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Let $A = (a_{ij})_{i,j=1,\dots,n}$ be a real symmetric square matrix. Suppose $A$ is positive definite. Are there sufficient conditions that guarantee $a_{ij} > 0$ for all $i,j = 1,\dots, n$?

I know that $a_{ii} > 0$ for all $i=1,\dots,n.$

One thing I found is that $a_{ij} > 0$ for all $i,j = 1, \dots, n$ if and only if $A^{-1}$ is monotone, i.e. $A^{-1}x \ge 0$ implies $x \ge 0$ for all $x \in \mathbb{R}^n.$ Is there a nice way to connect this with the fact that $A$ and $A^{-1}$ are positive definite?

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Sounds to me like you are describing the inverse of a symmetric M-matrix. Those matrices are extremely important in numerical analysis because they have stable behaviours with respect to several algorithms. There are PLENTY of characterizations of such matrices. Some of them are listed in the wikipedia page for M-matrices : https://en.wikipedia.org/wiki/M-matrix .

In your case, the easiest characterization is probably the following : every off-diagonal element of $A^{-1}$ (which exists since you assume positive definiteness) is negative.

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    $\begingroup$ For the sake of completeness, a symmetric M-matrix is called a Stieltjes matrix en.wikipedia.org/wiki/Stieltjes_matrix . $\endgroup$
    – G. Fougeron
    Feb 24, 2017 at 10:03
  • $\begingroup$ Thank you! Are there specific results on positive definite M-matrices? $\endgroup$
    – Elias Strehle
    Feb 26, 2017 at 21:46
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    $\begingroup$ I think the following reference might be helpful : Bendito, E. et al., 2009. Characterization of symmetric M-matrices as resistive inverses. Linear Algebra and Its Applications, 430(4), pp.1336–1349. $\endgroup$
    – G. Fougeron
    Feb 28, 2017 at 9:05

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