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Let A be a matrix with positive entries. The perron-frobenius-theorem states that A has a positive dominating simple eigenvalue, called the perron-frobenius-eigenvalue. I denote it with p(A). The Collatz-Wielandt-formula allows to estimate the eigenvalue. Using the vector $e_j$, which has a one one position j and 0 elsewhere, it can be shown that $a_{jj}\le p(A)$ for any j. So, p(A) is not smaller than any of the diagonalelements.

My question is whether equality is possible. Is there a positive matrix A, such that $a_{jj}=p(A)$ for some j ?

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The answer is NO, because there is a positive eigenvector $v$ associated to the eigenvalue $p(A)$ ; that implies that $Av=a_{j,j}v$ is impossible.

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