Eigenvalue rigidity for truncations of random unitary matrices

TL;DR

This paper studies the eigenvalue distribution of submatrices of random unitary matrices, showing that eigenvalues concentrate around a deterministic measure and providing bounds on their fluctuations.

Contribution

It extends previous results by establishing concentration inequalities for eigenvalues at a microscopic scale in large random unitary submatrices.

Findings

Eigenvalues are tightly concentrated around the deterministic measure.

Concentration inequalities are proved for eigenvalue counting functions.

Results apply to individual bulk eigenvalues on a microscopic scale.

Abstract

We consider the empirical eigenvalue distribution of an $m\times m$ principal submatrix of an $n\times n$ random unitary matrix distributed according to Haar measure. For $n$ and $m$ large with $\frac{m}{n}=\alpha$, the empirical spectral measure is well-approximated by a deterministic measure $\mu_\alpha$ supported on the unit disc. In earlier work, we showed that for fixed $n$ and $m$, the bounded-Lipschitz distance between the empirical spectral measure and the corresponding $\mu_\alpha$ is typically of order $\sqrt{\frac{\log(m)}{m}}$ or smaller. In this paper, we consider eigenvalues on a microscopic scale, proving concentration inequalities for the eigenvalue counting function and for individual bulk eigenvalues.

Equations

$${\mu }_m=\frac{1}{m}\sum _{p=1}^m{\delta }_{{\lambda }_p},$$

$$f_{\alpha }(z)=\{\begin{array}{ll}\frac{(1-\alpha )}{\pi \alpha (1-|z{|}^2{)}^2},& 0<|z|<\sqrt{\alpha };\\ 0,& \text{mbox}otherwise.\end{array}$$

$$g_{\alpha }(z)=\{\begin{array}{ll}\frac{(1-\alpha )}{\pi (1-\alpha |z{|}^2{)}^2},& 0<|z|<1;\\ 0,& \text{mbox}otherwise.\end{array}$$

$$\mathbb{P}_{n,m}\Big{[}d_{BL}\left(\mu_{m},\mu_{\alpha}\right)\geq r\Big{]}\leq e^{2}\exp\left\{-C_{\alpha}m^{2}r^{2}+2m\log(m)+C_{\alpha}^{\prime}m\right\}+\frac{e}{2\pi}\sqrt{\frac{m}{1-\alpha}}e^{-m},$$

$$A_{i,\theta }$$

$$\displaystyle=\left\{z\in\mathbb{C}\bigg{|}|z|<r_{i}\right\}\cup\left\{z\in\mathbb{C}\bigg{|}r_{i}\leq|z|<r_{i+1},0<\arg z<\theta\right\},$$

$$\mathbb{P}$$

$$2\le l\le \sqrt{m}{\left(1-\frac{{ϵ}_m}{1-\alpha (1-{ϵ}_m)}\right)}^{\frac{1}{2}},$$

$$\mathbb{P}\left[|{\lambda }_p-{\tilde{\lambda }}_p|\ge \frac{s}{\sqrt{n-m+(l-1{)}^2}}\right]\le 2\exp \left[-\frac{s^2}{C_{\alpha }l\sqrt{\log (l)}}\right];$$

$$\mathbb{P}\left[|{\lambda }_p-{\tilde{\lambda }}_p|\ge \frac{s}{\sqrt{n-m+(l-1{)}^2}}\right]\le 2\exp \left[-c_{\alpha }{s}^2\right];$$

$$\mathbb{P}\left[|{\lambda }_p-{\tilde{\lambda }}_p|\ge \frac{s}{\sqrt{n-m+(l-1{)}^2}}\right]=0.$$

$$\mathbb{P}\left[|{\lambda }_p-{\tilde{\lambda }}_p|\ge \sqrt{\frac{k\log (n)}{c_{\alpha }(n-m+(l-1{)}^2)}}\right]\le 2n^{-k},$$

$$\mathbb{P}\left[|{\lambda }_p-{\tilde{\lambda }}_p|\ge (\log (l){)}^{\frac{1}{4}}\sqrt{\frac{k{C}_{\alpha }l\log (n)}{n-m+(l-1{)}^2}}\right]\le 2n^{-k},$$

$$\mathrm{Var}\left({\lambda }_p\right)\le C_{\alpha }\frac{\sqrt{p\log (p+1)}}{n}.$$

$$K(z_1,z_2)=\sum _{j=1}^m\frac{1}{N_j}(z_1\overline{z_2}{)}^{j-1}(1-|z_1{|}^2{)}^{\frac{n-m-1}{2}}(1-|z_2{|}^2{)}^{\frac{n-m-1}{2}},$$

