Gaussian Fluctuations for Linear Eigenvalue Statistics of Products of Independent iid Random Matrices

Abstract

Consider the product $X = X_{1}\cdots X_{m}$ of m independent $n\times n$ iid random matrices. When m is fixed and the dimension n tends to infinity, we prove Gaussian limits for the centered linear spectral statistics of X for analytic test functions. We show that the limiting variance is universal in the sense that it does not depend on m (the number of factor matrices) or on the distribution of the entries of the matrices. The main result generalizes and improves upon previous limit statements for the linear spectral statistics of a single iid matrix by Rider and Silverstein as well as Renfrew and the second author.

Appendices

Appendix A. Truncation Arguments

This section is devoted to the proof of Lemma 4.3.

Proof of Lemma 4.3

First, we prove property (i). Observe that

$$\begin{aligned} 1&=\text {Var}(\xi )\\&=\mathbb {E}[\xi ^{2}{\mathbf {1}_{\{|\xi |\le n^{1/2-\varepsilon }\}}}]+\mathbb {E}[\xi ^{2}{\mathbf {1}_{\{|\xi |> n^{1/2-\varepsilon }\}}}]\\&=\text {Var}(\tilde{\xi })+\left( \mathbb {E}[\xi {\mathbf {1}_{\{|\xi |\le n^{1/2-\varepsilon }\}}}]\right) ^{2}+\mathbb {E}[\xi ^{2}{\mathbf {1}_{\{|\xi |> n^{1/2-\varepsilon }\}}}]. \end{aligned}$$

Also observe that

$$\begin{aligned} 0=\mathbb {E}[\xi ]=\mathbb {E}[\xi {\mathbf {1}_{\{|\xi |\le n^{1/2-\varepsilon }\}}}]+\mathbb {E}[\xi {\mathbf {1}_{\{|\xi |> n^{1/2-\varepsilon }\}}}] \end{aligned}$$

which implies $\left| \mathbb {E}[\xi {\mathbf {1}_{\{|\xi |\le n^{1/2-\varepsilon }\}}}]\right| =\left| \mathbb {E}[\xi {\mathbf {1}_{\{|\xi |> n^{1/2-\varepsilon }\}}}]\right| .$ Hence

$$\begin{aligned} |1-\text {Var}(\tilde{\xi })|&=\left( \mathbb {E}[\xi {\mathbf {1}_{\{|\xi |\le n^{1/2-\varepsilon }\}}}]\right) ^{2}+\mathbb {E}[\xi ^{2}{\mathbf {1}_{\{|\xi |> n^{1/2-\varepsilon }\}}}]\\&=\left| \mathbb {E}[\xi {\mathbf {1}_{\{|\xi |> n^{1/2-\varepsilon }\}}}]\right| ^{2}+\mathbb {E}[\xi ^{2}{\mathbf {1}_{\{|\xi |> n^{1/2-\varepsilon }\}}}]\\&\le 2\mathbb {E}\left[ \frac{|\xi |^{4}}{n^{1-2\varepsilon }}{\mathbf {1}_{\{|\xi |> n^{1/2-\varepsilon }\}}}\right] \\&=o(n^{-1-2\varepsilon }). \end{aligned}$$

Next we move onto (ii). By construction, $\mathbb {E}[\hat{\xi }]=0$ and $\text {Var}(\hat{\xi })=1$ provided n is sufficiently large. By part (i),

$$\begin{aligned} 1-\frac{C}{n^{1+2\varepsilon }}\le {{\,\mathrm{Var}\,}}(\tilde{\xi }) \end{aligned}$$

for some constant $C>0$ so choosing $N_{0}>\left( \frac{4C}{3}\right) ^{1/(1+2\varepsilon )}$ ensures that $\frac{1}{4}\le \text {Var}(\tilde{\xi })$, which gives $2\ge \left( \text {Var}(\tilde{\xi })\right) ^{-1/2}$ for $n>N_{0}$. With such an $n>N_{0}$,

$$\begin{aligned} \left| \hat{\xi }\right|&=\left| \frac{\xi {\mathbf {1}_{\{|\xi |\le n^{1/2-\varepsilon }\}}}-\mathbb {E}\left[ \xi {\mathbf {1}_{\{|\xi |\le n^{1/2-\varepsilon }\}}}\right] }{\sqrt{\text {Var}(\tilde{\xi })}}\right| \\&\le 2\left| \xi {\mathbf {1}_{\{|\xi |\le n^{1/2-\varepsilon }\}}}\right| +2\left| \mathbb {E}\left[ \xi {\mathbf {1}_{\{|\xi |\le n^{1/2-\varepsilon }\}}}\right] \right| \\&\le 4n^{1/2-\varepsilon } \end{aligned}$$

