Non-negative matrices with prescribed Perron-Frobenius eigenvectors - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:55:06Z http://mathoverflow.net/feeds/question/16670 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16670/non-negative-matrices-with-prescribed-perron-frobenius-eigenvectors Non-negative matrices with prescribed Perron-Frobenius eigenvectors SIB 2010-02-28T11:13:31Z 2010-03-01T14:39:30Z <p>In my research I came across the following question. </p> <p>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 $A$?</p> <p>I'll be thankfull for any comments or suggestions. </p> http://mathoverflow.net/questions/16670/non-negative-matrices-with-prescribed-perron-frobenius-eigenvectors/16759#16759 Answer by Pavel Etingof for Non-negative matrices with prescribed Perron-Frobenius eigenvectors Pavel Etingof 2010-03-01T14:39:30Z 2010-03-01T14:39:30Z <p>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.)</p>