The Circular Law (2)

Gaussian Case

Let \(X:=(X_{ij})_{1\leqslant i,j\leqslant n}\) be an i.i.d. random matrix on \(\mathbb{C}\) with variance \(1\). We consider \(X\) as a random variable in \(\mathcal{M}_{n}(\mathbb{C})\). Sometimes we denote \(G\) instead of \(X\) in order to distinguish the Gaussian case from the general case.^[1]

Mean Circular Law

Recall that \(\mu_{G}=\frac{1}{n}\sum_{k=1}^{n} \delta_{\lambda_k(G)}\). The first result relies on the fact that \(\mathbb{E}\mu_{G}\) has density \(\varphi_{n,1}\).

Theorem \(\mathbb{E}\mu_{n^{-1/2}G}\rightsquigarrow \mathcal{C}_{1}\) as \(n\to \infty\), where \(\mathcal{C}_{1}\) is the uniform law on the unit disc of \(\mathbb{C}\) with density \(z\mapsto \pi^{-1}\mathbf{1}_{\{z\in \mathbb{C}\colon \lvert z \rvert \leqslant 1\}}\).

proof From (2.35) with \(k=1\), we get that the density of \(\mathbb{E}\mu_{G}\) is \[ \varphi_{n,1}\colon \mapsto \gamma(z)\left( \frac{1}{n}\sum_{l=0} ^{n-1} \lvert H_l(z) \rvert ^{2} \right)=\frac{1}{n\pi}\mathrm{e}^{-\lvert z \rvert ^{2}} \sum_{l=0}^{n-1} \frac{\lvert z \rvert ^{2l}}{l!}. \tag{3.1} \]

For \(r^{2}<n\), \[ \mathrm{e}^{r^{2}} -\sum_{l=0}^{n-1} \frac{r^{2l}}{l!}=\sum_{l=n}^{\infty}\frac{r^{2l}}{l!}\leqslant \frac{r^{2n}}{n!}\sum_{l=0}^{\infty} \frac{r^{2l}}{(n+1)^{l}}=\frac{r^{2n}}{n!}\frac{n+1}{n+1-r^{2}} \tag{3.2} \]

while for \(r^{2}>n\), \[ \sum_{l=0}^{n-1} \frac{r^{2l}}{l!}\leqslant \frac{r^{2(n-1)}}{(n-1)!}\sum_{l=0}^{n-1} \left( \frac{n-1}{r^{2}} \right) ^{l}\leqslant \frac{r^{2(n-1)}}{(n-1)!}\frac{r^{2}}{r^{2}-n+1}. \tag{3.3} \]

By taking \(r^{2}=\lvert \sqrt{n}z \rvert ^{2}\) we obtain that for every compact \(C\subset \mathbb{C}\) \[ \lim_{n \to \infty}\sup _{z\in \mathbb{C}} \lvert n\varphi_{n,1}(\sqrt{n}z)-\pi^{-1} \mathbf{1}_{[0,1]}(\lvert z \rvert ) \rvert =0. \tag{3.4} \]

The \(n\) in front of \(\varphi_{n,1}\) is due to the fact that we are on \(\mathbb{C}\): \(\mathrm{d}\sqrt{n}x\mathrm{d}\sqrt{n}y=n\mathrm{d}x\mathrm{d}y\).

Strong Circular Law

Theorem (Silverstein, 1984) a.s. \(\mu_{n^{-1/2}G}\rightsquigarrow \mathcal{C}_{1}\) as \(n\to \infty\), where \(\mathcal{C}_{1}\) is the uniform law on the unit disc of \(\mathbb{C}\) with density \(z\mapsto \pi^{-1}\mathbf{1}_{\{z\in \mathbb{C}\colon \lvert z \rvert \leqslant 1\}}\)

proof Let us pick a compactly supported continuous bounded function \(f\) and set \[ S_n:=\int_{\mathbb{C}}^{} f \mathrm{d}\mu_{n^{-1/2}G}, \quad S_{\infty}:=\pi^{-1}\int_{\lvert z \rvert \leqslant 1}^{} f(z) \mathrm{d}x\mathrm{d}y. \tag{3.5} \]

