## You are here

Homeproperties of diagonally dominant matrix

## Primary tabs

# 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:

$\sum_{{j\neq i}}\left|a_{{ij}}\right|\geq\left|\lambda-a_{{ii}}\right|=\left|a% _{{ii}}\right|,$ |

which is in contrast with strictly diagonally dominance definition. ∎

2)(Ostrowski theorem) $\left|\det(A)\right|\geq\prod_{{i=1}}^{{n}}\left(\left|a_{{ii}}\right|-\sum_{{% j=1,j\neq i}}\left|a_{{ij}}\right|\right)$ (See here 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:

$\lambda\in[a_{{ii}}-\sum_{{j\neq i}}\left|a_{{ij}}\right|,a_{{ii}}+\sum_{{i% \neq j}}\left|a_{{ij}}\right|],$ |

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

## Mathematics Subject Classification

15-00*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff

## Recent Activity

new problem: Geometry by parag

Aug 24

new question: Scheduling Algorithm by ncovella

new question: Scheduling Algorithm by ncovella