Timeline for Probability for a random positive-semidefinite matrix to not be positive-definite?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 20, 2013 at 21:40 | vote | accept | Evgeni Sergeev | ||
| Nov 20, 2013 at 8:08 | answer | added | Dirk | timeline score: 3 | |
| Nov 20, 2013 at 2:19 | comment | added | Evgeni Sergeev | @NateEldredge That's interesting. I'd never appreciated that that theory had such limitations. | |
| Nov 19, 2013 at 22:36 | comment | added | Nate Eldredge | @evgeni: That question is more about the foundations of probability than matrices, and I'm not sure it's an appropriate one for MO. But in brief, as far as I know, one can't make sense of your question in measure-theoretic probability. "Number of successful trials" is not a measurable function on uncountable product space, and its expected value is not well defined. | |
| Nov 19, 2013 at 22:02 | comment | added | Evgeni Sergeev | @NateEldredge what if you run an uncountably infinite number of trials in parallel, where the infinity is of high enough cardinality, isn't it possible that the expected value of trials with an exact zero eigenvalue will be some non-zero number, for example 3 or 4? | |
| Nov 19, 2013 at 17:34 | comment | added | Nate Eldredge | @CarloBeenakker: The probability is not just "vanishingly small", it is zero. | |
| Nov 19, 2013 at 14:18 | answer | added | Igor Rivin | timeline score: 0 | |
| Nov 19, 2013 at 10:18 | comment | added | loup blanc | There is a robust Cholesky's method. math.berkeley.edu/~cinnawu/hss.pdf | |
| Nov 19, 2013 at 7:29 | review | First posts | |||
| Nov 19, 2013 at 8:24 | |||||
| Nov 19, 2013 at 7:28 | comment | added | Evgeni Sergeev | @CarloBeenakker Is it then typical to apply Cholesky decomposition without fearing numerical instability? I mean, given the above information alone? | |
| Nov 19, 2013 at 7:24 | comment | added | Carlo Beenakker | you're asking for the probability that $A^T A$ has an eigenvalue identical to zero; this probability is vanishingly small. | |
| Nov 19, 2013 at 7:11 | history | asked | Evgeni Sergeev | CC BY-SA 3.0 |