Perron-Frobenius theorem

Let $A$ be a nonnegative matrix. Denote its spectrum by $\sigma (A)$. Then the spectral radius $\rho (A)$ is an eigenvalue, that is, $\rho (A)\in \sigma (A)$, and is associated to a nonnegative eigenvector.

If, in addition, $A$ is an irreducible matrix, then $|\rho (A)|\ge |\lambda |$, for all $\lambda \in \sigma (A)$, $\lambda \ne \rho (A)$, and $\rho (A)$ is a simple eigenvalue associated to a positive eigenvector.

If, in addition, $A$ is a primitive matrix, then $\rho (A)>|\lambda |$ for all $\lambda \in \sigma (A)$, $\lambda \ne \rho (A)$.

Title	Perron-Frobenius theorem
Canonical name	PerronFrobeniusTheorem
Date of creation	2013-03-22 13:18:26
Last modified on	2013-03-22 13:18:26
Owner	jarino (552)
Last modified by	jarino (552)
Numerical id	5
Author	jarino (552)
Entry type	Theorem
Classification	msc 15A18
Related topic	FundamentalTheoremOfDemography