# Spectrum of random matrices has dominant eigenvalues

I am looking for pointers/references to results of the following kind:

For $M$ a real or integer square matrix drawn at random from some "reasonably" nice set of square matrices (possibly infinite/uncountable), with a "reasonably" nice probability distribution, then with "reasonably" high probability the spectrum of $M$ either has exactly two dominant conjugate eigenvalues, or exactly one dominant real eigenvalue. And in either case (with high probability) these eigenvalues have algebraic multiplicity 1.

(By "dominant" I mean of modulus strictly larger than the other eigenvalues.)

• Try looking up "eigenvalue repulsion" or "eigenvalue spacing." – Qiaochu Yuan May 12 '17 at 0:12
• Is $M$ symmetric ? What do you mean by "reasonably nice"? – Henry.L May 12 '17 at 0:18
• If the matrix were symmetric, I wouldn't expect complex conjugate eigenvalues. – Anthony Quas May 12 '17 at 0:36
• "Reasonably nice" is anything that will make the assertion true... :) $M$ is not necessarily symmetric (otherwise, as pointed out by Anthony Quas above, the spectrum would be entirely real). – Joël Ouaknine May 12 '17 at 11:25

The "good" matrices, those whose eigenvalues of largest absolute value are either a single simple real eigenvalue or a conjugate pair of simple eigenvalues, form a dense open set in the $n \times n$ real matrices. In fact, if $B$ is any "good" $n \times n$ matrix and $A$ any $n \times n$ matrix, $A + t B$ will be "good" for almost every real $t$. So the "good" matrices form a set of full $n^2$-dimensional Lebesgue measure, and a random matrix from any absolutely continuous probability distribution will be "good" with probability $1$.

Insofar as, after scaling, a "reasonably nice" distribution of integer matrices approximates an absolutely continuous distribution, it should produce "good" matrices with high probability.

• Thanks very much Robert, this is exactly the kind of result I was looking for (but I lack the reputation points to upvote your answer). I think I can just about prove these statements using elementary arguments, plus some facts about real algebraic geometry -- but is this already written up or mentioned somewhere that could be cited directly by any chance? – Joël Ouaknine May 12 '17 at 11:34

Yes. Take your favorite generating set of, say, $SL(n, \mathbb{R})$ (or a Zariski-dense subgroup thereof), and look at long random products of its elements. Amazingly, the entire spectrum of such a product is almost surely real and there is a unique dominant eigenvalue. In fact, the reality (realism? realty?) of the spectrum is essentially no harder than the existence of the dominant eigenvalue - because there is a unique dominant eigenvalue, it must be real (otherwise its conjugate would have the same modulus), and now when you look at the actions on the exterior powers, the same argument gives you all the eigenvalues. I am not sure where the first source for this is, but the papers by Yves Benoist

B\'enoist, Y., Asymptotic properties of linear groups, Geom. Funct. Anal. 7, No.1, 1-47 (1997). ZBL0947.22003.

and Breuillard and Gelander

Breuillard, E.; Gelander, T., On dense free subgroups of Lie groups, J. Algebra 261, No.2, 448-467 (2003). ZBL1014.22007.

should prove enlightening.