# Diagonal dominance versus positive semi-definiteness

I know that for a symmetric matrix $A$, diagonal dominance, i.e. $$A_{ii} \ge \sum\limits_{j \ne i} |A_{ij}|$$ implies positive semi-definiteness.

How about the other way? Does positive semi-definiteness imply diagonal dominance? Could you point to a proof or a counter example?

• I think you want $A_{ii} \ge \sum_{j \ne i} |A_{ij}|$ (without absolute values on the $A_{ii}$). – Robert Israel Sep 20 '12 at 0:29
• Whats wrong with the square matrix $J$ i.e., having all its entries $1$ ? – Tapu Sep 20 '12 at 0:49
• Yes, Robert. Thanks. – user25004 Sep 20 '12 at 6:11

## 4 Answers

Quick counter example

>>> a=2*ones(3,3)+eye(3)
a =

3   2   2
2   3   2
2   2   3

>>> eig(a)
ans =

1.00000
1.00000
7.00000

• $A=\left[\begin{smallmatrix}3&2\\2& 3\end{smallmatrix}\right]$ will also do - its eigenvalues are $1$ and $5$. – Pantelis Sopasakis Sep 20 '12 at 0:10
• @PantelisSopasakis: but that one does have diagonal dominance. – Robert Israel Sep 20 '12 at 0:30
• @RobertIsrael Oh, my bad! Sorry - blunder! – Pantelis Sopasakis Sep 20 '12 at 0:32

A $2 \times 2$ counterexample is $\pmatrix{a^2 & a\cr a & 1\cr}$ for $|a| \ne 1$.

Take the following matrix:

$$A=\left[ {\begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} } \right]$$

Notice that $A$ is positive semidefinite (it has a double eigenvalue at $\lambda=0$). But it is not diagonally dominant since $0=|A_{11}|<|A_{12}|=1$

• Did the same mistake, $A$ is assumed to be symmetric :) – Long Sep 20 '12 at 0:07
• The Question was: "Does positive semi-definiteness imply diagonal dominance?" But, you're right... – Pantelis Sopasakis Sep 20 '12 at 0:09

You can see link http://www.win.tue.nl/~aeb/srgbk/node16.html In that link, the author said that if ma trix symmetric and stricly diagonal dominant then A positve define.