# Gershgorin’s circle theorem result

Since the eigenvalues^{} of $A$ and $A$ transpose^{} are the same, you can get an additional set of discs which has the same centers, ${a}_{ii}$, but a radius calculated by the column ${\sum}_{j\ne i}|{a}_{ji}|$ (instead of the rows). If a disc is isolated it must contain an eigenvalue. The eigenvalues must lie in the intersection^{} of these circles. Hence, by comparing the row and column discs, the eigenvalues may be located efficiently.

Title | Gershgorin’s circle theorem result |
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Canonical name | GershgorinsCircleTheoremResult |

Date of creation | 2013-03-22 13:48:47 |

Last modified on | 2013-03-22 13:48:47 |

Owner | saki (2816) |

Last modified by | saki (2816) |

Numerical id | 11 |

Author | saki (2816) |

Entry type | Result |

Classification | msc 15A42 |