# matrix resolvent properties

The matrix resolvent norm for a complex-valued $s$ is related to the proximity of such value to the spectrum of $A$; more precisely, the following simple yet meaningful property holds:

$$\parallel {R}_{A}(s)\parallel \ge \frac{1}{\mathrm{dist}(s,{\sigma}_{A})},$$ |

where $\parallel .\parallel $ is any self consistent matrix norm^{}, ${\sigma}_{A}$ is the spectrum of $A$ and the distance between a complex point and the discrete set of the eigenvalues^{} ${\lambda}_{i}$ is defined as $\mathrm{dist}(s,{\sigma}_{A})=\underset{1\le i\le n}{\mathrm{min}}|s-{\lambda}_{i}|$.

From this fact it comes immediately, for any $1\le i\le n$,

$$\underset{s\to {\lambda}_{i}}{lim}\parallel {R}_{A}(s)\parallel =+\mathrm{\infty}.$$ |

###### Proof.

Let (${\lambda}_{i}$,$\mathbf{v}$) be an eigenvalue-eigenvector pair of $A$; then

$$(sI-A)v=sv-Av=(s-{\lambda}_{i})v$$ |

which shows $(s-{\lambda}_{i})$ to be an eigenvalue of $(sI-A)$; ${(s-{\lambda}_{i})}^{-1}$ is then an eigenvalue of ${(sI-A)}^{-1}$ and , since for any self consistent norm $|\lambda |\le \parallel A\parallel $, we have:

$$\underset{1\le i\le n}{\mathrm{max}}\frac{1}{|s-{\lambda}_{i}|}\le \parallel {(sI-A)}^{-1}\parallel $$ |

whence

$$\parallel {(sI-A)}^{-1}\parallel \ge \frac{1}{\underset{1\le i\le n}{\mathrm{min}}|s-{\lambda}_{i}|}=\frac{1}{\mathrm{dist}(s,{\sigma}_{A})}.$$ |

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Title | matrix resolvent properties |
---|---|

Canonical name | MatrixResolventProperties |

Date of creation | 2013-03-22 15:33:52 |

Last modified on | 2013-03-22 15:33:52 |

Owner | Andrea Ambrosio (7332) |

Last modified by | Andrea Ambrosio (7332) |

Numerical id | 15 |

Author | Andrea Ambrosio (7332) |

Entry type | Result |

Classification | msc 15A15 |