Timeline for Distribution of the spectrum of large non-negative matrices
Current License: CC BY-SA 3.0
13 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Mar 15, 2013 at 10:00 | history | edited | Denis Serre | CC BY-SA 3.0 |
edited title
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| Apr 21, 2011 at 20:41 | vote | accept | Denis Serre | ||
| Apr 21, 2011 at 17:13 | answer | added | Terry Tao | timeline score: 11 | |
| Apr 8, 2011 at 20:08 | comment | added | Alekk | isn't it the case that the circular law (Tao-Vu) shows that after rescaling by $\sqrt{n}$ and substracting the mean, the empirical spectral measures converge to the uniform measure on the unit disk ? | |
| Jan 12, 2011 at 21:49 | answer | added | Carlo Beenakker | timeline score: 1 | |
| Jan 12, 2011 at 21:25 | comment | added | Suvrit | Is the following chapter any useful: books.google.com/… | |
| Jan 12, 2011 at 20:46 | comment | added | Denis Serre | @Michael. According to Wikipedia, Wishart distribution is used in particular for positive definite Hermitian matrices. This is completely different from pointwise positive (and not symmetric) matrices. A completely different landscape. | |
| Jan 12, 2011 at 20:22 | comment | added | Michael | Another distribution on nonnegative matrices which people use is the Wishart distribution. Its eigenvalues are then asymptotically distributed according to the Marčenko–Pastur distribution. | |
| Jan 12, 2011 at 20:22 | comment | added | Jeff Schenker | This may help. Write $M == avg(M) + (M -avg(M)).$ Then $M - avg(M)$ has centered entries while $avg(M)$ is very simple -- it is $N \times $ the rank one projection onto the unit vector with all entries $1/\sqrt{N}$. So what you have is a "spiked matrix ensemble," most likely with one large eigenvalue (of order $N$) and a sea of smaller eigenvalues (of order $\sqrt{N}$). I think similar ensembles have been studied by Baik. I don't know the details. Anyway you can certainly get something by relating the resolvent of $M$ to the resolvent of $M-avg(M)$ with rank 1 perturbation formula. | |
| Jan 12, 2011 at 18:50 | answer | added | ndronen | timeline score: -2 | |
| Jan 12, 2011 at 15:20 | comment | added | Mark Meckes | I don't know whether anyone has already considered the distribution you suggest (although I think it's a natural and interesting one). People have considered restricting further to stochastic or doubly stochastic matrices, in which case it's natural to use a uniform distribution, e.g. arxiv.org/abs/1010.6136 | |
| Jan 12, 2011 at 14:52 | history | asked | Denis Serre | CC BY-SA 2.5 |