Suppose for now that we have \[ \mathbb{E}[(S_n-\mathbb{E}S_n)^{4}]=O(n^{-2}). \tag{3.6} \]

By monotone convergence (or by the Fubini-Tonelli theorem), \[ \mathbb{E}\sum_{n=1}^{\infty} (S_n-\mathbb{E}S_n)^{4}=\sum_{n=1}^{\infty} \mathbb{E}[(S_n-\mathbb{E}S_n)^{4}]<\infty \tag{3.7} \]

and consequently \(\sum_{n=1}^{\infty} (S_n-\mathbb{E}S_n)^{4}<\infty\) a.s. which implies \(\lim_{n \to \infty}S_n-\mathbb{E}S_n=0\) a.s. Since \(\lim_{n \to \infty}\mathbb{E}S_n=S_{\infty}\) by the mean circular law, we get that a.s. \[ \lim_{n \to \infty}S_n=S_{\infty}. \]

To establish (3.6), we set \[ S_n-\mathbb{E}S_n=\frac{1}{n}\sum_{i=1}^{n} Z_i \]

with \[ Z_i:=f(\lambda_i(n^{-1/2}G))-\mathbb{E}f(\lambda_i(n^{-1/2}G)) \tag{3.8} \]

Now \[ \begin{aligned} \mathbb{E}[(S_n-\mathbb{E}S_{n})^{4}]&= \frac{1}{n^{4}}\sum_{i_1}^{} \mathbb{E}[Z_{i_1}^{4}] \\ &+\frac{4}{n^{4}}\sum_{i_1,i_2}^{} \mathbb{E}[Z_{i_1}Z_{i_2}^{3}] \\ &+\frac{3}{n^{4}}\sum_{i_1,i_2}^{} \mathbb{E}[Z_{i_1}^{2}Z_{i_2}^{2}]\\ &+\frac{6}{n^{4}}\sum_{i_1,i_2,i_3}^{} \mathbb{E}[Z_{i_1}Z_{i_2}Z_{i_3}^{2}] \\ &+\frac{1}{n^{4}}\sum_{i_1,i_2,i_3,i_4}^{} \mathbb{E}[Z_{i_1}Z_{i_2}Z_{i_3}Z_{i_4}]. \end{aligned} \]

The first three terms of the RHS are \(O(n^{-2})\) since \(\max_{1\leqslant i\leqslant n}\lvert Z_i \rvert \leqslant 2\left\| f \right\|_{\infty}\).

Note that the \(k\)-correlation of spectral density for \(n^{-1/2}G\) is \[ \tilde{\varphi}_{n,k}(z_1,\cdots ,z_k)=\frac{(n-k)!}{n!}n^{k}\pi^{-k}\exp \left( -\sum_{i=1}^{k} n\lvert z_i \rvert^{2} \right) \det \left[\sum_{l=0}^{n-1} n^{l}\frac{(z_i \bar{z_j})^{l}}{l!}\right]_{1\leqslant i,j\leqslant k}. \tag{3.9} \]

For the fourth term, rewrite \[ \tilde{\varphi}_{n,3}(z_1,z_2,z_3)=\frac{(n-3)!}{n!}n^{3}\prod_{i=1}^{3} \tilde{\varphi}_{n,1}(z_i)-\left(\frac{(n-3)!}{n!}n^{3}\prod_{i=1}^{3} \tilde{\varphi}_{n,1}(z_i)- \tilde{\varphi}_{n,3}(z_1,z_2,z_3) \right) \]

The function in the parenthesis is a density and has mass \(\displaystyle \frac{(n-3)!}{n!}n^{3}-1=\frac{3n-2}{(n-1)(n-2)}=O(n^{-1})\). And \(Z_1,Z_2,Z_3\) are independent with respect to \(\prod_{i=1}^{3} \tilde{\varphi}_{n,1}(z_i)\). Hence the contribution of the fourth term is \(O(n^{-2})\).

