Let $G$ be a finitely generated subgroup of $GL(n,\mathbb{C})$. Assume that there exists a number field $k$ (i.e. a finite extension of $\mathbb{Q}$) such that for all $g \in G$, the eigenvalues of $g$ all lie in $k$. This implies that $g$ is conjugate to an element of $GL(n,k)$.


Question: must it be the case that some conjugate of $G$ lies in $GL(n,k)$? Or at least $GL(n,k')$ for some finite extension $k'$ of $k$? If this is not true, what kinds of assumptions can I put on $G$ to ensure that it is?



Finitely generated matrix groups whose eigenvalues are all algebraic
At the positive side, if $G$ acts irreducibly on $\mathbf{C}^n$ and $k$ is an arbitrary subfield of $\mathbf{C}$, then the answer is yes (allowing some field extension $k'$ of degree dividing $n$). This even works assuming that $G$ is a multiplicative submonoid of $M\_n(\mathbf{C})$ (keeping the irreducibility assumption). 


See for instance Proposition 2.2 in H. Bass, Groups of integral representation type. Pacific J. Math. 86, Number 1 (1980), 15-51. ([ProjectEuclid link](https://projecteuclid.org/euclid.pjm/1102780613), unrestricted access)


Robert Israel's simple example shows that some assumption such as irreducibility has to be done.


