Rational left eigenvector in Perron-Frobenius theorem with nonnegative integer matrix

Question

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.

First, working over $\mathbb Q$, look at $\dim \ker (A^T-I)$ then working over $\mathbb R$ look at $\dim \ker (A^T-I)$. They are the same dimension, so the eigenvector lives in $\mathbb Q$. ref https://math.stackexchange.com/questions/3777401/does-real-dimension-equal-rational-dimension/ — user8675309, Dec 19 '21 at 17:56
Does this answer your question? Does real dimension equal rational dimension? — Jacob Manaker, Dec 20 '21 at 05:15
Yes, or even simpler the boneheadedly obvious point that I put in my answer below. The question has been closed so I guess I'll add a bit more about it. — Sophie M, Dec 20 '21 at 18:09

score 1 · Answer 1 · answered Dec 18 '21 at 18:51

1

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.

answered Dec 18 '21 at 18:51

Sophie M

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Rational left eigenvector in Perron-Frobenius theorem with nonnegative integer matrix

1 Answers1