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Let $ A = (a_{ij})$ be a matrix of size $ n^2 $ such that $ a_{ij} > 0$. Let $x = (x_1,\ldots,x_n ) $ and$ y = (y_1,\ldots,y_n) = Ax$. Show that there exist an eigenvalue $\lambda $ of A, such that: $ \lambda \in \left[ {\min \left\{ {\frac{{y_i }} {{x_i }}} \right\}_{i = 1}^n ,\max \left\{ {\frac{{y_i }} {{x_i }}} \right\}_{i = 1}^n } \right]. $ I have no idea how to do it :S!!

Someone knows this result or related? or some solution?

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    You may want to look into the Perron-Frobenius theorem. I believe this problem is an exercise in Meyer's "Matrix Analysis."2011-11-19

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This sounds very like the Frobenius-Perron Theorem. The standard Frobenius-Perron Theorem can be thought of as asserting the result you want, but only for the special case $(x_1,\dots,x_n)=(1,\dots,1)$. (Multiplying by this vector gives us the row sums of the Theorem.) However, we can use Frobenius-Perron to obtain information from $(x_1,\dots,x_n)$ with positive $x_i$.

If the $x_i$ are positive, we can modify your matrix $A$ by multiplying the $i$-th column of $A$ by $x_i$. Then if A' is the modified matrix, and $e$ is the vector of all $1$'s, we have Ax= A'e. Now standard Frobenius-Perron, applied to A', should give what you want.

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    As a start, do look at the Wikipedia article. Look carefully at the inequality for the largest eigenvalue.2011-11-20