Exercise 18, § 6

Let $A$ be a real square matrix of type $(I,I)$, positive and irreducible. Let $\varrho$ be the spectral radius of A. For all $x\in \mathbf{R}^I$ such that $y\succcurlyeq 0$ and $x\not = 0$, we put

$\varrho^x=$ sup$_{i\in I}\frac{(Ax)_i}{x_i}$,

(with the convention $\varrho^x= +\infty$ if there exists an index $i$ such that $x_i= 0$ and ($Ax$)$_i\not = 0$). Let

$\varrho '=$ inf$_{xx\succ\not=00}\varrho^x$.

Following a procedure similar to that of exercise 12, now prove that $\varrho '$ is an eigenvalue of A and that $\varrho '=\varrho$.