Timeline for The probability for a symmetric matrix to be positive definite
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Nov 15, 2016 at 14:42 | answer | added | Nick Jones | timeline score: 9 | |
| Oct 7, 2014 at 8:27 | history | edited | Denis Serre | CC BY-SA 3.0 |
an equivalent definition
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| Oct 1, 2014 at 8:27 | vote | accept | Denis Serre | ||
| Mar 15, 2013 at 21:52 | history | edited | Denis Serre | CC BY-SA 3.0 |
added 17 characters in body
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| Mar 15, 2013 at 21:52 | comment | added | Denis Serre | @Federico. Right! I meant "among the Euclidian norms". | |
| Mar 15, 2013 at 13:47 | comment | added | Federico Poloni | "this is the most natural norm, because it is invariant under unitary conjugation": actually there are lots of norms invariant under unitary conjugation: the Frobenius norm, the Euclidean induced norm, Schatten norms, Ky Fan norms... | |
| Jan 10, 2013 at 20:53 | history | edited | Denis Serre | CC BY-SA 3.0 |
added 37 characters in body
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| Jan 10, 2013 at 17:46 | answer | added | Mikael de la Salle | timeline score: 27 | |
| Jan 10, 2013 at 16:32 | comment | added | Denis Serre | @Mark. Given $n$, you may plot $p_{n,k}$ as a function of $\frac{k}n$. Then, as $n\rightarrow+\infty$, the plots must tend to some graph, and I am interested in this graph. | |
| Jan 10, 2013 at 15:13 | comment | added | Mark Meckes | I don't understand your second question. The $p_{n,k}$ are deterministic, so what do you mean by their distribution? | |
| Jan 10, 2013 at 15:11 | comment | added | Mark Meckes | I don't know about exact values or precise asymptotics, but from Mikael's observation it's not too much work to see that $p_n$ goes to 0 exponentially fast. | |
| Jan 9, 2013 at 23:24 | comment | added | Mikael de la Salle | If you want to use a more standard terminology (to look for references for examples), you should say that the probability is with respect to the GOE random matrix model (=gaussian vector in $Sym(\mathbb R)$ with covariance $Id$). Hence, $p_n$ is the probability that a GOE matrix is positive. There are exact formulas for the eigenvalue distribution of a GOE. You can perhaps derive from them close formulas for $p_n$. The asymptotic behaviour of $p_n$ is also probably known (look for large deviations results for GOE). | |
| Jan 9, 2013 at 23:03 | history | edited | Dima Pasechnik |
more tags
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| Jan 9, 2013 at 22:57 | answer | added | Robert Bryant | timeline score: 10 | |
| Jan 9, 2013 at 21:34 | history | asked | Denis Serre | CC BY-SA 3.0 |