Non-singular matrices and if they can be unipotent.

Question

If a matrix W is non-singular and all its elements are integers and its not 'periodic' ( that is there exists no t such that W^t = W) then let the maximum element of W be c ( i.e., max(W) = c). If for any u , max(W^u) < = c then W would be 'periodic'. This implies if W is non-singular and non-periodic then given u is an element of positive integers max(W^u) can be as large as one likes. So for any large value T there exists u such that some element of W^u is > T. Is this true?

Answer 1

You have to assume that the sequence $W, W^2,W^3,\dots $ does not get periodic at some point.

Otherwise $W=\pmatrix{0&1\\0&0}$ is a counter-example to your claim.

You cannot prove the claim for the maximal element of $W^k$, but only for the elements of maximal absolute value: For $W=\pmatrix{1&-1&0\\0&1&0\\0&0&0}$, it holds $$ W^k = \pmatrix{1&-k&0\\0&1&0\\0&0&0}, $$ hence the entries of $W^k$ never exceed $1$, however the maximum of the absolute value goes to infinity.

Let me suppose now that for $j\ne k$ it holds $W^j \ne W^k$. Now take a positive integer $N$. Then the following set $$ M_N:=\{ A\in \mathbb Z^{n,n}: \ |a_{ij}|\le N \ \forall i,j=1\dots n\} $$ is finite.

All matrices in the sequence $W,W^2,W^3,\dots$ are integer matrices. Since this sequence contains an infinite number of matrices, there must be an infinite number of matrices $W^k \not \in M_N$. Thus for any large $N$ there is always some $k$ such that $W^k \not\in M_N$.