# Spectrum of random centrosymmetric matrices; CLT and Circular law

Indrajit Jana  Indian Institute of Technology, Bhubaneswar
ijana[at]iitbbs.ac.in and  Sunita Rani  Indian Institute of Technology,
Bhubaneswar  s21ma09007[at]iitbbs.ac.in

###### Abstract.

We analyze the asymptotic fluctuations of linear eigenvalue statistics of
random centrosymmetric matrices with i.i.d. entries. We prove that for a
complex analytic test function, the centered and normalized linear eigenvalue
statistics of random centrosymmetric matrices converge to a normal
distribution. We find the exact expression of the variance of the limiting
normal distribution via combinatorial arguments. Moreover, we also argue that
the limiting spectral distribution of properly scaled centrosymmetric matrices
follows the circular law.

keywords: Centrosymmetric matrix, Linear eigenvalue statistics, Central limit
theorem, Circular law.

Indrajit Jana’s research is partially supported by INSPIRE Fellowship
DST/INSPIRE/04/2019/000015, Dept. of Science and Technology, Govt. of India.

Sunita Rani’s research is fully supported by the University Grant Commission
(UGC), New Delhi.

##  1\. Introduction

Let $M$ be an $n\times n$ matrix with real or complex entries. The empirical
spectral distribution (ESD) is defined by the measure

$\displaystyle\mu_{n}(\cdot)=\frac{1}{n}\sum_{i=1}^{n}\delta_{\lambda_{i}}(\cdot),$

where $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ are the eigenvalues of the
matrix $M$, and $\delta_{x}(\cdot)$ is the point measure at $x\in\mathbb{C}$.
The linear eigenvalue statistics (LES) of $M$ corresponding to a test function
$f$ is defined by

$\operatorname{L}(f)=\frac{1}{n}\sum_{k=1}^{n}f\left(\lambda_{k}\right).$

The function $f$ is referred to as the test function. The asymptotic behaviour
of $\mu_{n}(\cdot)$ and $a_{n}(\operatorname{L}(f)-b_{n})$ for some sequences
$\langle a_{n}\rangle$ and $\langle b_{n}\rangle$ have been studied in depth
for different types of random matrices. We give a brief overview below.

Jonsson [20] established the central limit theorem (CLT) for LES in Wishart
matrices. Subsequently, researchers extensively studied the fluctuation of
eigenvalues of various random matrices. Some of the key contributions include
Johansson [19], Sinai and Soshnikov [32], Bai and Silverstein [5], Lytova and
Pastur [24], Shcherbina [29]. Furthermore, the same has been studied for
several structured matrices, such as band and sparse matrices [4, 22, 30, 18],
Toepliz and band Toeplitz matrices [10, 23], circulant matrices [1, 2, 25].

The CLT of LES for non-Hermitian matrices have been studied in different set
ups as well; such as when the test function is analytic [27, 26, 17], the test
function is non-analytic but the moments of the matrix entries match with that
of a Ginibre ensemble up to fourth moment [21]. Later, it was generalized for
continuous bounded test functions [13, 11, 12]. A more comprehensive list of
recent results can be found on the review article by Forrester [16].

In this article, we study the spectral properties of random centrosymmetric
matrices. A centrosymmetric matrix is symmetric around its center which is
formally defined in the Definition 3.1. Centrosymmetric matrices appeared in
many different contexts in mathematics, such as solutions to chessboard
separation problems [9], stochastic matrices [8], and in various other
contexts [15, 33, 28, 6]. However, to the best of our knowledge the spectral
properties of random centrosymmetric matrices have not been studied yet. In
this paper, we investigate the limiting ESD of a random centrosymmetric matrix
and the fluctuations of the LES.

##  2\. Notations and layout

This article is organized as follows. We state the main theorem in Section 3.
The proof of the CLT is shown in Section 6, and the variance calculation of
the limiting normal distribution is delegated to Section 7. In addition, it
was also argued that the limiting spectral distribution follows the circular
law, which was demonstrated in Sectoin 5.

Throughout this article, the identity matrix is denoted by $I$, and the
counter-identity matrix is denoted by $J$, as defined in Definition 4.1. The
dimensions of such matrices should be understood from the context. The
eigenvalues of an $n\times n$ matrix $M$ are denoted by $\lambda_{1}(M),$
$\lambda_{2}(M),$ $\ldots,$ $\lambda_{n}(M).$ The complex conjugate transpose
of a matrix $Y$ is denoted by $Y{\operatorname{{}^{*}}}$, whereas $\bar{Y}$
stands for only the complex conjugate of $Y$ without taking transpose. The
notation $\mathcal{R}_{z}(M)=(zI-M)^{-1}$ represents the resolvent of the
square matrix $M$. The notations $\zeta_{n}\stackrel{{\scriptstyle
d}}{{\to}}\zeta$, $\zeta_{n}\stackrel{{\scriptstyle p}}{{\to}}\zeta$, and
$\zeta_{n}\stackrel{{\scriptstyle a.s.}}{{\to}}\zeta$ are used to indicate
that the sequence of random variables $\\{\zeta_{n}\\}$ converges to $\zeta$
in distribution, in probability, and almost surely, respectively.

For any random variable $\xi$, $\xi^{\circ}$ denotes the centered random
variable $\xi-\mathbb{E}[\xi].$ Moreover, if $\xi$ depends on a random matrix,
then $\xi_{k}^{\circ}:=\xi-\Xi_{k}[\xi]$, where $\Xi_{k}[\cdot]$ signifies
averaging with respect to the $k$-th column of the underlying random matrix.

The vectors $e_{1},e_{2},\ldots,e_{n}$ denote the canonical basis vectors of
$\mathbb{R}^{n}.$ The disc of radius $r$, denoted by $\mathbb{D}_{r}$, is the
set $\\{z\in\mathbb{C}:|z|\leq r\\}$, and its boundary is denoted by
$\partial\mathbb{D}_{r}$. The notations $K$ or $C$ are used to denote a
universal constant whose exact value may change from line to line.

##  3\. Main result

Let us first define the centrosymmetric matrix.

######  Definition 3.1 (Centrosymmetric Matrix).

A random Centrosymmetric matrix is a random matrix which is symmetric about
its center. Formally, let $M=[m_{i,j}]_{n\times n}$ be a random matrix, where
$m_{ij}$ are i.i.d. random variables subject to the following constraint

$m_{i,j}=m_{n-i+1,n-j+1}\text{ for }i,j\in\\{1,\ldots,n\\}.$

Then $M$ is called a random centrosymmetric matrix.

Figure 1. Symmetry pattern of a centrosymmetric 5$\times$5 matrix.

######  Definition 3.2 (Poincaré inequality).

A complex random variable $\zeta$ is said to satisfy the Poincaré inequality
with a positive constant $\alpha$ if for any differentiable function
$g:\mathbb{C}\to\mathbb{C}$, we have
$\operatorname{Var}(g(\zeta))\leq\frac{1}{\alpha}\mathbb{E}[|g^{\prime}(\zeta)|^{2}].$
Here, $\mathbb{C}$ is identified with $\mathbb{R}^{2}$ and
$|g^{\prime}(\zeta)|$ represents the $\ell^{2}$ norm of the two dimensional
vector $g^{\prime}$. Here are some characteristics of the Poincaré inequality.

  1. (1)

If $\zeta$ satisfies the Poincaré inequality with constant $\alpha$, then
$\kappa\zeta$ also satisfies the Poincaré inequality with constant
$\alpha/\kappa^{2}$ for any nonzero real constant $\kappa.$

  2. (2)

If two independent random variables $\zeta_{1},\zeta_{2}$ satisfy the Poincaré
inequality with the same constant $\alpha$, then $\zeta=(\zeta_{1},\zeta_{2})$
also satisfies the Poincaré inequality with the same constant $\alpha$. Here
$\zeta$ is treated as a random vector and, accordingly, $g^{\prime}$ in the
definition of Poincaré inequality will be replaced by $\nabla g$.

  3. (3)

[3, Lemma 4.4.3] If $\zeta\in\mathbb{C}^{n}$ satisfies the Poincaré inequality
with a constant $\alpha$, then for any differentiable function
$g:\mathbb{C}^{n}\to\mathbb{C}$,

$\displaystyle\mathbb{P}(|g(\zeta)-\mathbb{E}[g(\zeta)]|>t)\leq
C\exp\left\\{-\frac{\sqrt{\alpha}t}{\sqrt{2}\|\|\nabla
g\|_{2}\|_{\infty}}\right\\},$

where $C=-2\sum_{k=1}^{\infty}2^{k}\log(1-2^{-2k-1}).$ Here $\mathbb{C}^{n}$
is identified with $\mathbb{R}^{2n}$. Moreover, $\|\nabla g(x)\|_{2}$ is the
$\ell^{2}$ norm of the $2n$ dimensional vector $\nabla g(x)$ at
$x\in\mathbb{C}^{n}$, and $\|\|\nabla
g\|_{2}\|_{\infty}=\sup_{x\in\mathbb{C}^{n}}\|\nabla g(x)\|_{2}.$

For instance, a Gaussian random variable satisfies the Poincaré inequality.

######  Condition 3.3.

Let $M=[m_{ij}]_{n\times n}$ be a centrosymmetric matrix, where

$\displaystyle m_{ij}=\frac{1}{\sqrt{n}}x_{ij}.$

Assume that $\\{x_{ij}:1\leq i,j\leq n\\}$ are i.i.d. complex random variables
subject to the constraint $x_{ij}=x_{n+1-i,n+1-j}$. In addition, $x_{ij}$s
satisfy the following conditions.

  1. (i)

$\mathbb{E}[x_{ij}]=0,\mathbb{E}[x_{ij}^{2}]=0$ and
$\mathbb{E}[|x_{ij}|^{2}]=1$ for all $1\leq i,j\leq n,$

  2. (ii)

$x_{ij}$s are continuous random variables with bounded density and satisfy the
Poincaré inequality with some universal constant $\alpha$.

Before stating our main theorem, let us define the centered LES as

$\displaystyle\operatorname{L}_{f}^{\circ}(M)=\sum_{k=1}^{n}f(\lambda_{k}(M))-\sum_{k=1}^{n}\mathbb{E}[f(\lambda_{k}(M))].$

We are now ready to state our main theorem.

######  Theorem 3.4.

Suppose $M$ is a centrosymmetric random matrix satisfying Condition 3.3. Let
$f:\mathbb{C}\to\mathbb{C}$ be an analytic function. Then
$\operatorname{L}_{f}^{\circ}(M)/\sqrt{n}\stackrel{{\scriptstyle
d}}{{\to}}N(0,\sigma_{f}^{2})$, where

$\displaystyle\sigma_{f}^{2}=\frac{2}{\pi}\int_{\mathbb{D}_{1}}f(w)\overline{f^{\prime}(w)}\;d\mathfrak{R}(w)d\mathfrak{I}(w).$

In particular, if $f$ is a polynomial given by
$P_{d}(z)=\sum_{k=0}^{d}a_{k}z^{k}$, then
$\sigma_{p_{d}}^{2}=\sum_{k=1}^{d}2ka_{k}^{2}$.

##  4\. Truncation and reduction

Before proceeding to the technical proof of the main theorem, we need to
recall one important theorem regarding centrosymmetric matrices.

######  Definition 4.1 (Counter-identity Matrix).

It is defined as the square matrix whose counter-diagonal elements are $1$ and
all the other elements are zero.

Throughout this article, we shall denote the counter-identity matrix by $J$.
The size of $J$ should be understood from the context. Notice that

$\displaystyle J=J^{T}$ $\displaystyle J^{2}=I,$

where $I$ the identity matrix of the same dimension. Now, we invoke the
following theorem which reduces the centrosymmetric matrix to block diagonal
matrix.

######  Theorem 4.2.

[35, Theorem 9]

  1. (a)

If $M=\left(\begin{array}[]{ll}A&B\\\ C&D\end{array}\right)$ is an $n\times n$
centrosymmetric matrix with $n=2s$ and $A,B,C$, and $D\text{ are }s\times$ s
matrices, then $M$ is orthogonally similar to

$\displaystyle Q^{T}MQ=\left(\begin{array}[]{cc}A+JC&0\\\
0&A-JC\end{array}\right),$

where

$\displaystyle Q=\sqrt{\frac{1}{2}}\left(\begin{array}[]{ll}I&-J\\\
J&I\end{array}\right).$

  2. (b)

If $M=\left(\begin{array}[]{lll}A&x&B\\\ y&q&yJ\\\ C&Jx&D\end{array}\right)$
is an $n\times n$ centrosymmetric matrix with $n=2s+1$, and $A,B,C,D$ are
$s\times s$ matrices, $x$ is $s\times 1$ matrix, $y$ is $1\times s$ matrix,
and $q$ is a scalar, then $M$ is orthogonally similar to

$\displaystyle Q^{T}MQ=\left(\begin{array}[]{ccc}A+JC&\sqrt{2}x&0\\\
\sqrt{2}y&q&0\\\ 0&0&A-JC\end{array}\right),$

where

$\displaystyle Q=\sqrt{\frac{1}{2}}\left(\begin{array}[]{lll}I&0&-J\\\
0&\sqrt{2}&0\\\ J&0&I\end{array}\right).$

In the definitions of $Q$ above, $I$ is an $s\times s$ identity matrix and $J$
is an $s\times s$ counter-identity matrix. Moreover, notice that in both the
cases $Q^{T}Q=I=QQ^{T}.$

Since the $Q^{T}MQ$ is orthogonally similar to the matrix $M$, the eigenvalues
of $Q^{T}MQ$ are the same as those of $M$. The remaining part of this article
analyses the eigenvalues of $Q^{T}MQ$. For convenience, we adopt the following
simplified notation for $Q^{T}MQ$.

