#  Repeated erfc statistics for deformed GinUEs

Dang-Zheng Liu111CAS Key Laboratory of Wu Wen-Tsun Mathematics, School of
Mathematical Sciences, University of Science and Technology of China, Hefei
230026, P.R. China. E-mail<EMAIL_ADDRESS>and Lu Zhang222School of
Mathematical Sciences, University of Science and Technology of China, Hefei
230026, P.R. China. E-mail<EMAIL_ADDRESS>
###### Abstract

For an additive perturbation of the complex Ginibre ensemble under a
deterministic matrix $X_{0}$, under certain assumption on $X_{0}$, we observe
that there are only two kinds of local statistical patterns at the spectral
edge: GinUE statistics and critical statistics, which corresponds to regular
or quadratic vanishing spectral points. As a continuation of our previous
study on critical statistics [25], in this paper we establish the local
statistics of GinUE type at the regular spectral edge, which is characterized
by a repeated erfc integral found in [24].

##  1 Introduction and main results

###  1.1 Introduction

We begin our study by specifying the deformed complex Ginibre ensembles (GinUE
for short)

$X:=X_{0}+\sqrt{\frac{\tau}{N}}G,$ (1.1)

where $X_{0}$ is a deterministic matrix, and $G=[g_{ij}]_{i,j=1}^{N}$ is a
random matrix whose entries are i.i.d. complex normal variables with mean 0
and variance 1. Or, to be more exact, we have

######  Definition 1.1.

A random complex $N\times N$ matrix $X$, is said to belong to the deformed
complex Ginibre ensemble with mean $X_{0}={\rm
diag}\left(A_{0},0_{N-r}\right)$ and time $\tau>0$, where $A_{0}$ is a
$r\times r$ constant matrix, denoted by GinUE${}_{N}(A_{0})$, if the joint
probability density function for matrix entries is given by

$P_{N}(A_{0};X)=\Big{(}\frac{N}{\pi\tau}\Big{)}^{N^{2}}\
e^{-\frac{N}{\tau}{\rm Tr}(X-X_{0})(X-X_{0})^{*}}.$ (1.2)

One motivation behind the study of the deformed model (1.1) comes mostly from
the general effort toward the understanding of the effect of a perturbation on
the spectrum of a large-dimensional random matrix; see [7] and [24] for a more
detailed explanation. This is a continuation of our previous paper [25], where
a kind of critical statistics has been observed at quadratic vanishing
spectral edge points, and now focuses on local statistics at the regular
spectral edge points.

In the non-perturbative case, the study of non-Hermitian random matrices dates
back to the work of Ginibre [15] for i.i.d. real/complex Gaussian random
matrices, and then was extended to i.i.d. case. At a macroscopic level, the
limit spectral measure was the famous circular law after a long list of works
[2, 16, 17, 30, 33]; see an excellent survey [9] and references therein.
However, at a microscopic level, local eigenvalue statistics in the bulk and
at the soft edge of the spectrum were revealed first for the Ginibre ensembles
[10, 14, 23], with the help of exact eigenvalue correlations. Later in [32],
Tao and Vu established a four moment match theorem in the bulk and at the soft
edge, both in the real and complex cases. Recently, the four moment matching
condition has been removed, so the edge and bulk universality have been proved
for any random matrix with i.i.d. entries, respectively by Cipolloni, Erdős
and Schrőder [11] and by Maltsev and Osman [26, 29].

For the deformed model (1.1), the circular law still holds under a small low
rank perturbation from $X_{0}$. see e.g. [33, Corollary 1.12]. Particularly
when the perturbation strength of $X_{0}$ is above a certain threshold
(supercritical regime), such a perturbation can create outliers outside the
unit disk [31]. Since then, this outlier phenomenon has been extensively
studied in [7, 4, 6, 12, 28]. Furthermore, the fluctuations of outlier
eigenvalues are investigated respectively by Benaych-Georges and Rochet for
deformed invariantly matrix ensembles [6], and by Bordenave and Capitaine for
deformed i.i.d. random matrices [7]. These fluctuations, due to non-Hermitian
structure, become much more complicated, and indeed highly depend on the shape
of the Jordan canonical form of the perturbation. Meanwhile, when the
perturbation strength is near or below a certain threshold (subcritical and
critical regimes), the finite-rank perturbation effect has been investigated
in our previous paper [24]and it has been proved that the edge statistics
forms a determinantal point processes with correlation kernels characterized
by repeated integrals of the erfc function, depending only on geometric
multiplicity of eigenvalue. Together, all these results establish a non-
Hermitian analogue of the BBP phase transition for the largest eigenvalue of
spiked random matrices [3].

Our main goal is to identify all the possible local eigenvalue statistics at
the edge and in the bulk of the spectrum, for the deformed model
GinUE${}_{N}(A_{0})$ defined in 1.1 under certain restriction (1.3) on
$A_{0}$. Related to this is the limit spectral measure of the deformed model.
Its characterization can be given by using tools from free probability, the
Brown measure of free circular Brownian motion $x_{0}+\sqrt{\tau}c$, where $c$
is a a circular variable, freely independent of a general operator $x_{0}$.
The Brown measure and corresponding density formula has been extensively
studied in [5, 8, 13, 21, 34]. The exact density will play a crucial role in
the study of local eigenvalue statistics for non-Hermitian random matrices.
Recently, [13] Erdós and Ji further observe a remarkable phenomenon: “The
density of the Brown measure has one of the following two types of behavior
around each point on the boundary of its support-either (i) sharp cut, or (ii)
quadratic decay at certain critical points on the boundary”. This dichotomy
can also be observed in the deformed model (1.1) when $X_{0}$ is normal under
certain restriction on $A_{0}$. In the previous paper [25] we have proved that
the critical edge point belongs to the quadratic edge and in this paper we are
devoted to the regular (non-critical) edge point belongs to the sharp edge.

###  1.2 Main results

Throughout the present paper, we select $A_{0}$ as a normal matrix with the
following form

$A_{0}={\rm
diag}\left(a_{1}\mathbb{I}_{r_{1}},\cdots,a_{t}\mathbb{I}_{r_{t}},z_{0}\mathbb{I}_{r_{0}},A_{t+1}\right),$
(1.3)

where $t$ is a given non-negative integer, $z_{0}$ is a spectral parameter and
$A_{t+1}$ is a $r_{t+1}\times r_{t+1}$ normal matrix with $z_{0}$ not an
eigenvalue. Put $R_{0}=N-r$ where $r=\sum_{\alpha=0}^{t+1}r_{\alpha}$, we
assume that $r_{0}$, $r_{t+1}$ and $R_{0}$ are finite, independent of $N$, and
as $N\to\infty$

$r_{\alpha}=c_{\alpha}N+R_{\alpha,N},\quad
R_{\alpha,N}=O(1),\quad\alpha=1,\cdots,t.$ (1.4)

Obviously,

$\sum_{\alpha=1}^{t}c_{\alpha}=1.$ (1.5)

For given complex numbers $a_{1},\cdots,a_{t}$, introduce a probability
measure on the complex plane

$d\nu(z)=\sum_{\alpha=1}^{t}c_{\alpha}\delta(z-a_{\alpha}),$ (1.6)

and several relevant expectations

$P_{00}(z_{0}):=\int\frac{1}{|z-z_{0}|^{2}}d\nu(z),\quad
P_{0}:=P_{0}(z_{0})=\int\frac{z-z_{0}}{|z-z_{0}|^{4}}d\nu(z),$ (1.7)

and

$P_{1}:=P_{1}(z_{0})=\int\frac{1}{|z-z_{0}|^{4}}d\nu(z).$ (1.8)

These quantities are crucial to characterize exact functional form of the
limit spectral measure and to distinguish spectral properties. For instances,
the support of limit spectral measure $\mu_{\infty}$ is

$\mathrm{Supp}(\mu_{\infty}):=\Big{\\{}z_{0}\in\mathbb{C}:P_{00}(z_{0})\geq\frac{1}{\tau}\Big{\\}};$
(1.9)

with the boundary curve characterized by the equation $P_{00}(z_{0})==1/\tau$;
see e.g. [7, Proposition 1.2]. On the other hand, any point on the boundary
curve that is critical (quadratic) or not (regular) edge the boundary points
is completely characterized by the relation $P_{0}(z_{0})\neq 0$ or not.

We also need to define a function, depending on a real parameter $n>-1$, by

$\mathrm{IE}_{n}(z)=\frac{1}{\sqrt{2\pi}\Gamma(n+1)}\int_{0}^{\infty}v^{n}e^{-\frac{1}{2}(v+z)^{2}}{\rm
d}v.$ (1.10)

while as a limit of $n>-1$ from above

$\mathrm{IE}_{-1}(z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^{2}}.$

At the regular edge point of the spectrum, a kernel emerges and has been
introduced in [24] via

$K_{n}\left(z,w\right)=\sqrt{\frac{2}{\pi}}\Gamma(n+1)e^{\frac{1}{2}(z+\overline{w})^{2}}\sqrt{\mathrm{IE}_{n-1}(-z-\overline{z})\mathrm{IE}_{n-1}(-w-\overline{w})}\,\mathrm{IE}_{n}\big{(}{z}+\overline{w}\big{)}.$
(1.11)

In the special case of $n=0$, $\mathrm{IE}_{0}(z)$ reduces to the erfc
function and the above kernel is none other than the well-known GinUE edge
kernel.

Now, we are ready to formulate the main result concerning edge limits of the
$n$-point correlation functions of eigenvalues, denoted by
$R_{N}^{(n)}(A_{0};z_{1},\ldots,z_{n})$ as shown in [24], for the
${\mathrm{GinUE}}_{N}(A_{0})$ ensemble.

######  Theorem 1.2.

For the ${\mathrm{GinUE}}_{N}(A_{0})$ ensemble with $R_{0}\geq n$ and the
assumptions (1.3) on $A_{0}$, if $z_{0}$ is a regular edge point that
satisfies

$P_{00}(z_{0})=1/\tau\quad\mathrm{and}\quad P_{0}(z_{0})\neq 0,$ (1.12)

then as $N\to\infty$ the scaled correlation functions

$\Big{(}\frac{\sqrt{P_{1}}}{|P_{0}|\sqrt{N}}\Big{)}^{2n}R_{N}^{(n)}\Big{(}A_{0};z_{0}-\frac{\sqrt{P_{1}}\hat{z}_{1}}{\overline{P_{0}}\sqrt{N}},\cdots,z_{0}-\frac{\sqrt{P_{1}}\hat{z}_{n}}{\overline{P_{0}}\sqrt{N}}\Big{)}=\det\left(\big{[}K_{\widetilde{r}_{0}}\left(\hat{z}_{i},\hat{z}_{j}\big{)}\right]_{i,j=1}^{n}\right)+O\big{(}N^{-\frac{1}{4}}\big{)},$
(1.13)

hold uniformly for all $\hat{z}_{1},\ldots,\hat{z}_{n}$ in a compact subset of
$\mathbb{C}$, where $\widetilde{r}_{0}=r_{0}+R_{0}$ if $z_{0}=0,$ and
$\widetilde{r}_{0}=r_{0}$ otherwise.

######  Remark 1.3.

The microscopic limit is completely governed by the GinUE edge statistics in
the absence of the finite-rank perturbation effect, so the combination of the
critical statistics proved in [25] probably shows that there are only two
types of statistics at the edge of the spectrum. Besides, although we assume
that the limit spectral measure $\nu$ of the normal matrix $A_{0}$ takes a
special form as in (1.6), it is expected that the edge universality holds for
the general $\nu$.

Structure of this paper. Section 2 is devoted to the proof of Theorem 1.2:
$z_{0}\not=0$, and Section 3 is for the postponed proofs of two propositions
and also for the proof of Theorem 1.2: $z_{0}=0$.

##  2 Proof of Theorem 1.2

###  2.1 Notation and matrix variables

By convention, the conjugate, transpose, and conjugate transpose of a complex
matrix $A$ are denoted by $\overline{A}$, $A^{t}$ and $A^{*}$, respectively.
Besides, the following symbols may be used in this and subsequent sections.

  * •

Tensor The tensor product, or Kronecker product, of an $m\times n$ matrix
$A=[a_{i,j}]$ and a $p\times q$ matrix $B$ is a block matrix

$A\otimes B=\left[\begin{smallmatrix}a_{11}B&\cdots&a_{1n}B\\\
\vdots&\ddots&\vdots\\\ a_{m1}B&\cdots&a_{mn}B\end{smallmatrix}\right].$

When $m=n$ and $p=q$, $A\otimes B$ is similar to $B\otimes A$ by a permutation
matrix.

  * •

Norm For a complex matrix $M$, define the Hilbert-Schmidt norm as
$\|M\|:=\sqrt{{\rm{Tr}}(MM^{*})}$. $M=O(A_{N})$ for some (matrix) sequence
$A_{N}$ means that each element has the same order as that of $A_{N}$.

  * •

Constants $\gamma_{N}=N/(N-n)$, and $f_{\alpha}=|z_{0}-a_{\alpha}|^{2}$
($\alpha=1,\cdots,t$).

