# 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^{} |
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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 |