$$N_j=\frac{\pi (j-1)!(n-m-1)!}{(n-m+j-1)!}.$$

$$f_{\alpha }(z)=\{\begin{array}{ll}\frac{(1-\alpha )}{\pi \alpha (1-|z{|}^2{)}^2},& 0<|z|<\sqrt{\alpha };\\ 0,& \text{mbox}otherwise.\end{array}$$

$$m{\mu }_{\alpha }(A)-\frac{6\sqrt{2m\log (m+1)}}{1-\alpha }\le \mathbb{E}{\mathcal{N}}_A\le m{\mu }_{\alpha }.$$

$$m{\mu }_{\alpha }(A)-4\le \mathbb{E}{\mathcal{N}}_A\le m{\mu }_{\alpha }(A).$$

$$\mathbb{E}\mathcal{N}(A)={\int }_{A}K(x,x)d\mu (x).$$

$$K(z,z)$$

$$=\frac{(n-m)(1-|z{|}^2{)}^{n-m-1}}{\pi }\sum _{p=0}^{m-1}\frac{(n-m+1{)}_p}{p!}|z{|}^{2p},$$

$${}_2{F}_1(a,b;c;z)=\sum _{n=0}^{\infty }\frac{(a{)}_n(b{)}_n}{(c{)}_n}\frac{z^n}{n!}$$

$$\sum _{p=0}^{\infty }\frac{(n-m+1{)}_p}{p!}|z{|}^{2p}={}_2{F}_1(n-m+1,1;1;|z{|}^2)=(1-|z{|}^2{)}^{-n+m-1}.$$

$$\begin{array}{rl}K(z,z)& =\frac{n-m}{\pi }\left[\frac{1}{(1-|z{|}^2{)}^2}-(1-|z{|}^2{)}^{n-m-1}\sum _{p=m}^{\infty }\frac{(n-m+1{)}_p}{p!}|z{|}^{2p}\right]\\ & =m{f}_{\alpha }(z)\left[1-(1-|z{|}^2{)}^{n-m+1}\sum _{p=m}^{\infty }\frac{(n-m+1{)}_p}{p!}|z{|}^{2p}\right],\end{array}$$

$$\mathbb{E}{\mathcal{N}}_A\le {\int }_{A}m{f}_{\alpha }(z)d\lambda (z)=m{\mu }_{\alpha }(A).$$

$$A\subseteq \left\{|z{|}^2\le \alpha \left(1-\sqrt{\frac{2\log (m+1)}{m}}\right)\right\}.$$

$$\mathbb{P}[Y_k(x)=p]=\frac{(k{)}_p}{p!}(1-x{)}^k{x}^p.$$

$$\mathbb{E}[e^{t{Y}_k(x)}]=\sum _{p=0}^{\infty }\frac{(k{)}_p}{p!}(1-x{)}^k(e^{t}x{)}^p={\left[\frac{1-x}{1-x{e}^t}\right]}^k.$$

$$(1-|z{|}^2{)}^{n-m+1}$$

$$=\mathbb{P}\left[Y_{n-m+1}(|z{|}^2)\ge m\right]\le e^{-tm}{\left[\frac{1-|z{|}^2}{1-|z{|}^2{e}^t}\right]}^{n-m+1},$$

$$(1-|z{|}^2{)}^{n-m+1}$$

Full text

11footnotemark: 1† Supported in part by NSF DMS 1612589.

1. Introduction

Let $U$ be an $n\times n$ Haar-distributed unitary matrix and let $U_{m}$ be the $m\times m$ top-left block of $U$, where $m<n$. We refer to $U_{m}$ as a truncation of $U$. The eigenvalues of the truncation are all located within the unit disc and the asymptotic distribution of the eigenvalues can be described quite explicitly. Let $\mu_{m}$ denote the empirical spectral measure of $U_{m}$, that is,

[TABLE]

where $\lambda_{1},\ldots,\lambda_{m}$ are the eigenvalues of $U_{m}$. Petz and Réffy [5] proved that if $\frac{m}{n}\to\alpha\in(0,1)$, then $\mu_{m}$ converges almost surely to a limiting spectral measure $\mu_{\alpha}$; it has radial density with respect to Lebesgue measure on $\mathbb{C}$ given by

[TABLE]

In [4], we proved the following non-asymptotic, quantitative version of this result. The rescaling was chosen so that the support of the limiting measure is the full unit disc, independent of $\alpha$.

