# Non-negative matrices with prescribed Perron-Frobenius eigenvectors

In my research I came across the following question.

Let $A$ be an integer non-negative matrix (every entry of $A$ is non-negative) and $x = (x_1,...,x_n)^T$ the probability Perron-Frobenius eigenvecor, i.e., $Ax = \lambda x$ and all $x_i \geq 0$. Denote by $H(x_1,...,x_n)$ the additive abelian group generated by $x_1,...,x_n$. Consider the set $\mathcal M(A)$ of all integer non-negative matrices $B$ such that $By = \lambda y$ where $y= (y_1,...,y_k)^T$ is the probability Perron-Frobenius eigenvecor. Suppose additionally that $H(x_1,...,x_n) =H(y_1,...,y_k)$. Is the set $\mathcal M(A)$ finite for every fixed $A$?

I'll be thankfull for any comments or suggestions.

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Unless I misunderstood something, there is an obvious possibility of adding zeros, so your question should probably be slightly changed. E.g. let $A=1$ (1 by 1 matrix), and $B=E_{11}$ ($k$ by $k$ matrix). Then $x_1=1$, $\lambda=1$, so $H(x_1)=\Bbb Z$, and one can use $y_1=1$, $y_2=...y_k=0$. Do you want to assume that the matrices are irreducible? indecomposable? That $x_i,y_j>0$? –  Pavel Etingof Feb 28 '10 at 17:17
Thank you for your comments. The case of irreducible matrices is the most interesting for me. But in more general context, I would admit that some entries of eigenvectors are zero. They of course do not contribute to the group $H$ but extend the set of possible matrices $B$. I'm not sure I understand your second comment. The point is that $\lambda$ and the entries of eigenvectors might be irrational algebraic numbers. –  SIB Mar 1 '10 at 8:25
Sorry, you are right. I deleted this comment. –  Pavel Etingof Mar 1 '10 at 13:03
Could you put the notation in the question in $..$ latex form ? it's hard to read this way. –  Suresh Venkat Mar 1 '10 at 21:51
Without the assumption of irreducibility of $B$, there seems to be the following counterexample, even if we restrict to $x_i,y_j>0$. Let $A=E_{12}+E_{21}+E_{22}$, a 2 by 2 matrix. Then $\lambda$ is the golden ratio, $x_1=\lambda^{-2}$, $x_2=\lambda^{-1}$, and so the group $H$ generated by $x_1$, $x_2$ is the ring of integers ${\mathcal O}$ of the real quadratic field ${\Bbb Q}(\lambda)$. The ring ${\mathcal O}$ is dense in the real line, so for any $n$ there are $n$-tuples $(a_1,...,a_n)\in (0,1)\cap {\mathcal O}$ such that $a_1+...+a_n=1$. Then the vectors $(a_1x_1,a_1x_2,a_2x_1,a_2x_2,...,a_nx_1,a_nx_2)^T$ work as vectors $y$ for $B=A\oplus...\oplus A$ ($n$ times). (As I understand, you allow $k$ to vary. Otherwise there are finitely many $B$ without any assumption on $H$ since the norm of $B$ is $\lambda$, where I use the norm $||z||:=\sum y_i|z_i|$ on row vectors. Indeed, the set of such $B$ is both discrete and compact.)
Yes, you are right: $B$ must be irreducible to avoid some trivialities. Thank you for pointing out this fact. The other remark is also true: the size of $B$ can vary. –  SIB Mar 2 '10 at 13:04