$\displaystyle Q^{T}MQ=\left[\begin{array}[]{cc}T_{1}&0\\\
0&T_{2}\end{array}\right],$

where $T_{2}=A-JC$ and

$\displaystyle T_{1}=\left\\{\begin{array}[]{ll}A+JC&\text{if $n=2s$}\\\
\left(\begin{array}[]{cc}A+JC&\sqrt{2}x\\\
\sqrt{2}y&q\end{array}\right)&\text{if $n=2s+1$}.\end{array}\right.$

Though the above theorem reduces the centrosymmetric matrix to a block
diagonal matrix, the block matrices are not independent. Therefore, we need to
take care of the dependence structure in the course of the proof.

##  5\. Circular Law for Random Centrosymmetric Matrix

Let us first state the general circular law.

######  Theorem 5.1 (Circular Law, [34]).

Consider a random matrix $M=\frac{1}{\sqrt{n}}[x_{ij}]_{i,j=1}^{n},$ where
$x_{ij}$s are i.i.d. random variables with mean zero and variance $1$. Then,
as $n\to\infty$, the ESD of $M$ converges almost surely to the uniform
distribution on the unit disc of $\mathbb{C}$ with density
$\frac{1}{\pi}1_{\\{z\in\mathbb{C}\;|\;|z|\leq 1\\}}$.

Now, let us state the circular law for random centrosymmetric matrix.

######  Theorem 5.2 (Circular Law for Random Centrosymmetric Matrix).

Consider a random centrosymmetric matrix
$M=\frac{1}{\sqrt{n}}[x_{ij}]_{i,j=1}^{n},$ where $x_{ij}$s are i.i.d. random
variables with mean zero and variance $1.$ Then, as $n\to\infty$, the ESD of
$M$ converges almost surely to the uniform distribution on the unit disc of
$\mathbb{C}$ with density $\frac{1}{\pi}1_{\\{z\in\mathbb{C}\;|\;|z|\leq
1\\}}$.

Figure 2. Scatter plot of real and imaginary parts of the eigenvalues of a
$5000\times 5000$ random centrosymmetric matrix having standard Gaussian
entries.

###### Proof.

We have a centrosymmetric matrix $M=\frac{1}{\sqrt{n}}[x_{ij}]_{i,j=1}^{n},$
where $x_{ij}$s are i.i.d. centered random variables with variance $1$.

From Theorem 4.2, it is evident that the eigenvalues of $M$ are the
eigenvalues of $T_{1}$ and $T_{2}.$ The entries of $T_{1}$ and $T_{2}$ are
i.i.d. centered random variables having variance $2$. However, sizes of
$T_{1}$ and $T_{2}$ is $\lceil n/2\rceil\times\lceil n/2\rceil$, and they are
scaled by $1/\sqrt{n}.$ Therefore, by Theorem 5.1 the ESD of $T_{1}$ converges
to the circular law. The same applies to $T_{2}$ as well. Hence, ESD of the
matrix $M$ converges to the circular law.

∎

##  6\. Proof of the central limit theorem

In Section 5, we have seen that the limiting ESD follows the circular law.
However, for any finite $n$, there might be some eigenvalues outside the unit
disc. To avoid such situations, we shall work on the event where the
eigenvalues are confined within a disc of finite radius. Keeping this in mind,
we define the event

(6.1) $\Omega_{n}=\\{\|T_{1}\|\leq\rho_{1}\\}\cap\\{\|T_{2}\|\leq\rho_{2}\\},$

where $\rho_{1}$ and $\rho_{2}$ are as defined in Lemma A.2. From the Lemma
A.2, we also have

(6.2) $P\left(\Omega_{n}^{c}\right)\leq K\exp\left(-\sqrt{\frac{\alpha
n}{2}}\frac{3\rho_{m}}{4}\right),\quad\text{where
}\rho_{m}=\min\\{\rho_{1},\rho_{2}\\}.$

Let us denote $\rho=\max\\{\rho_{1},\rho_{2}\\}.$ If
$f:\mathbb{C}\rightarrow\mathbb{C}$ is analytic on $\mathbb{D}_{\rho+\tau}$,
then

$\displaystyle\operatorname{L}_{f}(M)$
$\displaystyle=\sum_{i=1}^{n}f(\lambda_{i}(M))$ $\displaystyle=\frac{1}{2\pi
i}\oint_{\partial\mathbb{D}_{\rho+\tau}}\sum_{i=1}^{n}\frac{f(z)}{z-\lambda_{i}(M)}\,dz$
$\displaystyle=\frac{1}{2\pi
i}\oint_{\partial\mathbb{D}_{\rho+\tau}}f(z)\operatorname{Tr}\mathcal{R}_{z}(M)\,dz.$

We can express

$\operatorname{L}_{f}^{\circ}(M)=\frac{1}{2\pi
i}\oint_{\partial\mathbb{D}_{\rho+\tau}}f(z)\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(M)\,dz+o(1).$

where $\hat{\xi}=\xi 1_{\Omega_{n}},$ and $\xi^{\circ}=\xi-\mathbb{E}[\xi].$
Furthermore, standard use of complex analysis shows that

$\displaystyle\frac{1}{\pi}\int_{\mathbb{D}_{1}}f^{\prime}(w)\overline{f^{\prime}(w)}\;d\mathfrak{R}(w)d\mathfrak{I}(w)$
$\displaystyle=\frac{1}{\pi}\int_{\mathbb{D}_{1}}\left[\frac{1}{2\pi
i}\oint_{\partial\mathbb{D}_{1}}\frac{f(z)\;dz}{(w-z)^{2}}\frac{1}{2\pi
i}\oint_{\partial\mathbb{D}_{1}}\frac{f(\bar{\eta})\;d\bar{\eta}}{(\bar{w}-\bar{\eta})^{2}}\right]\;d\mathfrak{R}(w)d\mathfrak{I}(w)$
$\displaystyle=-\frac{1}{4\pi^{2}}\oint_{\partial\mathbb{D}_{1}}\oint_{\partial\mathbb{D}_{1}}f(z)\overline{f(\eta)}\left[\frac{1}{\pi}\int_{\mathbb{D}_{1}}\frac{d\mathfrak{R}(w)d\mathfrak{I}(w)}{(w-z)^{2}(\bar{w}-\bar{\eta})^{2}}\right]\;dz\;d\bar{\eta}$
$\displaystyle=-\frac{1}{4\pi^{2}}\oint_{\partial\mathbb{D}_{1}}\oint_{\partial\mathbb{D}_{1}}\frac{f(z)\overline{f(\eta)}}{(1-z\bar{\eta})^{2}}\;dz\;d\bar{\eta}.$

Therefore, proving the main Theorem 3.4 is equivalent to proving Proposition
6.1.

######  Proposition 6.1.

The sequence $\\{\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(M)\\}_{n}$ is
tight in the space of continuous functions on $\partial\mathbb{D}_{\rho+\tau}$
and converges in distribution to a Gaussian process with covariance kernel
given by $2(1-z\bar{\eta})^{-2}$.

Our first goal is to reduce the above proposition to (6.5), (6.6), (6.7).
First of all, by Cramer-Wold device, it suffices to show that for any
$\left\\{z_{l};1\leq l\leq q\right\\}\subset\partial\mathbb{D}_{\rho+\tau},$
and $\left\\{\alpha_{l},\beta_{l};1\leq l\leq q\right\\}\subset\mathbb{C}$ for
which

(6.3)
$\sum_{l=1}^{q}\left[\alpha_{l}\operatorname{Tr}\hat{\mathcal{R}}_{z_{l}}^{\circ}(M)+\beta_{l}\overline{\operatorname{Tr}\hat{\mathcal{R}}_{z_{l}}^{\circ}(M)}\right]\mathbf{1}_{\Omega_{n}}$

is real; converges to a Gaussian random variable with mean zero. In order to
show this, we use the martingale CLT, which is stated in Lemma A.3. Let
$\Xi_{k}[\xi]$ be the averaging of the random variable $\xi$ over the $k$-th
columns of matrices $T_{1}$ and $T_{2}$, and
$\mathbb{E}_{k}[\cdot]=\Xi_{k+1}\Xi_{k+2}\ldots\Xi_{\lceil\frac{n}{2}\rceil}[\cdot]=\Xi_{k+1}\mathbb{E}_{k+1}[\cdot]$.
Clearly, $\mathbb{E}_{0}[\cdot]=\mathbb{E}[\cdot]$ and
$\mathbb{E}_{\lceil\frac{n}{2}\rceil}[\cdot]=[\cdot]$. Therefore, we can
express

$\displaystyle\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(M)$
$\displaystyle=\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{1})+\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{2})$
$\displaystyle=\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{1})-\mathbb{E}[\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{1})]+\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{2})-\mathbb{E}[\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{2})]$
$\displaystyle=\sum_{k=1}^{\lceil
n/2\rceil}(\mathbb{E}_{k}-\mathbb{E}_{k-1})\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{1})+\sum_{k=1}^{\lceil
n/2\rceil}(\mathbb{E}_{k}-\mathbb{E}_{k-1})\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{2})$
$\displaystyle=\sum_{k=1}^{\lceil
n/2\rceil}\mathbb{E}_{k}\left\\{\left(\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{1})\right)_{k}^{\circ}\right\\}+\sum_{k=1}^{\lceil
n/2\rceil}\mathbb{E}_{k}\left\\{\left(\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{2})\right)_{k}^{\circ}\right\\}$
$\displaystyle=:\sum_{k=1}^{\lceil n/2\rceil}W_{z,k}(T_{1})+\sum_{k=1}^{\lceil
n/2\rceil}W_{z,k}(T_{2})$ (6.4) $\displaystyle=\sum_{k=1}^{\lceil
n/2\rceil}\left(W_{z,k}(T_{1})+W_{z,k}(T_{2})\right),$

where $\xi_{k}^{\circ}=\xi-\Xi_{k}[\xi].$ Clearly, the sequences
$\left\\{W_{z,k}(T_{1})\right\\}_{1\leq k\leq\lceil n/2\rceil},$ and
$\left\\{W_{z,k}(T_{2})\right\\}_{1\leq k\leq\lceil n/2\rceil}$ are martingale
difference sequences with respect to the filtration
$\mathcal{F}_{n,k}=\sigma\\{t^{1}_{j},t^{2}_{j}:0\leq j\leq k\\}$, where
$t^{1}_{j}$ and $t^{2}_{j}$ are the $j$th columns of matrices $T_{1}$ and
$T_{2}$ respectively. Therefore, the sum $W_{z,k}(T_{1})+W_{z,k}(T_{2})$ also
forms a martingale difference sequence with respect to the filtration
$\mathcal{F}_{n,k}.$

Now, we can rewrite

$\displaystyle\sum_{i=1}^{q}\left[\alpha_{i}\operatorname{Tr}\hat{\mathcal{R}}_{z_{i}}^{\circ}(M)+\beta_{i}\overline{\operatorname{Tr}\hat{\mathcal{R}}_{z_{i}}^{\circ}(M)}\right]$
$\displaystyle=\sum_{k=1}^{\lceil
n/2\rceil}\left\\{\sum_{i=1}^{q}\left(\alpha_{i}(W_{z_{i},k}(T_{1})+W_{z_{i},k}(T_{2}))+\beta_{i}(W_{\bar{z_{i}},k}(T_{1}{\operatorname{{}^{*}}})+W_{\bar{z_{i}},k}(T_{2}{\operatorname{{}^{*}}}))\right)\right\\}$
$\displaystyle=\sum_{k=1}^{\lceil
n/2\rceil}\left\\{\sum_{i=1}^{q}\alpha_{i}W_{z_{i},k}(T_{1})+\beta_{i}W_{\bar{z_{i}},k}(T_{1}{\operatorname{{}^{*}}})+\sum_{i=1}^{q}\alpha_{i}W_{z_{i},k}(T_{2})+\beta_{i}W_{\bar{z_{i}},k}(T_{2}{\operatorname{{}^{*}}})\right\\}$
$\displaystyle=:\sum_{k=1}^{\lceil
n/2\rceil}\left(\psi_{n,k}^{(1)}+\psi_{n,k}^{(2)}\right)$
$\displaystyle=:\sum_{k=1}^{\lceil n/2\rceil}\psi_{n,k},$

where $X{\operatorname{{}^{*}}}$ denotes the complex conjugate of the matrix
$X$. Consequently, $\left\\{\psi_{n,k}\right\\}_{1\leq k\leq{\lceil
n/2\rceil}}$ forms a martingale difference sequence with respect to the same
filtration $\mathcal{F}_{n,k}$. Also, note that
$\mathbb{E}[\cdot|\mathcal{F}_{n,k-1}]=\mathbb{E}_{k-1}[\cdot].$ Therefore,
condition (i) of Lemma A.3 can be expressed as

(6.5) $\displaystyle\sum_{k=1}^{\lceil
n/2\rceil}\mathbb{E}\left[\psi_{n,k}^{2}\mathbf{1}_{\left|\psi_{n,k}\right|>\theta}\right]\leq\frac{1}{\theta^{2}}\sum_{k=1}^{\lceil
n/2\rceil}\mathbb{E}\left[\psi_{n,k}^{4}\right]\rightarrow 0.$

Furthermore, the condition (ii) of Lemma A.3 can be expressed as

(6.6) $\displaystyle\sum_{k=1}^{\lceil
n/2\rceil}\mathbb{E}_{k-1}[(W_{z_{i},k}(T_{1})+W_{z_{i},k}(T_{2}))(W_{z_{j},k}(T_{1})+W_{z_{j},k}(T_{2}))]\stackrel{{\scriptstyle
p}}{{\rightarrow}}0,$

and

(6.7)
$\displaystyle\mathbb{E}_{k-1}[(W_{z_{i},k}(T_{1})+W_{z_{i},k}(T_{2}))(W_{\bar{z_{j}},k}(T_{1}{\operatorname{{}^{*}}})+W_{\bar{z_{j}},k}(T_{2}{\operatorname{{}^{*}}}))]\stackrel{{\scriptstyle
p}}{{\rightarrow}}\sigma\left(z_{i},z_{j}\right),$

where $\sigma$ is a covariance kernel to be found later. We defer the
calculation of the covariance kernel to Sectoin 7.4. Therefore, proving the
Proposition 6.1 is equivalent to showing the tightness of
$\\{\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{1})+\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{2})\\}_{n}$
and (6.5), (6.6) and (6.7). Before proceeding any further, let us do some
reductions.

