I've been trying to go through this paper for the past month, but I'm stuck on Lemma 4.4
So, the argument goes as follows: g is an integer $n{\times}n$ matrix with irreducible characteristic polynomial (over $\mathbb Z[x]$). and $\psi$ is any real $n{\times} n$ matrix that commutes with $g$ (so it has to be diagonal)
Now, the author says
"The space of complex solutions, $L_c \subset M_n(\mathbb C)$ of the equation $\psi g=g \psi$ with unknown vector $\psi \in M_n(\mathbb C)$ has also dimension $n$. It follows that the space of real solutions $L_r = L_c \cap M_n( \mathbb R)$ is $n$-dimensional over $\mathbb R$ since the coefficients of the equation are reals."
I know that they're simultaneously diagonalizable and that the columns of $g$ span $\mathbb R^n$ i.e. an n-dimensional vector space, but I don't understand why the same must be true for $\psi$, because, for all I know, it could very well be the $0$ matrix.
Now, I think I'm missing something obvious here, but it's beaten me for more than a week now, so I've thrown in the towel, and here I come seeking some help.
The characteristic polynomial of $g$ is irreducible over $\mathbb C$, so its roots are all distinct and it is diagonalizable, and is diagonal in a basis of eigenvectors. The system $\psi g - g \psi = 0$ has kernel of dimension $n$ (over $\mathbb C$), the solutions being all matrices that are diagonal in this basis. Now the entries of $g$ are real, so that system (as an $n^2 \times n^2$ system for the entries $\psi_{ij}$ of $\psi$) has real coefficients. A real $n^2 \times n^2$ system with kernel of dimension $n$ has a basis with real entries, which could e.g. be obtained by row reduction.