Timeline for Is the componentwise square-root of a positive-definite matrix also pos.-def.?
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 27, 2016 at 11:56 | answer | added | Andreas Thom | timeline score: 1 | |
| Jan 23, 2011 at 17:32 | comment | added | darij grinberg | @Johannes: ok, if you are talking Zariski topology, then yes. | |
| Jan 21, 2011 at 14:08 | vote | accept | Christian Stahlhut | ||
| Jan 21, 2011 at 14:05 | vote | accept | Christian Stahlhut | ||
| Jan 21, 2011 at 14:05 | |||||
| Jan 21, 2011 at 14:05 | vote | accept | Christian Stahlhut | ||
| Jan 21, 2011 at 14:05 | |||||
| Jan 21, 2011 at 11:30 | comment | added | Johannes Hahn | @darij: The set where your "some identity" does not hold is a null set, open and dense subset of $\lbrace sym.matrices\rbrace = \mathbb{R}^{n(n+1)/2}$. If you intersect it with the space of all matrices with the required properties it is still a null set which is open and dense in this subspace (positive definite + positive entries is an open set in $\mathbb{R}^{n(n+1)/2}$ so there are no problems with dimensions etc.). | |
| Jan 21, 2011 at 2:05 | answer | added | Igor Rivin | timeline score: 8 | |
| Jan 20, 2011 at 21:50 | comment | added | Yemon Choi | Hmm, I didn't realise (before doing the calculation) that this is true for $n=2$ | |
| Jan 20, 2011 at 18:09 | comment | added | darij grinberg | Uhm, I don't think so. | |
| Jan 20, 2011 at 17:14 | answer | added | Suvrit | timeline score: 17 | |
| Jan 20, 2011 at 17:12 | comment | added | Johannes Hahn | And as a corollary to darij's observation: Not only are there counterexamples, but almost every matrix A with the required properties gives a counterexample. | |
| Jan 20, 2011 at 16:17 | answer | added | darij grinberg | timeline score: 8 | |
| Jan 20, 2011 at 16:12 | comment | added | darij grinberg | It would be very strange. It is known that $A$ is positive-definite whenever $B$ is (more generally, the componentwise product of two positive-definite matrices is positive definite, because it is a submatrix of the Kronecker product), so we would get an if-and-only-if assertion, which would (by real algebraic geometry or something like that) mean some identities between dets, which almost certainly don't hold. | |
| Jan 20, 2011 at 16:09 | history | asked | Christian Stahlhut | CC BY-SA 2.5 |