Timeline for Seeking proof for linear algebra constraint problem.
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Aug 31, 2011 at 15:55 | vote | accept | Jeremy | ||
| Aug 31, 2011 at 15:55 | history | bounty ended | Jeremy | ||
| Aug 30, 2011 at 14:34 | answer | added | Mikael de la Salle | timeline score: 5 | |
| Aug 29, 2011 at 21:30 | answer | added | Will Jagy | timeline score: 1 | |
| Aug 29, 2011 at 20:39 | history | bounty started | Jeremy | ||
| Aug 29, 2011 at 20:39 | history | edited | Jeremy | CC BY-SA 3.0 |
Remove literature ref stuff before starting bounty
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| Aug 29, 2011 at 19:29 | history | edited | Jeremy | CC BY-SA 3.0 |
added 578 characters in body
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| Aug 28, 2011 at 19:41 | comment | added | Noah Stein | It may be worth noting that you can write the condition that the diagonal elements of $X^{-1}$ are at most $1$ using semidefinite constraints. Suppose $X$ is positive definite and let $e_i$ be the $i^{\text{th}}$ unit column vector. Taking Schur complements, $\begin{bmatrix} 1 & e_i' \\ e_i & X\end{bmatrix}\succeq 0$ if and only if $e_i'X^{-1}e_i\leq 1$. Imposing these constraints for all $i$ gives the claimed condition. The question, then, is whether there is a suitable objective function which would encourage all of these Schur complement conditions to be tight simultaneously. | |
| Aug 28, 2011 at 17:49 | comment | added | Suvrit | Sorry; it seems that in my speed, all I proved was that $e^TX^{-1}e-1$ is convex, not its square as I claimed---so this idea gets deleted. Indeed, only for $e^TX^{-1}e \ge 1$ is the claimed function guaranteed to be convex. | |
| Aug 28, 2011 at 1:46 | history | edited | Jeremy | CC BY-SA 3.0 |
added 188 characters in body
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| Aug 27, 2011 at 22:59 | history | edited | Jeremy | CC BY-SA 3.0 |
Added description of current approach to proof
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| Aug 27, 2011 at 22:16 | comment | added | Jeremy | From a computational perspecive, I've run thousands of random examples on 10 by 10 matrices and all of them had a solution, although that doesn't say much about uniqueness clearly. | |
| Aug 27, 2011 at 21:25 | comment | added | Will Jagy | unique for 2 by 2, assuming you mean all diagonal entries of $M$ are 0. Seems worth a good symbolic working over in the 3 by 3 case, at the same time a pretty full test with randomized entries. This kind of thing, fairly often, is either true or dies in dimension no larger than 4. | |
| Aug 27, 2011 at 20:20 | history | asked | Jeremy | CC BY-SA 3.0 |