I have the following question which I think should be simple to show but I have not succeeded so far in. I have a primitive matrix $A\in \big(\mathbb{N}_0\big) _{n\times n}$, with $\mathbb{N}_0=\{0,1,2,...\}$. I know by the Perron-Frobenius theory that $A$ has a simple leading eigenvalue $\lambda$, with a corresponding eigenvector $v=(v_1,...,v_n)$ such that $\sum v_j=1$ and $v_j>0$ for all $j$. If I assume that $\lambda\in \mathbb{Q}$, does this imply that $v_1,...,v_n\in \mathbb{Q}$ ?
I've numerically computed this for some examples, but I was wondering whether this has to be the case or is there possibly a counter-example?
Later edit: I also know that this does not work when I assume that the Perron-Frobenius eigenvector $\lambda$, is not rational. For example, for $A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$, the Perron-Frobenius eigenvetor is $\lambda = \frac{1+\sqrt{5}}{2}$. Its normalized Perron-Frobenius eigenvector is $(\frac{1}{\lambda},\frac{1}{\lambda^2})$.
Yes. For any singular matrix with entries in a field $\mathbb F$, the standard row reduction methods to find vectors in the null space produce a vector with entries in $\mathbb F$.