Theorem 1** (E. Meckes and K. Stewart).**

Let $n,m\in\mathbb{N}$ with $1\leq m<n$. Let $U\in\mathbb{U}\left(n\right)$ be distributed according to Haar measure, and let $\lambda_{1},\ldots,\lambda_{m}$ denote the eigenvalues of the top-left $m\times m$ block of $\sqrt{\frac{n}{m}}U$. The joint law of $\lambda_{1},\ldots,\lambda_{m}$ is denoted $\mathbb{P}_{n,m}$. Let $\mu_{m}$ be the random measure with mass $\frac{1}{m}$ at each of the $\lambda_{p}$, and let $\alpha=\frac{m}{n}$. Let $\mu_{\alpha}$ be the probability measure on the unit disc with the density $g_{\alpha}$ defined by

[TABLE]

For any $r>0$,

[TABLE]

*where $C_{\alpha}=\frac{1}{128\pi(1+\sqrt{3+\log(\alpha^{-1})})^{2}}$ and $C_{\alpha}^{\prime}=6+3\log(\alpha^{-1})$.

The result above is essentially macroscopic; it says that with high probability, $d_{BL}(\mu_{m},\mu_{\alpha})$ is of order $\sqrt{\frac{\log(m)}{m}}$. The purpose of this paper is to examine the microscopic level, by considering the eigenvalue counting function on small sets. Throughout the paper, we assume that $\alpha=\frac{m}{n}$ is bounded away from 0 and 1; i.e., that there is a fixed $\delta>0$ such that $\alpha\in(\delta,1-\delta)$. Throughout the statements and proofs, there are constants $C_{\alpha}$ depending only on $\alpha$; their exact values may vary from one line to the next.

We begin by ordering the eigenvalues $\left\{\lambda_{p}\right\}_{p=1}^{m}$ in the spiral fashion introduced in [3]. Define a linear order $\prec$ on $\mathbb{C}$ by making 0 initial, and for nonzero $w,z\in\mathbb{C},$ declare $w\prec z$ if either of the following hold:

• •

$|w|<|z|$

• •

$|w|=|z|$ and $\arg w<\arg z$

We divide the disc of radius $\sqrt{\frac{m}{n}}$ (i.e., the support of the limiting eigenvalue density) into annuli with radii $r_{i}=\frac{i}{\sqrt{n-m+i^{2}}}$; it is verified below that the expected number of eigenvalues in the annulus from radius $r_{i-1}$ to $r_{i}$ is approximately $2i-1$.

More generally, for $\theta\in(0,2\pi]$, define

[TABLE]

with $r_{i}=\frac{i}{\sqrt{n-m+i^{2}}}$ and $1\leq i\leq\sqrt{m}$ (see Figure 1).

Our first main result is on the concentration of the eigenvalue counting function for the sets $A_{i,\theta}$.

Theorem 2**.**

Let $\mathcal{N}_{i,\theta}$ denote the number of eigenvalues of an $m\times m$ truncation of a Haar-distributed matrix in $\mathbb{U}\left(n\right)$ which lie in $A_{i,\theta}$. If $\epsilon_{m}=\sqrt{\frac{2\log(m+1)}{m}}$, then for each $1\leq i\leq\sqrt{m}\left(1-\frac{\epsilon_{m}}{1-\alpha\left(1-\epsilon_{m}\right)}\right)^{\frac{1}{2}}$, $0\leq\theta\leq 2\pi$, and $t>0$,

[TABLE]

If $t>\frac{12}{1-\alpha}\sqrt{2m\log(m+1)},$ then this estimate is also valid for those $i$ with $\sqrt{m}\left(1-\frac{\epsilon_{m}}{1-\alpha\left(1-\epsilon_{m}\right)}\right)^{\frac{1}{2}}\leq i\leq\sqrt{m}$.

We next define predicted locations $\{\tilde{\lambda}_{p}\}_{p=1}^{m}$ for the eigenvalues by choosing $2i-1$ equally spaced points in the annulus with inner radius $r_{i-1}$ and outer radius $r_{i}$. The concentration inequalities in Theorem 2 for the counting function lead to the following concentration inequality for bulk eigenvalues about their predicted locations.

