Showing that the eigenvectors (when eigenvalue is 1) can be chosen to be integer valued Suppose $A$ is an $d \times d$ matrix with integer entries. 
If there exists $\underline{n} \neq 0 \in \mathbb{R}^d$ such that $(A^T)^k \underline{n}= \underline{n}$.
How can you show/justify that you can choose $\underline{n}\in \mathbb{Z}^d$?

If you have a rational matrix with rational eigenvalues, then it is not too hard to see that the eigenvectors can be chosen to be rational. (Just get $A-\lambda I$ to reduced echelon form and then take all free variables to be rational.) Now if $v$ is a rational vector, what's an integer vector which is a scalar multiple of $v$?
