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This question is related to that of Thurston. However, I am not interested in algebraic integers, and I wish to focus on random matrices instead of random polynomials.

When considering (entrywise) non-negative matrices $M$, a natural probability measure seems to be $$\prod_{i,j=1}^ne^{-m_{ij}}dm_{ij}.$$ By Perron-Frobenius theorem, the spectral radius $\rho(M)$ is an eigenvalue, associated to a non-negative eigenvector. Almost surely, $M$ is positive and therefore this eigenvalue is simple and its eigenvector is positive.

What is the distribution of the eigenvalues of $M$ as the size $n$ goes to infinity ? What is the relevant normalization ? Should we consider $\lambda/\rho(M)$ or $\lambda/\sqrt{\rho(M)}$ or something else ? Is it the same asymptotics as in the case of the conjugates of the algebraic Perron numbers considered by Thurston ?

Note that because of the constraint $m_{ij}\ge0$, an exponential law looked more natural to me than a Gaussian. Has anyone an other suggestion of probability over non-negative matrices ?

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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. – Mark Meckes Jan 12 '11 at 15:20
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. – Jeff Schenker Jan 12 '11 at 20:22
Another distribution on nonnegative matrices which people use is the Wishart distribution. Its eigenvalues are then asymptotically distributed according to the Marčenko–Pastur distribution. – Michael Jan 12 '11 at 20:22
@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. – Denis Serre Jan 12 '11 at 20:46
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 ? – Alekk Apr 8 '11 at 20:08
up vote 9 down vote accepted

This is a non-centered iid random matrix whose entries have mean one and variance one (and decay exponentially at infinity), and as such, is subject to the circular law with one outlier. Thus, there will be one eigenvalue roughly near n, and the rest will be more or less uniformly distributed in the complex disk of radius $\sqrt{n}$. The latter result is due to Chafai (at least for almost all of the eigenvalues), and the former is due to Silverstein. I discuss some finer aspects of the outlier eigenvalue (and show that all the other eigenvalues are nearly contained in the disk) in a more recent paper. (See in particular Phillip Wood's Figure 3 in that paper for an example of the eigenvalue distribution of a similar matrix model to the one you propose.)

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This is indeed suggested by the following very naive approach: Supposing that the Frobenius vector of such a matrix has more or less constant coordinates, it gets multiplied by $n$ times the expectancy (which is $1$) of the probability with density $e^{-x}$. – Roland Bacher Apr 21 '11 at 17:41
I am impressed that you have a paper that gives a definitive answer to this specific question! In particular, Il ike the fact that there is precisely one outlier. You will be amazed that Alice's office is next to mine... – Denis Serre Apr 22 '11 at 6:04
Your result is at the opposite side of the so-called `circulant case', where the matrix is assumed to be in block form $(M_{k\ell})_{1\le k,\ell\le p}$ with $M_{k\ell}=0$ unless $\ell=k+1$ ($\ell=1$ when $k=p$). Such a matrix has $p$ dominant eigenvalues at the vertices of a regular $p$-agon. – Denis Serre Apr 22 '11 at 6:41

starting from the Gaussian probability measure $\prod_{i,j=1}^{n}e^{-m_{ij}^{2}}$ constrained by $m_{i,j}\geq 0$, I would surmise that the positivity constraint would become irrelevant for eigenvalue correlation functions in the large-$n$ limit; such large-$n$ universality of spectral correlations holds for real symmetric matrices (the Gaussian orthogonal ensemble), and I would expect it to hold for real asymmetric matrices as well.

without the constraint $m_{i,j}\geq 0$ we have the socalled real Ginibre ensemble of random real asymmetric matrices; the literature on this ensemble is extensive, an entry point could be

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The given reference does not deal with entriwise positive matrices. It seems irrelevant. – Denis Serre Jan 13 '11 at 7:23
added explanation why I would expect the Ginibre ensemble to be relevant – Carlo Beenakker Jan 13 '11 at 8:08
I agree with Dennis. Without the constraint you qualitatively change the family of matrices you are looking at, from a spectral point of view. Entrywise positive matrices have significantly more structure than arbitrary matrices -- see Perron-Frobenius -- and in the large $n$ limit the collection of e.p. matrices constitutes an ever smaller proportion of the Ginibre ensemble, so trying to use statistics for the latter to get at the former doesn't seem promising to me – Yemon Choi Jan 13 '11 at 16:55

I believe the distribution of eigenvalues of a random symmetric matrix tends towards the Wigner semicircle distribution as $n \rightarrow \infty$.

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true (once you specify what scaling and convergence you mean), but I think not terribly relevant to the present question – Yemon Choi Jan 12 '11 at 19:09
Clearly not relevant. – Denis Serre Jan 12 '11 at 20:42

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