Theorem 3**.**

Let $\{\lambda_{p}\}_{p=1}^{m}$ denote the eigenvalues of an $m\times m$ truncation of a Haar-distributed matrix in $\mathbb{U}\left(n\right)$, ordered according to $\prec$. Let $l=\lceil\sqrt{p}\rceil$. There are constants $c_{\alpha},C_{\alpha}$ depending only on $\alpha=\frac{m}{n}$ such that, if $\epsilon_{m}=\sqrt{\frac{2\log(m+1)}{m}}$, then for those $p$ with

[TABLE]

when $s\leq 2\pi(l-1)$,

[TABLE]

when $2\pi(l-1)<s\leq 2\sqrt{n-m+(l-1)^{2}}$,

[TABLE]

and when $s>2\sqrt{n-m+(l-1)^{2}}$,

[TABLE]

By way of example, if $2\pi(l-1)\leq\sqrt{\frac{k}{c_{\alpha}}\log(n)}$, then

[TABLE]

whereas if, e.g., $\log(n)\leq\frac{4\pi^{2}(l-1)^{2}}{kC_{\alpha}l\sqrt{\log(l)}}$, then

[TABLE]

For reference, spacing of predicted locations around $\tilde{\lambda}_{p}$ is about $\frac{1}{\sqrt{n-m+(l-1)^{2}}}$.

The concentration inequalities of Theorem 3 also easily imply the following variance bound for bulk eigenvalues.

Corollary 4**.**

Let $\epsilon_{m}=\sqrt{\frac{2\log(m+1)}{m}}$ and $p$ be such that $2\leq\lceil\sqrt{p}\rceil\leq\sqrt{m}\left(1-\frac{\epsilon_{m}}{1-\alpha(1-\epsilon_{m})}\right)^{\frac{1}{2}}.$

There is a constant $C_{\alpha}$ depending only on $\alpha=\frac{m}{n}$ such that

[TABLE]

2. Means and Variances

Throughout the proofs, we will make heavy use of the fact that the eigenvalues we consider are a determinantal point process on $\{|z|\leq 1\}$ with kernel (with respect to Lebesgue measure) given by

[TABLE]

with

[TABLE]

See, e.g., [6] or [5].

Recall that for large $n$ and $\frac{m}{n}=\alpha\in(0,1)$ the spectral measure of the truncation is approximately given by the measure $\mu_{\alpha}$, with density with respect to Lebesgue measure given by

[TABLE]

In particular, given a set $A\subseteq\{|z|\leq\sqrt{\alpha}\}$, the expected number $\mathcal{N}_{A}$ of eigenvalues inside $A$ is approximately $m\mu_{\alpha}(A)$. We begin by giving explicit estimates quantifying this approximation.

Lemma 5**.**

For any measurable $A\subseteq\{|z|\leq\sqrt{\alpha}\}$,

[TABLE]

If additionally $A\subseteq\left\{|z|^{2}\leq\alpha\left(1-\sqrt{\frac{2\log(m+1)}{m}}\right)\right\},$ then

[TABLE]

Proof.

For a determinantal point process on $(\Lambda,\mu)$ with kernel $K$, the expected number of points in a set $A$ is given by

[TABLE]

From the formula for the kernel given in equation (1),

[TABLE]

where $(a)_{p}=a(a+1)\cdots(a+p-1)$ is the rising Pochhammer symbol. Letting

[TABLE]

denote the hypergeometric function with parameters $a,b,c$,

[TABLE]

It follows that

[TABLE]

and as an immediate consequence,

[TABLE]

For the lower bound, we first treat the more restrictive case of

[TABLE]

Consider the random variable $Y_{k}(x)$ on $\mathbb{N}\cup\{0\}$ with mass function

[TABLE]

The moment generating function of $Y_{k}(x)$ is given by

[TABLE]

Now,

[TABLE]

for any $t>0$. Since $|z|^{2}<\alpha\left(1-\sqrt{\frac{2\log(m+1)}{m}}\right)<\frac{\alpha n}{n+1}$, we may choose $t=\log\left(\frac{m}{|z|^{2}(n+1)}\right)>0.$ Then

[TABLE]

For $|z|^{2}\leq\frac{\alpha n}{n+1},$ this last quantity is increasing in $|z|$; if we further assume that $|z|^{2}\leq\alpha(1-\epsilon_{n})$, we thus have that

[TABLE]

The claimed estimate follows by taking $\epsilon_{n}=\sqrt{\frac{2\log(m+1)}{m}}=\sqrt{\frac{2\log(\alpha n+1)}{\alpha n}}$ (the constant $4$ in the statement is for concreteness; the actual estimate resulting from this choice of $\epsilon_{n}$ is $e^{\sqrt{\frac{2\alpha\log(\alpha n+1)}{n}}}$).

Returning to the more general case, using the expression for $K(z,z)$ in (2)

[TABLE]

making use of the analysis above. To estimate the remaining integral, we reconsider the quantity

[TABLE]

this time simply estimating via Markov’s inequality. Given $k$ and $x$,

[TABLE]

and so

[TABLE]

It follows that

[TABLE]

Now,

[TABLE]

for $|z|^{2}\leq\alpha$, and so

[TABLE]

It follows that

[TABLE]

Observing that $\max\left\{4,\frac{\alpha}{(1-\alpha)^{2}}\sqrt{\frac{2\log(m+1)}{m}}\right\}\leq\frac{1}{1-\alpha}\sqrt{2m\log(m+1)}$ completes the proof. ∎

The following is an immediate consequence of Lemma 5.

