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This is a tricky problem I encountered in my research. $A\in \mathbb{R}^{n\times n}_+, x,y\in \mathbb{R}^{n}_+$, i.e. $\forall 1\leq i \leq n, 1 \leq j\leq n, A(i, j)>0, x(i), y(i)>0$.

As known, since $A$ is a positive definite matrix, $A^{-1}$ is a positive definite matrix.

  • If $x=y$, then $x^TA^{-1}y>0$ always holds.
  • Since $A(i, j)>0$, then $x^TAy>0$ always holds. However, $A^{-1}(i, j)> 0$ doesn't hold and $x^TA^{-1}y>0$ as well.

Based on my experiments, I have reached the following views:

  1. Since $x(i), y(i)>0$, $x^TA^{-1}y>0$ is likely to hold when $y(i)>x(i)$.
  2. The diagonal elements of $𝐴$ are all positive, and their magnitudes are nearly $n$ times larger than those of the off-diagonal elements.

I want to figure out under what conditions on $A, x, y$ that $x^TA^{-1}y>0$ will always hold.

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    $\begingroup$ Might be worth noting that for any given $A$ meeting your conditions, there will be at least some $x, y$ meeting your conditions, such that $x^T A ^{-1}y < 0$. So looking for a condition purely on $A$ won't work. $\endgroup$
    – Martin M. W.
    Commented Sep 16 at 11:04
  • $\begingroup$ $A$ is subject to certain conditions: each element of $A$ is greater than 0, and $A$ is a symmetric positive definite matrix. In fact, what I am more interested in are the conditions regarding $x$ and $y$, such as how much larger the magnitude of $y$ is compared to $x$, or in what range their angle lies. $\endgroup$
    – Songqiao Hu
    Commented Sep 16 at 12:07
  • $\begingroup$ I did not understand the statement "Since $A(i,j)>0$, $x^TAy>0$ always holds. This is not so. $\endgroup$
    – Alexandre Eremenko
    Commented Sep 16 at 12:09
  • $\begingroup$ Thank you for your attention. Since each element of $x$, $y$, and $A$ is greater than 0, it follows directly from the rules of matrix multiplication that $x^TAy>0$. $\endgroup$
    – Songqiao Hu
    Commented Sep 16 at 12:12
  • $\begingroup$ I may not have expressed it clearly. I have already revised the question content. $\endgroup$
    – Songqiao Hu
    Commented Sep 16 at 12:16

1 Answer 1

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Just a general comment: you might be interested in checking out the theory of M-matrices. An M-matrix is a matrix such that $M_{ij} \leq 0$ for $i\neq j$, with the additional property that all its eigenvalues are in the right half-plane (hence, in particular, the diagonal elements $M_{ii}$ usually are positive and quite large with respect to the off-diagonal ones).

The inverse of an M-matrix has non-negative entries; hence it is possible that your $A$ is the inverse of an M-matrix, especially since you mention that it is diagonally dominant (or almost). This might help shed some more light on your conditions; as the question is now, it seems difficult to give a general answer that is not a tautology ("the condition is that $x^T A^{-1}y > 0$").

For instance, for every invertible M-matrix there exists a vector $v>0$ such that $Mv \geq 0$; this seems related to your property.

A possible starting point is the classical book

  • Berman, Abraham; Plemmons, Robert J., Nonnegative matrices in the mathematical sciences, Classics in Applied Mathematics. 9. Philadelphia, PA: SIAM,. xx, 340 p. (1994). ZBL0815.15016.
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  • $\begingroup$ Thank you very much for your answer; it is very valuable to me. I will try it! $\endgroup$
    – Songqiao Hu
    Commented Sep 16 at 13:20

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