For $i=1,2$, let $T_{i}^{k}$ be the matrix obtained by zeroing out the $k$th
column of $T_{i}$, and let
$\hat{\mathcal{R}}_{z}(T_{i}^{k}):=\mathcal{R}_{z}(T_{i}^{k})\mathbf{1}_{\Omega_{n,k}}$,
where $\Omega_{n,k}=\left\\{\left\lVert
T_{1}^{k}\right\rVert\leq\rho_{1}\right\\}\cap\left\\{\left\lVert
T_{2}^{k}\right\rVert\leq\rho_{2}\right\\}.$ The space $\Omega_{n,k}$ is
different from $\Omega_{n},$ but asymptotically both the spaces are the same.
In fact, we can check that

$\displaystyle\mathbb{E}\left|\sum_{i=1}^{\lceil
n/2\rceil}\Xi_{i}\left[\operatorname{Tr}\mathcal{R}_{z}(T_{1}^{k})\mathbf{1}_{\Omega_{n}}-\operatorname{Tr}\mathcal{R}_{z}(T_{1}^{k})\mathbf{1}_{\Omega_{n,i}}\right]_{i}^{\circ}\right|$
$\displaystyle\leq 2\mathbb{E}\left|\sum_{i=1}^{\lceil
n/2\rceil}\Xi_{i}\left[\operatorname{Tr}\mathcal{R}_{z}(T_{1}^{k})\mathbf{1}_{\Omega_{n}^{c}\cap\Omega_{n,i}}\right]\right|$
$\displaystyle\leq\frac{n^{2}}{\tau}\mathbb{P}(\Omega_{n}^{c})=o(1).$

The same is true for $T_{2}^{k}$ as well. The important difference between
$\Omega_{n}$ and $\Omega_{n,k}$ is that the latter is independent of the
$k$-th columns of $T_{1}$ and $T_{2}$, which will allow us to decouple some
expectation calculations in future calculations. Consequently, we shall
proceed with the space $\Omega_{n,k}$.

By using resolvent identity and Lemma A.4, we have

$\displaystyle\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{1})$
$\displaystyle=\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{1}^{k})+e_{k}^{t}\hat{\mathcal{R}}_{z}(T_{1}^{k})\mathcal{R}_{z}(T_{1})t_{k}^{1}$
$\displaystyle=\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{1}^{k})+\frac{e_{k}^{t}(\hat{\mathcal{R}}_{z}(T_{1}^{k}))^{2}t_{k}^{1}}{1+e_{k}^{t}\hat{\mathcal{R}}_{z}(T_{1}^{k})t_{k}^{1}}.$

Similarly,

$\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{2})=\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{2}^{k})+\frac{e_{k}^{t}(\hat{\mathcal{R}}_{z}(T_{2}^{k}))^{2}t_{k}^{2}}{1+e_{k}^{t}\hat{\mathcal{R}}_{z}(T_{2}^{k})t_{k}^{2}},$

where $t_{k}^{1}$ and $t_{k}^{2}$ are the $k$th columns of $T_{1}$ and $T_{2}$
respectively. Thus,

$\displaystyle\operatorname{Tr}\hat{\mathcal{R}}_{z}(M)$
$\displaystyle=\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{1}^{k})+\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{2}^{k})+\frac{e_{k}^{t}(\hat{\mathcal{R}}_{z}(T_{1}^{k}))^{2}t_{k}^{1}}{1+e_{k}^{t}\hat{\mathcal{R}}_{z}(T_{1}^{k})t_{k}^{1}}+\frac{e_{k}^{t}(\hat{\mathcal{R}}_{z}(T_{2}^{k}))^{2}t_{k}^{2}}{1+e_{k}^{t}\hat{\mathcal{R}}_{z}(T_{2}^{k})t_{k}^{2}}$
$\displaystyle=\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{1}^{k})+\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{2}^{k})-\frac{\partial}{\partial
z}\log(1+\gamma_{1k}(z))-\frac{\partial}{\partial z}\log(1+\gamma_{2k}(z)),$

where $\gamma_{1k}(z):=e_{k}^{t}\hat{\mathcal{R}}_{z}(T_{1}^{k})t_{k}^{1}$,
and $\gamma_{2k}(z):=e_{k}^{t}\hat{\mathcal{R}}_{z}(T_{2}^{k})t_{k}^{2}$. Now,
let us find the tail bounds of $\gamma_{1k}(z)$ and $\gamma_{2k}(z).$ Note
that $\gamma_{1k}(z)$ is product of two independent random variables.
Intuitively, $\gamma_{1k}(z)\stackrel{{\scriptstyle a.s.}}{{\to}}0$ by
conditioning on $\\{t^{1}_{j},t^{2}_{j}:j\in\\{1,2,...,\lceil
n/2\rceil\\}\backslash\\{k\\}\\}$.

Define
$h(t_{k}^{1})=e_{k}^{t}\hat{\mathcal{R}}_{z}(T_{1}^{k})t_{k}^{1}=\gamma_{1k}(z).$
Then by using property $3$ in Definition 3.2 and the fact that
$\sum_{j=1}^{\lceil
n/2\rceil}|\hat{\mathcal{R}_{z}}(T_{1}^{k})_{ij}|\leq\|\hat{\mathcal{R}_{z}}(T_{1}^{k})\|^{2}\leq\tau^{-2}$
for $z\in\partial{\mathbb{D}}_{\rho+\tau}$, we have

(6.8)
$\mathbb{P}\left(|\gamma_{1k}(z)|>t\;\;|\;\;\hat{\mathcal{R}_{z}}(T_{1}^{k})\right)\leq
K\operatorname{exp}\left(-\tau\sqrt{\frac{\alpha n}{2}}t\right).$

In addition,

(6.9) $\mathbb{E}[|\gamma_{1k}(z)|^{p}]\leq Kp!\left(\frac{2}{\tau^{2}\alpha
n}\right)^{p/2}\text{ for all }z\in\partial{\mathbb{D}}_{\rho+\tau}.$

Similarly, we have bound for $\gamma_{2k}$ as follows.

(6.10) $\mathbb{E}[|\gamma_{2k}(z)|^{p}]\leq Kp!\left(\frac{2}{\tau^{2}\alpha
n}\right)^{p/2}\text{ for all }z\in\partial{\mathbb{D}}_{\rho+\tau}.$

Since $T_{1}^{k},T_{2}^{k}$ are independent of the $k$th column of
$T_{1},T_{2}$ respectively, we can easily see that

$\displaystyle\left[\operatorname{Tr}\hat{\mathcal{R}_{z}}(T_{1}^{k})+\operatorname{Tr}\hat{\mathcal{R}_{z}}(T_{2}^{k})\right]_{k}^{\circ}$
(6.11)
$\displaystyle=\operatorname{Tr}\hat{\mathcal{R}_{z}}(T_{1}^{k})+\operatorname{Tr}\hat{\mathcal{R}_{z}}(T_{2}^{k})-\Xi_{k}[\operatorname{Tr}\hat{\mathcal{R}_{z}}(T_{1}^{k})+\operatorname{Tr}\hat{\mathcal{R}_{z}}(T_{2}^{k})]=0.$

Hence,

(6.12) $\displaystyle W_{z,k}(T_{1})+W_{z,k}(T_{2})$
$\displaystyle=\mathbb{E}_{k}\left[\left(\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{1})+\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{2})\right)_{k}^{\circ}\right]$
$\displaystyle=-\mathbb{E}_{k}\left[\left(\frac{\partial}{\partial
z}\operatorname{\log}(1+\gamma_{1k}(z))+\frac{\partial}{\partial
z}\operatorname{\log}(1+\gamma_{2k}(z))\right)_{k}^{\circ}\right]$
$\displaystyle=-\frac{\partial}{\partial
z}\mathbb{E}_{k}[\operatorname{\log}(1+\gamma_{1k}(z))+\operatorname{\log}(1+\gamma_{2k}(z))]_{k}^{\circ}$
(6.13) $\displaystyle=-\frac{\partial}{\partial
z}\mathbb{E}_{k}[\operatorname{\log}\\{(1+\gamma_{1k}(z))(1+\gamma_{2k}(z))\\}]_{k}^{\circ}.$

In the above, $\mathbb{E}_{k}$ and $\partial/\partial z$ are interchanged
using the dominated convergence theorem. We can estimate
$W_{z,k}(T_{1})+W_{z,k}(T_{2})$ by the following fact from complex analysis.
Let $v$ be an analytic function on $z+\mathbb{D}_{\tau/2}.$ Then,

(6.14) $|v^{\prime}(z)|=\left|\frac{1}{2\pi
i}\oint_{z+\partial{\mathbb{D}_{\tau/3}}}\frac{v(t)dt}{(t-z)^{2}}\right|\leq\frac{5}{\pi\tau^{2}}\oint_{z+\partial{\mathbb{D}_{\tau/3}}}|v(t)|dt.$

###  6.1. Proof of tightness

In this subsection, we show the tightness of
$\\{\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(M)\\}_{n}$ in the space
$\mathcal{C}(\partial{\mathbb{D}}_{\rho+\tau})$. Since
$\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(M)=\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{1})+\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{2})$,
it is sufficient to show that both
$\\{\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{1})\\}_{n}$ and
$\\{\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{2})\\}_{n}$ are tight.
To prove the tightness of
$\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{1})$ we need to show that
for any $\epsilon>0$ there exit a compact set
$\mathcal{K}(\epsilon)\subset\mathcal{C}(\partial{\mathbb{D}}_{\rho+\tau})$
such that

(6.15)
$\mathbb{P}\left(\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{1})\in\mathcal{K}(\epsilon)^{c}\right)<\epsilon.$

However by Arzela-Ascoli theorem, the compact sets in space of continuous
functions are space of equicontinuous functions. Let us define,

$\Phi_{n}:=\\{|\gamma_{1k}(z)|<n^{-1/8},\text{ for all }1\leq k\leq\lceil
n/2\rceil\\}.$

Therefore from (6.8),

$\mathbb{P}(\Phi_{n}^{c})\leq
K\exp{\left(\log\left(\frac{n}{2}\right)-\tau\sqrt{\frac{\alpha}{2}}n^{\frac{3}{8}}\right)}=o(1).$

Consequently,

$\displaystyle\mathbb{P}\left(\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{1})\in\mathcal{K}(\epsilon)^{c}\right)$
$\displaystyle\leq\mathbb{P}\left(\frac{|\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{1})-\operatorname{Tr}\hat{\mathcal{R}}_{\eta}^{\circ}(T_{1})|}{|z-\eta|}>K\right)$
(6.16)
$\displaystyle\leq\mathbb{P}\left(\frac{|\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{1})-\operatorname{Tr}\hat{\mathcal{R}}_{\eta}^{\circ}(T_{1})|}{|z-\eta|}1_{\Phi_{n}}>K\right)+o(1),$

for some $K>0$. Now, the following equation (6.17) is sufficient to prove
(6.15);

(6.17)
$\mathbb{E}\left\\{\left|\frac{|\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{1})-\operatorname{Tr}\hat{\mathcal{R}}_{\eta}^{\circ}(T_{1})}{z-\eta}\right|^{2}1_{\Phi_{n}}\right\\}\leq
C,$

uniformly for all $z,\eta\in\partial{\mathbb{D}}_{\rho+\tau}$ and
$n\in\mathbb{N}$. Using the resolvent identity, we can write