Corollary 6**.**

Let $\theta\in(0,2\pi]$ and let $r_{i}=\frac{i}{\sqrt{n-m+i^{2}}}$. Let $\mathcal{N}_{i,\theta}$ be defined as above, and suppose that $m\geq 3$ and $1\leq i\leq\sqrt{m}\left(1-\frac{\sqrt{\frac{2\log(m+1)}{m}}}{1-\alpha+\alpha\sqrt{\frac{2\log(m+1)}{m}}}\right)$. Then

[TABLE]

Proof.

The condition on $i$ guarantees that $A_{i,\theta}\subseteq\left\{|z|^{2}\leq\alpha\left(1-\sqrt{\frac{2\log(m+1)}{m}}\right)\right\},$ so that the sharper estimate from Lemma 5 applies.

For $A_{i,\theta}$ defined as above,

[TABLE]

so that

[TABLE]

∎

We next estimate the variance of $\mathcal{N}_{i,\theta}$.

Lemma 7**.**

Let $A_{i,\theta}$ be as above. There is a constant $C_{\alpha}$ depending only on $\alpha=\frac{m}{n}$ such that

[TABLE]

Proof.

By an argument similar to the one in [1, Appendix B],

[TABLE]

Observe that for $r_{1},r_{2}\leq 1$,

[TABLE]

Integrating in polar coordinates gives that

[TABLE]

since the angular integrals vanish unless $j=k$. By repeated applications of integration by parts,

[TABLE]

which is exactly $\mathbb{P}[Y_{j}>j]$ for $Y_{j}\sim\mathrm{Binom}(n-m+j,r_{i}^{2}).$ Similarly,

[TABLE]

which is $\mathbb{P}[X_{j}\leq j]$ for $X_{j}\sim\mathrm{Binom}(n-m+j,r_{i+1}^{2}).$

It thus follows from (4) that

[TABLE]

For the first sum, observe that $\mathbb{E}X_{j}=\frac{(n-m+j)(i+1)^{2}}{n-m+(i+1)^{2}}>j$ for $j\leq i^{2}-1$. By Bernstein’s inequality,

[TABLE]

The first term of the minimum is smaller exactly when $j\geq j_{0}:=\frac{(i+1)^{4}}{n-m+2(i+1)^{2}}$. Note that $j_{0}\leq\frac{(i+1)^{2}}{2}$, so that

[TABLE]

which is bounded independent of $i$.

Now consider

[TABLE]

where we have used the fact that the summand in the second line is increasing in $j$ and bounded by $\frac{1}{i+1}$ at the upper limit of the sum.

For the second sum of Equation (5), we again apply Bernstein’s inequality:

[TABLE]

The change in behavior of the bound is at $j=j_{1}:=\frac{i^{2}[2(n-m)+i^{2}]}{n-m}$. Note that $j_{1}\leq i^{2}\left(2+\frac{\alpha}{1-\alpha}\right)$ since $i^{2}\leq m$. Decomposing as before,

[TABLE]

This last sum is

[TABLE]

which is bounded independent of $i$. Collecting terms, we have

[TABLE]

for a constant $C_{\alpha}$ depending only on $\alpha$.

The remaining terms of (3) are estimated similarly. For $V_{2}$, integrating in polar coordinates gives that

[TABLE]

Proceeding exactly as for $V_{1}$,

[TABLE]

where $Y_{j}\sim$ Binom$(n-m+j,r_{i}^{2})$.

Integrating in polar coordinates and proceeding as above,

[TABLE]

where $X_{j}\sim$ Binom$(n-m+j,r_{i+1}^{2})$. The final integral in (3) is

[TABLE]

For $j\neq k$,

[TABLE]

Therefore, if $j\neq k$ in the sum the term is negative. Thus

[TABLE]

All together then,

[TABLE]

for a constant $C_{\alpha}$ depending only on $\alpha$. ∎

3. Concentration

We now move on to concentration for the counting functions $\mathcal{N}_{i,\theta}$. The key ingredient is the following general result on determinantal point processes.

