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I have read that the spectral density of an NxN random matrix consisting of iid random variables with zero mean and unit variance converges as N goes to infinity to the uniform distribution on the unit disc. I was wondering if there is an intuitive way of understanding why this should be the case.

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This nice post of Terry Tao might help… – Liviu Nicolaescu Apr 10 '13 at 9:37
up vote 12 down vote accepted

I don't know of a fully intuitive derivation, but there are some informal arguments that give the circular law with a relatively small amount of calculation.

Let $M$ be a matrix where the entries are iid with mean zero and variance one. One can begin with the determinant formula

$$ \log |\det( M - z )| = \sum_{j=1}^n \log |\lambda_j - z|.$$

The circular law suggests that the eigenvalues $\lambda_j$ should be uniformly distributed in the disk of radius $\sqrt{n}$, so one should be proving something like

$$ \log |\det( M - z )| \approx \frac{1}{\pi} \int_{|w| \leq \sqrt{n}} \log |w-z|\ dw.$$

A routine calculation (e.g. using Jensen's formula, or the fundamental solution for the Laplacian) reveals that the RHS is equal to $n \log |z|$ when $|z| \geq \sqrt{n}$ and $\frac{1}{2} n \log n - \frac{1}{2} n + \frac{1}{2} |z|^2$ for $|z| \leq \sqrt{n}$. So heuristically, the circular law is equivalent to the approximations

$$ |\det(M-z)| \approx |z|^n$$

for $|z| \geq \sqrt{n}$ and

$$ |\det(M-z)| \approx n^{n/2} e^{-n/2} e^{|z|^2/2}$$

for $|z| \leq \sqrt{n}$. Here one should interpret the $\approx$ symbol rather loosely (in particular, polynomial factors in $n$ should be considered negligible).

However, by the Leibniz formula for determinants and the iid mean zero variance one nature of the entries (which makes all the covariances between the terms in the Leibniz formula vanish), one can easily compute that

$$ {\bf E} |\det(M-z)|^2 = \sum_{j=0}^n |z|^{2j} \frac{n!}{j!}.$$

(This type of calculation goes back to an old paper of Turan.)

For $|z| \gg \sqrt{n}$, the $|z|^{2n}$ term on the RHS dominates, while for $|z| \ll \sqrt{n}$, the RHS is most of the Taylor series for $n! e^{|z|^2}$. The claim then morally follows from Stirling's approximation.

(Incidentally, my recent paper with Van Vu on local versions of the circular law basically proceeds by making the above argument rigorous; see also recent work of Bourgarde, Yau, and Yin. The idea of controlling the spectrum of an iid matrix through its log-determinant goes back to the early work of Girko.)


Note also that without too much calculation, one can see that the limiting law of the spectrum should be invariant with respect to rotations around the origin (especially if one assumes that the entries of the iid matrix are similarly invariant, e.g. they are complex gaussian). From the matrix inequality $\sum_{j=1}^n |\lambda_j|^2 \leq \hbox{tr}(M M^*)$ and the law of large numbers we also see that the typical size of an eigenvalue $\lambda_j$ should be of the order of $\sqrt{n}$. These facts fall well short of the full circular law but are certainly consistent with that law.

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Nice answer! I think you have a typo in the first displayed equation: $\lambda_j-n$ should be $\lambda_j-z$. – David Speyer Apr 10 '13 at 17:23
Corrected, thanks! – Terry Tao Apr 10 '13 at 17:59
Thank you for the nice response. I have verified most of the computations, but am still having some trouble verifying the inequality $\sum_{j=1}^n |\lambda_j|^2 \leq tr(MM^*)$ for a non-normal matrix $M$. Is there a way to do this using the inner product $\langle A,B \rangle=tr(AB^*)$? with $A=M$ and an appropriate choice of $B$? Thanks again. – Eric Foxall Apr 13 '13 at 18:25
If M is upper-triangular, one can write $\sum_{j=1}^n |\lambda_j|^2$ as $\hbox{tr}(M \overline{M})$. For the general case, one can use the Schur decomposition, – Terry Tao Apr 13 '13 at 21:59

Minor initial comment: You probably mean "uniform distribution in the unit circle", rather than on the unit circle --- the eigenvalues $\lambda_{k}$ ($i=1,2,... n$), scaled by $\sqrt{n}$, uniformly fill the interior of the unit circle in the complex plane. So what is commonly called the "circle law" is perhaps more appropriately referred to as the "disc law".

You ask for a simple/intuitive derivation. The easiest I know is given by Eric Kostlan in On the spectra of Gaussian matrices. When the real and imaginary parts of each matrix element are independent normally distributed, then it is rather easy to show that the absolute values squared of the eigenvalues, $\mu_{k}=|\lambda_{k}|^{2}$, have independent $\chi^2$ distributions with $k=2,4\ldots 2n$ degrees of freedom.

The large-$n$ asymptotics of the $\chi^2$ distribution then implies that $\mu$ is uniformly distributed in the interval $(0,n)$. It thus follows that the eigenvalues uniformly fill a disc in the complex plane of radius $\sqrt{n}$.

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Yes, thank you -- i have corrected it to read "on the unit disc". – Eric Foxall Apr 10 '13 at 17:48
This as well as the other is a good "intuitive" answer - I apologize that I cannot check both of them. – Eric Foxall Apr 10 '13 at 18:13
By "check", I mean accept (that is, put a check-mark). – Eric Foxall Apr 10 '13 at 18:14
thank you @Eric, you're welcome; I bow respectfully for the Matthew Principle. – Carlo Beenakker Apr 10 '13 at 19:44

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