In view of (1.3), we need to introduce several matrix variables $Y$, $Q_{0}$,
$Q_{t+1}$ and upper triangular $T_{\alpha}$ with non-negative diagonal
elements for $\alpha=1,\cdots,t$, which are of sizes $n\times n$, $n\times
r_{0}$, $n\times r_{t+1}$ and $n\times n$ respectively. Introduce scaled
variables

$Z=z_{0}I_{n}+N^{-\frac{1}{2}}\hat{Z},\quad
Z=\mathrm{diag}(z_{1},\ldots,z_{n}),\
\hat{Z}=\mathrm{diag}(\hat{z}_{1},\ldots,\hat{z}_{n}).$ (2.1)

Meanwhile, for $\alpha=1,\cdots,t$, setting block matrices as follows

$A_{\alpha}=\begin{bmatrix}\sqrt{\gamma_{N}}(Z-a_{\alpha}\mathbb{I}_{n})&-Y^{*}\\\
Y&\sqrt{\gamma_{N}}\big{(}Z^{*}-\overline{a}_{\alpha}\mathbb{I}_{n}\big{)}\end{bmatrix},$
(2.2) $\widehat{L}_{1}={\rm
diag}\big{(}A_{1}\otimes\mathbb{I}_{n},\cdots,A_{t}\otimes\mathbb{I}_{n},\widehat{B}_{t+1}\big{)},$
(2.3)
$\widehat{B}_{t+1}=\begin{bmatrix}\sqrt{\gamma_{N}}\big{(}Z\otimes\mathbb{I}_{\widetilde{r}_{t+1}}-\mathbb{I}_{n}\otimes\widetilde{A}_{t+1}\big{)}&-Y^{*}\otimes\mathbb{I}_{\widetilde{r}_{t+1}}\\\
Y\otimes\mathbb{I}_{\widetilde{r}_{t+1}}&\sqrt{\gamma_{N}}\big{(}Z^{*}\otimes\mathbb{I}_{\widetilde{r}_{t+1}}-\mathbb{I}_{n}\otimes\widetilde{A}_{t+1}^{*}\big{)}\end{bmatrix},$
(2.4)
$\widehat{L}_{2}=\left[\begin{smallmatrix}\left[\left[\begin{smallmatrix}a_{\alpha}\mathbb{I}_{n}&\\\
&\overline{a}_{\alpha}\mathbb{I}_{n}\end{smallmatrix}\right]\otimes(T_{\alpha}^{*}T_{\beta})\right]_{\alpha,\beta=1}^{t}&\left[\left[\begin{smallmatrix}a_{\alpha}\mathbb{I}_{n}&\\\
&\overline{a}_{\alpha}\mathbb{I}_{n}\end{smallmatrix}\right]\otimes(T_{\alpha}^{*}\widetilde{Q}_{t+1})\right]_{\alpha=1}^{t}\\\
\left[\begin{smallmatrix}\mathbb{I}_{n}\otimes(\widetilde{A}_{t+1}\widetilde{Q}_{t+1}^{*}T_{\beta})&\\\
&\mathbb{I}_{n}\otimes(\widetilde{A}_{t+1}^{*}\widetilde{Q}_{t+1}^{*}T_{\beta})\end{smallmatrix}\right]_{\beta=1}^{t}&\left[\begin{smallmatrix}\mathbb{I}_{n}\otimes(\widetilde{A}_{t+1}\widetilde{Q}_{t+1}^{*}\widetilde{Q}_{t+1})&\\\
&\mathbb{I}_{n}\otimes(\widetilde{A}_{t+1}^{*}\widetilde{Q}_{t+1}^{*}\widetilde{Q}_{t+1})\end{smallmatrix}\right]\end{smallmatrix}\right],$
(2.5)

where

$\widetilde{Q}_{t+1}=\left[Q_{0},Q_{t+1}\right],\quad\widetilde{A}_{t+1}={\rm
diag}\left(z_{0}\mathbb{I}_{r_{0}},A_{t+1}\right),\quad\widetilde{r}_{t+1}=r_{0}+r_{t+1},$
(2.6)

whenever $r_{0}>0$.

On the other hand, rewrite $T_{\alpha}$ as sum of a diagonal matrix
$\sqrt{T_{{\rm d},\alpha}}$ and a strictly upper triangular matrix ${T_{{\rm
u},\alpha}}$

$T_{\alpha}=\sqrt{T_{{\rm d},\alpha}}+T_{{\rm
u},\alpha}:=\begin{bmatrix}\sqrt{t_{1,1}^{(\alpha)}}&\cdots&t_{i,j}^{(\alpha)}\\\
&\ddots&\vdots\\\ &&\sqrt{t_{n,n}^{(\alpha)}}\end{bmatrix},$ (2.7)

where

$T_{{\rm d},\alpha}={\rm
diag}\big{(}t_{1,1}^{(\alpha)},\cdots,t_{n,n}^{(\alpha)}\big{)}.$ (2.8)

###  2.2 Concentration reduction

Combining integral representations for the $n$-point correlations given in our
previous paper [24, Proposition 1.3] and duality formulas for auto-correlation
functions of characteristic polynomials in [18, 24], we can write the
$n$-point correlation function for the GinUE${}_{N}(A_{0})$ ensemble in
Definition 1.1 in an appropriate form which is ready for asymptotic analysis
by the Laplace method argument. This has already been done in [25, Proposition
2.2], but for convenience we restate it as follows.

######  Proposition 2.1.

For the ${\mathrm{GinUE}}_{N}(A_{0})$ ensemble with $R_{0}\geq n$, with the
assumptions (1.3) on $A_{0}$ and the above notations, the $n$-point
correlation function can be written as

$\displaystyle
R_{N}^{(n)}(A_{0};z_{1},\cdots,z_{n})=\widetilde{C}_{N,n}\int\cdots\int
g(T,Y,Q)\exp\\{Nf(T,Y,Q)\\}{\rm d}Y{\rm d}Q_{0}{\rm
d}Q_{t+1}\prod_{\alpha=1}^{t}{\rm d}T_{\alpha},$ (2.9)

where

$\displaystyle g(T,Y,Q)=$
$\displaystyle\Big{(}\det\big{(}\mathbb{I}_{n}-\sum_{\alpha=1}^{t}T_{\alpha}T_{\alpha}^{*}-Q_{0}Q_{0}^{*}-Q_{t+1}Q_{t+1}^{*}\big{)}\Big{)}^{R_{0}-n}\Bigg{(}\det\\!\begin{bmatrix}\sqrt{\frac{N}{N-n}}Z&-Y^{*}\\\
Y&\sqrt{\frac{N}{N-n}}Z^{*}\end{bmatrix}\Bigg{)}^{R_{0}-n}$ (2.10)
$\displaystyle\times\Big{(}\prod_{\alpha=1}^{t}\prod_{j=1}^{n}\big{(}t_{j,j}^{(\alpha)}\big{)}^{R_{\alpha,N}-j}\Big{)}\Big{(}\prod_{\alpha=1}^{t}\big{(}\det(A_{\alpha})\big{)}^{R_{\alpha,N}-n}\Big{)}\det\\!\bigg{(}\widehat{L}_{1}+\sqrt{\frac{N}{N-n}}\widehat{L}_{2}\bigg{)},$

and

$\displaystyle
f(T,Y,Q)=\sum_{\alpha=1}^{t}c_{\alpha}\log\det(T_{\alpha}T_{\alpha}^{*})+\frac{1}{\tau}h(T,Q)-\frac{N-n}{\tau
N}{\rm
Tr}\big{(}YY^{*}\big{)}+\sum_{\alpha=1}^{t}c_{\alpha}\log\det(A_{\alpha}),$
(2.11)

with

$\displaystyle h(T,Q)=\sum_{\alpha=1}^{t}\left(-|z_{0}-a_{\alpha}|^{2}{\rm
Tr}\big{(}T_{\alpha}T_{\alpha}^{*}\big{)}+N^{-\frac{1}{2}}\left(\overline{a}_{\alpha}{\rm
Tr}\big{(}\hat{Z}T_{\alpha}T_{\alpha}^{*}\big{)}+a_{\alpha}{\rm
Tr}\big{(}\hat{Z}^{*}T_{\alpha}T_{\alpha}^{*}\big{)}\right)\right)$ (2.12)
$\displaystyle+|z_{0}|^{2}\Big{(}\sum_{\alpha=1}^{t}{\rm
Tr}\big{(}T_{\alpha}T_{\alpha}^{*}\big{)}+{\rm
Tr}\big{(}Q_{0}Q_{0}^{*}\big{)}+{\rm
Tr}\big{(}Q_{t+1}Q_{t+1}^{*}\big{)}\Big{)}$
$\displaystyle+N^{-\frac{1}{2}}\Big{(}\overline{z}_{0}{\rm
Tr}\big{(}\hat{Z}Q_{0}Q_{0}^{*}\big{)}+z_{0}{\rm
Tr}\big{(}\hat{Z}^{*}Q_{0}Q_{0}^{*}\big{)}+{\rm
Tr}\big{(}\hat{Z}Q_{t+1}A_{t+1}^{*}Q_{t+1}^{*}\big{)}+{\rm
Tr}\big{(}\hat{Z}^{*}Q_{t+1}A_{t+1}Q_{t+1}^{*}\big{)}\Big{)}$
$\displaystyle+\sum_{1\leq i<j\leq
n}\Big{|}\Big{(}\sum_{\alpha=1}^{t}a_{\alpha}T_{\alpha}T_{\alpha}^{*}+z_{0}Q_{0}Q_{0}^{*}+Q_{t+1}A_{t+1}Q_{t+1}^{*}\Big{)}_{i,j}\Big{|}^{2}$
$\displaystyle-{\rm
Tr}Q_{t+1}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}^{*}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}Q_{t+1}^{*}.$

Here

$\widetilde{C}_{N,n,\tau}=\Big{(}\frac{N-n}{\pi\tau}\Big{)}^{n^{2}}\frac{1}{C_{N,\tau}}e^{-\frac{N}{\tau}\sum_{k=1}^{n}|z_{k}|^{2}}\prod_{1\leq
i<j\leq
n}|z_{i}-z_{j}|^{2}\prod_{\alpha=1}^{t}\frac{\pi^{nr_{\alpha}-\frac{n(n-1)}{2}}}{\prod_{j=1}^{n}(r_{\alpha}-j)!},$
(2.13)
$C_{N,\tau}=\tau^{nN-\frac{n(n-1)}{2}}\pi^{n(r+1)}N^{-\frac{1}{2}n(n+1)}(N-n)^{-n(N-n)}\prod_{k=N-n-r}^{N-r-1}k!,$

and matrix variables are restricted to the integration region

$\sum_{\alpha=1}^{t}T_{\alpha}T_{\alpha}^{*}+Q_{0}Q_{0}^{*}+Q_{t+1}Q_{t+1}^{*}\leq\mathbb{I}_{n}.$
(2.14)

In order to find the exact approximation of the integral representation in
(2.9), we first give a concentration reduction via restricting the integration
region to a small domain. For this, it’s easy to see from (2.2) that

$A_{\alpha}=\begin{bmatrix}\sqrt{\gamma_{N}}(z_{0}-a_{\alpha})\mathbb{I}_{n}&-Y^{*}\\\
Y&\sqrt{\gamma_{N}}\overline{z_{0}-a_{\alpha}}\mathbb{I}_{n}\end{bmatrix}+\sqrt{\gamma_{N}}N^{-\frac{1}{2}}\begin{bmatrix}\hat{Z}&\\\
&\hat{Z}^{*}\end{bmatrix},$

and

$\log\det(A_{\alpha})=\log\det\left(\gamma_{N}f_{\alpha}\mathbb{I}_{n}+YY^{*}\right)+\log\det\left(\mathbb{I}_{2n}+\sqrt{\gamma_{N}}N^{-\frac{1}{2}}\widehat{A}_{\alpha}\right),$
(2.15)

where

$f_{\alpha}:=|z_{0}-a_{\alpha}|^{2},\quad\alpha=1,\cdots,t,$ (2.16)

and

$\displaystyle\widehat{A}_{\alpha}=\left[\begin{smallmatrix}\sqrt{\gamma_{N}}\overline{z_{0}-a_{\alpha}}\left(\gamma_{N}f_{\alpha}\mathbb{I}_{n}+Y^{*}Y\right)^{-1}\hat{Z}&Y^{*}\left(\gamma_{N}f_{\alpha}\mathbb{I}_{n}+YY^{*}\right)^{-1}\hat{Z}^{*}\\\
-Y\left(\gamma_{N}f_{\alpha}\mathbb{I}_{n}+Y^{*}Y\right)^{-1}\hat{Z}&\sqrt{\gamma_{N}}(z_{0}-a_{\alpha})\left(\gamma_{N}f_{\alpha}\mathbb{I}_{n}+YY^{*}\right)^{-1}\hat{Z}^{*}\end{smallmatrix}\right].$
(2.17)

Take all the leading terms of $f(T,Y,Q)$ in (2.11) and put

$\displaystyle
f_{0}(\tau;T,Y,Q)=\sum_{\alpha=1}^{t}\left(c_{\alpha}\log\det(T_{\alpha}T_{\alpha}^{*})+c_{\alpha}\log\det\big{(}f_{\alpha}\mathbb{I}_{n}+YY^{*}\big{)}\right)+\frac{1}{\tau}\Big{(}h_{0}(T,Q)-{\rm
Tr}\big{(}YY^{*})\Big{)}$ (2.18)

with

$\displaystyle h_{0}(T,Q)=-\sum_{\alpha=1}^{t}f_{\alpha}{\rm
Tr}\big{(}T_{\alpha}T_{\alpha}^{*}\big{)}-{\rm
Tr}Q_{t+1}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}^{*}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}Q_{t+1}^{*}$
$\displaystyle+|z_{0}|^{2}\bigg{(}\sum_{\alpha=1}^{t}{\rm
Tr}\big{(}T_{\alpha}T_{\alpha}^{*}\big{)}+{\rm
Tr}\big{(}Q_{0}Q_{0}^{*}\big{)}+{\rm
Tr}\big{(}Q_{t+1}Q_{t+1}^{*}\big{)}\bigg{)}$ $\displaystyle+\sum_{1\leq
i<j\leq
n}\bigg{|}\bigg{(}\sum_{\alpha=1}^{t}a_{\alpha}T_{\alpha}T_{\alpha}^{*}+z_{0}Q_{0}Q_{0}^{*}+Q_{t+1}A_{t+1}Q_{t+1}^{*}\bigg{)}_{i,j}\bigg{|}^{2}.$

Note that the two sets of matrix variables $\\{Y\\}$ and
$\\{Q_{0},Q_{t+1},T_{1},\ldots,T_{t}\\}$ in $f_{0}(\tau;T,Y)$ are separate, we
can do Taylor expansions near the maximum points respectively, according to
two maximum lemmas about $f_{0}(\tau;T,Y)$ shown in [25, Lemma 2.4 and Lemma
2.5]. They play a crucial role in verifying local eigenvalue statistics of the
deformed GinUEs.

