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