I am interested in infinite order elements $A\in SL(3, {\mathbb Z})$ whose spectra are not contained in ${\mathbb R}$ (i.e. such $A$ has two distinct complex-conjugate eigenvalues which are not roots of unity); I will refer to them as NRS matrices. 


**Question.** Is there a pair of commuting NRS matrices $A, B\in SL(3, {\mathbb Z})$ whose product is again an NRS matrix, such that $A, B$ generate a non-cyclic subgroup of $SL(3, {\mathbb Z})$? 


As the last resort, one can look for such matrices by computer-search, but I would prefer to avoid doing this. 



integer matrices with non-real spectra
There are no such pairs.


Let $\lambda\_1$, $\lambda\_2$, $\lambda\_3$ be the eigenvalues of $A$, and $\mu\_1,\mu\_2,\mu\_3$ be those of $B$ (with $\lambda\_1,\mu\_1\in\mathbb R$). Since the eigenvalues are distinct, $A$ and $B$ are diagonalizable; since they commute, they are simultaneously diagonalizable, i.e., an eigenbasis for $A$ is also that for $B$. We assume that in a common eigenbasis, $\lambda\_i$ corresponds to $\mu\_i$ (clearly, $\lambda\_1$ corresponds to $\mu\_1$ --- because only these eigenvectors may be chosen real).


Now let $e\_2$ be an eigenvector for $A$ corresponding to $\lambda\_2$. The linear system defining it has coefficients in $K=\mathbb Q[\lambda\_2]$, so we may assume that the elements of $e\_2$ are also in $K$. Writing down the condition that $e\_2$ is an eigenvector of $B$ (with eigenvalue $\mu\_2$) we get that $\mu\_2\in K$. Thus $\mathbb Q[\mu\_2]=K=\mathbb Q[\lambda\_2]$. Moreover, $\lambda\_2$ and $\mu\_2$ belong to the units group of $K$ --- which, [as is known](https://en.wikipedia.org/wiki/Cubic_field#Unit_group), is cyclic. 


Thus $\lambda\_2^k=\mu\_2^\ell$ for some $k$ and $\ell$, which yields also $\lambda\_i^k=\mu\_i^\ell$, and hence $A^k=B^\ell$. Thus the subgroup generated by $A$ and $B$ is cyclic.