almost surely. For part 4.3, we have

$$\begin{aligned} \mathbb {E}|\hat{\xi }|^{4}&=\mathbb {E}\left| \frac{\xi {\mathbf {1}_{\{|\xi |\le n^{1/2-\varepsilon }\}}}-\mathbb {E}\left[ \xi {\mathbf {1}_{\{|\xi |\le n^{1/2-\varepsilon }\}}}\right] }{\sqrt{\text {Var}(\tilde{\xi })}}\right| ^{4}\\&\le 2^{4}\mathbb {E}\left| \xi {\mathbf {1}_{\{|\xi |\le n^{1/2-\varepsilon }\}}}-\mathbb {E}\left[ \xi {\mathbf {1}_{\{|\xi |\le n^{1/2-\varepsilon }\}}}\right] \right| ^{4}\\&\le 2^{8}\mathbb {E}\left| \xi \right| ^{4}, \end{aligned}$$

completing the proof of the claim. $\square$

Appendix B. Largest and Smallest Singular Values

In this section, we consider events concerning the largest and smallest singular values for the random matrices appearing in this paper. These results are included as an appendix because the methods used to prove them are slight modifications of those in [23, 48, 52]. In order to prove these results, we need to introduce an intermediate truncation of the matrices. Specifically, let $\xi _{1},\xi _{2},\dots \xi _{m}$ be real-valued random variables each having mean zero, variance one, and finite $4+\tau$ moment for some $\tau >0$. Let $X_{n,1}X_{n,2},\dots X_{n,m}$ be independent iid $n\times n$ random matrices with atom random variables $\xi _{1},\xi _{2},\dots \xi _{m}$, respectively. For a fixed $\varepsilon >0$, and for each $1\le k\le m$, define truncated random variables (at $n^{1/2-\varepsilon }$) $\tilde{\xi }_{k}$ and $\hat{\xi }_{k}$ as in (19). Also define truncated matrices $\tilde{X}_{n,k}$ and $\hat{X}_{n,k}$ as in (21) and (22), respectively. Define the linearized truncated matrix $\mathcal {Y}_{n}$ as in (31). Also recall that $P_{n}=n^{-m/2}X_{n,1}X_{n,2}\cdots X_{n,m}$ and $\hat{P}_{n}=n^{-m/2}\hat{X}_{n,1}\hat{X}_{n,2}\cdots \hat{X}_{n,m}.$

Let X be an $n\times n$ random matrix filled with iid copies of a random variable $\xi$ which has mean zero, unit variance, and finite $4+\tau$ moment. For a fixed constant $L>0$, define matrices $\mathring{X}$ and $\check{X}$ to be the $n\times n$ matrices with entries defined by

$$\begin{aligned} \mathring{X}_{(i,j)}:=X_{(i,j)}{\mathbf {1}_{\{|X_{(i,j)}|\le L/\sqrt{2}\}}}-\mathbb {E}\left[ X_{(i,j)}{\mathbf {1}_{\{|X_{(i,j)}|\le L/\sqrt{2}\}}}\right] \end{aligned}$$

(109)

and

$$\begin{aligned} \check{X}_{(i,j)}:= \frac{\mathring{X}_{(i,j)}}{\sqrt{\text {Var}(\mathring{X}_{(i,j)})}} \end{aligned}$$

(110)

for $1 \le i,j \le n$. Define $\mathring{X}_{n,1},\mathring{X}_{n,2},\dots \mathring{X}_{n,m}$ and $\check{X}_{n,1},\check{X}_{n,2},\dots \check{X}_{n,m}$ as in (109) and (110), respectively. Finally, define the linearized truncated matrix

$$\begin{aligned} \check{\mathcal {Y}}_{n} :=n^{-1/2}\left[ \begin{array}{ccccc} 0 &{}\quad \check{X}_{n,1} &{}\quad 0 &{}\quad \cdots &{}\quad 0\\ 0 &{}\quad 0 &{}\quad \check{X}_{n,2} &{}\quad \dots &{}\quad 0\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad \check{X}_{n,m-1}\\ \check{X}_{n,m} &{}\quad 0 &{}\quad 0 &{}\quad \dots &{}\quad 0 \end{array}\right] . \end{aligned}$$

(111)

Lemma B.1

Fix $\varepsilon >0$. For a fixed integer $m>0$, let $\xi _{1},\xi _{2},\dots \xi _{m}$ be real-valued random variables each mean zero, variance one, and finite $4+\tau$ moment for some $\tau >0$. Let $\hat{X}_{n,1},\hat{X}_{n,2},\dots ,\hat{X}_{n,m}$ be independent iid random matrices with atom variables as defined in (22), and define $\mathcal {Y}_{n}$ as in (31). For every $\delta >0$, there exists a constant $c>0$ depending only on $\delta$ such that

$$\begin{aligned} \inf _{|z|> 1+\delta /2}s_{mn}\left( \mathcal {Y}_{n}-zI\right) \ge c \end{aligned}$$

with overwhelming probability.