By restricting the integration region to

$\Omega_{N,\delta}=\Omega\cap A_{N,\delta},$ (2.19)

where

$\Omega=\bigg{\\{}\big{(}\\{T_{\alpha}\\},Y,Q_{0},Q_{t+1}\big{)}\Big{|}\sum_{\alpha=1}^{t}T_{\alpha}T_{\alpha}^{*}+Q_{0}Q_{0}^{*}+Q_{t+1}Q_{t+1}^{*}\leq\mathbb{I}_{n}\bigg{\\}}$
(2.20)

and

$A_{N,\delta}=\left\\{\big{(}\\{T_{\alpha}\\},Y,Q_{0},Q_{t+1}\big{)}\Big{|}S(T,Y,Q)\leq\delta\right\\},$
(2.21)

with positive constant $\delta$ and

$\displaystyle S(T,Y,Q)$ $\displaystyle=\sum_{\alpha=1}^{t}\left({\rm
Tr}\big{(}T_{{\rm d},\alpha}-\tau
c_{\alpha}|z_{0}-a_{\alpha}|^{-2}\mathbb{I}_{n}\big{)}\big{(}T_{{\rm
d},\alpha}-\tau
c_{\alpha}|z_{0}-a_{\alpha}|^{-2}\mathbb{I}_{n}\big{)}^{*}+{\rm Tr}(T_{{\rm
u},\alpha}T_{{\rm u},\alpha}^{*})\right)$ (2.22) $\displaystyle+{\rm
Tr}\big{(}YY^{*}+Q_{0}Q_{0}^{*}+Q_{t+1}Q_{t+1}^{*}\big{)},$

we have

######  Proposition 2.2.

With the same notation and assumptions as in Proposition 2.1, if
$P_{00}(z_{0})=1/\tau$, then for any positive constant $\delta$ there exists
$\Delta>0$ such that

$R_{N}^{(n)}(A_{0};z_{1},\cdots,z_{n})=D_{N,n}\,\Big{(}I_{N,\delta}+O\big{(}e^{-\frac{1}{2}N\Delta}\big{)}\Big{)},$
(2.23)

where

$I_{N,\delta}=\int_{\Omega_{N,\delta}}g(T,Y,Q)\exp\Big{\\{}N\big{(}f(T,Y,Q)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}}\mathbb{I}_{n}\big{\\}},0,0\big{)}\big{)}\Big{\\}}{\rm
d}V,$ (2.24)

and

${\rm d}V={\rm d}Y{\rm d}Q_{0}{\rm d}Q_{t+1}\prod_{\alpha=1}^{t}{\rm
d}T_{\alpha},\quad
D_{N,n}=\widetilde{C}_{N,n,\tau}e^{nN(|z_{0}|^{2}/\tau-1)}\prod_{\alpha=1}^{t}(\tau
c_{\alpha})^{nNc_{\alpha}}.$ (2.25)

###### Proof of Proposition 2.2..

Using the two maximum lemmas about $f_{0}(\tau;T,Y)$ in [25, Lemma 2.4 and
Lemma 2.5], we see that $f_{0}(\tau;T,Y,Q)$ in (2.18) attains its maximum only
at

$Y=0,\ Q_{0}=0,\ Q_{t+1}=0,\quad T_{\alpha}=\sqrt{\tau
c_{\alpha}/f_{\alpha}}\mathbb{I}_{n},\quad\alpha=1,\cdots,t.$ (2.26)

So there exists $\Delta>0$ such that

$f_{0}(\tau;T,Y,Q)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}}\mathbb{I}_{n}\big{\\}},0,0\big{)}\leq-\Delta$

in the domain $\Omega\cap A_{N,\delta}^{c}$.

On one hand, noting that

$\sum_{\alpha=1}^{t}T_{\alpha}T_{\alpha}^{*}+Q_{0}Q_{0}^{*}+Q_{t+1}Q_{t+1}^{*}\leq\mathbb{I}_{n}$

in the domain $\Omega$ defined by (2.20), we know for $\hat{Z}$ in a compact
subset of $\mathbb{C}^{n}$ that the partial sub-leading terms of $f(T,Y,Q)$ in
(2.11) can be bounded by $O(N^{-\frac{1}{2}})$, that is,

$\displaystyle
N^{-\frac{1}{2}}\sum_{\alpha=1}^{t}\left(\overline{a}_{\alpha}{\rm
Tr}\big{(}\hat{Z}T_{\alpha}T_{\alpha}^{*}\big{)}+a_{\alpha}{\rm
Tr}\big{(}\hat{Z}^{*}T_{\alpha}T_{\alpha}^{*}\big{)}\right)+N^{-\frac{1}{2}}\left(\overline{z}_{0}{\rm
Tr}\big{(}\hat{Z}Q_{0}Q_{0}^{*}\big{)}+z_{0}{\rm
Tr}\big{(}\hat{Z}^{*}Q_{0}Q_{0}^{*}\big{)}\right)$ (2.27)
$\displaystyle+N^{-\frac{1}{2}}\left({\rm
Tr}\big{(}\hat{Z}Q_{t+1}A_{t+1}^{*}Q_{t+1}^{*}\big{)}+{\rm
Tr}\big{(}\hat{Z}^{*}Q_{t+1}A_{t+1}Q_{t+1}^{*}\big{)}\right)=O(N^{-\frac{1}{2}}).$

On the other hand, apply the singular value decomposition (2.47) for $Y$ and
note that all $f_{\alpha}\not=0$ because of the restriction $P_{00}(z_{0})=1$,
we obtain

$\displaystyle(\sqrt{\gamma_{N}}f_{\alpha}\mathbb{I}_{n}+Y^{*}Y)^{-1}=O(1),\quad
Y(\sqrt{\gamma_{N}}f_{\alpha}\mathbb{I}_{n}+Y^{*}Y)^{-1}=O(1),$

and

$\log\det\left(\mathbb{I}_{2n}+\sqrt{\gamma_{N}}N^{-\frac{1}{2}}\widehat{A}_{\alpha}\right)=O(N^{-\frac{1}{2}})$
(2.28)

since $\widehat{A}_{\alpha}=O(1)$. Meanwhile,

$\log\det\left(\gamma_{N}f_{\alpha}\mathbb{I}_{n}+YY^{*}\right)=\log\det\left(f_{\alpha}\mathbb{I}_{n}+YY^{*}\right)+O\big{(}N^{-1}\big{)}.$

Combination of (2.11), (2.15), (2.18), (2.27) and (2.28) give rise to

$N\Big{(}f(T,Y,Q)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}}\mathbb{I}_{n}\big{\\}},0,0\big{)}\Big{)}=N\Big{(}f_{0}(\tau;T,Y,Q)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}}\mathbb{I}_{n}\big{\\}},0,0\big{)}\Big{)}+\frac{n}{\tau}{\rm
Tr}(YY^{*})+O\big{(}\sqrt{N}\big{)}.$

For $g(T,Y,Q)$ in (2.10), all factors can be controlled by a polynomial
function, so by taking a sufficiently large
$N_{0}>\sum_{\alpha=1}^{t}(n-R_{\alpha,N})/c_{\alpha}$ we know

$\int_{A_{N,\delta}^{c}\cap\Omega}|g(T,Y,Q)|\exp\left\\{N_{0}\Big{(}f_{0}(\tau;T,Y,Q)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}}\mathbb{I}_{n}\big{\\}},0,0\big{)}\Big{)}+\frac{n}{\tau}{\rm
Tr}(YY^{*})\right\\}{\rm d}V=O(1).$

Hence,

$\displaystyle\left|\int_{A_{N,\delta}^{\complement}\cap\Omega}g(T,Y,Q)\exp\left\\{N\Big{(}f(T,Y,Q)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}}\mathbb{I}_{n}\big{\\}},0,0\big{)}\Big{)}\right\\}{\rm
d}V\right|$ $\displaystyle\leq
e^{-(N-N_{0})\Delta+O(\sqrt{N})}O(1)=O\big{(}e^{-\frac{1}{2}N\Delta}\big{)}.$

This thus completes the proof. ∎

The remaining task is to analyze the matrix integral $I_{N,\delta}$ with
sufficiently small positive $\delta$, which follows the argument of Laplace
method. To be precise, the proof of Theorem 1.2 will be divided into three
steps: Step 1 deals with Taylor expansions of $f(T,Y,Q)$ and of $g(T,Y,Q)$.
According to the integration region in (2.14), the most relevant variables are
triangular matrices $\big{\\{}T_{\alpha}\big{\\}}$, equivalently, diagonal
matrices $\big{\\{}T_{{\rm d},\alpha}\big{\\}}$ and strictly upper triangular
matrices $\big{\\{}T_{{\rm u},\alpha}\big{\\}}$ as in (2.7). With the coming
of the change of variables

$T_{{\rm d},1},T_{{\rm d},2},\cdots,T_{{\rm
d},t}\longmapsto\widetilde{T}_{{\rm d},1},\widetilde{T}_{{\rm
d},2},\cdots,\widetilde{T}_{{\rm
d},t}\longmapsto\Gamma_{1},\widetilde{T}_{{\rm
d},2},\cdots,\widetilde{T}_{{\rm d},t}$ (2.29)

and

$T_{{\rm u},1},T_{{\rm u},2},\cdots,T_{{\rm u},t}\longmapsto G_{1},T_{{\rm
u},2},\cdots,T_{{\rm u},t};$ (2.30)

Step 2 is a standard procedure in Laplace method to remove the error terms
resulted from the expansions in first two steps; In Step 3, after dealing with
several matrix integrals, we can complete the proof.

###  2.3 Proof of Theorem 1.2: $z_{0}\neq 0$

The proof of Theorem is divided into two cases, one deals with the case when
$z_{0}\not=0$ while the other deals with the case when $z_{0}=0$. With Section
2.1 in mind, we still need some extra notations. For a family of matrices
$\\{T_{{\rm u},\alpha}|\alpha=1,\cdots,t\\}$, define $\|T_{\rm
u}\|:=\sum_{\alpha=1}^{t}\|T_{{\rm u},\alpha}\|,\quad\|T_{\rm
u}\|_{2}:=\sum_{\alpha=2}^{t}\|T_{{\rm u},\alpha}\|.$ Similarly, for a family
of matrices $\\{\widetilde{T}_{{\rm d},\alpha}|\alpha=1,\cdots,t\\}$, define
$\|\widetilde{T}_{{\rm d}}\|:=\sum_{\alpha=1}^{t}\|\widetilde{T}_{{\rm
d},\alpha}\|,\ \|\widetilde{T}_{{\rm
d}}\|_{2}:=\sum_{\alpha=2}^{t}\|\widetilde{T}_{{\rm d},\alpha}\|.$ For a
complex matrix $M$, denote $M^{({\rm d})}$ the diagonal part of M, and
$M^{({\rm off},{\rm u})}$ the strictly upper triangular part of M; in
particular, for $M_{\alpha}$ with index $\alpha$, the corresponding quantity
is denoted by $M^{(\alpha,{\rm d})}$ and $M^{(\alpha,{\rm off},{\rm u})}$.

We need the following two propositions concerning the Taylor expansions, whose
proofs are postponed until next section.

######  Proposition 2.3.

With $f(T,Y,Q)$ in (2.11), we have

$f(T,Y,Q)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}}\mathbb{I}_{n}\big{\\}},0,0\big{)}=F_{0}+F_{1}+O(F_{2}),$
(2.31)

where

$\displaystyle F_{0}$
$\displaystyle=\frac{N^{-\frac{1}{2}}}{\tau}\left(\overline{z_{0}}{\rm
Tr}\big{(}\hat{Z}\big{)}+z_{0}{\rm
Tr}\big{(}\hat{Z}^{*}\big{)}\right)+\frac{n^{2}}{N}$
$\displaystyle-\frac{N^{-1}}{2}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}\left(\overline{z_{0}-a_{\alpha}}^{2}{\rm
Tr}\big{(}\hat{Z}^{2}\big{)}+(z_{0}-a_{\alpha})^{2}{\rm
Tr}\big{(}\hat{Z}^{*}\big{)}^{2}\right),$ $\displaystyle
F_{1}=\frac{|z_{0}|^{2}}{\tau}{\rm Tr}(\Gamma_{1})-\frac{1}{\tau}{\rm
Tr}Q_{t+1}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}^{*}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}Q_{t+1}^{*}-\sum_{\alpha=2}^{t}\frac{f_{\alpha}^{2}}{2\tau^{2}c_{\alpha}}{\rm
Tr}\big{(}\widetilde{T}_{{\rm d},\alpha}^{2}\big{)}$ (2.32)
$\displaystyle-\frac{f_{1}^{2}}{2\tau^{2}c_{1}}{\rm
Tr}\left((Q_{0}Q_{0}^{*})^{({\rm d})}+\sum_{\alpha=2}^{t}\widetilde{T}_{{\rm
d},\alpha}\right)^{2}-\frac{N^{-\frac{1}{2}}}{\tau}{\rm
Tr}\Big{(}(\overline{a}_{1}\hat{Z}+a_{1}\hat{Z}^{*})\big{(}(Q_{0}Q_{0}^{*})^{({\rm
d})}+\sum_{\alpha=2}^{t}\widetilde{T}_{{\rm d},\alpha}\big{)}\Big{)}$
$\displaystyle+\frac{N^{-\frac{1}{2}}}{\tau}\sum_{\alpha=2}^{t}{\rm
Tr}\big{(}(\overline{a}_{\alpha}\hat{Z}+a_{\alpha}\hat{Z}^{*})\widetilde{T}_{{\rm
d},\alpha}\big{)}+\frac{N^{-\frac{1}{2}}}{\tau}\left(\overline{z}_{0}{\rm
Tr}\big{(}\hat{Z}Q_{0}Q_{0}^{*}\big{)}+z_{0}{\rm
Tr}\big{(}\hat{Z}^{*}Q_{0}Q_{0}^{*}\big{)}\right)$
$\displaystyle-\frac{f_{1}^{2}}{\tau^{2}c_{1}}{\rm
Tr}\left((Q_{0}Q_{0}^{*})^{({\rm off},{\rm
u})}+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau c_{\alpha}}{f_{\alpha}}}T_{{\rm
u},\alpha}\right)\left((Q_{0}Q_{0}^{*})^{({\rm off},{\rm
u})}+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau c_{\alpha}}{f_{\alpha}}}T_{{\rm
u},\alpha}\right)^{*}$
$\displaystyle-\sum_{\alpha=2}^{t}\frac{f_{\alpha}}{\tau}{\rm
Tr}\big{(}T_{{\rm u},\alpha}T_{{\rm
u},\alpha}^{*}\big{)}+\frac{1}{\tau}\sum_{1\leq i<j\leq
n}\left|(z_{0}-a_{1})(Q_{0}Q_{0}^{*})^{({\rm off},{\rm
u})}_{i,j}+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}(a_{\alpha}-a_{1})t_{i,j}^{(\alpha)}\right|^{2}$
$\displaystyle-\frac{P_{1}}{2}{\rm
Tr}(YY^{*})^{2}+N^{-\frac{1}{2}}\left(\overline{P_{0}}{\rm
Tr}\big{(}Y^{*}Y\hat{Z}\big{)}+P_{0}{\rm
Tr}\big{(}YY^{*}\hat{Z}^{*}\big{)}\right),$

and

$\displaystyle F_{2}$
$\displaystyle=N^{-\frac{3}{2}}+\|Y\|^{6}+N^{-1}\|Y\|^{2}+N^{-\frac{1}{2}}\|Y\|^{4}$
$\displaystyle+\Omega_{{\rm
error}}^{3}+N^{-\frac{1}{2}}\big{(}\|\Gamma_{1}\|+\Omega_{{\rm
error}}^{2}\big{)}+\big{(}\|\Gamma_{1}\|+\|G_{1}\|\big{)}\Omega_{{\rm
error}}.$

