# Weyl’s inequality

Let $A$ and $E$ be two $n\times n$ Hermitian matrices^{}, with $E$ positive semidefinite^{}.

Let ${\lambda}_{i}(A)$, ${\lambda}_{i}(A+E)$, $1\le i\le n$ be the eigenvalues^{} of $A$ and $A+E$ respectively, ordered in such a way that

$$|{\lambda}_{1}|\le |{\lambda}_{2}|\le \mathrm{\cdots}\le |{\lambda}_{n}|.$$ |

Then

$${\lambda}_{i}(A)\le {\lambda}_{i}(A+E).$$ |

Title | Weyl’s inequality |
---|---|

Canonical name | WeylsInequality |

Date of creation | 2013-03-22 15:33:37 |

Last modified on | 2013-03-22 15:33:37 |

Owner | Andrea Ambrosio (7332) |

Last modified by | Andrea Ambrosio (7332) |

Numerical id | 10 |

Author | Andrea Ambrosio (7332) |

Entry type | Theorem |

Classification | msc 15A42 |