Proof

Fix $\delta >0$ and define $\check{\mathcal {Y}}_{n}$ as in (111). By [23, Lemma 8.1], which is based on techniques in [48, 49], we know that there exists a constant $c'>0$ which depends only on $\delta$ such that $\inf _{|z|> 1+\delta /2}s_{mn}\left( \check{\mathcal {Y}}_{n}-zI\right) \ge c'$ with overwhelming probability. Note that by Weyl’s inequality (13),

$$\begin{aligned} \sup _{z\in \mathcal {C}}\left| s_{mn}\left( \check{\mathcal {Y}}_{n}-zI\right) -s_{mn}\left( \mathcal {Y}_{n}-zI\right) \right| \le \left\| \check{\mathcal {Y}}_{n}-\mathcal {Y}_{n}\right\| \le \max _{1 \le k \le m}\frac{1}{\sqrt{n}}\left\| \check{X}_{n,k}-\hat{X}_{n,k}\right\| . \end{aligned}$$

(112)

Focusing on an arbitrary value of k, we have

$$\begin{aligned} \frac{1}{\sqrt{n}}\left\| \check{X}_{n,k}-\hat{X}_{n,k}\right\| \le \frac{1}{\sqrt{n}}\left\| \frac{\mathring{X}_{n,k}}{\sqrt{\text {Var}((\mathring{X}_{n,k})_{(i,j)})}}-\frac{\tilde{X}_{n,k}}{\sqrt{\text {Var}((\tilde{X}_{n,k})_{(i,j)})}}\right\| \end{aligned}$$

for any $1\le i,j\le n$. Observe that

$$\begin{aligned} \frac{1}{\sqrt{n}}\left\| \frac{\mathring{X}_{n,k}}{\sqrt{\text {Var}((\mathring{X}_{n,k})_{(i,j)})}}- \mathring{X}_{n,k}\right\| =\frac{1}{\sqrt{n}}\left\| \frac{\mathring{X}_{n,k}\left( 1-\sqrt{\text {Var}((\mathring{X}_{n,k})_{(i,j)})}\right) }{\sqrt{\text {Var}((\mathring{X}_{n,k})_{(i,j)})}}\right\| . \end{aligned}$$

By [23, Lemma 7.1], $\left( \text {Var}((\mathring{X}_{n,k})_{(i,j)})\right) ^{-1/2}\le 2$ for L sufficiently large. Additionally, an argument similar to that of [23, Lemma 7.1] shows that $\left| 1-\sqrt{\text {Var}((\mathring{X}_{n,k})_{(i,j)})}\right| \le \frac{C}{L^{2}}$ for any $1\le i,j\le n$ and some constant $C>0$. Therefore, by [62, Theorem 1.4], for L sufficiently large,

$$\begin{aligned} \frac{1}{\sqrt{n}}\left\| \frac{\mathring{X}_{n,k}}{\sqrt{\text {Var}((\mathring{X}_{n,k})_{(i,j)})}}- \mathring{X}_{n,k}\right\| \le \frac{C}{L^{2}\sqrt{n}}\left\| \frac{\mathring{X}_{n,k}}{\sqrt{\text {Var}((\mathring{X}_{n,k})_{(i,j)})}}\right\| \le \frac{c'}{16} \end{aligned}$$

with overwhelming probability. Similarly,

$$\begin{aligned} \frac{1}{\sqrt{n}}\left\| \frac{\tilde{X}_{n,k}}{\sqrt{\text {Var}((\tilde{X}_{n,k})_{(i,j)})}}- \tilde{X}_{n,k}\right\| =\frac{1}{\sqrt{n}}\left\| \frac{\tilde{X}_{n,k}\left( 1-\sqrt{\text {Var}((\tilde{X}_{n,k})_{(i,j)})}\right) }{\sqrt{\text {Var}((\tilde{X}_{n,k})_{(i,j)})}}\right\| . \end{aligned}$$

By the arguments to prove part (ii) of Lemma 4.3, $\left( \text {Var}((\tilde{X}_{n,k})_{(i,j)})\right) ^{-1/2}\le 2$ for n sufficiently large. Also, by part (i) of Lemma 4.3, we can show that $\left| 1-\sqrt{\text {Var}((\tilde{X}_{n,k})_{(i,j)})}\right| =o(n^{-1+2\varepsilon })$. Therefore, by [13, Theorem 5.9],

$$\begin{aligned} \frac{1}{\sqrt{n}}\left\| \frac{\tilde{X}_{n,k}}{\sqrt{\text {Var}((\tilde{X}_{n,k})_{(i,j)})}}- \tilde{X}_{n,k}\right\| =o(n^{-1-2\varepsilon }) \frac{1}{\sqrt{n}}\left\| \tilde{X}_{n,k}\right\| \le \frac{c'}{16} \end{aligned}$$

with overwhelming probability. Ergo, by the triangle inequality, for L sufficiently large,

$$\begin{aligned} \frac{1}{\sqrt{n}}\left\| \check{X}_{n,k}-\hat{X}_{n,k}\right\|&\le \frac{1}{\sqrt{n}}\left\| \frac{\mathring{X}_{n,k}}{\sqrt{\text {Var}((\mathring{X}_{n,k})_{(i,j)})}}-\frac{\tilde{X}_{n,k}}{\sqrt{\text {Var}((\tilde{X}_{n,k})_{(i,j)})}}\right\| \nonumber \\&\le \frac{c'}{8}+ \frac{1}{\sqrt{n}}\left\| \mathring{X}_{n,k}-\tilde{X}_{n,k}\right\| \end{aligned}$$

(113)

with overwhelming probability.