Here the changes of variables in (3.1) (3.6) and (3.7) have been used, and
$\Omega_{{\rm error}}$ is defined in (3.11), in Section 3.

######  Proposition 2.4.

With $g(T,Y,Q)$ in (2.10), we have

$\displaystyle g(T,Y,Q)$
$\displaystyle=\Big{(}\big{(}|z_{0}|^{2n}+O\big{(}N^{-\frac{1}{2}}+\|Y\|\big{)}\big{)}\det\big{(}-\Gamma_{1}-G_{1}-G_{1}^{*}\big{)}\Big{)}^{R_{0}-n}$
$\displaystyle\times\prod_{\alpha=1}^{t}\prod_{j=1}^{n}\Big{(}\frac{\tau
c_{\alpha}}{f_{\alpha}}+O\big{(}\Omega_{\rm
error}\big{)}\Big{)}^{R_{\alpha,N}-j}.\prod_{\alpha=1}^{t}\Big{(}f_{\alpha}^{n}+O\big{(}N^{-\frac{1}{2}}+\|Y\|\big{)}\Big{)}^{R_{\alpha,N}-n}$
$\displaystyle\times\big{|}\det\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}\big{|}^{2n}\Big{(}\big{(}\det(YY^{*})\big{)}^{r_{0}}+O\Big{(}\sum_{\alpha,\beta}\|Y\|^{\alpha}\Omega_{\rm
error}^{\beta}\Big{)}\Big{)}$
$\displaystyle\times\Big{(}\tau^{2n^{2}}|z_{0}|^{2n^{2}}|P_{0}|^{2n^{2}}\prod_{\alpha=1}^{t}f_{\alpha}^{n^{2}}+O\big{(}\Omega_{\rm
error}+N^{-\frac{1}{2}}+\|Y\|\big{)}\Big{)}$

where the finite non-negative integer pair $(\alpha,\beta)$ satisfies (3.24)
and $\Omega_{{\rm error}}$ is defined in (3.11), in Section 3.

###### Proof of Theorem 1.2: $z_{0}\neq 0$.

We proceed in three steps.

Step 1: Taylor expansions of $f(T,Y,Q)$ and $g(T,Y,Q)$.

For $z_{0}\not=0$, by Proposition 2.2 we can restrict all the matrix variables
in the region $\Omega_{N,\delta}$ defined in (2.19) for sufficiently small
$\delta>0$. Hence, we can do Taylor expansions of the relevant matrix
variables as stated in Proposition 2.3 and Proposition 2.4.

Step 2: Removal of error terms.

Recalling (2.24), (3.12) and (2.31), we rewrite

$e^{-NF_{0}}I_{N,\delta}=J_{1,N}+J_{2,N},$ (2.33)

where

$J_{1,N}=\Big{(}\frac{f_{1}}{\tau
c_{1}}\Big{)}^{\frac{n(n-1)}{2}}\int_{\hat{\Omega}_{N,\delta}}g(T,Y,Q)\big{(}1+O\big{(}\Omega_{\rm
error}\big{)}\big{)}e^{NF_{1}}{\rm d}\widetilde{V},$

and

$J_{2,N}=\Big{(}\frac{f_{1}}{\tau
c_{1}}\Big{)}^{\frac{n(n-1)}{2}}\int_{\hat{\Omega}_{N,\delta}}g(T,Y,Q)e^{NF_{1}}\big{(}e^{O(NF_{2})}-1\big{)}\big{(}1+O\big{(}\Omega_{\rm
error}\big{)}\big{)}{\rm d}\widetilde{V},$

with

${\rm d}\widetilde{V}={\rm d}Y{\rm d}Q_{0}{\rm d}Q_{t+1}{\rm d}\Gamma_{1}{\rm
d}G_{1}\prod_{\alpha=2}^{t}{\rm d}\widetilde{T}_{{\rm d},\alpha}{\rm d}T_{{\rm
u},\alpha}$

and

$\hat{\Omega}_{N,\delta}=A_{N,\delta}\cap\big{\\{}\Gamma_{1}+G_{1}+G_{1}^{*}\leq
0\big{\\}},$

cf. (2.21) for definition of $A_{N,\delta}$.

Under the restriction condition of $\Gamma_{1}+G_{1}+G_{1}^{*}\leq 0$, every
principal minor of order 2 is non-positive definite, so all diagonal entries
of $\Gamma_{1}$ are zero or negative and

${\rm Tr}\big{(}G_{1}G_{1}^{*}\big{)}=\sum_{1\leq i<j\leq
n}\big{|}(G_{1})_{i,j}\big{|}^{2}\leq\sum_{1\leq i<j\leq
n}(\Gamma_{1})_{i,i}(\Gamma_{1})_{j,j}\leq\big{(}{\rm
Tr}(\Gamma_{1})\big{)}^{2}.$ (2.34)

Noticing the absence of $G_{1}$ in $F_{1}$ given in (2.32), for convergence we
need to control $G_{1}$ by $\Gamma_{1}$ in $F_{2}$, just as shown in (2.34).
Since $\|G_{1}\|=O(\|\Gamma_{1}\|)$ from (2.34), also from (2.19) and (2.21)
for sufficiently large $N$ and small $\delta$, there exists some $C>0$ such
that

$\displaystyle\frac{1}{C}NF_{2}\leq
N^{-\frac{1}{2}}+\sqrt{\delta}+\sqrt{N\delta}\big{(}\|Y\|^{2}+\|\widetilde{T}_{{\rm
d}}\|_{2}+\|Q_{0}\|^{2}+\|Q_{t+1}\|+\|T_{{\rm u}}\|_{2}\big{)}$
$\displaystyle+N\sqrt{\delta}\big{(}\|Q_{0}\|^{4}+\|\widetilde{T}_{{\rm
d}}\|_{2}^{2}+\|T_{{\rm
u}}\|_{2}^{2}+\|Y\|^{4}+\|Q_{t+1}\|^{2}+\|\Gamma_{1}\|\big{)}.$

Using the inequality

$\big{|}e^{O(NF_{2})}-1\big{|}\leq O(N|F_{2}|)e^{O(N|F_{2}|)},$ (2.35)

after change of variables

$\displaystyle(G_{1},\Gamma_{1})\rightarrow
N^{-1}(G_{1},\Gamma_{1}),\quad(Q_{0},Y)\rightarrow
N^{-\frac{1}{4}}(Q_{0},Y),\quad Q_{t+1}\rightarrow N^{-\frac{1}{2}}Q_{t+1},$
(2.36) $\displaystyle\big{(}\widetilde{T}_{{\rm d},\alpha},T_{{\rm
u},\alpha}\big{)}\rightarrow N^{-\frac{1}{2}}\big{(}\widetilde{T}_{{\rm
d},\alpha},T_{{\rm u},\alpha}\big{)}\quad\alpha=2,\cdots,t,$

the term $O(NF_{2})$ in (2.35) has an upper bound by
$N^{-\frac{1}{4}}P(\tilde{V})$ for some polynomial of variables. since $F_{1}$
can control $F_{2}$ for sufficiently small $\delta$, by the argument of
Laplace method and the dominant convergence theorem we know that $J_{2,N}$ is
typically of order $N^{-\frac{1}{4}}$ compared with $J_{1,N}$, that is,

$J_{2,N}=O\big{(}N^{-\frac{1}{4}}\big{)}J_{1,N}.$ (2.37)

For $J_{1,N}$, take a large $M_{0}>0$ such that

$\displaystyle A_{N,\delta}^{\complement}\subseteq$
$\displaystyle\bigcup_{\alpha=2}^{t}\Big{\\{}{\rm Tr}\big{(}T_{{\rm
u},\alpha}T_{{\rm
u},\alpha}^{*}\big{)}>\frac{\delta}{M_{0}}\Big{\\}}\bigcup\bigcup_{\alpha=2}^{t}\Big{\\{}{\rm
Tr}\big{(}\widetilde{T}_{{\rm d},\alpha}\widetilde{T}_{{\rm
d},\alpha}^{*}\big{)}>\frac{\delta}{M_{0}}\Big{\\}}\bigcup\Big{\\{}{\rm
Tr}\big{(}\Gamma_{1}\Gamma_{1}^{*}\big{)}>\frac{\delta}{M_{0}}\Big{\\}}$
$\displaystyle\bigcup\Big{\\{}{\rm
Tr}\big{(}Q_{0}Q_{0}^{*}\big{)}>\frac{\delta}{M_{0}}\Big{\\}}\bigcup\Big{\\{}{\rm
Tr}\big{(}YY^{*}\big{)}>\frac{\delta}{M_{0}}\Big{\\}}\bigcup\Big{\\{}{\rm
Tr}\big{(}Q_{t+1}Q_{t+1}^{*}\big{)}>\frac{\delta}{M_{0}}\Big{\\}}.$

Here we have used (2.34) to drop out the domain $\\{{\rm
Tr}\big{(}G_{1}G_{1}^{*}\big{)}>\delta/M_{0}\\}$. For each piece of domain,
only keep the restricted matrix variable and let the others free, it’s easy to
prove that the corresponding matrix integral is exponentially small, that is,
$O\big{(}e^{-\delta_{1}N}\big{)}$ for some $\delta_{1}>0$.

So we can extend the integration region from $\hat{\Omega}_{N,\delta}$ to
$\big{\\{}\Gamma_{1}+G_{1}+G_{1}^{*}\leq 0\big{\\}}$. From (1.4) and (1.5), we
have $\sum_{\alpha=1}^{t}R_{\alpha,N}=-r_{0}-r_{t+1}-R_{0}$, and by the change
of variables (2.36) we have

$\displaystyle J_{1,N}$
$\displaystyle=N^{-\frac{n^{2}t}{2}-n(R_{0}+r_{0}+r_{t+1})}\Big{(}\frac{f_{1}}{\tau
c_{1}}\Big{)}^{\frac{n(n-1)}{2}}|z_{0}|^{2nR_{0}}|P_{0}|^{2n^{2}}\tau^{2n^{2}-n(r_{0}+r_{t+1}+R_{0})-\frac{n(n+1)}{2}t}$
(2.38)
$\displaystyle\times\Big{|}\det\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}\Big{|}^{2n}\prod_{\alpha=1}^{t}c_{\alpha}^{nR_{\alpha,N}-\frac{n(n+1)}{2}}f_{\alpha}^{\frac{n(n+1)}{2}}\Big{(}I_{0}+O\big{(}N^{-\frac{1}{4}}\big{)}\Big{)},$

where with $\widehat{H}=\Gamma_{1}+G_{1}+G_{1}^{*}$,

$I_{0}=\int_{\widehat{H}\leq
0}\big{(}\det(-\widehat{H})\big{)}^{R_{0}-n}\big{(}\det(YY^{*})\big{)}^{r_{0}}e^{F}{\rm
d}\widetilde{V},$ (2.39) $\displaystyle F=\frac{|z_{0}|^{2}}{\tau}{\rm
Tr}(\Gamma_{1})-\frac{1}{\tau}{\rm
Tr}Q_{t+1}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}^{*}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}Q_{t+1}^{*}-\sum_{\alpha=2}^{t}\frac{f_{\alpha}^{2}}{2\tau^{2}c_{\alpha}}{\rm
Tr}\big{(}\widetilde{T}_{{\rm d},\alpha}^{2}\big{)}$
$\displaystyle-\frac{f_{1}^{2}}{2\tau^{2}c_{1}}{\rm
Tr}\Big{(}(Q_{0}Q_{0}^{*})^{({\rm d})}+\sum_{\alpha=2}^{t}\widetilde{T}_{{\rm
d},\alpha}\Big{)}^{2}-\frac{1}{\tau}{\rm
Tr}\Big{(}\big{(}\overline{a}_{1}\hat{Z}+a_{1}\hat{Z}^{*}\big{)}\big{(}(Q_{0}Q_{0}^{*})^{({\rm
d})}+\sum_{\alpha=2}^{t}\widetilde{T}_{{\rm d},\alpha}\big{)}\Big{)}$
$\displaystyle+\frac{1}{\tau}\sum_{\alpha=2}^{t}{\rm
Tr}\Big{(}\big{(}\overline{a}_{\alpha}\hat{Z}+a_{\alpha}\hat{Z}^{*}\big{)}\widetilde{T}_{{\rm
d},\alpha}\Big{)}+\frac{1}{\tau}\Big{(}\overline{z}_{0}{\rm
Tr}\big{(}\hat{Z}Q_{0}Q_{0}^{*}\big{)}+z_{0}{\rm
Tr}\big{(}\hat{Z}^{*}Q_{0}Q_{0}^{*}\big{)}\Big{)}$
$\displaystyle-\frac{f_{1}^{2}}{\tau^{2}c_{1}}{\rm
Tr}\Big{(}(Q_{0}Q_{0}^{*})^{({\rm off},{\rm
u})}+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau c_{\alpha}}{f_{\alpha}}}T_{{\rm
u},\alpha}\Big{)}\Big{(}(Q_{0}Q_{0}^{*})^{({\rm off},{\rm
u})}+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau c_{\alpha}}{f_{\alpha}}}T_{{\rm
u},\alpha}\Big{)}^{*}$
$\displaystyle-\sum_{\alpha=2}^{t}\frac{1}{\tau}f_{\alpha}{\rm
Tr}\big{(}T_{{\rm u},\alpha}T_{{\rm
u},\alpha}^{*}\big{)}+\frac{1}{\tau}\sum_{1\leq i<j\leq
n}\left|(z_{0}-a_{1})(Q_{0}Q_{0}^{*})^{({\rm off},{\rm
u})}_{i,j}+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}(a_{\alpha}-a_{1})t_{i,j}^{(\alpha)}\right|^{2}$
$\displaystyle-\frac{P_{1}}{2}{\rm Tr}(YY^{*})^{2}+\overline{P_{0}}{\rm
Tr}\big{(}Y^{*}Y\hat{Z}\big{)}+P_{0}{\rm Tr}\big{(}YY^{*}\hat{Z}^{*}\big{)}$

Step 3: Matrix integrals and final proof.

