $V\_i$ is $(1/12)^d$. So $(12^d+1)/12^d\le 1$, that is, $1+12^{-d}\le 1$, contradiction.


I conclude with the following claim. Consider a matrix in $A=\mathrm{GL}\_d(\mathbf{R})$ with $d\ge 3$, $|\det(A)|=1$, and all eigenvalues having argument in $[-\pi/6,\pi/6]$. Then the trace of $A$ is $>2$. In particular, if $A$ is integral then its trace is $\ge 3$.


Indeed, let $\lambda\_1,\dots,\lambda\_d$ be the eigenvalues, of modulus $r\_1\ge r\_2\ge \dots \ge r\_d$ (we can suppose that conjugate eigenvalues are written consecutively). If $r\_3\ge 1$, then the trace is $\ge (\sqrt{3}/2)(r\_1+r\_2+r\_3)\ge 3\sqrt{3}/2> 2$. 


Now suppose $r\_3<1$.
Note that $\prod r\_i=1$, so $r\_1r\_2=\prod\_{i\ge 3}r\_i^{-1}>1$, so $r\_2>1/r\_1$. If $\lambda\_1$ and $\lambda\_2$ are real, then the trace is $>r\_1+r\_2>r\_1+r\_1^{-1}\ge 2$. If $\lambda\_1$ is real and $\lambda\_2$ is nonreal, then $r\_2=r\_3$ and hence the trace is $\ge r\_1+2(\sqrt{3}/2)r\_2>r\_1+r\_2>r\_1+r\_1^{-1}\ge 2$. If $\lambda\_1$ is non-real and $\lambda\_3$ is real, then $r\_1=r\_2$ and the trace is $>\sqrt{3}r\_1+r\_2\ge r\_1+r\_1^{-1}\ge 2$. Finally if both $\lambda\_1$ and $\lambda\_3$ are non-real, then $k\ge 4$, $r\_1=r\_2$, $r\_3=r\_4$ the trace is $\ge \sqrt{3}(r\_1+r\_3)$; we have $r\_1^2r\_3^2\ge 1$ implying $r\_3\ge r\_1^{-1}$ and hence the trace is $\ge \sqrt{3}(r\_1+r\_1^{-1})\ge 2\sqrt{3}>3$.