Now, recall that the entries of $\mathring{X}_{n,k}$ are truncated at level L for a fixed $L>0$ so for sufficiently large n, $L\le n^{1/2-\varepsilon }$. Note that if all entries are less than L in absolute value, then the entries in $\mathring{X}_{n,k}$ and $\tilde{X}_{n}$ agree. Similarly, if all entries are greater than $n^{1/2-\varepsilon }$, then the entries in $\mathring{X}_{n,k}$ and $\tilde{X}_{n}$ agree. Ergo, we need only consider the case when there exists some entries $1\le i,j\le n$ such that $L\le |(\tilde{X}_{n,k})_{i,j}|\le n^{1/2-\varepsilon }$. For each $1\le k\le m$, define the random variables

$$\begin{aligned} \dot{\xi }_{k} := \xi _{k}{\mathbf {1}_{\{L\le |\xi _{k}|\le n^{1/2-\varepsilon }\}}}-\mathbb {E}\left[ \xi _{k}{\mathbf {1}_{\{L\le |\xi _{k}|\le n^{1/2-\varepsilon }\}}}\right] \end{aligned}$$

and define $\dot{X}_{n,k}$ to be the matrix with entries

$$\begin{aligned} (\dot{X}_{n,k})_{(i,j)} := (X_{n,k})_{(i,j)}{\mathbf {1}_{\{L\le |(X_{n,k})_{(i,j)}|\le n^{1/2-\varepsilon }\}}}-\mathbb {E}\left[ (X_{n,k})_{(i,j)}{\mathbf {1}_{\{L\le |(X_{n,k})_{(i,j)}|\le n^{1/2-\varepsilon }\}}}\right] , \end{aligned}$$

for $1\le i,j \le n$. Note that the definitions of $\dot{\xi }$ and $\dot{X}_{n,k}$ differ from the definitions in Sect. 4. We will use the definition given in this appendix for the remainder of this proof. We can write

$$\begin{aligned} \frac{1}{\sqrt{n}}\left\| \mathring{X}_{n,k}-\tilde{X}_{n,k}\right\| = \frac{1}{\sqrt{n}}\left\| \dot{X}_{n,k}\right\| . \end{aligned}$$

By [13, Lemma 5.9], for L sufficiently large

$$\begin{aligned} \frac{1}{\sqrt{n}}\left\| \dot{X}_{n,k}\right\| \le \frac{c'}{8} \end{aligned}$$

(114)

with overwhelming probability. Thus, by choosing L large enough to satisfy both conditions, by (113) and (114),

$$\begin{aligned} \max _{1 \le k \le m}\frac{1}{\sqrt{n}}\left\| \check{X}_{n,k}-\hat{X}_{n,k}\right\| <\frac{c'}{4} \end{aligned}$$

with overwhelming probability. By recalling (112), this implies that, for L sufficiently large,

$$\begin{aligned} \inf _{|z|>1+\delta /2}s_{mn}\left( \mathcal {Y}_{n}-zI\right) \ge c \end{aligned}$$

with overwhelming probability where $c=\frac{c'}{2}$. $\square$

Lemma B.2

Fix $\varepsilon >0$. For a fixed integer $m>0$, let $\xi _{1},\xi _{2},\dots \xi _{m}$ be real-valued random variables each mean zero, variance one, and finite $4+\tau$ moment for some $\tau >0$. Let ${X}_{n,1},{X}_{n,2},\dots ,{X}_{n,m}$ be independent iid random matrices with atom variables $\xi _{1},\xi _{2},\dots ,\xi _{m}$, respectively. Define $\hat{X}_{n,1},\hat{X}_{n,2},\dots \hat{X}_{n,m}$ as in (22), and define $\hat{P}_{n}$ as in (24). For any $\delta >0$, there exists a constant $c>0$ depending only on $\delta$ such that

$$\begin{aligned} \inf _{|z|> 1+\delta /2}s_{mn}\left( \hat{P}_{n}-zI\right) \ge c \end{aligned}$$

with overwhelming probability.