To simplify $I_{0}$ further, we first integrate out $Q_{t+1},T_{{\rm
u},2},\cdots,T_{{\rm u},t}$,

$\int e^{-\frac{1}{\tau}{\rm
Tr}Q_{t+1}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}^{*}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}Q_{t+1}^{*}}{\rm
d}Q_{t+1}=\Big{|}\det\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}\Big{|}^{-2n}(\pi\tau)^{nr_{t+1}}.$

From Proposition A.3, we have

$\displaystyle\int\exp\Big{\\{}-\frac{f_{1}^{2}}{\tau^{2}c_{1}}{\rm
Tr}\Big{(}(Q_{0}Q_{0}^{*})^{({\rm off},{\rm
u})}+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau c_{\alpha}}{f_{\alpha}}}T_{{\rm
u},\alpha}\Big{)}\Big{(}(Q_{0}Q_{0}^{*})^{({\rm off},{\rm
u})}+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau c_{\alpha}}{f_{\alpha}}}T_{{\rm
u},\alpha}\Big{)}^{*}$
$\displaystyle-\sum_{\alpha=2}^{t}\frac{1}{\tau}f_{\alpha}{\rm
Tr}\big{(}T_{{\rm u},\alpha}T_{{\rm
u},\alpha}^{*}\big{)}+\frac{1}{\tau}\sum_{1\leq i<j\leq
n}\left|(z_{0}-a_{1})(Q_{0}Q_{0}^{*})^{({\rm off},{\rm
u})}_{i,j}+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}(a_{\alpha}-a_{1})t_{i,j}^{(\alpha)}\right|^{2}\Big{\\}}$
$\displaystyle\times\prod_{\alpha=2}^{t}{\rm d}T_{{\rm
u},\alpha}=\tau^{\frac{n(n-1)}{2}(t-2)}\Big{(}\frac{f_{1}}{c_{1}}\big{(}\prod_{\alpha=1}^{t}f_{\alpha}\big{)}|P_{0}|^{2}\Big{)}^{-\frac{n(n-1)}{2}}\pi^{\frac{n(n-1)}{2}(t-1)},$

which is independent of $Q_{0}$.

Secondly, to integrate out $G_{1}$ and $\Gamma_{1}$, by Proposition A.2, we
have

$\int_{\widehat{H}\leq 0}\big{(}\det(-\widehat{H})\big{)}^{R_{0}-n}{\rm
d}G_{1}=(-1)^{n(R_{0}-n)}\pi^{\frac{n(n-1)}{2}}\frac{\prod_{i=1}^{n}(R_{0}-i)!\big{(}\det(\Gamma_{1})\big{)}^{R_{0}-1}}{\big{(}(R_{0}-1)!\big{)}^{n}},$

from which

$(-1)^{n(R_{0}-n)}\int_{\Gamma_{1}\leq
0}\big{(}\det(\Gamma_{1})\big{)}^{R_{0}-1}e^{\frac{1}{\tau}|z_{0}|^{2}{\rm
Tr}(\Gamma_{1})}{\rm
d}\Gamma_{1}=\tau^{nR_{0}}|z_{0}|^{-2nR_{0}}\big{(}(R_{0}-1)!\big{)}^{n}.$
(2.40)

Thirdly, we need to integrate out $\widetilde{T}_{{\rm d},\alpha}$ for
$\alpha=2,\cdots,t$. For $i=1,\cdots,n$, let

$\vec{b}_{i}=\big{(}b_{i}^{(2)},b_{i}^{(3)},\cdots,b_{i}^{(t)}\big{)},\quad
b_{i}^{(\alpha)}=\Big{(}\overline{a_{\alpha}-a_{1}}\hat{Z}+(a_{\alpha}-a_{1})\hat{Z}^{*}-\frac{f_{1}^{2}}{\tau
c_{1}}Q_{0}Q_{0}^{*}\Big{)}_{i,i},$

and introduce a square matrix

$\Sigma=[\Sigma_{\alpha,\beta}]_{\alpha,\beta=2}^{t},\quad\Sigma_{\alpha,\beta}=\frac{f_{\alpha}^{2}}{2c_{\alpha}}\delta_{\alpha,\beta}+\frac{f_{1}^{2}}{2c_{1}},$

Noting

$\det(\Sigma)=2P_{1}\prod_{\alpha=1}^{t}\frac{f_{\alpha}^{2}}{2c_{\alpha}},\quad(\Sigma^{-1})_{\alpha,\beta}=\frac{2c_{\alpha}\delta_{\alpha,\beta}}{f_{\alpha}^{2}}-\frac{2c_{\alpha}c_{\beta}}{f_{\alpha}^{2}f_{\beta}^{2}P_{1}},\quad\alpha,\beta=2,\cdots,t,$
(2.41)

where $P_{1}$ is given in (1.8). Calculate the gaussian integrals and we
obtain

$\displaystyle\int\exp\Big{\\{}-\frac{f_{1}^{2}}{2\tau^{2}c_{1}}{\rm
Tr}\big{(}(Q_{0}Q_{0}^{*})^{({\rm d})}+\sum_{\alpha=2}^{t}\widetilde{T}_{{\rm
d},\alpha}\big{)}^{2}-\sum_{\alpha=2}^{t}\frac{f_{\alpha}^{2}}{2\tau^{2}c_{\alpha}}{\rm
Tr}\big{(}\widetilde{T}_{{\rm d},\alpha}^{2}\big{)}$ (2.42)
$\displaystyle+\frac{1}{\tau}\sum_{\alpha=2}^{t}{\rm
Tr}\Big{(}\big{(}\overline{a_{\alpha}-a_{1}}\hat{Z}+(a_{\alpha}-a_{1})\hat{Z}^{*}\big{)}\widetilde{T}_{{\rm
d},\alpha}\Big{)}\Big{\\}}\prod_{\alpha=2}^{t}{\rm d}\widetilde{T}_{{\rm
d},\alpha}$
$\displaystyle=\Big{(}2P_{1}\prod_{\alpha=1}^{t}\frac{f_{\alpha}^{2}}{2c_{\alpha}}\Big{)}^{-\frac{n}{2}}\pi^{\frac{t-1}{2}n}\tau^{(t-1)n}e^{-\frac{f_{1}^{2}}{2\tau^{2}c_{1}}{\rm
Tr}\big{(}(Q_{0}Q_{0}^{*})^{({\rm
d})}\big{)}^{2}+\frac{1}{4}\sum_{i=1}^{n}\vec{b}_{i}\Sigma^{-1}\vec{b}_{i}^{t}}$

Obviously,

$\displaystyle\sum_{i=1}^{n}\sum_{\alpha=2}^{t}\frac{c_{\alpha}\big{(}b_{i}^{(\alpha)}\big{)}^{2}}{f_{\alpha}^{2}}$
$\displaystyle=\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}{\rm
Tr}\Big{(}\big{(}\overline{a}_{\alpha}\hat{Z}+a_{\alpha}\hat{Z}^{*}\big{)}^{2}-2\big{(}\overline{a}_{\alpha}\hat{Z}+a_{\alpha}\hat{Z}^{*}\big{)}$
$\displaystyle\times\big{(}\overline{a}_{1}\hat{Z}+a_{1}\hat{Z}^{*}+\frac{f_{1}^{2}}{\tau
c_{1}}(Q_{0}Q_{0}^{*})^{({\rm
d})}\big{)}+\big{(}\overline{a}_{1}\hat{Z}+a_{1}\hat{Z}^{*}+\frac{f_{1}^{2}}{\tau
c_{1}}(Q_{0}Q_{0}^{*})^{({\rm d})}\big{)}^{2}\Big{)},$
$\displaystyle\sum_{i=1}^{n}\Big{(}\sum_{\alpha=2}^{t}\frac{c_{\alpha}b_{i}^{(\alpha)}}{f_{\alpha}^{2}}\Big{)}^{2}={\rm
Tr}\Big{(}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}\big{(}\overline{a}_{\alpha}\hat{Z}+a_{\alpha}\hat{Z}^{*}\big{)}-P_{1}\big{(}\overline{a}_{1}\hat{Z}+a_{1}\hat{Z}^{*}+\frac{f_{1}^{2}}{\tau
c_{1}}(Q_{0}Q_{0}^{*})^{({\rm
d})}\big{)}+\frac{1}{\tau}(Q_{0}Q_{0}^{*})^{({\rm d})}\Big{)}^{2}.$

Use (2.41), elementary calculation gives us

$\frac{1}{4}\vec{b}_{i}\Sigma^{-1}\vec{b}_{i}^{t}=\frac{1}{2}\sum_{\alpha=2}^{t}\frac{c_{\alpha}\big{(}b_{i}^{(\alpha)}\big{)}^{2}}{f_{\alpha}^{2}}-\frac{1}{2P_{1}}\Big{(}\sum_{\alpha=2}^{t}\frac{c_{\alpha}b_{i}^{(\alpha)}}{f_{\alpha}^{2}}\Big{)}^{2},$

and

$\displaystyle-\frac{f_{1}^{2}}{2\tau^{2}c_{1}}{\rm Tr}$
$\displaystyle\big{(}(Q_{0}Q_{0}^{*})^{({\rm
d})}\big{)}^{2}+\frac{1}{\tau}{\rm
Tr}\Big{(}\big{(}\overline{z_{0}-a_{1}}\hat{Z}+(z_{0}-a_{1})\hat{Z}^{*}\big{)}(Q_{0}Q_{0}^{*})^{({\rm
d})}\Big{)}+\frac{1}{4}\sum_{i=1}^{n}\vec{b}_{i}\Sigma^{-1}\vec{b}_{i}^{t}$
(2.43) $\displaystyle=-\frac{1}{2P_{1}}$ $\displaystyle{\rm
Tr}\Big{(}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}\big{(}\overline{a}_{\alpha}\hat{Z}+a_{\alpha}\hat{Z}^{*}\big{)}\Big{)}^{2}+\frac{1}{2}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}{\rm
Tr}\big{(}\overline{a}_{\alpha}\hat{Z}+a_{\alpha}\hat{Z}^{*}\big{)}^{2}$
$\displaystyle-\sum_{i=1}^{n}\Big{(}\frac{1}{2P_{1}\tau^{2}}\big{(}(Q_{0}Q_{0}^{*})^{({\rm
d})}_{i,i}\big{)}^{2}-\frac{1}{P_{1}\tau}(Q_{0}Q_{0}^{*})^{({\rm
d})}_{i,i}\big{(}\overline{P}_{0}\hat{z}_{i}+P_{0}\overline{\hat{z}_{i}}\big{)}\Big{)},$

Let $Q_{0}=[q_{i,j}^{(0)}]$, and apply the spherical coordinates to
$(Q_{0}Q_{0}^{*})^{({\rm d})}_{i,i}$:

$\big{(}q_{i,1}^{(0)},\cdots,q_{i,r_{0}}^{(0)}\big{)}=\sqrt{q_{i}}\vec{u}_{i},\quad\|\vec{u}_{i}\|=1,$

with Jacobian

$\prod_{j=1}^{r_{0}}{\rm
d}q_{i,j}^{(0)}=\frac{\pi^{r_{0}}}{\Gamma(r_{0})}q_{i}^{r_{0}-1}{\rm
d}q_{i}{\rm d}\vec{u}_{i},$

where ${\rm d}\vec{u}_{i}$ is the Haar measure on the sphere $S^{r_{0}}.$ Then

$\displaystyle\prod_{i=1}^{n}\int
e^{-\frac{q_{i}^{2}}{2\tau^{2}P_{1}}-\frac{q_{i}}{P_{1}\tau}\big{(}\overline{P}_{0}\hat{z}_{i}+P_{0}\overline{\hat{z}_{i}}\big{)}}\prod_{j=1}^{r_{0}}{\rm
d}q_{i,j}^{(0)}=\frac{(\pi\tau)^{nr_{0}}}{\big{(}\Gamma(r_{0})\big{)}^{n}}\prod_{i=1}^{n}\int_{0}^{+\infty}x^{r_{0}-1}e^{-\frac{x^{2}}{2P_{1}}-\frac{x}{P_{1}}\big{(}\overline{P}_{0}\hat{z}_{i}+P_{0}\overline{\hat{z}_{i}}\big{)}}{\rm
d}x.$ (2.44)

