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$B$ is a non-negative irreducible block matrix as follows:

$B= \left[ \begin{array}{c|c|c} 0 &B_{12}&B_{13}\\ \hline B_{21}& 0& B_{23}\\ \hline B_{31}& B_{32}&0 \end{array} \right]$.

Is there any relation between the maximum positive eigenvalue of $B$ and the maximum positive eigenvalue of the following matrix:

$M= \left[ \begin{array}{c|c} B_{21}B_{12}& B_{21}B_{13}+B_{23}\\ \hline B_{31}B_{12}+B_{32}B_{21}B_{12}&B_{31}B_{13}+B_{32}B_{21}B_{13}+B_{32}B_{23} \end{array} \right]$?

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  • $\begingroup$ Are you maybe in the special case in which the Perron eigenvalue of $B$ is $1$? Because $M$ is obtained from $B$ by censoring the first block of states, in the language of Markov chains, i.e., $I-M$ is a Schur complement of $I-B$; in particular, if $B$ is stochastic then $M$ is, too, and the Perron eigenvector of $M$ is given by chopping off the last first block of that of $B$. $\endgroup$ – Federico Poloni Nov 20 '18 at 7:49
  • $\begingroup$ In fact, the maximum eigenvalue of $B$ is positive and greater than 1. Also $M$ is coming from the following multiplication:$B_3B_2B_1 in which $\\ $ B_1=\left[ \begin{array}{c|c|c} 0 &B_{12}&B_{13}\\ \hline 0& I& 0\\ \hline 0& 0&I \end{array} \right]$ \ $ B_2=\left[ \begin{array}{c|c|c} I &0&0\\ \hline B_{21}& 0& B_{23}\\ \hline 0& 0&I \end{array} \right]$ \ $ B_3=\left[ \begin{array}{c|c|c} I &0&0\\ \hline 0& I& 0\\ \hline B_{31}& B_{32}&0 \end{array} \right]$ $\endgroup$ – afra Nov 20 '18 at 13:04

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