$\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{1})-\operatorname{Tr}\hat{\mathcal{R}}_{\eta}(T_{1})=(z-\eta)\operatorname{Tr}\left[\hat{\mathcal{R}}_{z}(T_{1})\hat{\mathcal{R}}_{\eta}(T_{1})\right].$

Therefore, using (6.4) we have

(6.18)
$\displaystyle(z-\eta)^{-1}\left[\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{1})-\operatorname{Tr}\hat{\mathcal{R}}_{\eta}^{\circ}(T_{1})\right]=\sum_{k=1}^{\lceil
n/2\rceil}\mathbb{E}_{k}\left\\{\left(\operatorname{Tr}\left[\hat{\mathcal{R}}_{k}(T_{1})\hat{\mathcal{R}}_{\eta}(T_{1})\right)\right]_{k}^{\circ}\right\\}.$

Using resolvent identity and Lemma A.4,

$\displaystyle\operatorname{Tr}$
$\displaystyle\left[\hat{\mathcal{R}}_{z}(T_{1})\hat{\mathcal{R}}_{\eta}(T_{1})-\hat{\mathcal{R}}_{z}\left(T_{1}^{k}\right)\hat{\mathcal{R}}_{\eta}\left(T_{1}^{k}\right)\right]$
$\displaystyle=\operatorname{Tr}\left[\left\\{\hat{\mathcal{R}}_{z}(T_{1})-\hat{\mathcal{R}}_{z}(T_{1}^{k})\right\\}\left\\{\hat{\mathcal{R}}_{\eta}(T_{1})-\hat{\mathcal{R}}_{\eta}(T_{1}^{k})\right\\}\right]$
$\displaystyle\quad+\operatorname{Tr}\left[\hat{\mathcal{R}}_{z}\left(T_{1}^{k}\right)\left\\{\hat{\mathcal{R}}_{\eta}(T_{1})-\hat{\mathcal{R}}_{\eta}\left(T_{1}^{k}\right)\right\\}\right]+\operatorname{Tr}\left[\left\\{\hat{\mathcal{R}}_{z}(T_{1})-\hat{\mathcal{R}}_{z}\left(T_{1}^{k}\right)\right\\}\hat{\mathcal{R}}_{\eta}\left(T_{1}^{k}\right)\right]$
$\displaystyle=\operatorname{Tr}\frac{\hat{\mathcal{R}}_{z}\left(T_{1}^{k}\right)t_{k}^{1}e_{k}^{t}\hat{\mathcal{R}}_{z}\left(T_{1}^{k}\right)}{1+\gamma_{k1}(z)}\frac{\hat{\mathcal{R}}_{\eta}\left(T_{1}^{k}\right)t_{k}^{1}e_{k}^{t}\hat{\mathcal{R}}_{\eta}\left(T_{1}^{k}\right)}{1+\gamma_{k1}(\eta)}$
$\displaystyle\quad+\operatorname{Tr}\frac{\hat{\mathcal{R}}_{z}\left(T_{1}^{k}\right)\hat{\mathcal{R}}_{\eta}\left(T_{1}^{k}\right)t_{k}^{1}e_{k}^{t}\hat{\mathcal{R}}_{\eta}\left(T_{1}^{k}\right)}{1+\gamma_{k1}(\eta)}+\operatorname{Tr}\frac{\hat{\mathcal{R}}_{z}\left(T_{1}^{k}\right)t_{k}^{1}e_{k}^{t}\hat{\mathcal{R}}_{z}\left(T_{1}^{k}\right)\hat{\mathcal{R}}_{\eta}\left(T_{1}^{k}\right)}{1+\gamma_{k1}(z)}$
$\displaystyle=\frac{\left(e_{k}^{t}\hat{\mathcal{R}}_{z}\left(T_{1}^{k}\right)\hat{\mathcal{R}}_{\eta}\left(T_{1}^{k}\right)t_{k}^{1}\right)^{2}}{\left(1+\gamma_{k1}(z)\right)\left(1+\gamma_{k1}(\eta)\right)}+\frac{e_{k}^{t}\hat{\mathcal{R}}_{\eta}\left(T_{1}^{k}\right)\hat{\mathcal{R}}_{z}\left(T_{1}^{k}\right)\hat{\mathcal{R}}_{\eta}\left(T_{1}^{k}\right)t_{k}^{1}}{1+\gamma_{k1}(\eta)}$
$\displaystyle\quad+\frac{e_{k}^{t}\hat{\mathcal{R}}_{z}\left(T_{1}^{k}\right)\hat{\mathcal{R}}_{\eta}\left(T_{1}^{k}\right)\hat{\mathcal{R}}_{z}\left(T_{1}^{k}\right)t_{k}^{1}}{1+\gamma_{k1}(z)}$
$\displaystyle=:\vartheta_{1}(k)+\vartheta_{2}(k)+\vartheta_{3}(k).$

By (6.11), we have
$\left[\operatorname{Tr}\left(\hat{\mathcal{R}}_{z}\left(T_{1}^{k}\right)\hat{\mathcal{R}}_{\eta}\left(T_{1}^{k}\right)\right)\right]_{k}^{\circ}=0$.
Therefore, we can express $\eqref{eqn: tightness: difference of tr R as a
product of R}$ as

$(z-\eta)^{-1}\left[\operatorname{Tr}\hat{\mathcal{R}}_{z}(T_{1})-\operatorname{Tr}\hat{\mathcal{R}}_{\eta}^{\circ}(T_{1})\right]=\sum_{k=1}^{\lceil
n/2\rceil}\mathbb{E}_{k}\left\\{\left[\vartheta_{1}(k)+\vartheta_{2}(k)+\vartheta_{3}(k)\right]_{k}^{\circ}\right\\}.$

Since $\left\\{\left[\vartheta_{i}(k)\right]_{k}^{\circ}\right\\}$ is a
martingale difference sequence,

$\mathbb{E}\,\left|\,\sum_{k=1}^{\lceil
n/2\rceil}\mathbb{E}_{k}\left\\{\left[\vartheta_{1}(k)+\vartheta_{2}(k)+\vartheta_{3}(k)\right]_{k}^{\circ}\right\\}\right|^{2}=\sum_{k=1}^{\lceil
n/2\rceil}\mathbb{E}\left|\mathbb{E}_{k}\left\\{\left[\vartheta_{1}(k)+\vartheta_{2}(k)+\vartheta_{3}(k)\right]_{k}^{\circ}\right\\}\right|^{2}.$

Hence, proving (6.17) is equivalent to verifying the validity of

(6.19) $\displaystyle
n\mathbb{E}[|\vartheta_{i}(k)|^{2}1_{\mathrm{\Phi_{n}}}]\leq C,\text{ for all
}1\leq k\leq\lceil n/2\rceil,\;\;i=1,2,\dots.$

uniformly for all $n\in\mathbb{N}$ and
$z,\eta\in\partial{\mathbb{D}}_{\rho+\tau}$. By adopting the same approach as
demonstrated in (6.9), similar tail estimates for
$e_{k}^{t}\hat{\mathcal{R}}_{z}\left(T_{1}^{k}\right)\hat{\mathcal{R}}_{\eta}\left(T_{1}^{k}\right)t_{k}^{1}$
can be obtained possibly with $\tau^{2}$ or $\tau^{3}$ in the rhs of (6.9).
Noting that $|\gamma_{1k}(z)|<n^{-1/8}$ on $\Phi_{n}$, leveraging the estimate
$|(1+\gamma_{1k}(z))|^{-1}\leq 2$ in $\mathbb{E}[|\vartheta_{i}(k)|^{2}],$ we
have (6.19). Similarly, we can prove the tightness of
$\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{2})$.

###  6.2. Proof of (6.5)

Expanding $\operatorname{\log}(1+\gamma_{k1}(z))$ and
$\operatorname{\log}(1+\gamma_{k2}(z))$ up to two terms and using
(6.9),(6.10),(6.13), and (6.14) we have

$\mathbb{E}[|W_{z,k}(T_{1})+W_{z,k}(T_{2})|^{4}]=O(n^{-2}).$

Substituting the above into (6.5), we obtain

$\displaystyle\sum_{k=1}^{\lceil
n/2\rceil}\mathbb{E}[\psi_{n,k}^{2}1_{|\psi_{n,k}|>\theta}]$
$\displaystyle\leq\frac{(2q)^{3}}{\theta^{2}}\sum_{k=1}^{\lceil
n/2\rceil}\sum_{i=1}^{q}(|\alpha_{i}|^{4}\mathbb{E}[|W_{z_{i},k}(T_{1})+W_{z_{i},k}(T_{2})|^{4}]$
$\displaystyle+|\beta_{i}|^{4}\mathbb{E}[|W_{\bar{z}_{i},k}(T_{1}{\operatorname{{}^{*}}})+W_{\bar{z}_{i},k}(T_{2}{\operatorname{{}^{*}}})|^{4}])$
$\displaystyle=\theta^{-2}O(n^{-1}),$

which proves (6.5).

###  6.3. Proof of (6.6)

Expanding $\operatorname{\log}(1+\gamma_{k1}(z))$ and
$\operatorname{\log}(1+\gamma_{k2}(z))$ up to two terms and using (6.9) and
(6.10) we have

$\Xi_{k}\\{\mathbb{E}_{k}[(\log(1+\gamma_{1k}(z))+\log(1+\gamma_{2k}(z)))_{k}^{\circ})]\mathbb{E}_{k}[(\log(1+\gamma_{1k}(\eta))+\log(1+\gamma_{2k}(\eta)))_{k}^{\circ})]\\}=O(n^{-2}).$

Thus, using (6.13),(6.14) and the above we have,

$\sum_{k=1}^{\lceil
n/2\rceil}\mathbb{E}_{k-1}[(W_{z_{i},k}(T_{1})+W_{z_{i},k}(T_{2}))(W_{z_{j},k}(T_{1})+W_{z_{j},k}(T_{2}))]=O(n^{-1}),$

which proves (6.6).

###  6.4. Proof of (6.7)

Expanding $\operatorname{\log}(1+\gamma_{k1}(z))$ and
$\operatorname{\log}(1+\gamma_{k2}(z))$ up to two terms and using (6.9),(6.10)
and (6.13) we have

$\mathbb{E}_{k-1}[(W_{z,k}(T_{1})+W_{z,k}(T_{2}))(W_{\bar{\eta},k}(T_{1}{\operatorname{{}^{*}}})+W_{\bar{\eta},k}(T_{2}{\operatorname{{}^{*}}}))]=\frac{\partial^{2}}{\partial
z\partial\bar{\eta}}\Theta_{k}(z,\bar{\eta}),$

where

$\displaystyle\Theta_{k}(z,\bar{\eta})$
$\displaystyle=\Xi_{k}\\{\mathbb{E}_{k}[(\log(1+\gamma_{1k}(z))+\log(1+\gamma_{2k}(z)))_{k}^{\circ}]\mathbb{E}_{k}[(\log(1+\overline{\gamma_{1k}(\eta)})+\log(1+\overline{\gamma_{2k}(\eta)}))_{k}^{\circ}]\\}$
$\displaystyle=\frac{1}{n}\left(e_{k}^{t}\mathbb{E}_{k}[\hat{\mathcal{R}}_{z}(T_{1}^{k})]\operatorname{Var}((T_{1}^{k})_{ki})I\mathbb{E}_{k}[\hat{\mathcal{R}}_{\bar{\eta}}(T_{1}^{k}{\operatorname{{}^{*}}})e_{k}]\right)$
$\displaystyle+\frac{1}{n}\left(e_{k}^{t}\mathbb{E}_{k}[\hat{\mathcal{R}}_{z}(T_{2}^{k})]\operatorname{Var}((T_{2}^{k})_{ki})I\mathbb{E}_{k}[\hat{\mathcal{R}}_{\bar{\eta}}(T_{2}^{k}{\operatorname{{}^{*}}})e_{k}]\right)+O(n^{-2})$
$\displaystyle=:\frac{1}{n}\Gamma_{k}(z,\bar{\eta})+O(n^{-2}),$

where $(T_{1}^{k})_{ij}$ and $(T_{2}^{k})_{ij}$ are the $ij$th element of
$T_{1}^{k}$ and $T_{2}^{k}$ respectively.

As a result, (6.7) becomes

$\displaystyle\sum_{k=1}^{\lceil
n/2\rceil}\mathbb{E}_{k-1}[(W_{z,k}(T_{1})+W_{z,k}(T_{2}))(W_{\bar{\eta},k}(T_{1})+W_{\bar{\eta},k}(T_{2}))]$
$\displaystyle=\frac{\partial^{2}}{\partial
z\partial\bar{\eta}}\left[\frac{1}{n}\sum_{k=1}^{\lceil
n/2\rceil}\Gamma_{k}(z,\bar{\eta})\right]+O(n^{-1})$ (6.20)
$\displaystyle=:\frac{\partial^{2}}{\partial
z\partial\bar{\eta}}\chi_{k}(z,\bar{\eta})+O(n^{-1}).$

Since
$\|\hat{\mathcal{R}}_{z}(T_{1}^{k})+\hat{\mathcal{R}}_{z}(T_{2}^{k})\|\leq
4\tau^{-1}$ on $\mathbb{D}_{\rho+\tau/2}^{c}$, $\chi_{k}(z,\bar{\eta})$ is a
sequence of uniformly bounded analytic functions on
$\mathbb{D}_{\rho+\tau/2}^{c}$. Therefore, by Vitali’s theorem (see [14,
Section 2.9]), proving (6.7) equivalent to showing that
$\chi_{k}(z,\bar{\eta})$ converges in probability.