Proof

Fix $\delta >0$. By Lemma B.1, we know that there exists some $c'>0$ such that $\inf _{|z|>1+\delta /2}s_{mn}\left( \mathcal {Y}_{n}-zI\right) \ge c'$ with overwhelming probability as well. Recall that $s_{mn}\left( \mathcal {Y}_{n}-zI\right) =s_{1}\left( \left( \mathcal {Y}_{n}-zI\right) ^{-1}\right)$ provided z is not an eigenvalue of $\mathcal {Y}_{n}$. A block inverse matrix calculation reveals that

$$\begin{aligned} \left( \left( \mathcal {Y}_{n}-zI\right) ^{-1}\right) ^{[1,1]}=z^{m-1}\left( \hat{P}_{n}-z^{m}I\right) ^{-1}, \end{aligned}$$

where the notation $A^{[1,1]}$ denotes the upper left $n\times n$ block of A. Therefore,

$$\begin{aligned} \frac{1}{c'} \ge \sup _{|z|> 1+\delta /2}s_{1}\left( \left( \mathcal {Y}_{n}-zI\right) ^{-1}\right) \ge \sup _{|z|> 1+\delta /2}|z|^{m-1}\left\| \left( \hat{P}_{n}-z^{m}I\right) ^{-1}\right\| . \end{aligned}$$

This implies that there exists a constant $c>0$ such that

$$\begin{aligned} \frac{1}{c}\ge \sup _{|z|> 1+\delta /2}s_{1}\left( \left( \hat{P}_{n}-zI\right) ^{-1}\right) \end{aligned}$$

with overwhelming probability. This gives $\inf _{|z| > 1+\delta /2}s_{n}\left( \hat{P}_{n}-zI\right) \ge c$ with overwhelming probability. $\square$

Lemma B.3

For a fixed integer $m>0$, let $\xi _{1},\xi _{2},\dots \xi _{m}$ be real-valued random variables each satisfying Assumption 2.1. Fix $\delta >0$ and let $X_{n,1},X_{n,2},\dots X_{n,m}$ be independent iid random matrices with atom variables $\xi _{1},\xi _{2},\dots \xi _{m}$, respectively. Then there exists a constant $c>0$ depending only on $\delta$ such that

$$\begin{aligned} \inf _{|z|> 1+\delta /2}s_{n}\left( P_{n}/\sigma -zI\right) \ge c \end{aligned}$$

with probability $1-o(1)$ where $\sigma = \sigma _{1}\cdots \sigma _{m}$.

Proof

By a simple rescaling, it is sufficient to assume that the variance of each random variable is 1 so that $\sigma =1$. Let $\delta >0$ and recall by Lemma B.2 there exists a $c'>0$ depending only on $\delta$ such that $\inf _{|z|> 1+\delta /2}s_{n}\left( \hat{P}_{n}-zI\right) \ge c'$ with overwhelming probability. Then by Lemma 4.10,

$$\begin{aligned}&\mathbb {P}\left( \inf _{|z|> 1+\delta /2}s_{n}\left( P_{n}-zI\right)<\frac{c'}{2}\right) \\&\quad = \mathbb {P}\left( \inf _{|z|> 1+\delta /2}s_{n}\left( P_{n}-zI\right)<\frac{c'}{2}\;\;\text { and }\;\;\left\| P_{n}-\hat{P}_{n}\right\| \le n^{-\varepsilon }\right) \\&\qquad + \mathbb {P}\left( \inf _{|z|> 1+\delta /2}s_{n}\left( P_{n}-zI\right)<\frac{c'}{2}\;\;\text { and }\;\;\left\| P_{n}-\hat{P}_{n}\right\|> n^{-\varepsilon }\right) \\&\quad \le \mathbb {P}\left( \inf _{|z|> 1+\delta /2}s_{n}\left( P_{n}-zI\right)<\frac{c'}{2}\;\;\text { and }\;\;\left\| P_{n}-\hat{P}_{n}\right\| \le n^{-\varepsilon }\right) \\&\qquad + \mathbb {P}\left( \left\| P_{n}-\hat{P}_{n}\right\|> n^{-\varepsilon }\right) \\&\quad \le \mathbb {P}\left( \inf _{|z|> 1+\delta /2}s_{n}\left( P_{n}-zI\right) <\frac{c'}{2}\;\;\text { and }\;\;\left\| P_{n}-\hat{P}_{n}\right\| \le n^{-\varepsilon }\right) +o(1). \end{aligned}$$

Suppose that there exists a $z_{0}\in \mathbb {C}$ with $|z_{0}|\ge 1+\delta /2$ such that $s_{n}\left( P_{n}-z_{0}I\right) <\frac{c'}{2}$ and $\left\| P_{n}-\hat{P}_{n}\right\|<n^{-\varepsilon }<\frac{c'}{2}$. Then, by Weyl’s inequality (13), $\Big |s_{n}(P_{n}-z_{0}I)-s_{n}(\hat{P}_{n}-z_{0}I)\Big |<\frac{c'}{2}$ which implies $s_{n}(\hat{P}_{n}-z_{0}I)<c'$. Thus, for n sufficiently large to ensure that $n^{-\varepsilon }<\frac{c'}{2}$, by Lemma 4.10

$$\begin{aligned} \mathbb {P}\left( \inf _{|z|> 1+\delta /2}s_{n}\left( P_{n}-zI\right)<\frac{c'}{2}\right) \le \mathbb {P}\left( \inf _{|z|> 1+\delta /2}s_{n}\left( \hat{P}_{n}-zI\right) <c'\right) +o(1). \end{aligned}$$

