# properties of diagonally dominant matrix

###### 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)() $\left|\det(A)\right|\geq\prod_{i=1}^{n}\left(\left|a_{ii}\right|-\sum_{j=1,j% \neq i}\left|a_{ij}\right|\right)$ (See here (http://planetmath.org/ProofOfDeterminantLowerBoundOfAStrictDiagonallyDominantMatrix) for a proof.)

###### 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,$\lambda\geq 0.$

Title properties of diagonally dominant matrix PropertiesOfDiagonallyDominantMatrix 2013-03-22 15:34:32 2013-03-22 15:34:32 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 15 Andrea Ambrosio (7332) Result msc 15-00