######  Lemma 6.2.

Let $\chi_{k}(z,\bar{\eta})$ be the same as defined in (6.20), that is,
$\chi_{k}(z,\bar{\eta})=\frac{1}{n}\sum_{k=1}^{\lceil
n/2\rceil}\Gamma_{k}(z,\bar{\eta}).$ Then
$\operatorname{Var}(\chi_{k}(z,\bar{\eta}))\to 0$ as $n\to\infty.$

###### Proof.

Since $\\{W_{z,k}(T_{1})+W_{z,k}(T_{2})\\}_{n}$ is a martingale difference
sequence, we have

$\mathbb{E}[(W_{z,k}(T_{1})+W_{z,k}(T_{2}))(W_{\bar{\eta},l}(T_{1}{\operatorname{{}^{*}}})+W_{\bar{\eta},l}(T_{2}{\operatorname{{}^{*}}}))]=0\text{
if }k\neq l.$

This implies,

$\displaystyle\mathbb{E}\left\\{\sum_{k=1}^{\lceil
n/2\rceil}\mathbb{E}_{k-1}[(W_{z,k}(T_{1})+W_{z,k}(T_{2}))(W_{\bar{\eta},k}(T_{1}{\operatorname{{}^{*}}})+W_{\bar{\eta},k}(T_{2}{\operatorname{{}^{*}}}))]\right\\}$
$\displaystyle=\mathbb{E}\left\\{\left(\sum_{k=1}^{\lceil
n/2\rceil}(W_{z,k}(T_{1})+W_{z,k}(T_{2}))\right)\left(\sum_{k=1}^{\lceil
n/2\rceil}(W_{\bar{\eta},k}(T_{1}{\operatorname{{}^{*}}})+W_{\bar{\eta},k}(T_{2}{\operatorname{{}^{*}}}))\right)\right\\}$
$\displaystyle=\mathbb{E}\left\\{\left(\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{1})+\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{2})\right)\left(\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{1}{\operatorname{{}^{*}}})+\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(T_{2}{\operatorname{{}^{*}}})\right)\right\\}$
$\displaystyle=\mathbb{E}\left\\{\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(M)\operatorname{Tr}\hat{\mathcal{R}}_{\bar{\eta}}^{\circ}(M^{*})\right\\}.$

Let us rewrite $\Gamma_{k}(z,\bar{\eta})$ as

$\displaystyle\Gamma_{k}(z,\bar{\eta})$
$\displaystyle=4e_{k}^{t}\mathbb{E}_{k}[\hat{\mathcal{R}}_{z}(T_{1}^{k})]\mathbb{E}_{k}[\hat{\mathcal{R}}_{\bar{\eta}}(T_{1}^{k}{\operatorname{{}^{*}}})]e_{k}+4e_{k}^{t}\mathbb{E}_{k}[\hat{\mathcal{R}}_{z}(T_{2}^{k})]\mathbb{E}_{k}[\hat{\mathcal{R}}_{\bar{\eta}}(T_{2}^{k}{\operatorname{{}^{*}}})]e_{k}$
$\displaystyle:=\Gamma_{k}^{1}(z,\bar{\eta})+\Gamma_{k}^{2}(z,\bar{\eta}).$

Consequently, we may decompose

$\displaystyle\operatorname{Var}\left(\frac{1}{n}\Gamma_{k}(z,\bar{\eta})\right)$
$\displaystyle=\operatorname{Var}\left[\frac{1}{n}\Gamma_{k}^{1}(z,\bar{\eta})+\frac{1}{n}\Gamma_{k}^{2}(z,\bar{\eta})\right]$
$\displaystyle\leq
2\operatorname{Var}\left(\frac{1}{n}\Gamma_{k}^{1}(z,\bar{\eta})\right)+2\operatorname{Var}\left(\frac{1}{n}\Gamma_{k}^{2}(z,\bar{\eta})\right).$

Therefore, proving the lemma is equivalent to showing

$\displaystyle\operatorname{Var}\left(\frac{1}{n}\Gamma_{k}^{1}(z,\bar{\eta})\right)+\operatorname{Var}\left(\frac{1}{n}\Gamma_{k}^{2}(z,\bar{\eta})\right)\rightarrow
0.$

We show that
$\operatorname{Var}\left(\frac{1}{n}\Gamma_{k}^{1}(z,\bar{\eta})\right)\rightarrow
0.$ We shall use the Poincaré inequality to get estimates on
$\operatorname{Var}\left(\frac{1}{n}\Gamma_{k}^{1}(z,\bar{\eta})\right).$ This
is shown in the equation (6.22). However, since we are working on the event
$\Omega_{n,k}$, where norms of the matrices are bounded, some terms may not be
differentiable with respect to the matrix entries. To circumvent such a
problem, we define a smoothing function as follows.

Let $\epsilon_{n}=4/n,\text{ and }\omega_{\epsilon_{n},k}:\mathbb{R}^{\lceil
n^{2}/4\rceil}\rightarrow[0,1]$ be a smooth function such that

$\displaystyle\omega_{\epsilon_{n},k}|_{\\{\|T_{1}^{k}\|\leq\rho_{1}\\}}\equiv
1,$
$\displaystyle\omega_{\epsilon_{n},k}|_{\\{\|T_{1}^{k}\|\geq\rho_{1}+\epsilon_{n}\\}}\equiv
0,$ $\displaystyle\left|\frac{\partial}{\partial
x_{i}}\omega_{\epsilon_{n},k}\right|\leq\frac{2}{\epsilon_{n}},\;\;\;\;\forall\;\;1\leq
i\leq\lceil n^{2}/4\rceil.$

Let us define the smoothed version of $\mathcal{R}$ as

$\tilde{\mathcal{R}}_{z}(T_{1}^{k})=\mathcal{R}_{z}(T_{1}^{k})\omega_{\epsilon_{n},k},$

and

$\tilde{\Gamma}_{k}^{1}(z,\bar{\eta})=4e_{k}^{t}\mathbb{E}_{k}[\tilde{\mathcal{R}}_{z}(T_{1}^{k})]\mathbb{E}_{k}[\tilde{\mathcal{R}}_{\bar{\eta}}(T_{1}^{k}{\operatorname{{}^{*}}})]e_{k}.$

Since the smoothed version and the original one match everywhere except the
narrow annulus $\\{\rho_{1}<\|T_{1}^{k}\|<\rho_{1}+\epsilon_{n}\\}$, we can
easily obtain

$|\Gamma_{k}^{1}(z,\bar{\eta})-\tilde{\Gamma}_{k}^{1}(z,\bar{\eta})|\leq\frac{K}{(\tau-\epsilon_{n})^{2}}1_{\\{\rho_{1}<\|T_{1}^{k}\|<\rho_{1}+\epsilon_{n}\\}},$

where $K$ is some universal constant. Now, by Lemma A.2, we have,

(6.21)
$\displaystyle\mathbb{E}[|\Gamma_{k}^{1}(z,\bar{\eta})-\tilde{\Gamma}_{k}^{1}(z,\bar{\eta})|^{2}]\leq\frac{K}{(\tau-\epsilon_{n})^{4}}\exp{\left(-\tau\sqrt{\frac{\alpha
n}{2}}\frac{3\rho_{1}}{4}\right)}.$

Therefore the smoothing function $\omega$ did not alter the original
quantities by significant amount. For the notational convenience, let us
denote

$\displaystyle\Lambda$
$\displaystyle:=\mathbb{E}_{k}[\mathcal{R}_{z}(T_{1}^{k})],$
$\displaystyle\Upsilon$
$\displaystyle:=\mathbb{E}_{k}[\mathcal{R}_{\eta}(T_{2}^{k})^{\operatorname{{}^{*}}}],$
$\displaystyle\tilde{\Lambda}$
$\displaystyle:=\mathbb{E}_{k}[\tilde{\mathcal{R}}_{z}(T_{1}^{k})],$
$\displaystyle\tilde{\Upsilon}$
$\displaystyle:=\mathbb{E}_{k}[\tilde{\mathcal{R}}_{\eta}(T_{2}^{k})^{\operatorname{{}^{*}}}].$

So, $\tilde{\Gamma}_{k}^{1}(z,\bar{\eta})=\sum_{s=0}^{\lceil
n/2\rceil}\tilde{\Lambda}_{ks}\tilde{\Upsilon}_{sk}.$ Since $x_{ij}s$ satisfy
Poincaré inequality, we have

(6.22)
$\displaystyle\operatorname{Var}(\tilde{\Gamma}_{k}^{1}(z,\bar{\eta}))\leq\frac{1}{n\alpha}\sum_{i,j=1}^{k-1}\left[\mathbb{E}\left|\frac{\partial\tilde{\Gamma}_{k}^{1}(z,\bar{\eta})}{\partial
t^{1}_{ij}}\right|^{2}+\mathbb{E}\left|\frac{\partial\tilde{\Gamma}_{k}^{1}(z,\bar{\eta})}{\partial\bar{t^{1}}_{ij}}\right|^{2}\right].$

The sum is divided by $n$ because the matrix $T_{1}$ is scaled by
$1/\sqrt{n}$. Moreover, the sum stops at $k-1$ because
$\tilde{\Gamma}_{k}^{1}(z,\bar{\eta})$ is a constant for
$i,j=k,k+1,\dots,\lceil n/2\rceil$. On the other hand, we have

$\displaystyle\frac{\partial\mathcal{R}_{z}(T_{1}^{k})_{ks}}{\partial
t_{ij}^{1}}$
$\displaystyle=\mathcal{R}_{z}(T_{1}^{k})_{ki}\mathcal{R}_{z}(T_{1}^{k})_{js}\mathbf{1}_{\\{j\neq
k\\}},$
$\displaystyle\frac{\partial\overline{\mathcal{R}_{z}(T_{1}^{k})_{ks}}}{\partial
t_{ij}^{1}}$ $\displaystyle=0,$

As a result,

$\displaystyle\frac{\partial\tilde{\Lambda}_{ks}}{\partial
t^{1}_{ij}}=\tilde{\Lambda}_{ki}\tilde{\Lambda}_{js}\mathbf{1}_{\\{j\neq
k\\}}+\Lambda_{ks}\frac{\partial\omega_{\epsilon_{n},k}}{\partial
t^{1}_{ij}}1_{\left\\{\rho_{1}<\|T_{1}^{k}\|<\rho_{1}+\epsilon_{n}\right\\}},$
$\displaystyle\frac{\partial\tilde{\Upsilon}_{sk}}{\partial
t^{1}_{ij}}=\Upsilon_{sk}\frac{\partial\omega_{\epsilon_{n},k}}{\partial
t^{1}_{ij}}1_{\left\\{\rho_{1}<\|T_{1}^{k}\|<\rho_{1}+\epsilon_{n}\right\\}},$
$\displaystyle\frac{\partial\left(\tilde{\Gamma}_{k}^{1}(z,\bar{\eta})\right)}{\partial
t^{1}_{ij}}=\sum_{s=0}^{\lceil
n/2\rceil}\tilde{\Lambda}_{ki}\tilde{\Lambda}_{js}\tilde{\Upsilon}_{sk}+\sum_{s=0}^{n/2}\left(\Lambda_{ks}\tilde{\Upsilon}_{sk}+\tilde{\Lambda}_{ks}\Upsilon_{sk}\right)\frac{\partial\omega_{\epsilon_{n},k}}{\partial
t^{1}_{ij}}1_{\left\\{\rho_{1}<\|T_{1}^{k}\|<\rho_{1}+\epsilon_{n}\right\\}}.$

Let us denote $\tilde{y}_{jk}=\sum_{s=0}^{\lceil
n/2\rceil}\tilde{\Lambda}_{js}\tilde{\Upsilon}_{sk}$. Then using the facts
that
$\|\tilde{\Lambda}\|,\|\tilde{\Upsilon}\|\leq\left(\tau-\epsilon_{n}\right)^{-1}$,
we have

