TITLE: Rational left eigenvector in Perron-Frobenius theorem with nonnegative integer matrix
QUESTION [1 upvotes]: On pages 1133-1135 of Joel Friedman's On the road coloring problem, (PAMS 1990),  we have an $n \times n$ primitive integer matrix $A$ with all row sums equal to $d \geq 2$. Clearly the right Perron eigenvector $\mathbf{v}$ is constant, i.e. $v_i = v_j$ for all $i,j$. Friedman claims that the entries of the left Perron eigenvector $\mathbf{w}$ are commensurable, i.e. by scaling $\mathbf{w}$ appropriately we can take $w_i \in \mathbb{N}$ for all $i$. It's probably very easy to see why this is true, but I'm not seeing it.
Additional material, after the question was closed: this is an important paper in the literature on the road coloring problem. The divisibility properties of the sum of the entries of $\mathbf{w}$ (once those entries are normalized to be integers with gcd $1$) are the point of the paper, and because I forgot some high-school math, I wasn't seeing why that normalization was possible.

REPLY [0 votes]: Oh, wow, this ended up being really simple, and it doesn't use constant row sums. $D w_j = \sum_i w_i A_{ij}$, then Gaussian elimination.
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