Finally, with (2.13) and (2.25) in mind, by the Stirling’s formula we can
obtain

$\displaystyle
e^{NF_{0}}D_{N,n}=\frac{N^{\frac{n^{2}t}{2}+n(R_{0}+1+r_{0}+r_{t+1})}}{\pi^{n(n+1+r_{0}+r_{t+1})+\frac{n(n-1)}{2}t}}(2\pi)^{-\frac{nt}{2}}\frac{\prod_{\alpha=1}^{t}c_{\alpha}^{-nR_{\alpha,N}+\frac{n^{2}}{2}}}{\prod_{k=1}^{n}(R_{0}-k)!}\tau^{-\frac{n(n+1)}{2}}$
(2.45) $\displaystyle\times\prod_{1\leq i<j\leq
n}\big{|}\hat{z}_{i}-\hat{z}_{j}\big{|}^{2}e^{-\frac{1}{\tau}{\rm
Tr}\big{(}\hat{Z}\hat{Z}^{*}\big{)}-\frac{1}{2}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}\Big{(}\overline{z_{0}-a_{\alpha}}^{2}{\rm
Tr}\big{(}\hat{Z}^{2}\big{)}+(z_{0}-a_{\alpha})^{2}{\rm
Tr}\big{(}\hat{Z}^{*}\big{)}^{2}\Big{)}}\big{(}1+O(N^{-1})\big{)}.$

In view of (1.8), rewrite

$\displaystyle\overline{a}_{\alpha}\hat{Z}+a_{\alpha}\hat{Z}^{*}=\overline{a_{\alpha}-z_{0}}\hat{Z}+(a_{\alpha}-z_{0})\hat{Z}^{*}+\overline{z}_{0}\hat{Z}+z_{0}\hat{Z}^{*},$
$\displaystyle\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}\big{(}\overline{a}_{\alpha}\hat{Z}+a_{\alpha}\hat{Z}^{*}\big{)}=P_{1}\big{(}\overline{z}_{0}\hat{Z}+z_{0}\hat{Z}^{*}\big{)}+\overline{P}_{0}\hat{Z}+P_{0}\hat{Z}^{*},$

expand the square of above, we have

$\displaystyle-\frac{1}{2}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}\Big{(}\overline{z_{0}-a_{\alpha}}^{2}{\rm
Tr}\big{(}\hat{Z}^{2}\big{)}+(z_{0}-a_{\alpha})^{2}{\rm
Tr}\big{(}\hat{Z}^{*}\big{)}^{2}\Big{)}-\frac{1}{2P_{1}}{\rm
Tr}\Big{(}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}\big{(}\overline{a}_{\alpha}\hat{Z}+a_{\alpha}\hat{Z}^{*}\big{)}\Big{)}^{2}$
(2.46) $\displaystyle-\frac{1}{\tau}{\rm
Tr}\big{(}\hat{Z}\hat{Z}^{*}\big{)}+\frac{1}{2}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}{\rm
Tr}\big{(}\big{(}\overline{a}_{\alpha}\hat{Z}+a_{\alpha}\hat{Z}^{*}\big{)}^{2}\big{)}=-\frac{1}{2P_{1}}{\rm
Tr}\big{(}\overline{P}_{0}\hat{Z}+P_{0}\hat{Z}^{*}\big{)}^{2}.$

Changing variables like
$\hat{Z}\rightarrow-\frac{\sqrt{P_{1}}}{\overline{P}_{0}}\hat{Z}$,
$x\rightarrow\sqrt{P_{1}}x$ in (2.44), $Y\rightarrow P_{1}^{-\frac{1}{4}}Y$ in
(2.39), and combining (2.23), (2.33), (2.37), (2.38)-(2.40), (2.42), (2.43),
(2.44), (2.45) and (2.46), we can get

$\displaystyle\frac{1}{\big{(}N|P_{0}|^{2}P_{1}^{-1}\big{)}^{n}}R_{N}^{(n)}\big{(}A_{0};z_{0}-\frac{\sqrt{P_{1}}\hat{z}_{1}}{\overline{P_{0}}\sqrt{N}},\cdots,z_{0}-\frac{\sqrt{P_{1}}\hat{z}_{n}}{\overline{P_{0}}\sqrt{N}}\big{)}$
$\displaystyle=\frac{\prod_{1\leq i<j\leq
n}\big{|}\hat{z}_{i}-\hat{z}_{j}\big{|}^{2}}{\pi^{n^{2}+n}}e^{-\frac{1}{2}\sum_{i=1}^{n}\big{(}\hat{z}_{i}+\overline{\hat{z}}_{i}\big{)}^{2}}\prod_{i=1}^{n}\mathrm{IE}_{r_{0}-1}\big{(}-\hat{z}_{i}-\overline{\hat{z}_{i}}\big{)}$
$\displaystyle\times\int_{\mathbb{C}^{n^{2}}}\big{(}\det(YY^{*})\big{)}^{r_{0}}\exp\Big{\\{}-\frac{1}{2}{\rm
Tr}(YY^{*})^{2}-{\rm Tr}\big{(}Y^{*}Y\hat{Z}\big{)}-{\rm
Tr}\big{(}YY^{*}\hat{Z}^{*}\big{)}\Big{\\}}{\rm
d}Y+O\big{(}N^{-\frac{1}{4}}\big{)}.$

where $\mathrm{IE}_{n}(z)$ is defined in (1.10). By the singular value
decomposition:

$Y=U\sqrt{R}V,\quad R={\rm diag}\left(r_{1},\cdots,r_{n}\right),\quad
r_{1}\geq\cdots\geq r_{n}\geq 0,$ (2.47)

the Jacobian reads

${\rm d}Y=\pi^{n^{2}}\Big{(}\prod_{i=1}^{n-1}i!\Big{)}^{-2}\prod_{1\leq
i<j\leq n}(r_{j}-r_{i})^{2}{\rm d}R{\rm d}U{\rm d}V,$ (2.48)

where $U$ and $V$ are chosen from the unitary group $\mathcal{U}(n)$ with the
Haar measure. Use Harish-Chandra-Itzykson-Zuber integration formula [19, 22]
(see also [27])

$\int_{\mathcal{U}(n)}\exp\\!\left\\{{\rm
Tr}\left(AUBU^{*}\right)\right\\}{\rm
d}U=\frac{\det\big{(}[e^{a_{i}b_{j}}]_{i,j=1}^{n}\big{)}}{\prod_{1\leq i<j\leq
n}(a_{j}-a_{i})(b_{j}-b_{i})}\prod_{i=1}^{n-1}i!,$ (2.49)

with $A={\rm diag}\left(a_{1},\cdots,a_{n}\right)$ and $B={\rm
diag}\left(b_{1},\cdots,b_{n}\right)$, and also Andréief’s integration formula
[1]

$\int\det\big{(}[f_{i}(x_{j})]_{i,j=1}^{n}\big{)}\det\big{(}[g_{i}(x_{j})]_{i,j=1}^{n}\big{)}\prod_{i=1}^{n}{\rm
d}\mu(x_{i})=n!\det\Big{(}\Big{[}\int f_{i}(x)g_{j}(x){\rm
d}\mu(x)\Big{]}_{i,j=1}^{n}\Big{)},$ (2.50)

we obtain

$\displaystyle\frac{\prod_{1\leq i<j\leq
n}\big{|}\hat{z}_{i}-\hat{z}_{j}\big{|}^{2}}{\pi^{n^{2}+n}}\int_{\mathbb{C}^{n^{2}}}$
$\displaystyle\big{(}\det(YY^{*})\big{)}^{r_{0}}\exp\Big{\\{}-\frac{1}{2}{\rm
Tr}(YY^{*})^{2}-{\rm Tr}\big{(}Y^{*}Y\hat{Z}\big{)}-{\rm
Tr}\big{(}YY^{*}\hat{Z}^{*}\big{)}\Big{\\}}{\rm d}Y$
$\displaystyle=\det\left(\Big{[}\sqrt{\frac{2}{\pi}}\Gamma(r_{0}+1)e^{\frac{1}{2}\big{(}\hat{z}_{i}+\overline{\hat{z}}_{j}\big{)}^{2}}\mathrm{IE}_{r_{0}}\left(\hat{z}_{i}+\overline{\hat{z}}_{j}\right)\Big{]}_{i,j=1}^{n}\right).$

After Step 1, Step 2 and Step 3, we thus give a complete proof for
$z_{0}\not=0$ in Theorem 1.2. ∎

##  3 Proofs of Propositions 2.3, 2.4 and Theorem 1.2: $z_{0}=0$

###  3.1 Proofs of Propositions 2.3 and 2.4

###### Proof of Proposition 2.3.

With the notation $f_{\alpha}$ in (2.16), noticing the integration domain due
to (2.22) the decomposition of $T_{\alpha}$ as sum of a diagonal matrix
$\sqrt{T_{{\rm d},\alpha}}$ and a strictly upper triangular matrix ${T_{{\rm
u},\alpha}}$ as in (2.7), introduce new matrix variables

$T_{{\rm d},\alpha}=\frac{\tau
c_{\alpha}}{f_{\alpha}}\mathbb{I}_{n}+\widetilde{T}_{{\rm d},\alpha},$ (3.1)

It’s easy to obtain

$c_{\alpha}{\rm Tr}\log T_{{\rm d},\alpha}-\frac{f_{\alpha}}{\tau}{\rm
Tr}(T_{{\rm d},\alpha})=nc_{\alpha}\big{(}\log\frac{\tau
c_{\alpha}}{f_{\alpha}}-1\big{)}-\frac{f_{\alpha}^{2}}{2\tau^{2}c_{\alpha}}{\rm
Tr}\widetilde{T}_{{\rm d},\alpha}^{2}+O\big{(}\|\widetilde{T}_{{\rm
d},\alpha}\|^{3}\big{)},$ (3.2) $\sqrt{T_{{\rm d},\alpha}}=\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}\left(\mathbb{I}_{n}+\frac{f_{\alpha}}{2\tau
c_{\alpha}}\widetilde{T}_{{\rm d},\alpha}+O\big{(}\|\widetilde{T}_{{\rm
d},\alpha}\|^{2}\big{)}\right),$ $\displaystyle T_{\alpha}T_{\alpha}^{*}$
$\displaystyle=\frac{\tau
c_{\alpha}}{f_{\alpha}}\mathbb{I}_{n}+\widetilde{T}_{{\rm
d},\alpha}+\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}\left(\mathbb{I}_{n}+\frac{f_{\alpha}}{2\tau
c_{\alpha}}\widetilde{T}_{{\rm d},\alpha}+O\big{(}\|\widetilde{T}_{{\rm
d},\alpha}\|^{2}\big{)}\right)T_{{\rm u},\alpha}^{*}$
$\displaystyle+\sqrt{\frac{\tau c_{\alpha}}{f_{\alpha}}}T_{{\rm
u},\alpha}\left(\mathbb{I}_{n}+\frac{f_{\alpha}}{2\tau
c_{\alpha}}\widetilde{T}_{{\rm d},\alpha}+O\big{(}\|\widetilde{T}_{{\rm
d},\alpha}\|^{2}\big{)}\right)+T_{{\rm u},\alpha}T_{{\rm u},\alpha}^{*},$

from which we have for $i<j$,

$\left(\sum_{\alpha=1}^{t}a_{\alpha}T_{\alpha}T_{\alpha}^{*}\right)_{i,j}=\sum_{\alpha=1}^{t}a_{\alpha}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}t_{i,j}^{(\alpha)}+O\left(\|T_{{\rm
u},\alpha}\|^{2}+\|T_{{\rm u},\alpha}\|\|\widetilde{T}_{{\rm
d},\alpha}\|\right),$ (3.3)

where (2.7) has been used, and

${\rm Tr}\big{(}\hat{Z}T_{\alpha}T_{\alpha}^{*}\big{)}=\frac{\tau
c_{\alpha}}{f_{\alpha}}{\rm Tr}\big{(}\hat{Z}\big{)}+{\rm
Tr}\big{(}\hat{Z}\widetilde{T}_{{\rm d},\alpha}\big{)}+O\left(\|T_{{\rm
u},\alpha}\|^{2}\right).$ (3.4)

At the same time,

$\left(z_{0}Q_{0}Q_{0}^{*}+Q_{t+1}A_{t+1}Q_{t+1}^{*}\right)_{i,j}=z_{0}\left(Q_{0}Q_{0}^{*}\right)_{i,j}+O\left(\|Q_{t+1}\|^{2}\right).$

Moreover, we see from the boundary condition $P_{00}(z_{0})=1$ that the
restriction condition in (2.20) becomes

$\sum_{\alpha=1}^{t}\widetilde{T}_{{\rm
d},\alpha}+\sum_{\alpha=1}^{t}\sqrt{\frac{\tau c_{\alpha}}{f_{\alpha}}}T_{{\rm
u},\alpha}+\sum_{\alpha=1}^{t}\sqrt{\frac{\tau c_{\alpha}}{f_{\alpha}}}T_{{\rm
u},\alpha}^{*}+S\leq 0,$

where

$\displaystyle S$ $\displaystyle=\sum_{\alpha=1}^{t}T_{{\rm u},\alpha}T_{{\rm
u},\alpha}^{*}+Q_{0}Q_{0}^{*}+Q_{t+1}Q_{t+1}^{*}$ (3.5)
$\displaystyle+\sum_{\alpha=1}^{t}T_{{\rm u},\alpha}\Big{(}\sqrt{T_{{\rm
d},\alpha}}-\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}\mathbb{I}_{n}\Big{)}+\sum_{\alpha=1}^{t}\Big{(}\sqrt{T_{{\rm
d},\alpha}}-\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}\mathbb{I}_{n}\Big{)}T_{{\rm u},\alpha}^{*}.$

Here the first line contains all diagonal parts of $S$, while the second line
only contains the non-diagonal parts.