$\displaystyle\sum_{i,j=1}^{k-1}\left|\sum_{s=0}^{\lceil
n/2\rceil}\tilde{\Lambda}_{ki}\tilde{\Lambda}_{js}\tilde{\Upsilon}_{sk}\right|^{2}=\sum_{i,j=1}^{k-1}\left|\tilde{\Lambda}_{ki}\tilde{y}_{jk}\right|^{2}\leq\left\|\tilde{\Lambda}_{k}\right\|_{2}^{2}\left\|\tilde{y}_{k}\right\|_{2}^{2}\leq\left(\tau-\epsilon_{n}\right)^{-6},$
$\displaystyle\sum_{i,j=1}^{k-1}\left|\sum_{s=0}^{\lceil
n/2\rceil}\left(\Lambda_{ks}\tilde{\Upsilon}_{sk}+\tilde{\Lambda}_{ks}\Upsilon_{sk}\right)\frac{\partial\omega_{\epsilon_{n},k}}{\partial
a_{ij}}\mathbf{1}_{\left\\{\left(\rho_{1}<\|T_{1}^{k}\|<\rho_{1}+\epsilon_{n}\right\\}\right.}\right|^{2}$
$\displaystyle\leq\frac{10}{\epsilon_{n}^{2}}\sum_{i,j=1}^{k-1}\left\|\tilde{\Lambda}_{k}\right\|_{2}^{2}\left\|\tilde{\Upsilon}_{k}\right\|_{2}^{2}1_{\left\\{\rho<\|T_{1}^{k}\|<\rho+\epsilon_{n}\right\\}}\leq
10n^{4}\left(\tau-\epsilon_{n}\right)^{-4}1_{\left\\{\rho_{1}<\|T_{1}^{k}\|<\rho_{1}+\epsilon_{n}\right\\}}.$

Using the above estimates, (6.1), we have

$\displaystyle\operatorname{Var}\left(\frac{1}{n}\sum_{k=1}^{\lceil
n/2\rceil}\tilde{\Gamma}_{k}^{1}(z,\bar{\eta})\right)$
$\displaystyle\leq\frac{1}{n}\sum_{k=1}^{\lceil
n/2\rceil}\operatorname{Var}\left(\tilde{\Gamma}_{k}^{1}(z,\bar{\eta})\right)$
$\displaystyle\leq\frac{2}{\alpha
n\left(\tau-\epsilon_{n}\right)^{6}}+\frac{Kn^{3}}{\left(\tau-\epsilon_{n}\right)^{4}}\exp\left(-\sqrt{\frac{\alpha
n}{2}}\frac{3\rho_{1}}{4}\right).$

Therefore from (6.21), we obtain

$\displaystyle\operatorname{Var}\left(\frac{1}{n}\sum_{k=1}^{\lceil
n/2\rceil}\Gamma_{k}^{1}(z,\bar{\eta})\right)$
$\displaystyle\leq\frac{1}{n}\sum_{k=1}^{\lceil
n/2\rceil}\mathbb{E}\left[\left|\Gamma_{k}^{1}(z,\bar{\eta})-\tilde{\Gamma}_{k}^{1}(z,\bar{\eta})\right|^{2}\right]+\operatorname{Var}\left(\frac{1}{n}\sum_{k=1}^{\lceil
n/2\rceil}\tilde{\Gamma}_{k}^{1}(z,\bar{\eta})\right)$
$\displaystyle\leq\frac{2}{\alpha
n\left(\tau-\epsilon_{n}\right)^{6}}+\frac{Kn^{3}}{\left(\tau-\epsilon_{n}\right)^{4}}\exp\left(-\sqrt{\frac{\alpha
n}{2}}\frac{3\rho_{1}}{4}\right)\rightarrow 0\text{ as }n\rightarrow\infty.$

Similarly, $\operatorname{Var}\left(\frac{1}{n}\sum_{k=1}^{\lceil
n/2\rceil}\Gamma_{k}^{2}(z,\bar{\eta})\right)\rightarrow 0\text{ as
}n\rightarrow\infty.$ Thus,

$\displaystyle\operatorname{Var}(\chi_{k}(z,\bar{\eta}))$ $\displaystyle\leq
2\operatorname{Var}\left(\frac{1}{n}\sum_{k=1}^{\lceil
n/2\rceil}\Gamma_{k}^{1}(z,\bar{\eta})\right)+2\operatorname{Var}\left(\frac{1}{n}\sum_{k=1}^{\lceil
n/2\rceil}\Gamma_{k}^{2}(z,\bar{\eta})\right)\rightarrow 0,\text{ as
}n\rightarrow\infty.$

∎

This completes the proof of the central limit theorem. We now demonstrate the
calculation of the covariance kernel in Section 7. Before that let us verify
this result through numerical simulation; see Figure 3.

Figure 3. Histogram of $\operatorname{L}^{\circ}(P_{d})/\sqrt{n}$ of a
$4000\times 4000$ random centrosymmetric matrix sampled 1000 times. The
entries of the matrix are drawn from the Gaussian distribution, where
$P_{d}(x)=2x^{3}+x^{4}$.

##  7\. Calculation of variance

This section is dedicated to calculating the covariance kernel. We proceed
with the expression of the variance in the case of polynomial test functions.

######  Proposition 7.1.

Let $M$ be the random matrix as defined in Definition 3.1 satisfying the
Condition 3.3. Then we have

$\operatorname{Var}(\operatorname{Tr}(P_{d}(M)))=\sum_{k=1}^{d}2ka_{k}^{2}+o(1),$

where $P_{d}(x)=\sum_{k=1}^{d}a_{k}x^{k}$ is a polynomial of degree $d\geq 1$.

Notice that, the variance expression does not contain the constant term of the
polynomial $P_{d}(x)$. The constant term of a matrix polynomial will be a
constant multiplied by the identity matrix, and this term will not affect the
fluctuation result of the LES of a random matrix, as we center the LES by its
mean.

We recognize that the essential ingredients are
$\mathbb{E}[\operatorname{Tr}(M^{k})\operatorname{Tr}(\bar{M}^{l})]$ and
$\mathbb{E}[\operatorname{Tr}(M^{k})]$, where $\bar{M}$ is obtained by taking
complex conjugates of all the entries of $M$. We compute these quantities
using combinatorial arguments. Before proving the Proposition 7.1, we
introduce and analyze some required notions such as pair partitions and chain
merging in the following subsections.

###  7.1. Analyzing expected product

We evaluate $\mathbb{E}\left[x_{1}x_{2}\ldots x_{n}\right]$, where $x_{i}s$
are random variables from condition (3.3). Note that $\mathbb{E}[x_{i}^{2}]=0$
if $x_{i}$ satisfies condition (3.3).

In the next calculations, we need to evaluate the terms of the forms
$\mathbb{E}[\operatorname{Tr}(M^{k})]$ and
$\mathbb{E}[\operatorname{Tr}(M^{k})\operatorname{Tr}(\bar{M}^{l})]$. Upon
expansion, the term $\operatorname{Tr}(M^{k})$ is represented as
$\sum_{i_{1},i_{2},\ldots,i_{k}=1}^{n}x_{i_{1}i_{2}}x_{i_{1}i_{2}}$
$x_{i_{2}i_{3}}$ $\cdots$ $x_{i_{k}i_{1}}$. Notice that here the random
variables $x_{i_{1}i_{2}}$ $x_{i_{2}i_{3}}$ $\cdots$ $x_{i_{k}i_{1}}$ are
linked via their indices $i_{1},i_{2},\ldots,i_{n}$. The arrangement of these
indices plays an important role in evaluating expectations. We formalize this
using the notation of index chain as defined below.

######  Definition 7.2 (Index chains).

The index chain of the product of random variables $x_{i_{1}i_{2}}$
$x_{i_{2}i_{3}}$ $\cdots$ $x_{i_{k}i_{1}}$ is the unique sequence $i_{1}\to
i_{2}\to\cdots\to i_{k}\to i_{1}$ of indices.

  1. (a)

Single chain: In the calculation of $\mathbb{E}[\operatorname{Tr}(M^{k})]$,
the indices of the product are chained as $i_{1}\to i_{2}\to\cdots\to i_{k}\to
i_{1}$ which will be referred as single chain.

  2. (b)

Double chain: In the calculation of
$\mathbb{E}[\operatorname{Tr}(M^{k})\operatorname{Tr}(\bar{M}^{l})]$, there
are two chains of indices $i_{1}\to i_{2}\to\cdots\to i_{k}\to i_{1}$ and
$j_{1}\to j_{2}\to\cdots\to j_{l}\to j_{1}$ which will be referred as double
chain.

In addition, while calculating $\mathbb{E}[\operatorname{Tr}(M^{k})]$ and
$\mathbb{E}[\operatorname{Tr}(M^{k})\operatorname{Tr}(\bar{M}^{l})]$, we shall
use the terms ‘degree of freedom’ and ‘free variables’, which are defined in
Definition 7.3.

######  Definition 7.3 (Degrees of freedom and free variables).

The degree of freedom indicates the number of independent choices when
selecting indices, such as $i_{1},i_{2},\dots,i_{k}$ and
$j_{1},j_{2},\dots,j_{l}$, from a set $\\{1,2,3,\dots,n\\}$. Moreover,
$i_{1},i_{2},\dots$ are referred to as free variables.

Now, we proceed by calculating $\mathbb{E}[\operatorname{Tr}(M^{k})]$ and
$\mathbb{E}[\operatorname{Tr}(M^{k})\operatorname{Tr}(\bar{M}^{l})]$ in two
different scenarios such as single chain and double chain merging.

###  7.2. Single Chain Expectation

In this subsection, we focus on computing the expected value of the trace of a
matrix raised to the power of $k$, that is, the expectation of a single chain.
Specifically, we are interested in evaluating the expressions of the form.

$\mathbb{E}[\mathrm{Tr}(M^{k})]=\frac{1}{n^{{k/2}}}\sum_{i_{1},i_{2},\dots,i_{k}=1}^{n}\mathbb{E}[x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots
x_{i_{(k-1)}i_{k}}x_{i_{k}i_{1}}].$

We state the first lemma in this regard.

######  Lemma 7.4.

Let $M$ be the random matrix as defined in Definition 3.1 satisfying the
Condition 3.3. Then we have the following

$\mathbb{E}[\operatorname{Tr}(M^{k})]=O(k^{k}n^{-k/6}).$

In particular, if $k\leq n^{\alpha}$ for some $\alpha<1/6$, then
$\mathbb{E}[\operatorname{Tr}(M^{k})]=o(1).$

###### Proof.

This lemma evaluates the limit expectation of a single chain through intra
chain merging, which is defined in Section 7.3. The expectation of the trace
of $M^{k}$ can be expressed as

$\mathbb{E}[\text{Tr}(M^{k})]=\frac{1}{n^{k/2}}\sum_{i_{1},i_{2},\dots,i_{k}=1}^{n}\mathbb{E}[x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots
x_{i_{(k-1)}i_{k}}x_{i_{k}i_{1}}].$

To evaluate $\mathbb{E}[x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots
x_{i_{(k-1)}i_{k}}x_{i_{k}i_{1}}]$, we need to find out the number of ways by
which $x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots x_{i_{(k-1)}i_{k}}x_{i_{k}i_{1}}$
simplifies to a product where each $x_{ij}$ occurs with a power greater than
two. Note that in the product $x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots
x_{i_{(k-1)}i_{k}}x_{i_{k}i_{1}}$, if we make any two random variables equal,
then the expectation vanishes. So, we shall consider the cases where more than
two random variables are equal to each other. Let us consider the case where
we make at least three random variables equal. Although we cannot determine
the exact number of combinations, we can find an upper bound, which is
$n^{k/3}k^{k}$. In this case, we have a total of $k$ random variables
$x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots x_{i_{(k-1)}i_{k}}x_{i_{k}i_{1}}$ where
$x_{ij}$s satisfy Condition 3.3. Therefore, we can have at most $k^{k}$
arrangements by rearranging their positions for fixed indices
$i_{1},i_{2},\dots,i_{k}$; for example, $x_{i_{10}i_{11}}x_{i_{8}i_{9}}\dots
x_{i_{(6)}i_{7}}x_{i_{k}i_{1}}$ or $x_{i_{k}i_{1}}x_{i_{2}i_{3}}\dots
x_{i_{(k-1)}i_{k}}x_{i_{1}i_{2}}$ or $x_{i_{2}i_{3}}x_{i_{1}i_{2}}\dots
x_{i_{(k-1)}i_{k}}x_{i_{k}i_{1}}$ etc. In each specific arrangement, we can
have at most $k/3$ groups where each group is of size at least three. Each
group gives us at most one free variable, which can be explained as follows.

We want to make all three random variables equal within a group of size three.
Let us consider a group $(x_{i_{1}i_{2}}x_{i_{4}i_{5}}x_{i_{7}i_{8}})$ of size
three. To make all the random variables in it equal, we need to impose
constraints on the indices. We have various possibilities as follows,

  1. (1)

$(i_{1},i_{2})=(i_{4},i_{5})=(i_{7},i_{8})$ or

  2. (2)

$(i_{1},i_{2})=(i_{7},i_{8})$ and $(i_{4},i_{5})=((n+1)-i_{1},(n+1)-i_{2})$ or

  3. (3)

$(i_{1},i_{2})=(i_{4},i_{5})$ and $(i_{7},i_{8})=((n+1)-i_{1},(n+1)-i_{2}).$

Notice that in each case, we have only two free variables $i_{1}$ and $i_{2}$.
We must have another group that contains $x_{i_{8}i_{9}}$, a consecutive
element to one of the groups we have already examined, such as
$(x_{i_{8}i_{9}}x_{i_{10}i_{11}}x_{i_{12}i_{13}})$. In that group, we have two
remaining free variables, namely $i_{8}$ and $i_{9}$, once we make all
variables equal. However, $i_{8}$ has already been counted in a previous
group. So, here we are left with only one free variable. Moreover, we may have
a group $x_{i_{13}i_{14}}x_{i_{20}i_{21}}x_{i_{6}i_{7}}$. Here, we have two
free variables; $i_{13}$ and $i_{14}$. However, $i_{14}$ depends on $i_{7}$
because $x_{i_{13}i_{14}}=x_{i_{6}i_{7}}$, and both $i_{13}$ and $i_{7}$ are
already fixed in previous groups. Thus, in this group, we do not have any free
variables. Consequently, we can say that every $3$-sized group has at most one
free variable.

