# properties of diagonally dominant matrix

1)(Levy-Desplanques theorem) A strictly diagonally dominant matrix is non-singular.

###### Proof.

Let $A$ be a strictly diagonally dominant matrix and let’s assume $A$ is singular, that is, $\lambda =0\in \sigma (A)$. Then, by Gershgorin’s circle theorem, an index $i$ exists such that:

$$\sum _{j\ne i}\left|{a}_{ij}\right|\ge \left|\lambda -{a}_{ii}\right|=\left|{a}_{ii}\right|,$$ |

which is in contrast with strictly diagonally dominance definition. ∎

2)() $\left|det(A)\right|\ge {\prod}_{i=1}^{n}\left(\left|{a}_{ii}\right|-{\sum}_{j=1,j\ne i}\left|{a}_{ij}\right|\right)$ (See here (http://planetmath.org/ProofOfDeterminantLowerBoundOfAStrictDiagonallyDominantMatrix) for a proof.)

3) A Hermitian diagonally dominant matrix with real nonnegative diagonal entries is positive semidefinite^{}.

###### Proof.

Let $A$ be a Hermitian diagonally dominant matrix with real nonnegative diagonal entries; then its eigenvalues^{} are real and, by Gershgorin’s circle theorem, for each eigenvalue an index $i$ exists such that:

$$\lambda \in [{a}_{ii}-\sum _{j\ne i}\left|{a}_{ij}\right|,{a}_{ii}+\sum _{i\ne j}\left|{a}_{ij}\right|],$$ |

which implies, by definition of diagonally dominance,$\lambda \ge 0.$ ∎

Title | properties of diagonally dominant matrix |
---|---|

Canonical name | PropertiesOfDiagonallyDominantMatrix |

Date of creation | 2013-03-22 15:34:32 |

Last modified on | 2013-03-22 15:34:32 |

Owner | Andrea Ambrosio (7332) |

Last modified by | Andrea Ambrosio (7332) |

Numerical id | 15 |

Author | Andrea Ambrosio (7332) |

Entry type | Result |

Classification | msc 15-00 |