Theorem 8** (Hough–Krishnapur–Peres–Virág [2]).**

Let $\Lambda$ be a locally compact Polish space and $\mu$ a Radon measure on $\Lambda$. Suppose that $K:\Lambda\times\Lambda\to\mathbb{C}$ is the kernel of a determinantal point process, such that the corresponding integral operator $\mathcal{K}:L^{2}(\mu)\to L^{2}(\mu)$ defined by

[TABLE]

is self-adjoint, nonnegative, and locally trace-class. Let $D\subseteq\Lambda$ be such that the restriction $K_{D}(x,y)=\mathbbm{1}_{D}(x)K(x,y)\mathbbm{1}_{D}(y)$ defines a trace-class operator $\mathcal{K}_{D}$ on $L^{2}(\mu)$. Then the number of points $\mathcal{N}_{D}$ lying in $D$ of the process governed by $K$ is distributed as $\sum_{k}\xi_{k}$, where the $\xi_{k}$ are independent Bernoulli random variables whose means are given by the eigenvalues of the operator $\mathcal{K}_{D}$.

It is not hard to see that the kernel given in (1) has the properties required by Theorem 8, and so the random variable $\mathcal{N}_{i,\theta}$ is distributed as a sum of independent Bernoulli random variables. It is thus an immediate consequence of Bernstein’s inequality that

[TABLE]

This is the key observation underlying the proof of Theorem 2.

Proof of Theorem 2.

For the first claim, the assumption on $i$ implies that

[TABLE]

where $D$ is the unit disc, so that by lemmas 5 and 7 together with Bernstein’s inequality, if $t>4$, then

[TABLE]

If $t\geq 8$, then $t-4\geq t/2$, so that

[TABLE]

if $t<8$, then

[TABLE]

and the first claim follows. The proof of the second claim is an immediate consequence of the second estimate of Lemma 5 together with Lemma 7. ∎

We now focus our attention on individual eigenvalues. Given $1\leq p\leq m$, let $l=\lceil\sqrt{p}\rceil$ and $q=p-(l-1)^{2}$, so that $p=(l-1)^{2}+q$ and $1\leq q\leq 2l-1$. Let

[TABLE]

The predicted locations $\tilde{\lambda}_{p}$ for the eigenvalues are defined by

[TABLE]

To shed some light on this choice, consider the annulus $A_{l}$ with inner radius $r_{l-1}$ and outer radius $r_{l}.$ Then

[TABLE]

It follows from Lemma 5 that the expected number of eigenvalues in $A_{l}$ is approximately $2l-1.$

Proof of Theorem 3.

The essential idea of the proof is that if $\lambda_{p}$ is far from its predicted location $\tilde{\lambda}_{p}$, then there is either a set of the form $A_{\ell,\theta}$ with substantially more eigenvalues than predicted by the mean (if $\lambda_{p}$ comes early) or a set of the form $A_{\ell,\theta}$ with substantially fewer eigenvalues than predicted by the mean (if $\lambda_{p}$ comes late). Theorem 2 then gives control on the probabilities of such events.

To implement this strategy, several cases must be considered, which we first outline here.

1. (I)

$\lambda_{p}\prec\tilde{\lambda}_{p}$

1. (A)

$\frac{s}{2(l-1)}<\frac{2\pi q}{2l-1}$ 2. (B)

$\frac{2\pi q}{2l-1}\leq\frac{s}{2(l-1)}$ 2. (II)

$\lambda_{p}\succ\tilde{\lambda}_{p}$

1. (A)

$\frac{s}{2(l-1)}<2\pi-\frac{2\pi q}{2l-1}$ 2. (B)

$2\pi-\frac{2\pi q}{2l-1}\leq\frac{s}{2(l-1)}\leq\pi$ 3. (C)

$\pi<\frac{s}{2(l-1)}\leq\frac{\sqrt{m}\left(1-\frac{\epsilon_{m}}{1-\alpha(1-\epsilon_{m})}\right)^{\frac{1}{2}}+l-1}{2(l-1)}$ 4. (D)

$\frac{\sqrt{m}\left(1-\frac{\epsilon_{m}}{1-\alpha(1-\epsilon_{m})}\right)^{\frac{1}{2}}+l-1}{2(l-1)}<\frac{s}{2(l-1)}$

Combining cases (A) and (B) from both I and II yields the first part of the lemma (small $s$) and combining (C) and (D) of II gives the second part of the lemma (large $s$).

In most of the cases we will make use of the fact that

[TABLE]

where $a(\theta,\phi)$ denotes the length of the shorter arc on the unit circle between $e^{i\theta}$ and $e^{i\phi}$.