In any arrangement of $k$ random variables, such as
$x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots x_{i_{(k-1)}i_{k}}x_{i_{k}i_{1}}$, we have
at most $k/3$ many groups having size at least three. Therefore, in any
arrangement, we have at most $n^{k/3}$ degrees of freedom. And we can have at
most $k^{k}$ such arrangements. Thus, the total number of combinations when we
make three or more random variables equal in every group is at most
$n^{k/3}k^{k}$.

Therefore,

$\displaystyle\mathbb{E}[\text{Tr}(M^{k})]$
$\displaystyle=\frac{1}{n^{k/2}}\sum_{i_{1},i_{2},\dots,i_{k}=1}^{n}\mathbb{E}[x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots
x_{i_{(k-1)}i_{k}}x_{i_{k}i_{1}}]$
$\displaystyle=O\left(\frac{n^{k/3}k^{k}}{n^{k/2}}\right)$
$\displaystyle=O(k^{k}n^{-k/6}).$

∎

We shall now proceed with the calculation of the covariance between
$\operatorname{Tr}M^{k}$, $\operatorname{Tr}M^{l}.$

###  7.3. Double chain Expectation

In this section, our focus is on computing the expected value of the product
of traces of matrices raised to different powers, which requires the
computation of the expectation of a double chain. Specifically, we are
interested in expressions of the form

$\sum_{\begin{subarray}{c}k,l=1\end{subarray}}^{d}\frac{1}{n^{(k+l)/2}}\sum_{{i_{1},i_{2},\dots
i_{k},j_{1},j_{2},\dots
j_{l}=1}}^{n}\mathbb{E}\left[(x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots
x_{i_{k}i_{1}})(\bar{x}_{j_{1}j_{2}}\bar{x}_{j_{2}j_{3}}\dots\bar{x}_{j_{l}j_{1}})\right].$

Here, $k$ and $l$ are positive integers denoting the lengths of the respective
chains. As we know, trying to make more random variables equal reduces the
degree of freedom. Therefore, we aim to make fewer random variables equal,
preferably $2$, as in this case, we have the term $\mathbb{E}(|x_{ij}|^{2})$,
which is $1$. Thus, when we have a product of two chains such as
$(x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots
x_{i_{k}i_{1}})(\bar{x}_{j_{1}j_{2}}\bar{x}_{j_{2}j_{3}}\dots\bar{x}_{j_{l}j_{1}})$,
we can pair random variables from the first chain
$x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots x_{i_{k}i_{1}}$ with the second chain
$\bar{x}_{j_{1}j_{2}}\bar{x}_{j_{2}j_{3}}\dots\bar{x}_{j_{l}j_{1}}$ in the
following three different ways:

  1. (1)

All the random variables of one chain find their pair within the chain itself
- intra chain merging.

  2. (2)

All random variables in one chain find their pair in the other chain - cross
chain merging.

  3. (3)

Some of the random variables of one chain find their pairs in the chain
itself, and others find in the other chain - partial chain merging.

Let us explain each case as follows.

Case 1 (Intra chain merging): As we have seen in Section 7.2, the single chain
expectation is asymptotically zero. Therefore, intra chain merging gives us
zero asymptotic contribution.

Case 2 (Partial chain merging): If the chain lengths are equal, then we shall
have a cross chain merging, which is explained in the next case. Here the
chain lengths are not equal. Without loss of generality, let us assume that
$k<l$. We can divide the second chain into two parts - one with the length of
the first chain and one with the remaining elements, as follows;
$(x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots x_{i_{k}i_{1}})$
$(\bar{x}_{j_{1}j_{2}}\bar{x}_{j_{2}j_{3}}\dots\bar{x}_{j_{k}j_{(k+1)}})$
$(\bar{x}_{j_{(k+1)}j_{k+2}}\bar{x}_{j_{(k+2)}j_{k+3}}\dots\bar{x}_{j_{(l)}j_{1}})$.
To get the maximum contribution, the first and second chains will have cross
chain merging, and the third chain will have intra chain merging.

From the cross chain merging of the first and second chains, we have $2kn^{k}$
combinations; see case 3. Now, we have one option for the third chain, which
is intra chain merging. We have at most $n^{(l-k-1)/3}(l-k)^{(l-k)}$
combinations from intra chain merging. Note that the random variable $j_{k+1}$
has already been fixed in the previous case where we merged the first and
second chains. Additionally, we chose the second and third chains from a
single chain, and this can be done in $l$ many ways. Therefore, we have at
most $ln^{(l-k-1)/3}(l-k)^{(l-k)}$ combinations. Thus, the total number of
combinations is of order $O(lk(l-k)^{l-k}n^{(l+2k-1)/3})$. However, when
evaluating
$\mathbb{E}[\operatorname{Tr}(M^{k})\operatorname{Tr}(\bar{M}^{l})]$, the
expression $\mathbb{E}\left[(x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots
x_{i_{k}i_{1}})\right.$
$\left.(\bar{x}_{j_{1}j_{2}}\bar{x}_{j_{2}j_{3}}\dots\bar{x}_{j_{l}j_{1}})\right]$
is multiplied by $(1/n)^{(k+l)/2}$. Therefore, from this case we have a
contribution of $O((l-k)^{l-k}n^{-(l-k)/6})$.

Case 3 (Cross chain merging): The only nonzero asymptotic contribution we get
is from cross chain merging. Let us consider a simple example to evaluate
$\mathbb{E}[\operatorname{Tr}(M^{k})\operatorname{Tr}(\bar{M}^{l})]$ with
$k=l=5$. Thus,

$\mathbb{E}[\operatorname{Tr}(M^{5})\operatorname{Tr}(\bar{M}^{5})]=\frac{1}{n^{5}}\sum_{\begin{subarray}{c}i_{1},i_{2},\dots
i_{k},\\\ j_{1},j_{2},\dots
j_{5}=1\end{subarray}}^{n}\mathbb{E}\left[(x_{i_{1}i_{2}}x_{i_{2}i_{3}}x_{i_{3}i_{4}}x_{i_{4}i_{5}}x_{i_{5}i_{1}})(\bar{x}_{j_{1}j_{2}}\bar{x}_{j_{2}j_{3}}\bar{x}_{j_{3}j_{4}}\bar{x}_{j_{4}j_{5}}\bar{x}_{j_{5}j_{1}})\right].$

As we know, intra chain and partial mergings are not possible. The only
possibility is a cross chain merging. In cross chain merging, we get maximum
contribution if we make any two random variable equal, one from first chain
and other from second chain.

One way of cross chain merging is illustrated in Figure 4, and it gives us a
nonzero asymptotic contribution.

$x_{i_{1}i_{2}}$$x_{i_{2}i_{3}}$$x_{i_{3}i_{4}}$$x_{i_{4}i_{5}}$$x_{i_{5}i_{1}}$$\bar{x}_{j_{1}j_{2}}$$\bar{x}_{j_{2}j_{3}}$$\bar{x}_{j_{3}j_{4}}$$\bar{x}_{j_{4}j_{5}}$$\bar{x}_{j_{5}j_{1}}$
Figure 4. Crossing-case

For any pair of random variables $(x_{ij},\bar{x}_{kl})$, where $x_{ij}$ and
$\bar{x}_{kl}$ are from a chain $x_{i_{1}i_{2}}x_{i_{2}i_{3}}x_{i_{3}i_{4}}$
$x_{i_{4}i_{5}}x_{i_{5}i_{1}}$, and
$\bar{x}_{j_{1}j_{2}}\bar{x}_{j_{2}j_{3}}\bar{x}_{j_{3}j_{4}}\bar{x}_{j_{4}j_{5}}\bar{x}_{j_{5}j_{1}}$
respectively, $\mathbb{E}[x_{ij}\bar{x}_{kl}]\neq 0$ if and only if
$(i,j)=(k,l)$ or $(i,j)=(n+1-k,n+1-l)$. This leads to the following possible
pairings in the above example.

$\begin{array}[]{l}\left\\{\begin{array}[]{ l }{j_{1}=n+1-i_{1}}\\\
{j_{2}=n+1-i_{2}}\end{array}\quad\text{ or
}\quad\left\\{\begin{array}[]{l}j_{1}=i_{1}\\\
j_{2}=i_{2}\end{array}\right.\right.\\\ \\\ \left\\{\begin{array}[]{ l
}{j_{2}=n+1-i_{2}}\\\ {j_{3}=n+1-i_{3}}\end{array}\quad\text{ or
}\quad\left\\{\begin{array}[]{l}j_{2}=i_{2}\\\
j_{3}=i_{3}\end{array}\right.\right.\\\ \\\ \left\\{\begin{array}[]{ l
}{j_{3}=n+1-i_{3}}\\\ {j_{4}=n+1-i_{4}}\end{array}\quad\text{ or
}\quad\left\\{\begin{array}[]{l}j_{3}=i_{3}\\\
j_{4}=i_{4}\end{array}\right.\right.\\\ \\\ \left\\{\begin{array}[]{ l
}{j_{4}=n+1-i_{4}}\\\ {j_{5}=n+1-i_{5}}\end{array}\quad\text{ or
}\quad\left\\{\begin{array}[]{l}j_{4}=i_{4}\\\
j_{5}=i_{5}\end{array}\right.\right.\\\ \\\ \left\\{\begin{array}[]{ l
}{j_{5}=n+1-i_{5}}\\\ {j_{1}=n+1-i_{1}}\end{array}\quad\text{ or
}\quad\left\\{\begin{array}[]{l}j_{5}=i_{5}\\\
j_{1}=i_{1}\end{array}\right.\right.\\\ \end{array}$

If we start with $j_{1}=n+1-i_{1}$, then the chain has to follow the
constraints listed in the left column; not the ones which are listed on the
right column. Similarly, if we start with the condition $j_{1}=i_{1}$, then we
have to follow the constraints that are listed on the right column. Mixing the
conditions listed in the left and right column will reduce the number of free
variables, leading to an asymptotically zero contribution, as the full term is
divided by $n^{5}$.

For example, if we take $(i_{1},i_{2})=(n+1-j_{1},n+1-j_{2})$ and decide to
take $(i_{2},i_{3})=(j_{2},j_{3})$, then to keep the continuity of the chain,
we must have $j_{2}=i_{2}=n+1-j_{2}$ which enforces one extra constraint
$j_{2}=n+1-j_{2}$. Thus, we have lost a degree of freedom there. Now, if the
indices follow the constraints listed on the first column, then there are
$n^{5}$ many choices for choosing all the indices. For each such choice, the
chain will reduce to a product of the terms of the form
$\mathbb{E}[|x_{ij}|^{2}](=1).$

Thus, we conclude that when we do the cross chain merging for the first and
the second chain, namely $i_{1}\rightarrow i_{2}\rightarrow i_{3}\rightarrow
i_{4}\rightarrow i_{5}\to i_{1}$ and $j_{1}\rightarrow j_{2}\rightarrow
j_{3}\rightarrow j_{4}\rightarrow j_{5}\to j_{1}$, one way for cross chain
merging is to copy the first chain in place of the second chain, resulting in:
$i_{1}\rightarrow i_{2}\rightarrow i_{3}\rightarrow i_{4}\rightarrow i_{5}\to
i_{1}\rightarrow i_{2}\rightarrow i_{3}\rightarrow i_{4}\rightarrow i_{5}\to
i_{1}$. This represents one possibility with $5$ free variables. Additionally,
we can start the second chain with $i_{2}$ or $i_{3}$ or $\dots$ or $i_{5}$
like $i_{2}\rightarrow i_{3}\rightarrow\dots\rightarrow i_{5}\rightarrow
i_{1}\rightarrow i_{2}$, resulting in $4$ more choices. Therefore, we have
$2\times 5n^{5}$ combinations for cross chain merging of the first and second
chain so far. Moreover, since $x_{(i,j)}=x_{((n+1-i),(n+1-j))}$ , we can
perform a cross chain merging by changing the second chain to
$(n+1-i_{1})\rightarrow(n+1-i_{2})\rightarrow\dots\rightarrow(n+1-i_{5})\rightarrow(n+1-i_{1})$,
representing another one possibility with $5$ free variables. Additionally, we
have $4$ more choices by rotating the second sequence, similar to the previous
case. Therefore, the total number of combinations for cross chain merging of
the first and second chain is $2\times 5n^{5}$ from type of merging in Figure
4.

So far, we have observed that if $x_{i_{t}i_{l}}$ finds its pair
$\bar{x}_{j_{p}j_{q}}$, then the pairing for $x_{i_{t+1}i_{l+1}}$ is
$\bar{x}_{j_{p+1}j_{q+1}}$ for all $t$ and $l$ such that $1\leq t,l\leq 5$.
Now, let us consider the another type of case as demonstrated in Figure 5.