Let $S^{({\rm d})}$ be a diagonal matrix extracted from the diagonal part of
$S$ given in (3.5), introduce a new diagonal matrix $\Gamma_{1}$ and make
change of variables from $\widetilde{T}_{{\rm d},1},\widetilde{T}_{{\rm
d},2},\cdots,\widetilde{T}_{{\rm d},t}$ to

$\displaystyle\Gamma_{1}:=\sum_{\alpha=1}^{t}\widetilde{T}_{{\rm
d},\alpha}+S^{({\rm d})},\quad\widetilde{T}_{{\rm
d},2},\,\ldots,\widetilde{T}_{{\rm d},t}.$ (3.6)

Note that $S^{({\rm d})}$ doesn’t depend on $\widetilde{T}_{{\rm d},\alpha}$,
so the above transformation is linear and

$S^{({\rm d})}=(Q_{0}Q_{0}^{*})^{({\rm d})}+O\big{(}\|T_{{\rm
u}}\|^{2}+\|Q_{t+1}\|^{2}\big{)}.$

We will see that the typical size for $Q_{t+1},\widetilde{T}_{{\rm
d},2},\,\ldots,\widetilde{T}_{{\rm d},t}$ are $N^{-1/2}$ while it is $N^{-1}$
for $\Gamma_{1}$ and it is $N^{-\frac{1}{4}}$ for $Q_{0}$. So we will do
Taylor expansions up to some proper orders accordingly.

Also let $S^{({\rm off},{\rm u})}$ be a strictly upper triangular matrix
extracted from the strictly upper triangular part of $S$ given in (3.5),
introduce a new strictly upper triangular matrix $G_{1}$ and make change of
variables from $T_{{\rm u},1},T_{{\rm u},2},\cdots,T_{{\rm u},t}$ to

$\displaystyle G_{1}:=\sum_{\alpha=1}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}T_{{\rm u},\alpha}+S^{({\rm off},{\rm u})},\quad
T_{{\rm u},2},\,\ldots,T_{{\rm u},t}.$ (3.7)

We will see that the typical size for $T_{{\rm u},2},\cdots,T_{{\rm u},t}$ are
$N^{-1/2}$ while it is $N^{-1}$ for $G_{1}$. So we will do Taylor expansions
up to some proper orders accordingly.

As to the change of variables (3.7), the Jacobian determinant reads

$\det\Big{(}\frac{\partial T_{{\rm u},1}}{\partial
G_{1}}\Big{)}=\left(\frac{f_{1}}{\tau
c_{1}}\right)^{\frac{n(n-1)}{2}}+O\big{(}\|T_{{\rm
u},1}\|+\|\widetilde{T}_{{\rm d}}\|\big{)}.$

At the same time,

$\|\widetilde{T}_{{\rm d}}\|=O\big{(}\|\widetilde{T}_{{\rm
d}}\|_{2}+\|\Gamma_{1}\|+\|Q_{0}\|^{2}+\|T_{{\rm u}}\|+\|Q_{t+1}\|\big{)},$
(3.8)

where the last two terms can be replaced by $\|T_{{\rm
u}}\|^{2}+\|Q_{t+1}\|^{2}$ for exactness, and from

$S^{({\rm off},{\rm u})}=(Q_{0}Q_{0}^{*})^{({\rm off},{\rm
u})}+O\Big{(}\|T_{{\rm u}}\|\big{(}\|T_{{\rm u}}\|+\|\widetilde{T}_{{\rm
d}}\|\big{)}+\|Q_{t+1}\|^{2}\Big{)},$

we have from (3.6) that

$\|T_{{\rm u}}\|=O\big{(}\|G_{1}\|+\|T_{{\rm
u}}\|_{2}+\|Q_{0}\|^{2}+\|Q_{t+1}\|\big{)}+\|T_{{\rm u}}\|O\big{(}\|T_{{\rm
u}}\|+\|\widetilde{T}_{{\rm d}}\|\big{)}.$

Also from (2.21) we have

$\|T_{{\rm u}}\|+\|\widetilde{T}_{{\rm d}}\|=O\big{(}\sqrt{\delta}\big{)},$

as $\delta>0$ is sufficiently small, we have

$\|T_{{\rm u}}\|=O\big{(}\|G_{1}\|+\|T_{{\rm
u}}\|_{2}+\|Q_{0}\|^{2}+\|Q_{t+1}\|\big{)}.$ (3.9)

Substitute (3.9) into (3.8), we have

$\|\widetilde{T}_{{\rm d}}\|=O\big{(}\Omega_{{\rm error}}\big{)},$ (3.10)

where

$\Omega_{{\rm error}}:=\|\widetilde{T}_{{\rm
d}}\|_{2}+\|\Gamma_{1}\|+\|G_{1}\|+\|T_{{\rm
u}}\|_{2}+\|Q_{0}\|^{2}+\|Q_{t+1}\|,$ (3.11)

and further

$\det\Big{(}\frac{\partial T_{{\rm u},1}}{\partial
G_{1}}\Big{)}=\left(\frac{f_{1}}{\tau
c_{1}}\right)^{\frac{n(n-1)}{2}}+O(\Omega_{{\rm error}}).$ (3.12)

In view of (2.11) and (2.12), we get

$\frac{|z_{0}|^{2}}{\tau}\Big{(}\sum_{\alpha=1}^{t}{\rm
Tr}(T_{\alpha}T_{\alpha}^{*})+{\rm Tr}(Q_{0}Q_{0}^{*})+{\rm
Tr}(Q_{t+1}Q_{t+1}^{*})\Big{)}=\frac{n|z_{0}|^{2}}{\tau}+\frac{|z_{0}|^{2}}{\tau}{\rm
Tr}(\Gamma_{1}).$

In view of (2.11), (2.12) and (3.3), from transformations (3.6) and (3.7) we
have

$\displaystyle-\sum_{\alpha=1}^{t}f_{\alpha}{\rm Tr}\big{(}T_{{\rm
u},\alpha}T_{{\rm u},\alpha}^{*}\big{)}+\sum_{1\leq i<j\leq
n}\left|\left(\sum_{\alpha=1}^{t}a_{\alpha}T_{\alpha}T_{\alpha}^{*}+z_{0}Q_{0}Q_{0}^{*}+Q_{t+1}A_{t+1}Q_{t+1}^{*}\right)_{i,j}\right|^{2}$
$\displaystyle=-\frac{f_{1}^{2}}{\tau c_{1}}{\rm
Tr}\left((Q_{0}Q_{0}^{*})^{({\rm off},{\rm
u})}+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau c_{\alpha}}{f_{\alpha}}}T_{{\rm
u},\alpha}\right)\left((Q_{0}Q_{0}^{*})^{({\rm off},{\rm
u})}+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau c_{\alpha}}{f_{\alpha}}}T_{{\rm
u},\alpha}\right)^{*}-\sum_{\alpha=2}^{t}f_{\alpha}{\rm Tr}\big{(}T_{{\rm
u},\alpha}T_{{\rm u},\alpha}^{*}\big{)}$ $\displaystyle+\sum_{1\leq i<j\leq
n}\left|(z_{0}-a_{1})(Q_{0}Q_{0}^{*})^{({\rm off},{\rm
u})}_{i,j}+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}(a_{\alpha}-a_{1})t_{i,j}^{(\alpha)}\right|^{2}+O(\Omega_{{\rm
error}}^{3}+\|G_{1}\|\Omega_{{\rm error}}),$

and

$\displaystyle-\sum_{\alpha=1}^{t}\frac{f_{\alpha}^{2}}{2\tau^{2}c_{\alpha}}{\rm
Tr}\left(\widetilde{T}_{{\rm
d},\alpha}^{2}\right)+\frac{|z_{0}|^{2}}{\tau}{\rm
Tr}\left(\sum_{\alpha=1}^{t}\widetilde{T}_{{\rm d},\alpha}+S^{({\rm
d})}\right)+\frac{N^{-\frac{1}{2}}}{\tau}\sum_{\alpha=1}^{t}\left(\overline{a}_{\alpha}{\rm
Tr}\big{(}\hat{Z}\widetilde{T}_{{\rm d},\alpha}\big{)}+a_{\alpha}{\rm
Tr}\big{(}\hat{Z}^{*}\widetilde{T}_{{\rm d},\alpha}\big{)}\right)$ (3.13)
$\displaystyle=\frac{|z_{0}|^{2}}{\tau}{\rm
Tr}(\Gamma_{1})-\frac{f_{1}^{2}}{2\tau^{2}c_{1}}{\rm
Tr}\left((Q_{0}Q_{0}^{*})^{({\rm d})}+\sum_{\alpha=2}^{t}\widetilde{T}_{{\rm
d},\alpha}\right)^{2}-\sum_{\alpha=2}^{t}\frac{f_{\alpha}^{2}}{2\tau^{2}c_{\alpha}}{\rm
Tr}\big{(}\widetilde{T}_{{\rm
d},\alpha}^{2}\big{)}+\frac{N^{-\frac{1}{2}}}{\tau}\sum_{\alpha=2}^{t}{\rm
Tr}\big{(}(\overline{a}_{\alpha}\hat{Z}+a_{\alpha}\hat{Z}^{*})\widetilde{T}_{{\rm
d},\alpha}\big{)}$ $\displaystyle-\frac{N^{-\frac{1}{2}}}{\tau}{\rm
Tr}\Big{(}(\overline{a}_{1}\hat{Z}+a_{1}\hat{Z}^{*})\big{(}(Q_{0}Q_{0}^{*})^{({\rm
d})}+\sum_{\alpha=2}^{t}\widetilde{T}_{{\rm
d},\alpha}\big{)}\Big{)}+O\Big{(}\big{(}\Omega_{{\rm
error}}+N^{-\frac{1}{2}}\big{)}\big{(}\Omega_{{\rm
error}}^{2}+\|\Gamma_{1}\|\big{)}\Big{)}.$

Also, the integration region now becomes simply

$\Gamma_{1}+G_{1}+G_{1}^{*}\leq 0.$ (3.14)

In view of (2.15), noticing (2.17), we obtain

$\displaystyle\log\det\left(\mathbb{I}_{2n}+\sqrt{\gamma_{N}}N^{-\frac{1}{2}}\widehat{A}_{\alpha}\right)=N^{-\frac{1}{2}}{\rm
Tr}\big{(}\widehat{A}_{\alpha}\big{)}-\frac{1}{2}N^{-1}{\rm
Tr}\big{(}\widehat{A}_{\alpha}^{2}\big{)}+O\big{(}N^{-\frac{3}{2}}\big{)}.$
(3.15)

Combining (2.15), (2.17) and (3.15), and introducing $P_{0}$ defined in (1.7),
we have

$\displaystyle{\rm
Tr}(YY^{*})-\sum_{\alpha=1}^{t}c_{\alpha}\log\det(A_{\alpha})=-\frac{n^{2}}{N}-n\sum_{\alpha=1}^{t}c_{\alpha}\log
f_{\alpha}$ (3.16)
$\displaystyle+\frac{N^{-1}}{2}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}\left(\overline{z_{0}-a_{\alpha}}^{2}{\rm
Tr}\big{(}\hat{Z}^{2}\big{)}+(z_{0}-a_{\alpha})^{2}{\rm
Tr}\big{(}\hat{Z}^{*}\big{)}^{2}\right)$
$\displaystyle-N^{-\frac{1}{2}}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}}\left(\overline{z_{0}-a_{\alpha}}{\rm
Tr}\big{(}\hat{Z}\big{)}+(z_{0}-a_{\alpha}){\rm
Tr}\big{(}\hat{Z}^{*}\big{)}\right)+\frac{P_{1}}{2}{\rm Tr}(YY^{*})^{2}$
$\displaystyle-N^{-\frac{1}{2}}\left(\overline{P_{0}}{\rm
Tr}\big{(}Y^{*}Y\hat{Z}\big{)}+P_{0}{\rm
Tr}\big{(}YY^{*}\hat{Z}^{*}\big{)}\right)+O\big{(}N^{-\frac{3}{2}}+\|Y\|^{6}+N^{-1}\|Y\|^{2}+N^{-\frac{1}{2}}\|Y\|^{4}\big{)}.$

Combining (2.11), (3.2), (3.4), (3.9)-(3.13) and (3.16), we thus obtain the
desired result. ∎

###### Proof of Proposition 2.4.

For $g(T,Y,Q)$ defined in (2.10), we write down leading terms for five
relevant factors while the estimate for the last determinant in (2.10) is non-
trivial.

  * (1)

For the first factor in (2.10), by (3.6) and (3.7),

$\Big{(}\det\big{(}\mathbb{I}_{n}-\sum_{\alpha=1}^{t}T_{\alpha}T_{\alpha}^{*}-Q_{0}Q_{0}^{*}-Q_{t+1}Q_{t+1}^{*}\big{)}\Big{)}^{R_{0}-n}=\Big{(}\det\big{(}-\Gamma_{1}-G_{1}-G_{1}^{*}\big{)}\Big{)}^{R_{0}-n},$
(3.17)

  * (2)

For the second factor in (2.10), by (2.1),

$\det\begin{bmatrix}\sqrt{\gamma_{N}}Z&-Y^{*}\\\
Y&\sqrt{\gamma_{N}}Z^{*}\end{bmatrix}=|z_{0}|^{2n}+O\big{(}N^{-\frac{1}{2}}+\|Y\|\big{)}.$

  * (3)

For the third factor in (2.10) with $\alpha=1,\cdots,t,$, it is easy to see
from (3.1) and (3.11) that

$\prod_{\alpha=1}^{t}\prod_{j=1}^{n}\big{(}t_{j,j}^{(\alpha)}\big{)}^{R_{\alpha,N}-j}=\prod_{\alpha=1}^{t}\prod_{j=1}^{n}\Big{(}\frac{\tau
c_{\alpha}}{f_{\alpha}}+O\big{(}\Omega_{\rm
error}\big{)}\Big{)}^{R_{\alpha,N}-j}.$ (3.18)

  * (4)

For the fourth factor in (2.10), by (2.1) and (2.2)

$\prod_{\alpha=1}^{t}\big{(}\det(A_{\alpha})\big{)}^{R_{\alpha,N}-n}=\prod_{\alpha=1}^{t}\Big{(}f_{\alpha}^{n}+O\big{(}N^{-\frac{1}{2}}+\|Y\|\big{)}\Big{)}^{R_{\alpha,N}-n}.$
(3.19)

The remaining part of this subsection is devoted to the last determinant in
(2.10).