(I, A) Suppose that $|\lambda_{p}-\tilde{\lambda}_{p}|\geq\frac{s}{\sqrt{n-m+(l-1)^{2}}}$, that $\lambda_{p}\prec\tilde{\lambda}_{p}$, and that $\frac{s}{2(l-1)}<\frac{2\pi q}{2l-1}.$ We claim that

[TABLE]

Indeed, since $\lambda_{p}\prec\tilde{\lambda}_{p}$, either

1. (i)

$\begin{aligned} r_{l-1}\leq|\lambda_{p}|<r_{l}&\text{ and }\arg\lambda_{p}<\arg\tilde{\lambda}_{p}=\frac{2\pi q}{2l-1}\end{aligned}$ or 2. (ii)

$\begin{aligned} |\lambda_{p}|<|\tilde{\lambda}_{p}|=r_{l-1}.\end{aligned}$

If $|\lambda_{p}|<r_{l-1}$ holds, then the claim holds trivially. Otherwise, the estimate in (9) implies

[TABLE]

Therefore when condition (i) holds and $|\lambda_{p}-\tilde{\lambda}_{p}|\geq\frac{s}{\sqrt{n-m+(l-1)^{2}}}$, then

[TABLE]

and so $\arg\lambda_{p}<\frac{2\pi q}{2l-1}-\frac{s}{2(l-1)}.$ In this case as well, then,

[TABLE]

It follows from the claim that

[TABLE]

Now, the computation of $m\mu_{\alpha}\left(A_{i,\theta}\right)$ in the proof of Corollary 6 gives that

[TABLE]

Then Theorem 2 implies that

[TABLE]

since $s\leq 2\pi(l-1)$.

(I, B) Suppose that $|\lambda_{p}-\tilde{\lambda}_{p}|\geq\frac{s}{\sqrt{n-m+(l-1)^{2}}}$, $\lambda_{p}\prec\tilde{\lambda}_{p}$ and $\frac{2\pi q}{2l-1}\leq\frac{s}{2(l-1)}.$ We claim that

[TABLE]

The estimate (9) implies condition (ii) above must hold; that is, $|\lambda_{p}|<r_{l-1}$. If $|\lambda_{p}|\geq r_{l-1}-\frac{s}{2\sqrt{n-m+(l-1)^{2}}}$, then the estimate (9) again implies that

[TABLE]

In particular, when $r_{l-2}\leq|\lambda_{p}|<r_{l-1}$, $\arg\lambda_{p}<2\pi+\frac{2\pi q}{2l-1}-\frac{s}{2(l-1)}$. If $|\lambda_{p}|<r_{l-1}-\frac{s}{2\sqrt{n-m+(l-1)^{2}}}<r_{l-2}$, then the estimate (9) implies

[TABLE]

Then

[TABLE]

Either way, $\frac{s}{2(l-1)}\leq\pi$,

[TABLE]

and

[TABLE]

Now the computation in the proof of Corollary 6 yields

[TABLE]

for $l\geq 2.$ Therefore in this range of $s$, Theorem 2 implies that

[TABLE]

The estimates above cover the entire range of $s$ when $\lambda_{p}\prec\tilde{\lambda}_{p}$ and so

[TABLE]

for all $s>0$.

(II, A) Suppose that $|\lambda_{p}-\tilde{\lambda}_{p}|\geq\frac{s}{\sqrt{n-m+(l-1)^{2}}}$, that $\lambda_{p}\succ\tilde{\lambda}_{p}$, and that $\frac{s}{2(l-1)}<2\pi-\frac{2\pi q}{2l-1}.$ We claim that

[TABLE]

Indeed, since $\lambda_{p}\succ\tilde{\lambda}_{p}$, either

1. (i)

$\begin{aligned} r_{l-1}\leq|\lambda_{p}|<r_{l}\text{ and }\arg\lambda_{p}>\arg\tilde{\lambda}_{p}\end{aligned}$ or 2. (ii)

$\begin{aligned} |\lambda_{p}|\geq r_{l}=\frac{l}{\sqrt{n-m+l}}.\end{aligned}$

If $|\lambda_{p}|\geq r_{l}$ holds, then the claim is trivially true. Suppose that condition (i) holds. Then for $s\geq 2$, $|\lambda_{p}|<r_{l}<r_{l-1}+\frac{s}{2\sqrt{n-m+(l-1)^{2}}}$. As above, combining this observation with the fact that $|\lambda_{p}-\tilde{\lambda}_{p}|\geq\frac{s}{\sqrt{n-m+(l-1)^{2}}}$ and the estimate (9) implies $a\left(\arg\lambda_{p},\frac{2\pi q}{2l-1}\right)\geq\frac{s}{2(l-1)}.$ It follows that if condition (i) holds, then $\arg\lambda_{p}>\frac{2\pi q}{2l-1}+\frac{s}{2(l-1)}$, and so

[TABLE]

It follows from the claim that

[TABLE]

By the proof of Corollary (6),

[TABLE]

and so Theorem 2 implies that

[TABLE]

since $s\leq 2\pi(l-1)$.