Thus, selecting $c=\frac{c'}{2}$, we have $\inf _{|z|> 1+\delta /2}s_{n}\left( P_{n}-zI\right) \ge c$ with probability $1-o(1)$. $\square$

Lemma B.4

Let A be an $n\times n$ matrix. Let R be a subset of the integer set $\{1,2,\dots n\}$. Let $A^{(R)}$ denote the matrix A, but with the rth column replaced with zero for each $r\in R$. Then

$$\begin{aligned} s_{n}\left( A^{(R)}-zI\right) \ge \min \{s_{n}(A-zI),|z|\}. \end{aligned}$$

Proof

Let $A^{((R))}$ denote the matrix A with column r removed for all $r\in R$. Note that $A^{((R))}$ is an $n\times (n-|R|)$ matrix, which is distinct from the $n\times n$ matrix $A^{(R)}$. Also, let $I^{((R))}$ denote the $n\times n$ identity matrix with column r removed for all $r\in R$. In order to bound the least singular value of $(A^{(R)}-zI)$, we will consider the eigenvalues of $\left( A-zI\right) ^{*}\left( A-zI\right) ,$$\left( A^{(R)}-zI\right) ^{*}\left( A^{(R)}-zI\right) ,$ and $\left( A^{((R))}-zI^{((R))}\right) ^{*}\left( A^{((R))}-zI^{((R))}\right) .$

Now, observe that $\left( A^{((R))}-zI^{((R))}\right) ^{*}\left( A^{((R))}-zI^{((R))}\right)$ is an $(n-|R|)\times (n-|R|)$ matrix, and is a principle sub-matrix of the Hermitian matrix $(A-zI)^{*}(A-zI)$. Therefore, the eigenvalues of $\left( A^{((R))}-zI^{((R))}\right) ^{*}\left( A^{((R))}-zI^{((R))}\right)$ must interlace with the eigenvalues of $\left( A-zI\right) ^{*}\left( A-zI\right)$ by Cauchy’s interlacing theorem [38, Theorem 1]. This implies

$$\begin{aligned} s_{n}\left( A^{((R))}-zI^{((R))}\right) ^{2}\ge s_{n}\left( A-zI\right) ^{2}. \end{aligned}$$

Next, we compare the eigenvalues of $\left( A^{(R)}-zI\right) ^{*}\left( A^{(R)}-zI\right)$ to the eigenvalues of $\left( A^{((R))}-zI^{((R))}\right) ^{*}\left( A^{((R))}-zI^{((R))}\right)$. Note that, after a possible permutation of columns to move all zero columns of $A^{(R)}$ to be in the last |R| columns, the product $\left( A^{(R)}-zI\right) ^{*}\left( A^{(R)}-zI\right)$ becomes

$$\begin{aligned} \left[ \begin{array}{cc} \left( A^{((R))}-zI^{((R))}\right) ^{*}\left( A^{((R))}-zI^{((R))}\right) &{} 0\cdot I_{|R|\times (n-|R|)}\\ &{} \\ 0\cdot I_{(n-|R|)\times |R|} &{} |z|^{2}\cdot I_{|R|\times |R|} \end{array}\right] . \end{aligned}$$

Due to the block structure of the matrix above, if w is an eigenvalue of $\left( A^{(R)}-zI\right) ^{*}\left( A^{(R)}-zI\right)$, then either w is an eigenvalue of $\left( A^{((R))}-zI^{((R))}\right) ^{*}\left( A^{((R))}-zI^{((R))}\right)$ or w is $|z|^{2}$. Ergo,

$$\begin{aligned} s_{n}\left( A^{(R)}-zI\right) ^{2}&= \min \left\{ s_{n}\left( A^{((R))}-zI^{((R))}\right) ^{2},\;|z|^{2}\right\} \\&\ge \min \left\{ s_{n}\left( A-zI\right) ^{2},\;|z|^{2}\right\} \end{aligned}$$

which implies $s_{n}\left( A^{(R)}-zI\right) \ge \min \left\{ s_{n}\left( A-zI\right) ,\;|z|\right\}$ concluding the proof. $\square$

This lemma gives way to the following two corollaries.

Corollary B.5

Fix $\varepsilon >0$. For a fixed integer $m>0$, let $\xi _{1},\xi _{2},\dots \xi _{m}$ be real-valued random variables each mean zero, variance one, and finite $4+\tau$ moment for some $\tau >0$. Let ${X}_{n,1},{X}_{n,2},\dots ,{X}_{n,m}$ be independent iid random matrices with atom variables $\xi _{1},\xi _{2},\dots ,\xi _{m}$, respectively, and define $\hat{X}_{n,1},\hat{X}_{n,2},\dots \hat{X}_{n,m}$ as in (22). Define $\mathcal {Y}_{n}$ as in (31) and $\mathcal {Y}_{n}^{(k)}$ as $\mathcal {Y}_{n}$ with the columns $c_{k},c_{n+k},c_{2n+k},\dots ,c_{(m-1)n+k}$ replaced with zeros. For any $\delta >0$, there exists a constant $c>0$ depending only on $\delta$ such that

$$\begin{aligned} \inf _{|z|> 1+\delta /2}s_{mn}\left( \mathcal {Y}_{n}^{(k)}-zI\right) \ge c \end{aligned}$$

with overwhelming probability.