$x_{i_{1}i_{2}}$$x_{i_{2}i_{3}}$$x_{i_{3}i_{4}}$$x_{i_{4}i_{5}}$$x_{i_{5}i_{1}}$$\bar{x}_{j_{1}j_{2}}$$\bar{x}_{j_{2}j_{3}}$$\bar{x}_{j_{3}j_{4}}$$\bar{x}_{j_{4}j_{5}}$$\bar{x}_{j_{5}j_{1}}$
Figure 5. Crossing-case

For this case, we need to impose more constraints that will reduce the degree
of freedom and leads to zero asymptotic contribution. Here,
$x_{i_{1}i_{2}}=\bar{x}_{j_{3}j_{4}}$ if and only if
$(i_{1},i_{2})=(j_{3},j_{4})$ or $(i_{2},i_{3})=((n+1)-j_{3},(n+1)-j_{4})$.

If $j_{3}=i_{1}$, this implies $j_{4}=i_{2}$. Furthermore, since we are
pairing $x_{i_{2}i_{3}}$ and $\bar{x}_{j_{5}j_{1}}$, we shall have
$j_{5}=i_{2}$ and $i_{3}=j_{1}$, or the dual case. Now, look at the
consecutive random variables of those already paired up. We observe that to
maintain the continuity of the chain, we must impose restrictions that will
significantly reduce the number of free variables and lead to zero asymptotic
contribution.

Therefore, the total number of combinations that provide a nonzero asymptotic
contribution is $2\times 5n^{5}$, and we only obtain that from cases similar
to Figure 4. Consequently,

$\lim_{{n\to\infty}}\mathbb{E}[\operatorname{Tr}(M^{5})\operatorname{Tr}(\bar{M}^{5})]=\lim_{{n\to\infty}}\frac{2\times
5n^{5}+o(n^{5})}{n^{5}}+o(1)=2\times 5+o(1).$

Now, we generalize the above result in the following lemma.

######  Lemma 7.5.

Let $M$ the random matrix as defined in Definition 3.1 satisfying the
Condition 3.3. Then we have

$\displaystyle\mathbb{E}[\operatorname{Tr}(M^{k})\operatorname{Tr}(\bar{M}^{k})]$
$\displaystyle=2k+o(1).$

###### Proof.

The lemma estimates the limiting expectation of cross chain merging. Consider

$\displaystyle\mathbb{E}[\operatorname{Tr}(M^{k})\operatorname{Tr}(\bar{M}^{k})]$
$\displaystyle=\frac{1}{n^{k}}\mathbb{E}\left[\sum_{{i_{1},i_{2},\dots
i_{k}\atop j_{1},j_{2},\dots j_{k}=1}}^{n}(x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots
x_{i_{k}i_{1}})(\bar{x}_{j_{1}j_{2}}\bar{x}_{j_{2}j_{3}}\dots\bar{x}_{j_{k}j_{1}})\right]$
$\displaystyle=\frac{1}{n^{k}}\sum_{{i_{1},i_{2},\dots i_{k}\atop
j_{1},j_{2},\dots
j_{k}=1}}^{n}\mathbb{E}\left[(x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots
x_{i_{k}i_{1}})(\bar{x}_{j_{1}j_{2}}\bar{x}_{j_{2}j_{3}}\dots\bar{x}_{j_{k}j_{1}})\right].$

As explained in the Case 3 of Section 7.3 (page 7.3), the contribution will
come from two possible cases, such as (a) the chain $i_{1}\rightarrow
i_{2}\rightarrow\dots\rightarrow i_{k}\to i_{1}$ overlaps completely with the
chain $j_{1}\rightarrow j_{2}\rightarrow\dots\rightarrow j_{k}\to j_{1}$, or
(b) it overlaps with the dual version of the second chain, which is
$(n+1-j_{1})\rightarrow(n+1-j_{2})\rightarrow\dots\rightarrow(n+1-j_{k})\rightarrow(n+1-j_{1}).$
In either cases, we have $k$ many possible ways to start the second chain.
Once the chains overlap, it reduces to the product of terms of the form
$\mathbb{E}[|x_{ij}|^{2}]$ $(=1)$. Therefore, we have

$\displaystyle\mathbb{E}[\operatorname{Tr}(M^{k})\operatorname{Tr}(\bar{M}^{k})]$
$\displaystyle=\frac{1}{n^{k}}\sum_{\begin{subarray}{c}i_{1},i_{2},\dots
i_{k},\\\ j_{1},j_{2},\dots
j_{k}=1\end{subarray}}^{n}\mathbb{E}\left[(x_{i_{1}i_{2}}x_{i_{2}i_{3}}\dots
x_{i_{k}i_{1}})(\bar{x}_{j_{1}j_{2}}\bar{x}_{j_{2}j_{3}}\dots\bar{x}_{j_{k}j_{1}})\right]$
$\displaystyle=\frac{2kn^{k}}{n^{k}}+o(1)$ $\displaystyle=2k+o(1).$

∎

We summarise all the above cases as follows.

######  Remark 7.6.

While calculating
$\mathbb{E}[\operatorname{Tr}(M^{k})\operatorname{Tr}(\bar{M}^{l})]$, the
significant contribution comes from the Case 3 only. From Case 1 and Case 2
(page 7.3), we conclude that if $k\neq l$, though intra chain merging and
partial chain merging are possible strategies, asymptotically they provide
zero contributions. In particular,
$\mathbb{E}[\operatorname{Tr}(M^{k})\operatorname{Tr}(\bar{M}^{l})]=O((l-k)^{l-k}n^{-(l-k)/6})$

Now, we are ready to summarize the above results into the proof of Proposition
7.1.

###### Proof of Proposition 7.1.

We may write the variance as

$\displaystyle\operatorname{Var}(\operatorname{Tr}(P_{d}(M)))$
$\displaystyle=\operatorname{Var}\left(\sum_{k=1}^{d}a_{k}\operatorname{Tr}(M^{k})\right)$
$\displaystyle=\sum_{k,l=1}^{d}a_{k}a_{l}\operatorname{Cov}\left(\operatorname{Tr}(M^{k}),\operatorname{Tr}(M^{l})\right)$
$\displaystyle=\sum_{k,l=1}^{d}a_{k}a_{l}\left(\mathbb{E}\left[\operatorname{Tr}(M^{k})\operatorname{Tr}(\bar{M}^{l})\right]-\mathbb{E}\left[\operatorname{Tr}(M^{k})\right]\mathbb{E}\left[\operatorname{Tr}(\bar{M}^{l})\right]\right)$
(7.1) $\displaystyle=\sum_{\begin{subarray}{c}k\neq l\\\
k,l=1\end{subarray}}^{d}0+\sum_{\begin{subarray}{c}k=l\\\
k,l=1\end{subarray}}^{d}2ka_{k}^{2}+o(1)$
$\displaystyle=\sum_{k=1}^{d}2ka_{k}^{2}+o(1).$

The equality in (7.3) follows from the Lemmas 7.4, 7.5, and Remark 7.6. ∎

###  7.4. Covariance kernel of $\operatorname{Tr}\hat{\mathcal{R}}_{z}(M)$

In this subsection, we find the covariance kernel of the Gaussian process as
mentioned in Proposition 6.1. As defined at the beginning of Section 6, let
$\rho=\max\\{\rho_{1},\rho_{2}\\}$. Now, on the event $\Omega_{n}$, we can
expand $\operatorname{Tr}\hat{\mathcal{R}}_{z}(M)$ on the boundary
$\partial\mathbb{D}_{\rho+\tau}$ as follows

$\displaystyle\operatorname{Tr}\hat{\mathcal{R}}_{z}(M)=\frac{n}{z}+\sum_{k=1}^{\lfloor\log^{2}n\rfloor-1}z^{-k-1}\operatorname{Tr}M^{k}+\frac{1}{z^{\lfloor\log^{2}n\rfloor+1}}\operatorname{Tr}[\hat{\mathcal{R}}_{z}(M)M^{\lfloor\log^{2}n\rfloor}].$

We see that on the event $\Omega_{n}$, the last term is bounded by
$\frac{n}{\tau}(\rho/|z|)^{\lfloor\log^{2}n\rfloor}=o(1).$ Now using Remark
7.6, we have

$\displaystyle\mathbb{E}[\operatorname{Tr}M^{k}\operatorname{Tr}\bar{M}^{l}]=O(|k-l|^{|k-l|}n^{-|k-l|/6}).$

Furthermore, since $(xn^{-1/6})^{x}$ is a decreasing function of $x$ for
$1\leq x\leq n^{1/6}/e$, we can estimate

$\displaystyle\sum_{k\neq
l}^{\lfloor\log^{2}n\rfloor}|k-l|^{|k-l|}n^{-|k-l|/6}$ $\displaystyle\leq
2\log^{2}n\sum_{m=1}^{\lfloor\log^{2}n\rfloor}m^{m}n^{-m/6}$
$\displaystyle\leq\frac{2\log^{4}n}{n^{1/6}}=o(1).$

As a result, $\sum_{k\neq
l=1}^{\lfloor\log^{2}n\rfloor}z^{(-k-1)}\bar{z}^{(-l-1)}\mathbb{E}[\operatorname{Tr}M^{k}\operatorname{Tr}\bar{M}^{l}]=o(1).$
Therefore using Lemma 7.5, we obtain

$\displaystyle\mathbb{E}[\operatorname{Tr}\hat{\mathcal{R}}_{z}^{\circ}(M)\operatorname{Tr}\hat{\mathcal{R}}_{\bar{\eta}}^{\circ}(\bar{M})]$
$\displaystyle=\left\\{\mathbb{E}[\operatorname{Tr}\hat{\mathcal{R}}_{z}(M)\operatorname{Tr}\hat{\mathcal{R}}_{\bar{\eta}}(\bar{M})]-\frac{n^{2}}{z\bar{\eta}}\right\\}+o(1)$
$\displaystyle=\sum_{k=1}^{\lfloor\log^{2}n\rfloor}(z\bar{\eta})^{-k-1}\mathbb{E}[M^{k}\bar{M}^{k}]+o(1)$
$\displaystyle=2\sum_{k=1}^{\lfloor\log^{2}n\rfloor}k(z\bar{\eta})^{-k-1}+o(1)$
$\displaystyle=2\left(\frac{1}{z\bar{\eta}}\right)^{2}\left(1-\frac{1}{z\bar{\eta}}\right)^{-2}$
$\displaystyle=2\left(1-z\bar{\eta}\right)^{-2}.$

This completes the proof of the Proposition 6.1.

## Appendix A

######  Lemma A.1.

[17, Lemma A.3] Let $M$ be a random band matrix with bandwidth $b_{n}$ and
variance profile $\omega$. Assume that the entries are i.i.d. having mean zero
and variance one, and satisfy the Poincaré inequality with constant $\alpha.$
Then there exists $\rho\geq 1$ such that

$\mathbb{P}(\|M\|>\rho/4+t)\leq
K\exp\left(-\sqrt{\frac{\alpha(2b_{n}+1)}{2\omega}}t\right),\;\;\;\;\forall\;\;t>0.$

In our case, we have a matrix with i.i.d. entries whose entries satisfy the
Poincaré inequality. Therefore, using the above lemma we can have the
following.

######  Lemma A.2.

Let $T_{m\times m}$ be a random matrix with i.i.d. entries having mean zero
and variance one. Moreover, assume that the entries satisfy the Poincaré
inequality with constant $\alpha$. Then there exists $\rho\geq 1$ such that

$\mathbb{P}(\|T/\sqrt{m}\|>\rho/4+t)\leq K\exp\left(-\sqrt{\frac{\alpha
m}{2}}t\right),\;\;\;\;\forall\;\;t>0,$

where $K>0$ is a universal constant.

We note down the essential ingredient of this article, which martingale CLT as
follows.

######  Lemma A.3.

[7, Theorem 35.12] Let $\left\\{\psi_{n,k}\right\\}_{1\leq k\leq n}$ be a
martingale difference sequence with respect to a filtration
$\\{\mathcal{F}_{k,n}\\}_{1\leq k\leq n}$. Suppose for any $\gamma>0$,
(i)
$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\mathbb{E}[\psi_{n,k}^{2},\mathbf{1}_{|\psi_{n,k}|>\gamma}]=0$.
(ii)
$\sum_{k=1}^{n}\mathbb{E}[\psi_{n,k}^{2}\left|\mathcal{F}_{n,k-1}\right]\stackrel{{\scriptstyle
p}}{{\rightarrow}}\sigma^{2}$ as $n\rightarrow\infty$.

Then $\sum_{k=1}^{n}\psi_{n,k}\stackrel{{\scriptstyle
d}}{{\rightarrow}}\mathcal{N}_{1}\left(0,\sigma^{2}\right)$.

######  Lemma A.4 (Sherman-Morrison formula, [31]).

Let $M$ and $M+ve_{k}^{t}$ be two invertible matrices, where
$v\in\mathbb{C}^{n}$. Then

$\displaystyle(M+ve_{k}^{t})^{-1}v=\frac{M^{-1}v}{1+e_{k}^{t}M^{-1}v}.$

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