Expansiveness of automorphism of n-torus Let $A$ be an automorphism of the $n$-torus, and $[A]$ the corresponding matrix. Then $A$ is expansive iff the linear map induced by $[A]\in GL_n(\mathbb{Z})$ of $\;\mathbb{R^n}\;$ is expansive.
( A homeomorphism $T:X\to X$ of a metric space $(X,d)$ is said to be expansive if $\;\exists\delta>0$ such that if $x\neq y$ then $\;\exists n\in\mathbb{Z}$ s.t. $d(T^n(x),T^n(y))>\delta$ )
The forward implication is clear. I am stuck on the reverse implication.
This is one of the step required to prove the result that $A$ is expansive iff $[A]$ has no eigenvalues of modules $1$ as outlined in P.Walters text. Any hints or other references to proofs of the latter result would be appreciated.
 A: It easy to see that the map on $\mathbb{R}^n$ induced by $[A]$ is expansive iff for all $x\in\mathbb{R}^n$ the set $\{\|[A]^nx\|:n\in\mathbb{Z}\}$ is unbounded. Also note that, in our case, it is enough to find a constant satisfying the expansiveness condition only for pairs $(x,0)$.   Let $c=max\left\{\|A\|,\|A^{-1}\|\right\}$ and $\delta=min\left\{\frac{1}{2c},\frac{1}{4}\right\}$. I claim that $\delta$ is a expansive constant for $A$. To see this, let $x\in [0,1]^n$. If $\|x\|>\delta$, then $$d\left(A^0(x),A^0(0)\right)=\|x\|>\delta$$ So suppose $\|x\|\leq\delta$. Let $k=\inf\left\{|n|:\|[A]^nx\|>\delta\right\}$(its non empty since we are assuming the orbits are unbounded). Then (atleast) one of $$\|[A]^kx\|>\delta$$ $$\|[A]^{-k}x\|>\delta $$ holds. Suppose the first one holds. Then $$\delta<\|[A]^kx\|=\|[A]([A]^{k-1}x)\|\leq c\|[A]^{k-1}x\|\leq c\delta\leq \frac{1}{2} $$ which shows that $[A]^kx\in[-1,1]^n$ and hence $$ d\left(A^k(x),A^k(0)\right)=\|[A]^kx\|>\delta. $$ Similar for the case $\|[A]^{-k}x\|>\delta $. Since $x\in[0,1]^n$ was arbitrary, we are done.