Proof

Note that by Lemmas B.1 and B.4,

$$\begin{aligned} \inf _{|z|> 1+\delta /2}s_{mn}\left( \mathcal {Y}_{n}^{(k)}-zI\right)&\ge \inf _{|z|> 1+\delta /2} \min \left\{ s_{mn}\left( \mathcal {Y}_{n}-zI\right) ,\;|z|\right\} \\&\ge \inf _{|z|> 1+\delta /2} \min \left\{ s_{mn}\left( \mathcal {Y}_{n}-zI\right) ,\;1\right\} \\&\ge \min \left\{ c',\;1\right\} \end{aligned}$$

with overwhelming probability for some constant $c'>0$ depending only on $\delta$. The result follows by setting $c=\min \left\{ c',\;1\right\}$. $\square$

Corollary B.6

Fix $\varepsilon >0$. For a fixed integer $m>0$, let $\xi _{1},\xi _{2},\dots \xi _{m}$ be real-valued random variables each mean zero, variance one, and finite $4+\tau$ moment for some $\tau >0$. Let $\hat{X}_{n,1},\hat{X}_{n,2},\dots ,\hat{X}_{n,m}$ be independent iid random matrices with atom variables as defined in (22). Define $\mathcal {Y}_{n}$ as in (31) and $\mathcal {Y}_{n}^{(k,s)}$ as $\mathcal {Y}_{n}$ with the columns $c_{k},c_{n+k},c_{2n+k},\dots ,c_{(m-1)n+k}$ and $c_{s}$ replaced with zeros. For any $\delta >0$, there exists a constant $c>0$ depending only on $\delta$ such that

$$\begin{aligned} \inf _{|z|> 1+\delta /2}s_{mn}\left( \mathcal {Y}_{n}^{(k,s)}-zI\right) \ge c \end{aligned}$$

with overwhelming probability.

The proof of Corollary B.6 follows in exactly the same way as the proof of Corollary B.5.

Appendix C. Useful Lemmas

Lemma C.1

(Lemma 2.7 from [12]). For $X = (x_{1},x_{2},\ldots ,x_{N})^{T}$ iid standardized complex entries, B an $N\times N$ complex matrix, we have, for any $p\ge 2$,

$$\begin{aligned} \mathbb {E}\left| X^{*}BX-{{\,\mathrm{tr}\,}}(B)\right| ^{p}\le K_{p}\left( \left( \mathbb {E}\left| x_{1}\right| ^{4}{} tr B^{*}B\right) ^{p/2}+\mathbb {E}|x_{1}|^{2p}{} tr (B^{*}B)^{p/2}\right) , \end{aligned}$$

where the constant $K_{p}>0$ depends only on p.

Lemma C.2

Let A be an $N\times N$ complex-valued matrix. Suppose that $\xi$ is a complex-valued random variable with mean zero and unit variance. Let $S\subseteq [N]$, and let $w=(w_{i})_{i=1}^{N}$ be a vector with the following properties:

• (i) $\{w_i : i \in S \}$ is a collection of iid copies of $\xi$,

• (ii) $w_{i}=0$ for $i\not \in S$.

Additionally, $A_{S\times S}$ denote the $|S|\times |S|$ matrix which has entries $A_{(i,j)}$ for $i,j\in S$. Then for any even $p\ge 2$,

$$\begin{aligned} \mathbb {E}\left| w^{*}Aw-{{\,\mathrm{tr}\,}}(A_{S\times S})\right| ^{p}\ll _{p}\mathbb {E}\left| \xi \right| ^{2p}\left( {{\,\mathrm{tr}\,}}(A^{*}A)\right) ^{p/2}. \end{aligned}$$

Proof

Let $w_{S}$ denote the |S|-vector which contains entries $w_{i}$ for $i\in S$ and observe

$$\begin{aligned} w^{*}Aw = \sum _{i,j}\bar{w}_{i}A_{(i,j)}w_{j}= w_{S}^{*}A_{S\times S}w_{S}. \end{aligned}$$

Therefore, by Lemma C.1, for any even $p\ge 2$,

$$\begin{aligned} \mathbb {E}\left| w^{*}Aw-{{\,\mathrm{tr}\,}}(A_{S\times S})\right| ^{p}&= \mathbb {E}\left| w_{S}^{*}A_{S\times S}w_{S}-{{\,\mathrm{tr}\,}}(A_{S\times S})\right| ^{p}\\&\ll _{p}\left( \mathbb {E}\left| \xi \right| ^{4}{{\,\mathrm{tr}\,}}(A_{S\times S}^{*}A_{S\times S})\right) ^{p/2}+\mathbb {E}\left| \xi \right| ^{2p}{{\,\mathrm{tr}\,}}(A_{S\times S}^{*}A_{S\times S})^{p/2}\\&\ll _{p}\mathbb {E}\left| \xi \right| ^{2p}\left( {{\,\mathrm{tr}\,}}(A_{S\times S}^{*}A_{S\times S})\right) ^{p/2}. \end{aligned}$$