For the bound of the last term, note that \[ \begin{aligned} &\exp \left( -\sum_{i=1}^{k} n\lvert z_i \rvert ^{2} \right) \det\left[ \sum_{l=0}^{n-1} \frac{n^{l}(z_i \bar{z_j})^{l}}{l!}\right] \\ &=\det\left[ \exp (-\frac{n}{2}(\lvert z_i \rvert^{2}+\lvert z_j \rvert ^{2}))\sum_{l=0}^{n-1} \frac{n^{l}(z_i\bar{z_j})^{l}}{l!}\right] \\ &=\det\left[\exp (n(-\frac{1}{2}\lvert z_i \rvert ^{2}-\frac{1}{2}\lvert z_j \rvert ^{2}))\left( \exp (nz_i\bar{z_j})- \frac{n^{n}(z_i\bar{z_j})^{n}}{2\pi i}\int_{\xi=n\mathrm{e}^{i\theta} }^{} \frac{\mathrm{e}^{\xi} \mathrm{d}\xi}{(\xi-nz_i\bar{z_j})\xi^{n}}\right) \right]. \end{aligned} \] (3.10)

Hence, if $S {z <<1} $, then \[ \begin{aligned} &\operatorname{Pr}(z_1 \in S,\cdots ,z_k\in S)\\ &= \frac{(n-k)!}{n!}n^{k}\pi^{-k}\int_{S^{k}}^{} \det \exp n\left( -\frac{1}{2}\lvert z_i \rvert ^{2}-\frac{1}{2}\lvert z_j \rvert ^{2}+z_i\bar{z_j} \right) \mathrm{d}z_1 \cdots \mathrm{d}z_k+O(\alpha^{n}). \\ \end{aligned} \] (3.11)

The diagonal term in the determinant expansion gives \[ \frac{(n-k)!}{n!}n^{k}\pi^{-k}\mu^{k}(S), \]

and this is the leading term. Let \[ B= \begin{cases} 0, \quad k=1 \\ -\frac{(n-k)!}{n!}n^{k}\pi^{-k}\tbinom{k}{2}\mu^{k-2}(S)\int_{S^{2}}^{} \exp \left( -\lvert z_1-z_2 \rvert ^{2}n \right) \mathrm{d}z_1\mathrm{d}z_2, \quad k\geqslant 2. \end{cases} \]

For \(k\geqslant 2\), \(B\) is the contribution to (3.11) from all terms in the determinant containing all but two of the diagonal elements. The contributions from all other terms are \(O(n^{-2})\). (3.11) can be written as \[ \frac{(n-k)!}{n!}n^{k}\left( \frac{\mu(S)}{\pi} \right) ^{k}+B+O(n^{-2}). \]

Using this it is not hard to see the last term is also an \(O(n^{-2})\).

% note danger % I don't understand! % endnote %

Theorem (Layers) We have the following equality in distribution: \[ (\lvert \lambda_1(G)\rvert,\cdots ,\lvert \lambda_n(G) \rvert )=(Z_{(1)},\cdots ,Z_{(n)}) \]

where \(Z_{(1)},\cdots ,Z_{(n)}\) is the non-increasing reordering of a sequence \(Z_1,\cdots ,Z_n\) of independent random variabes with \(Z_k^{2}\sim \Gamma(k,1)\) for every \(1\leqslant k\leqslant n\). Here \(\Gamma(a,\lambda)\) stands for the law on \(\mathbb{R}_{+}\) with Lebesgue density \(x\mapsto \lambda^{a}\Gamma(a)^{-1}x^{a-1}e^{-\lambda x}\).