(II, B) Suppose that $|\lambda_{p}-\tilde{\lambda}_{p}|\geq\frac{s}{\sqrt{n-m+(l-1)^{2}}}$, that $\lambda_{p}\succ\tilde{\lambda}_{p}$ and that $2\pi-\frac{2\pi q}{2l-1}\leq\frac{s}{2(l-1)}\leq\pi.$ We claim that

[TABLE]

By the estimate (9) again, it must be the case that condition (ii) holds and $|\lambda_{p}|\geq r_{l}$. If $s\geq 2$ and $|\lambda_{p}|\geq r_{l-1}+\frac{s}{2\sqrt{n-m+(l-1)^{2}}}>r_{l+1}$, then the claim holds trivially. If $|\lambda_{p}|<r_{l-1}+\frac{s}{2\sqrt{n-m+(l-1)^{2}}}$, then the estimate (9) yields $\arg\lambda_{p}>\frac{2\pi q}{2l-1}+\frac{s}{2(l-1)}-2\pi.$ In particular, if $r_{l}\leq|\lambda_{p}|<r_{l+1}$, then $\arg\lambda_{p}>\frac{2\pi q}{2l-1}+\frac{s}{2(l-1)}-2\pi$ and so

[TABLE]

It follows from the claim that

[TABLE]

Since

[TABLE]

Theorem 2 implies that in this regime,

[TABLE]

Combining cases (I, A), (I,B), (II, A), and (II, B) thus yields

[TABLE]

when $s\leq 2\pi(l-1)$. This proves the first part of the Theorem (small $s$).

Finally, we consider cases for the larger values of $s$ based on which part of Theorem 2 applies.

(C) Let $\epsilon_{m}=\sqrt{\frac{2\log(m+1)}{m}}$ and suppose that $|\lambda_{p}-\tilde{\lambda}_{p}|\geq\frac{s}{\sqrt{n-m+(l-1)^{2}}}$, that $\lambda_{p}\succ\tilde{\lambda}_{p}$, and that

[TABLE]

(That is, $s-l+1\leq\sqrt{m}\left(1-\frac{\epsilon_{m}}{1-\alpha\left(1-\epsilon_{m}\right)}\right)^{\frac{1}{2}}$.) By the triangle inequality,

[TABLE]

Now, $\frac{s}{2(l-1)}>\pi$ implies $(l-1)^{2}\leq(s-l+1)^{2}$. It follows that $|\lambda_{p}|\geq\frac{s-l+1}{\sqrt{n-m+(s-l+1)^{2}}}$, and so

[TABLE]

Since $p=(l-1)^{2}+q$, $1\leq q\leq 2l-1$, and $l<\frac{s}{2\pi}+1$,

[TABLE]

since $s\geq 2\pi$. It follows from Theorem 2,

[TABLE]

since $s\geq 2\pi(l-1)$.

(D) Suppose that $|\lambda_{p}-\tilde{\lambda}_{p}|\geq\frac{s}{\sqrt{n-m+(l-1)^{2}}}$, that $\lambda_{p}\succ\tilde{\lambda}_{p}$, and that

[TABLE]

that is,

[TABLE]

As in the previous case,

[TABLE]

If $\frac{s-l+1}{\sqrt{n-m+(s-l+1)^{2}}}\geq 2$, then $\mathbb{P}\left[\mathcal{N}\left(A_{\lfloor s-l+1\rfloor,2\pi}\right)<p\right]=0$. Otherwise, by the second estimate in Lemma 5,

[TABLE]

In this range, $s^{2}\geq m\left(1-\frac{\epsilon_{m}}{1-\alpha(1-\epsilon_{m})}\right)$, so the lower bound can be replaced, for large enough $m$, by $cs^{2}$ by slightly reducing the value of $c$. Theorem 2 applied with $t=cs^{2}\geq\frac{12\sqrt{2m\log(m)}}{1-\alpha}$ (again for $m$ large enough) then yields

[TABLE]

Cases (C) and (D) thus yield

[TABLE]

for $s\geq 2\pi(l-1)$. Finally, the empirical spectral measure is supported on the disc of radius 1. It follows that if $s\geq 2\sqrt{n-m+(l-1)^{2}}$, then

[TABLE]

This completes the proof. ∎

Proof of Corollary 4.

Let $\epsilon_{m}=\sqrt{\frac{2\log(m+1)}{m}}$ and $p$ be such that

[TABLE]

Then by Fubini’s theorem and Theorem 3,

[TABLE]

since $(l-1)^{2}\leq m$. ∎