1
$\begingroup$

Consider a class of real symmetric random matrices,$M_{n\times n}=(X_{i,j})_{n\times n},$ whose off-diagonal elements follow an exchangeable distribution, and the diagonal elements follow another exchangeable distribution and the diagonal part and off-diagonal part are independent. My question is what's the eigenvalue density of this type of random matrices? Is the eigenvalue density universal?

I know that Chatterjee has a paper.

But my matrices are different from his consideration. I do have a specific example. If we choose the off-diagonal part follow Dirichlet distribution and the diagonal part follow another Dirichlet distribution, then the eigenvalue density is some Gamma distribution. To my opinion, this is actually trivial, for after we rescale the matrix, it's asymptotic matrix is an infinite dimensional identity matrix. What about other cases?

$\endgroup$

1 Answer 1

2
$\begingroup$

If you don't require anything else besides exchangeable, then I guess not much can be said. Since exchangeability includes the trivial case of completely coupled random variables. Say consider the following matrix: the diagonal entries are all the same variable $X$, and off diagonal ones the same variable $Y$, independent of $X$. Then each instance is a circulant matrix with eigenvectors given by $(\omega, \omega^2, \ldots, \omega^n)$ where $\omega$ is an $n$th root of unity and eigenvalues given by $y \sum_{j=1}^{n-1} \omega^j + x$: see this wiki page: http://en.wikipedia.org/wiki/Circulant_matrix Thus the spectrum when you restrict to $n$ being a prime will converge to a random delta mass at $X-Y$. When $n$ is composite it's a bit more complicated and I guess it's supported on $\phi(n)-1$ number of points. I haven't thought about your Dirichlet example but I don't know how to phrase it universally.

$\endgroup$
1
  • $\begingroup$ Thanks for your comments. Yes, you are right. Probably it is not a universal case! $\endgroup$ Commented Oct 16, 2011 at 0:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .