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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?

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    $\begingroup$ I think you want $A_{ii} \ge \sum_{j \ne i} |A_{ij}|$ (without absolute values on the $A_{ii}$). $\endgroup$ – Robert Israel Sep 20 '12 at 0:29
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    $\begingroup$ Whats wrong with the square matrix $J$ i.e., having all its entries $1$ ? $\endgroup$ – Tapu Sep 20 '12 at 0:49
  • $\begingroup$ Yes, Robert. Thanks. $\endgroup$ – user25004 Sep 20 '12 at 6:11
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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
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  • $\begingroup$ $A=\left[\begin{smallmatrix}3&2\\2& 3\end{smallmatrix}\right]$ will also do - its eigenvalues are $1$ and $5$. $\endgroup$ – Pantelis Sopasakis Sep 20 '12 at 0:10
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    $\begingroup$ @PantelisSopasakis: but that one does have diagonal dominance. $\endgroup$ – Robert Israel Sep 20 '12 at 0:30
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    $\begingroup$ @RobertIsrael Oh, my bad! Sorry - blunder! $\endgroup$ – Pantelis Sopasakis Sep 20 '12 at 0:32
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A $2 \times 2$ counterexample is $\pmatrix{a^2 & a\cr a & 1\cr}$ for $|a| \ne 1$.

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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 $0=|A_{11}|<|A_{12}|=1$

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

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