integer matrices with non-real spectra 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. 
 A: 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, 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.
A: The eigenvalues $\lambda_1,\lambda_2,\lambda_3$ of a matrix $M\in SL_3(\mathbb{Z})$ satisfy:
$\lambda_1\lambda_2\lambda_3=\det M=1$.
Let $f(x)\in\mathbb{Z}[x]$ be the characteristic polynomial of $M$, then $\lambda_i$ is a root of $f(x)$.
Let $f(x)=x^3+tx-1$ for some $t\in\mathbb{Z}$, then the discriminant $d_{f}=-4*t^3 - 27$. Now if you pick $t>0$, then $d_f<0$, then $f(x)$ has two complex root and one real root.   
To find a choice for $M$, let $K$ be the number field defined by $K(\alpha)=\mathbb{Q}[x]/(f(x))$.
Then the ring of integer $\mathcal{O}_K$ is a free $\mathbb{Z}$-module of rank $3$. Now multiplication by $\alpha$ induces an Endomorphism $\phi(\alpha)\in End_{\mathbb{Z}}(\mathcal{O}_K)$. Since $\mathcal{O}_K$ is free $\mathbb{Z}$ module of rank $3$, we have $\phi(\alpha)\in GL_{3}(\mathbb{Z})$ since $\alpha$ is invertible in $\mathcal{O}_K$. Choose the sign of $\alpha$ such that $\det(\phi(\alpha))=1$, then we have $\phi(\alpha)\in SL_3(\mathbb{Z})$. Note that the characteristic polynomial of $\phi(\alpha)$ has $\alpha$ as a root and hence an irreducible polynomial. Then the conjugate of $\alpha$ also satisfies it. Now you can take $M=\phi(\alpha)$. 
