Operatornorm of powers of Matricies with integer coefficient

Question

Let $A\in GL_n(\mathbb{Z})$ have infinite order, so $A^k\neq Id_n$ for all $k>0$. The operator norm is defined by $\lVert A \rVert=\max\{\lVert Av\rVert \mid v\in\mathbb{R}^n: \lVert v\rVert=1\}$. It´s easy to see, that $\lVert A\rVert > 1$. My question is:

Is $\lVert A^k \rVert\geq k$ for all (big enough) $k$? If not, is there a constant $C>0$ with $\lVert A^k \rVert\geq Ck$ for all (big enough) $k$?

It feels like there should be some easy solution, since there are only finitely many matrices with integer coefficients with operator norm at most $k$ (or $Ck$), and since $A$ has infinite order, I should eventually outgrow them, but I have no nice formal argument.

Answer 1

There are two cases:

1. $\rho(A)>1$. Then $\|A^k\|_2\ge\rho(A^k)=\rho(A)^k$.
2. $\rho(A)\le1$. Then $1=|\det(A)|=\prod_{i=1}^n|\lambda_i|\le1$, where the $\lambda_k$s denote the eigenvalues of $A$ over $\mathbb C$. Therefore $|\lambda_i|=1$ for every $i$. In turn, the characteristic polynomial of $A$ is a monic polynomial with integer coefficients and unimodular roots. It follows from a result by Kronecker that all $\lambda_i$s are roots of unity. Hence $A$ is not diagonalisable over $\mathbb C$, or else $A$ will have a finite order. Therefore, some eigenvalue $\lambda_i$ of $A$ is not semi-simple, i.e., the Jordan form $J$ of $A$ contains a principal submatrix $\pmatrix{\lambda_i&1\\ 0&\lambda_i}$. It follows that $$ \|J^k\|_F \ge\left\|\pmatrix{\lambda_i&1\\ 0&\lambda_i}^k\right\|_F =\left\|\pmatrix{\lambda_i^k&k\lambda_i\\ 0&\lambda_i^k}\right\|_F >k. $$

It follows from the two cases above that there exists come $C>0$ such that $\|A^k\|_2\ge Ck$ for every integer $k\ge0$.