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Timeline for Sum of eigenvalues is nonpositive

Current License: CC BY-SA 4.0

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Jan 30, 2020 at 17:20 comment added fedja Alas, so far I can do only the original one :-(. But I'll think a bit more of it :-)
Jan 30, 2020 at 5:18 comment added Victor Reis Hi fedja, I figured out how to do the original problem, but still don't know about the extension (for $m > 1$). I am still pretty interested in that, if you know how to do it (even just a sketch)
Jan 29, 2020 at 2:36 comment added fedja I believe I have figured it out, but are you still interested?
Jun 16, 2019 at 23:11 comment added Victor Reis Ah, yes, you’re right that these are equivalent. I’m not sure if we can also state the more general version with $m$ in terms of projections.
Jun 16, 2019 at 22:55 comment added Christian Remling To make this more concrete: If my $M$ is not positive definite and $\langle v, Mv\rangle <0$, then I believe $A=\langle v, \cdot \rangle v$ gives a counterexample to your claim, for $k=n-d$ (and test on vectors spanning $R(P)$).
Jun 16, 2019 at 22:52 comment added Christian Remling Yes, but note that I'm only considering projections in my reformulation, so there is no immediate contradiction to what you're saying. (The argument establishing the equivalence was quite simple, but of course I still may have made a mistake somewhere.)
Jun 15, 2019 at 23:50 comment added Victor Reis Hi Christian. It seems like while the sum of the largest eigenvalues is nonpositive, they are not all always nonpositive.
Jun 15, 2019 at 20:59 comment added Christian Remling I think a possible reformulation is: Let $P$ be a projection onto a $d$-dimensional subspace. Is it then true that $M\ge 0$, with $M=-P+(n-d+1)\textrm{diag}\: P$? Not sure though if this gets us any closer to a solution.
Jun 15, 2019 at 18:05 review First posts
Jun 15, 2019 at 18:26
Jun 15, 2019 at 18:03 history asked Victor Reis CC BY-SA 4.0