Let me call a matrix $\hat{A}$ row and column diagonally dominant if for all $i$ simultaneously $\left|A_{i,i}\right|\geq\sum_{j\neq i}{\left|A_{ij}\right|}$ and $\left|A_{i,i}\right|\geq\sum_{j\neq i}{\left|A_{ji}\right|}$. Is it possible to prove that inverse of a row and column diagonally dominant matrix $\hat{A}$ is also a column and row diagonally dominant matrix? What if the matrix $\hat{A}$ is also symmetric?

I have played around with a few numerical examples, and it seems that this is indeed the case for both symmetric and asymmetric matrices as long as they are row and column diagonally dominant. Any disproofs by counterexample are also welcome!

Edit: Faced with a successful counterexample $\left(\begin{matrix}2&1&0\\1&100&98\\0&98&99\end{matrix}\right)$, I decided to play around with some randomly generated matrices. The diagonal dominance is indeed usually NOT conserved for $3\times 3$ matrices or larger. The diagonal dominance is however conserved in the case of $2\times 2$ matrices. So why are $2\times 2$ matrices special?

  • $\begingroup$ Should there be absolute values in that sum? Or, is $\hat A$ a non-negatve matrix? $\endgroup$ – Omnomnomnom Apr 6 '16 at 15:17
  • $\begingroup$ If you mean what I think you mean, the answer is no. For example, here is a $3 \times 3$ symmetric, diagonally dominant matrix whose inverse is not diagonally dominant. The result will hold, however, for $2 \times 2$ matrices. $\endgroup$ – Omnomnomnom Apr 6 '16 at 15:34
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    $\begingroup$ Your definition of diagonal dominance is quite different from the usual one. In the usual definition, we take absolute values of the entries first before taking sums and doing comparisons. Do you really mean not to take absolute values? $\endgroup$ – user1551 Apr 6 '16 at 16:21
  • $\begingroup$ My bad, the sum should be over absolute values (see edit). @Omnomnomnom in your example the diagonal dominance criterion is not fulfilled for the third row (and column) therefore the initial matrix is not diagonally dominant! $\endgroup$ – Arturs C. Apr 6 '16 at 19:45
  • $\begingroup$ @ArtursC. Sorry; change the lower-right entry to a $99$ and you have a counterexample. $\endgroup$ – Omnomnomnom Apr 6 '16 at 20:16

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