A somewhat lengthy calculation, involving integrals, reveals that the probability an $n\times n$ Hermitian matrix, drawn from the Gaussian unitary ensemble, is positive definite, decays as $\left(\sqrt{3}\right)^{-n^2}$ for large $n$. I can’t say what decay constant I had expected, given the context, but it certainly wasn’t $\sqrt{3}$ ! Is there a deeper or more direct explanation for the appearance of this number?
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2$\begingroup$ The Hermitian property means there are only about $n^2/2$ independent entries, so it would make more sense to write it as $3^{-n^2/2}$ and the constant doesn't look strange any more. One can still ask for a heuristic explanation, of course. $\endgroup$– Brendan McKayCommented Apr 17, 2014 at 14:05
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3$\begingroup$ But $n^2$ independent real numbers ... Still, why should there even be a 3? $\endgroup$– Veit ElserCommented Apr 17, 2014 at 14:13
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1$\begingroup$ I also wasn't exactly expecting Euler's constant $\gamma$ to be so close to $\dfrac1{\sqrt3}$ either, but hey, that's life: it always takes you by surprise! :-) $\endgroup$– LucianCommented Apr 17, 2014 at 16:40
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$\begingroup$ Related : mathoverflow.net/questions/118481/… $\endgroup$– Mikael de la SalleCommented Apr 17, 2014 at 23:11
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1 Answer
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this is a limit of a more general result by Majumdar and company, How many eigenvalues of a Gaussian random matrix are positive? (2010), see also their earlier papers from 2006 and 2008.
The coefficient $\sqrt{3}$, or $\frac{1}{2}\log 3$ in the exponent, appears from a saddle point approximation, see Eq. 59 of the 2008 paper, without any particular numerological significance.
Concerning a possible heuristic argument for the number $\sqrt{3}$, here is one published argument that gives $\sqrt{e}$, close but not quite correct.
answered Apr 17, 2014 at 13:54
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2$\begingroup$ In fact, this constant comes from minimizing the rate function of BenArous and Guionnet math.nyu.edu/~benarous/Publications/benarous_30.pdf over measures which are supported over the positive part of the real axis. This is essentially what Majumdar and als do. $\endgroup$ Commented Apr 17, 2014 at 19:12