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Suppose $A \in \mathbb{R}^{n \times n}$ has positive entries on it main diagonal and $\mbox{rank}(A) =: d < n$. Then, what is the maximum number of many negative entries $A$ can contain?

If, in addition, $A$ is positive semidefinite (PSD), then using the fact that in $d$-dimensional space there are at most $d+1$ vectors with pairwise negative inner product and Turán's Theorem from graph theory we can bound the number of negative entries by roughly $$\left(\frac{d-1}{d}\right) n^2$$

But what if $A$ is not PSD?

I've been stuck with this problem for a while. Any helpful reference is greatly appreciated!

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    $\begingroup$ Do you have results for special d, e.g. d = 1 or d = N-1? (sorry, MathJax seems to be broken somehow...) $\endgroup$ – Dirk Liebhold Feb 1 '18 at 10:40
  • $\begingroup$ The case where $d=1$ is easy. When $d=N-1$, actually all the off-diagonal entries can be negative. $\endgroup$ – Richard D. Feb 2 '18 at 3:37
  • $\begingroup$ Sounds like d=N-2 might be the first “intermediate” case on the upper part of the range, then. Perhaps getting a formula for it would help. $\endgroup$ – Abel Molina Feb 5 '18 at 9:13

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