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A matrix is said to be positive if each entry in the matrix is positive. If $A$ is real, irreducible, diagonally dominant (or strictly dominant matrix) and has positive diagonal and non-positive off-diagonal elements. Then how to show that inverse of $A$ exists and is positive.?

I am able to show that inverse of $A$ exists, but don't know how to prove that it is positive.

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  • $\begingroup$ what do you mean by positive matrix ? do you know hadamard lemma ? $\endgroup$ – Gabriel Romon Apr 23 '13 at 16:41
  • $\begingroup$ Here positive matrix is a square matrix whose all entries are positive. $\endgroup$ – Sujeet Apr 23 '13 at 17:14
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I'm not sure your problem is formulated correctly. Consider $\left( \begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array} \right)$.

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  • $\begingroup$ I have edited my question. Earlier, I wrote wrong definiiton of diagonally dominant matrix. I am sorry. Thank a lot for correcting me. $\endgroup$ – Sujeet Apr 23 '13 at 17:15
  • $\begingroup$ I believe this counterexample still holds for the definition of diagonally dominant matrix given at en.wikipedia.org/wiki/Diagonally_dominant_matrix $\endgroup$ – Ross B. Apr 23 '13 at 17:18
  • $\begingroup$ But this is for irreducible diagonally dominant matrix which is equivalent to strictly diagonally dominant matrix. $\endgroup$ – Sujeet Apr 23 '13 at 17:25
  • $\begingroup$ It seems that the definition comes from the fact that the terms "irreducible and diagonally dominant" and "irreducibly diagonal dominant" are not always consistent. In my brief review of related articles (which I'm sure you have seen) the latter seems to refer to the case where at least one of the inequalities is strict. $\endgroup$ – Ross B. Apr 23 '13 at 19:06

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