# Distribution of eigenvalues of a random matrix

Given a random symmetric matrix A whose entries are Poisson distributed, can anything be said about the distribution of A's eigenvalues? Would be great if someone could link a paper citing such a result!

• Principal component analysis and factor analysis would be where I would look. Also, for a general symmetric matrix the eigenvalues can be positive, negative, or complex. – probabilityislogic Jan 10 '18 at 11:38
• But is there a result that talks about what the underlying distribution of these eigenvalues could be ? – cbro Jan 10 '18 at 13:24
• (1/2) I think quite a lot can be said although I don't have the time to post a full answer right now. I'd guess that the exact distribution is intractable since even in the 2x2 case $$A = \begin{bmatrix} X & Y \\ Y & Z\end{bmatrix}$$ the eigenvalues are $$\lambda = \frac{X + Z \pm \sqrt{(X - Z)^2 + 4Y^2}}{2}.$$ Maybe you can find an analytical form for that, but I'd be really surprised if you can find one in the general case (not saying it's impossible though!). More generally, I did a few simulations and it seems that if the eigenvalues are $\lambda_1 \geq \dots \geq \lambda_n$ then – jld Jan 11 '18 at 15:29
• (2/2) $\lambda_2$ through $\lambda_n$ follow a semi-circle distribution while $\lambda_1$ gets bigger, much like what happens with the eigenvalues of a random uniform matrix. See here for some terms to get started: mathworld.wolfram.com/WignersSemicircleLaw.html and more generally you'll want to look into various semicircle laws for random matrices. You've got a nice case in that everything's real, iid, and all moments are finite, so I bet you can work out an asymptotic distribution. I'd love to see this presented if you or someone else does it – jld Jan 11 '18 at 15:31
• Thanks for the interesting insight @Chaconne! I think I will try to think about this result some more later. I needed such a result for something else I was trying to prove, but I guess this seems to be an interesting problem that still remains open and you (and possibly others) are interested in the solution! :) – cbro Jan 12 '18 at 3:44