I am seeking the following example, maybe it is easy to construct, but I have no idea now besides aimless computation.
Could anyone give me a matrix $A\in SL_n(\mathbb{Z})$ for some $n$, such that $A$ has an eigenvalue $\lambda$ such that $|\lambda|\geq 2$ and the eigenspace correspondence to $\lambda$ contains a $n$-column vector with all entries belong to integers.
Thanks in advance!
Such $A$ doesn't exist.
If $A$ has an eigenvector whose components are all integers, the corresponding eigenvalue $\lambda$ will be a rational number. Since $A \in SL_{n}(\mathbb{Z})$, $\det(A) = 1$. This implies the constant term in $\det(\lambda I_n - A)$, the characteristic polynomial of $A$, is $(-1)^n$. By Rational Root theorem, $\lambda$ can only be $\pm 1$.