Now observe that

$$\begin{aligned} {{\,\mathrm{tr}\,}}(A_{S\times S}^{*}A_{S\times S})= \sum _{i,j\in S} A_{i,j}^{*}A_{j,i} \le \sum _{i,j=1}^{N} A_{i,j}^{*}A_{j,i}= {{\,\mathrm{tr}\,}}(A^{*}A). \end{aligned}$$

Therefore,

$$\begin{aligned} \mathbb {E}\left| w^{*}Aw-{{\,\mathrm{tr}\,}}(A_{S\times S})\right| ^{p}\ll _{p}\mathbb {E}\left| \xi \right| ^{2p}\left( {{\,\mathrm{tr}\,}}(A_{S\times S}^{*}A_{S\times S})\right) ^{p/2}\le \mathbb {E}\left| \xi \right| ^{2p}\left( {{\,\mathrm{tr}\,}}(A^{*}A)\right) ^{p/2}. \end{aligned}$$

$\square$

Lemma C.3

(Lemma A.1 from [12]). For $X = (x_{1},x_{2},\ldots ,x_{N})^{T}$ iid standardized complex entries, B an $N\times N$ complex-valued Hermitian nonnegative definite matrix, we have, for any $p\ge 1$,

$$\begin{aligned} \mathbb {E}\left| X^{*}BX\right| ^{p}\le K_{p}\left( \left( tr B\right) ^{p}+\mathbb {E}|x_{1}|^{2p}{} tr B^{p}\right) , \end{aligned}$$

where $K_{p}>0$ depends only on p.

Lemma C.4

Let A be an $N\times N$ Hermitian positive semidefinite matrix. Suppose that $\xi$ is a complex-valued random variable with mean zero and unit variance. Let $S\subseteq [N]$, and let $w = (w_i)_{i=1}^N$ be a vector with the following properties:

• (i) $\{w_i : i \in S \}$ is a collection of iid copies of $\xi$,

• (ii) $w_{i}=0$ for $i\not \in S$.

Then for any $p\ge 2$,

$$\begin{aligned} \mathbb {E}\left| w^{*}Aw\right| ^{p} \ll _{p}\mathbb {E}|\xi |^{2p}\left( {{\,\mathrm{tr}\,}}A\right) ^{p}. \end{aligned}$$

(115)

Proof

Let $w_{S}$ denote the |S|-vector which contains entries $w_{i}$ for $i\in S$, and let $A_{S\times S}$ denote the $|S|\times |S|$ matrix which has entries $A_{(i,j)}$ for $i,j\in S$. Then we have

$$\begin{aligned} w^{*}Aw = \sum _{i,j}\bar{w}_{i}A_{(i,j)}w_{j}= w_{S}^{*}A_{S\times S}w_{S}. \end{aligned}$$

By Lemma C.3, we get

$$\begin{aligned} \mathbb {E}\left| w^{*}Aw\right| ^{p} \ll _{p}\left( {{\,\mathrm{tr}\,}}A_{S\times S}\right) ^{p}+\mathbb {E}|\xi |^{2p}{{\,\mathrm{tr}\,}}A_{S\times S}^{p}. \end{aligned}$$

Since A is nonnegative definite, the diagonal elements are nonnegative so that ${{\,\mathrm{tr}\,}}(A_{S\times S}^{p})\le ({{\,\mathrm{tr}\,}}(A_{A\times A}))^{p}$. By this and the fact that for a Hermitian positive semidefinite matrix, the partial trace is less than or equal to the full trace, we observe that

$$\begin{aligned} \left( {{\,\mathrm{tr}\,}}A_{S\times S}\right) ^{p}+\mathbb {E}|\xi |^{2p}{{\,\mathrm{tr}\,}}A_{S\times S}^{p} \ll _{p} \mathbb {E}|\xi |^{2p}\left( {{\,\mathrm{tr}\,}}A_{S\times S}\right) ^{p} \ll _{p} \mathbb {E}|\xi |^{2p}\left( {{\,\mathrm{tr}\,}}A\right) ^{p}. \end{aligned}$$

$\square$

Lemma C.5

Let A and B be $n\times n$ matrices. Then

$$\begin{aligned} \left| {{\,\mathrm{tr}\,}}(AB)\right| \le \sqrt{n}\left\| AB\right\| _{2}\le \sqrt{n}\left\| A\right\| \cdot \left\| B\right\| _{2}. \end{aligned}$$

Proof

This follows by an application of the Cauchy–Schwarz inequality and an application of [13, Theorem A.10]. $\square$