proof According to (2.24) and (3.9), for every \(r>0\), \[ \begin{aligned} &\operatorname{Pr}(\lvert \lambda_1(G) \rvert \leqslant \sqrt{n}r) \\ &= \int_{\lvert z_1 \rvert \leqslant \sqrt{n}r}^{} \cdots \int_{\lvert z_n \rvert \leqslant \sqrt{n}r}^{} \frac{1}{n!\pi^{n}}\exp \left( -\sum_{i=1}^{n} \lvert z_i \rvert^{2} \right) \det \left[\sum_{l=0}^{n-1} \frac{(z_i \bar{z_j})^{l}}{l!}\right]_{1\leqslant i,j\leqslant n} \mathrm{d}z_n^{r}\mathrm{d}z_n^{i}\cdots \mathrm{d}z_1^{r}\mathrm{d}z_1^{i}\\ &=\frac{1}{\pi^{n}}\det \left[ \int_{\lvert z \rvert \leqslant \sqrt{n}r}^{} \mathrm{e}^{-\lvert z \rvert ^{2}} z^{j}\bar{z}^{j} \mathrm{d}z^{r}\mathrm{d}z^{i} \right]=\prod_{j=1}^{n} \int_{0}^{nr^{2}} \mathrm{e}^{-t} t^{j} \mathrm{d}t=\prod_{j=1}^{n} \operatorname{Pr}\left( \frac{E_1+\cdots +E_j}{n}\leqslant r^{2} \right) \end{aligned} \]

where \(E_1,\cdots ,E_k\) are i.i.d. exponential random variables of unit mean.

The law of large numbers suggests that \(r=1\) is a critical value. The central limit theorem suggests that \(\lvert \lambda_1(G) \rvert\) behaves when \(n\gg 1\) as the maximum of i.i.d. Gaussians, for which the fluctuations follow the Gumbel law.

Universal Case

Marchenko-Pastur quarter circular law a.s. \(\nu_{n^{-1/2}X}\rightsquigarrow \mathcal{Q}_{2}\) as \(n\to \infty\), where \(\mathcal{Q}_{2}\) is the quarter circular law on $[0,2]_{+} $ with density \[ x\mapsto \frac{\sqrt{4-x^{2}}}{\pi}\mathbf{1}_{[0,2]}(x). \]

The \(n^{-1/2}\) normalization is easily understood from the law of large numbers:

\[ \int_{}^{} s^{2} \mathrm{d}\nu_{n^{-1/2}X}(s)=\frac{1}{n^{2}}\sum_{i=1}^{n} s_i(X)^{2}=\frac{1}{n^{2}}\operatorname{tr} (XX^{*})=\frac{1}{n^{2}}\sum_{i,j=1}^{n} \lvert X_{i,j} \rvert ^{2}\rightarrow \mathbb{E}(\lvert X_{1,1} \rvert )^{2}. \tag{2.1} \]

Girko circular law a.s. \(\mu_{n^{-1/2}X}\rightsquigarrow \mathcal{C}_{1}\) as \(n\to \infty\), where \(\mathcal{C}_{1}\) is the circular law which is the uniform law on the unit disc of \(\mathbb{C}\) with density \[ z\mapsto \frac{1}{\pi}\mathbf{1}_{\{z\in \mathbb{C}\colon \lvert z \rvert \leqslant 1\}}. \]

The a.s. tightness of \(\mu_{n^{-1/2}X}\) is easily understood since Weyl's inequality give \[ \int_{}^{} \lvert \lambda^{2} \rvert \mathrm{d}\mu_{n^{-1/2}X}(\lambda)=\frac{1}{n^{2}}\sum_{i=1}^{n} \lvert \lambda_i(X) \rvert ^{2}\leqslant \frac{1}{n^{2}}\sum_{i=1}^{n} s_i(X)^{2}=\int_{}^{} s^{2} \mathrm{d}\nu_{n^{-1/2}X}(s) \]

Reference

1. Charles Bordenave and Djalil Chafaï, The circular law ↩︎
2. R. A. Horn and Ch. R. Johnson, Topics in matrix analysis, Cambridge University Press, Cambridge, 1994, Corrected reprint of the 1991 original. ↩︎
3. Madan Lal Mehta, Random matrices, third ed. Pure and Applied Mathematics, vol.142, Acad. Press, 2004. ↩︎
4. Rider, B. A limit theorem at the edge of a non-Hermitian random matrix ensemble. Random matrix theory. J. Phys. A 36 (2003), no. 12, 3401–3409. ↩︎