I came across this answer which claims that not every Perron number admits a primitive non-negative integral matrix representation. This seems to contradict Lind's theorem, which states:
If $\lambda$ is a Perron number, then there is an aperiodic non-negative integral matrix with spectral radius $\lambda$.
How is this possible?
The proposed counterexample is the polynomial: $$p(x)=x^6−4x^4−5x^3+10x^2−12x+6$$
The largest root of the polynomial is the Perron number 2.2508.
There is no contradiction, because Lind’s theorem has fewer restrictions on the primitive matrix.
Let $p\in\mathbb Z[x]$ be the minimal polynomial of a Perron number $\lambda$. In the cited counterexample, the question it addresses requires that the primitive integral matrix $A$ is similar to the companion matrix of $p$ (which is $x^6−13x^4−20x^3+x^2−x+2$ in that counterexample). In other words, the question requires not only that $p$ is the characteristic polynomial of $A$, but also that $A$ is non-derogatory. If $A$ is $n\times n$, the question also implicitly requires that $n=\deg p$.
In contrast, Lind’s theorem only requires $\lambda$ to be equal to the spectral radius of $A$. It allows $n>\deg p$. It does not require $p$ to be the characteristic polynomial of $A$ (but by the definitions of $p$ and $A$, $p$ must divide the characteristic polynomial of $A$). Nor does it require $A$ to be non-derogatory.
In fact, on p.289, Lind has mentioned that if $\lambda\approx3.8916$ is the Perron number of $p(x)=t^3+3t^2-15t-46$, his algorithm will generate a matrix $A$ with $n=10>3=\deg p$, and so $p$ is only a factor of, but not identical to, the characteristic polynomial of $A$.
