Timeline for Fast trace of inverse of a square matrix
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 19, 2010 at 17:56 | vote | accept | César | ||
| Nov 19, 2010 at 12:58 | comment | added | Federico Poloni | Maybe it is also positive definite? This would help (no pivoting) | |
| Nov 19, 2010 at 12:35 | answer | added | J. M. isn't a mathematician | timeline score: 7 | |
| Nov 19, 2010 at 12:25 | comment | added | J. M. isn't a mathematician | Aha. People, people, remember, when dealing with matrix problems, please mention if there's anything special about them. Symmetry is a big thing! | |
| Nov 19, 2010 at 12:17 | comment | added | César | I forgot to mention that under normal conditions, H will be symmetric. But it will be always dense. | |
| Nov 19, 2010 at 12:16 | history | edited | César | CC BY-SA 2.5 |
added 79 characters in body
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| Nov 19, 2010 at 3:31 | comment | added | J. M. isn't a mathematician | The pivoting inherent in LU decomposition complicates a lot of things in what you want to do. I'm going to spitball and suggest the slightly more expensive QR decomposition, but the payoff (hopefully) is that you don't have to do the required backsubstitution on the rows of the orthogonal matrix Q in full, but only up until the component needed for the running sum of the trace. You'll have to do your own testing to be sure, of course, since the speed can be environment-dependent. | |
| Nov 19, 2010 at 2:04 | answer | added | Ian Agol | timeline score: 1 | |
| Nov 19, 2010 at 0:00 | comment | added | J. M. isn't a mathematician | Is there any structure inherent in your $H$ (e.g. sparse, banded, symmetric, etc.)? | |
| Nov 18, 2010 at 22:52 | history | asked | César | CC BY-SA 2.5 |