# proof of Weyl’s inequality

Let ${\lambda}_{i}$ be the i-th eigenvalue^{} of $A+E$. Then, by the Courant-Fisher min-max theorem and being ${x}^{H}Ex\ge 0$ by hypothesis, we have:

${\lambda}_{i}(A+E)=\underset{S,dimS=i}{\mathrm{max}}\underset{\parallel x\parallel \ne 0}{\mathrm{min}}\frac{{x}^{H}(A+E)x}{{x}^{H}x}=$

$=\underset{S,dimS=i}{\mathrm{max}}\underset{\parallel x\parallel \ne 0}{\mathrm{min}}\left(\frac{{x}^{H}Ax}{{x}^{H}x}+\frac{{x}^{H}Ex}{{x}^{H}x}\right)\ge \underset{S,dimS=i}{\mathrm{max}}\underset{\parallel x\parallel \ne 0}{\mathrm{min}}\frac{{x}^{H}Ax}{{x}^{H}x}={\lambda}_{i}(A)$.

Title | proof of Weyl’s inequality |
---|---|

Canonical name | ProofOfWeylsInequality |

Date of creation | 2013-03-22 15:33:39 |

Last modified on | 2013-03-22 15:33:39 |

Owner | Andrea Ambrosio (7332) |

Last modified by | Andrea Ambrosio (7332) |

Numerical id | 9 |

Author | Andrea Ambrosio (7332) |

Entry type | Proof |

Classification | msc 15A42 |