Spectral radius of product of matrices with same left and right eigenvector

Question

Suppose $A, B\in \mathbf{R}^{n\times n}$ where $A\mathbf{1}=B\mathbf{1}=\mathbf{1}$, $\mathbf{w}^TA=\mathbf{w}^TB=\mathbf{w}^T$, $\mathbf{w}\geq 0$ and $\rho(A)=\rho(B)=1$. Prove that $\rho(AB)=1$?

Answer 1

No. E.g. when $w\ne0$, since it is also nonnegative, we have $w^T\mathbf1\ne0$. Therefore, there exists an invertible matrix $P$ such that $\mathbf1$ and $\frac{1}{w^T\mathbf1}w^T$ are respectively the first column of $P$ and the first row of $P^{-1}$. The three given conditions in your question thus implies that $$ P^{-1}AP=\pmatrix{1&0\\ 0&X} \quad\text{and}\quad P^{-1}BP=\pmatrix{1&0\\ 0&Y} $$ for some matrices $X$ and $Y$ such that $\rho(X),\rho(Y)\le1$. So, you are essentially asking whether $\rho(X),\rho(Y)\le1\implies\rho(XY)\le1$. The answer is clearly negative, as shown by counterexample below: $$ X=Y^T=\pmatrix{0&2\\ 0&0},\quad\rho(X)=\rho(Y)=0<1,\quad\rho(XY)=4>1. $$ To construct a complete counterexample, we may take $w^T=(1,0,0)$, $$ P=\pmatrix{1&0&0\\ 1&1&0\\ 1&0&1}, \quad P^{-1}=\pmatrix{1&0&0\\ -1&1&0\\ -1&0&1}, $$ $$ A=P(1\oplus X)P^{-1}=\pmatrix{1&0&0\\ -1&0&2\\ 1&0&0}, $$ $$ B=P(1\oplus Y)P^{-1}=\pmatrix{1&0&0\\ 1&0&0\\ -1&2&0}, $$ $$ AB=\pmatrix{1&0&0\\ -3&4&0\\ 1&0&0}. $$ However, your conjecture is true when $A$ and $B$ are nonnegative matrices. In this case, that $A\mathbf1=B\mathbf1=\mathbf1$ means that the two matrices are stochastic matrices. Hence $AB$ is also stochastic and $\rho(AB)=1$. The condition that $w^TA=w^TB=w^T$ is irrelevant in this case.

Remark. Another counterexample. Here we take $w=\mathbf 1$, $$ X=Y^T=\pmatrix{0&2\\ \frac14&0}, \quad\text{and}\quad P=\pmatrix{1&1&1\\ 1&-1&1\\ 1&0&-2} $$ so that each of $$ A=P(1\oplus X)P^{-1}=\frac{1}{24} \pmatrix{19&13&-8\\ 3&-3&24\\ 2&14&8}, $$ $$ B=P(1\oplus Y)P^{-1}=\frac{1}{24}\pmatrix{33&-15&6\\ 31&-17&10\\ -40&56&8} $$ and $$ AB=\frac{1}{288}\pmatrix{675&-477&90\\ -477&675&90\\ 90&90&108} $$ is an entrywise nonzero matrix with three distinct eigenvalues. The common spectrum of $A$ and $B$ is $\{1,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\}$ but the eigenvalues of $AB$ are $1,4$ and $\frac{1}{16}$.