Recalling $\hat{L}_{1}$ and $\hat{L}_{2}$ defined in (2.3) and (2.5), since
all $a_{\alpha}\neq z_{0}$, $A_{a_{\alpha}}$ is invertible when $N$ is
sufficiently large. Rewrite

$\det\\!\bigg{(}\widehat{L}_{1}+\sqrt{\gamma_{N}}\widehat{L}_{2}\bigg{)}=\left[\begin{smallmatrix}P_{1,1}&O\big{(}\|\widetilde{Q}_{t+1}\|\big{)}\\\
O\big{(}\|\widetilde{Q}_{t+1}\|\big{)}&\widehat{B}_{t+1}+O\big{(}\|\widetilde{Q}_{t+1}\|^{2}\big{)}\end{smallmatrix}\right],$

where

$P_{1,1}=\left[\begin{smallmatrix}A_{1}\otimes\mathbb{I}_{n}&&\\\ &\ldots&\\\
&&A_{t}\otimes\mathbb{I}_{n}\end{smallmatrix}\right]+\sqrt{\gamma_{N}}\left[\begin{smallmatrix}\mathbb{I}_{2n}\otimes
T_{1}^{*}\\\ \vdots\\\ \mathbb{I}_{2n}\otimes
T_{t}^{*}\end{smallmatrix}\right]\left[\begin{smallmatrix}\left[\begin{smallmatrix}a_{1}\mathbb{I}_{n}&\\\
&\overline{a}_{1}\mathbb{I}_{n}\end{smallmatrix}\right]\otimes
T_{1}&,\cdots,&\left[\begin{smallmatrix}a_{t}\mathbb{I}_{n}&\\\
&\overline{a}_{t}\mathbb{I}_{n}\end{smallmatrix}\right]\otimes
T_{t}\end{smallmatrix}\right],$

change the order of the above matrix product and we obtain

$\displaystyle\det(P_{1,1})=\Big{(}\prod_{\alpha=1}^{t}\det\big{(}A_{\alpha}\big{)}\Big{)}^{n}\det\bigg{(}\mathbb{I}_{2n^{2}}+\sqrt{\gamma_{N}}\sum_{\alpha=1}^{t}\left(A_{\alpha}^{-1}\left[\begin{smallmatrix}a_{\alpha}\mathbb{I}_{n}&\\\
&\overline{a}_{\alpha}\mathbb{I}_{n}\end{smallmatrix}\right]\right)\otimes(T_{\alpha}T_{\alpha}^{*})\bigg{)},$

Simple calculations show that

$\Big{(}Z_{\alpha}+Y^{*}\big{(}Z_{\alpha}^{*}\big{)}^{-1}Y\Big{)}^{-1}=(z_{0}-a_{\alpha})^{-1}\mathbb{I}_{n}+O\big{(}N^{-\frac{1}{2}}+\|Y\|^{2}\big{)},$
(3.20) $T_{\alpha}T_{\alpha}^{*}=\frac{\tau
c_{\alpha}}{f_{\alpha}}\mathbb{I}_{n}+O\big{(}\Omega_{\rm error}\big{)}.$
(3.21)

Rewrite
$\frac{a_{\alpha}}{z_{0}-a_{\alpha}}=\frac{a_{\alpha}\overline{z_{0}-a_{\alpha}}}{f_{\alpha}}$,
we have

$\displaystyle\mathbb{I}_{2n^{2}}$
$\displaystyle+\sqrt{\gamma_{N}}\sum_{\alpha=1}^{t}\left(A_{\alpha}^{-1}\left[\begin{smallmatrix}a_{\alpha}\mathbb{I}_{n}&\\\
&\overline{a}_{\alpha}\mathbb{I}_{n}\end{smallmatrix}\right]\right)\otimes(T_{\alpha}T_{\alpha}^{*})=\mathbb{I}_{2n^{2}}+$
$\displaystyle\sum_{\alpha=1}^{t}\begin{bmatrix}\frac{\tau
c_{\alpha}a_{\alpha}\overline{z_{0}-a_{\alpha}}}{f_{\alpha}^{2}}\mathbb{I}_{n^{2}}&\\\
&\frac{\tau
c_{\alpha}\overline{a}_{\alpha}(z_{0}-a_{\alpha})}{f_{\alpha}^{2}}\mathbb{I}_{n^{2}}\end{bmatrix}+O\big{(}\Omega_{\rm
error}+N^{-\frac{1}{2}}+\|Y\|\big{)}.$

While from $P_{00}(z_{0})=1$

$\displaystyle 1+\sum_{\alpha=1}^{t}\frac{\tau
c_{\alpha}a_{\alpha}\overline{z_{0}-a_{\alpha}}}{f_{\alpha}^{2}}$
$\displaystyle=1+\tau\sum_{\alpha=1}^{t}\frac{c_{\alpha}(a_{\alpha}-z_{0}+z_{0})\overline{z_{0}-a_{\alpha}}}{f_{\alpha}^{2}}$
$\displaystyle=1-\tau\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}}+\tau
z_{0}\sum_{\alpha=1}^{t}\frac{c_{\alpha}\overline{z_{0}-a_{\alpha}}}{f_{\alpha}^{2}}=-\tau
z_{0}\overline{P}_{0},$

where $P_{0}$ is defined in (1.7). Combining (3.19) we have

$\det\big{(}P_{1,1}\big{)}=\tau^{2n^{2}}|z_{0}|^{2n^{2}}|P_{0}|^{2n^{2}}\prod_{\alpha=1}^{t}f_{\alpha}^{n^{2}}+O\big{(}\Omega_{\rm
error}+N^{-\frac{1}{2}}+\|Y\|\big{)}.$ (3.22)

As $P_{0}\not=0$ and $z_{0}\not=0$, we have $\det(P_{1,1})\not=0.$ and from
$\widetilde{Q}_{t+1}=[Q_{0},Q_{t+1}]$,
$\|\widetilde{Q}_{t+1}\|^{2}=O\big{(}\|Q_{0}\|^{2}+\|Q_{t+1}\|\big{)}=O\big{(}\Omega_{\rm
error}\big{)}.$ Therefore,

$\det\\!\bigg{(}\widehat{L}_{1}+\sqrt{\gamma_{N}}\widehat{L}_{2}\bigg{)}=\det(P_{1,1})\det\Big{(}\widehat{B}_{t+1}+O\big{(}\|\widetilde{Q}_{t+1}\|^{2}\big{)}\Big{)}.$

In view of (2.4) and (2.6), from elementary computations,

$\det\Big{(}\widehat{B}_{t+1}+O\big{(}\|\widetilde{Q}_{t+1}\|^{2}\big{)}\Big{)}=\big{|}\det\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}\big{|}^{2n}\Big{(}\big{(}\det(YY^{*})\big{)}^{r_{0}}+O\Big{(}\sum_{\alpha,\beta}\|Y\|^{\alpha}\Omega_{\rm
error}^{\beta}\Big{)}\Big{)},$ (3.23)

where the finite non-negative integer pair $(\alpha,\beta)$ is such that

$\frac{1}{4}\alpha+\frac{1}{2}\beta\geq\frac{1}{2}nr_{0}+\frac{1}{4}.$ (3.24)

Combing (3.22)-(3.23), we have

$\displaystyle\det\\!\bigg{(}\widehat{L}_{1}+\sqrt{\gamma_{N}}\widehat{L}_{2}\bigg{)}$
$\displaystyle=\big{|}\det\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}\big{|}^{2n}\Big{(}\big{(}\det(YY^{*})\big{)}^{r_{0}}+O\Big{(}\sum_{\alpha,\beta}\|Y\|^{\alpha}\Omega_{\rm
error}^{\beta}\Big{)}\Big{)}$
$\displaystyle\times\Big{(}\tau^{2n^{2}}|z_{0}|^{2n^{2}}|P_{0}|^{2n^{2}}\prod_{\alpha=1}^{t}f_{\alpha}^{n^{2}}+O\big{(}\Omega_{\rm
error}+N^{-\frac{1}{2}}+\|Y\|\big{)}\Big{)},$

where the finite non-negative integer pair $(\alpha,\beta)$ is such that
(3.24). ∎

###  3.2 Proof of Theorem 1.2: $z_{0}=0$

The proof in the case of $z_{0}=0$ is similar, we only point out the different
places from the case of $z_{0}\not=0$.

As $z_{0}=0$, in view of $X_{0}={\rm diag}\left(A_{0},0_{R_{0}}\right)$ in
Definition 1.1, $R_{0}=N-r$ and
$A_{0}={\rm
diag}\left(a_{1}\mathbb{I}_{r_{1}},\cdots,a_{t}\mathbb{I}_{r_{t}},0_{r_{0}},A_{t+1}\right)$
below and in (1.3), there are $R_{0}+r_{0}$ 0s. Hence, without loss of
generality we can simply set $r_{0}=0$, and further $Q_{0}=0$ in Section 2.1.
For Taylor expansion of $f(T,Y,Q)$, We still have (3.1)-(3.10) and (3.14) with
$z_{0}=0$ and $Q_{0}=0$. But note that as $z_{0}=0$, linear term of
$\Gamma_{1}$ does not exist on the exponent. This means that the typical size
of $\Gamma_{1}$ is $N^{-\frac{1}{2}}$. Correspondingly, the typical size of
$G_{1}$ is also $N^{-\frac{1}{2}}$, this is different from the case of
$z_{0}\not=0$.

Hence, we have

$\displaystyle-\sum_{\alpha=1}^{t}f_{\alpha}{\rm Tr}\big{(}T_{{\rm
u},\alpha}T_{{\rm u},\alpha}^{*}\big{)}+\sum_{1\leq i<j\leq
n}\left|\left(\sum_{\alpha=1}^{t}a_{\alpha}T_{\alpha}T_{\alpha}^{*}+Q_{t+1}A_{t+1}Q_{t+1}^{*}\right)_{i,j}\right|^{2}$
(3.25) $\displaystyle=-\frac{f_{1}^{2}}{\tau c_{1}}{\rm
Tr}\left(G_{1}-\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}T_{{\rm
u},\alpha}\right)\left(G_{1}-\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}T_{{\rm
u},\alpha}\right)^{*}-\sum_{\alpha=2}^{t}f_{\alpha}{\rm Tr}\big{(}T_{{\rm
u},\alpha}T_{{\rm u},\alpha}^{*}\big{)}$ $\displaystyle+\sum_{1\leq i<j\leq
n}\left|a_{1}(G_{1})_{i,j}+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}(a_{\alpha}-a_{1})t_{i,j}^{(\alpha)}\right|^{2}+O\left(\Omega_{\rm
error}^{3}\right),$

where

$\Omega_{\rm error}=\|G_{1}\|+\|T_{\rm
u}\|_{2}+\|\Gamma_{1}\|+\|\widetilde{T}_{\rm d}\|_{2}+\|Q_{t+1}\|.$

In view of (3.1) and (3.4),

$\displaystyle-\sum_{\alpha=1}^{t}\frac{f_{\alpha}^{2}}{2\tau^{2}c_{\alpha}}{\rm
Tr}\big{(}\widetilde{T}_{{\rm
d},\alpha}^{2}\big{)}+\frac{1}{\tau}N^{-\frac{1}{2}}\sum_{\alpha=1}^{t}\Big{(}\overline{a}_{\alpha}{\rm
Tr}\big{(}\hat{Z}\widetilde{T}_{{\rm d},\alpha}\big{)}+a_{\alpha}{\rm
Tr}\big{(}\hat{Z}^{*}\widetilde{T}_{{\rm d},\alpha}\big{)}\Big{)}$ (3.26)
$\displaystyle=-\frac{f_{1}^{2}}{2\tau^{2}c_{1}}{\rm
Tr}\Big{(}\Gamma_{1}-\sum_{\alpha=2}^{t}\widetilde{T}_{{\rm
d},\alpha}\Big{)}^{2}-\sum_{\alpha=2}^{t}\frac{f_{\alpha}^{2}}{2\tau^{2}c_{\alpha}}{\rm
Tr}\big{(}\widetilde{T}_{{\rm
d},\alpha}^{2}\big{)}+\frac{1}{\tau}N^{-\frac{1}{2}}\sum_{\alpha=2}^{t}{\rm
Tr}\Big{(}\big{(}\overline{a}_{\alpha}\hat{Z}+a_{\alpha}\hat{Z}^{*}\big{)}\widetilde{T}_{{\rm
d},\alpha}\Big{)}$ $\displaystyle+\frac{1}{\tau}N^{-\frac{1}{2}}{\rm
Tr}\Big{(}\big{(}\overline{a}_{1}\hat{Z}+a_{1}\hat{Z}^{*}\big{)}\big{(}\Gamma_{1}-\sum_{\alpha=2}^{t}\widetilde{T}_{{\rm
d},\alpha}\big{)}\Big{)}+O\Big{(}\big{(}N^{-\frac{1}{2}}+\Omega_{\rm
error}\big{)}^{3}\Big{)}.$

Also we still have (3.16) with $z_{0}=0$. Combining all together,

$f(T,Y)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}}\mathbb{I}_{n}\big{\\}},0,0\big{)}=F_{0}+F_{1}+O(F_{2}),$

where

$F_{0}=\frac{n^{2}}{N}-\frac{N^{-1}}{2}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}\left(\overline{a_{\alpha}}^{2}{\rm
Tr}\big{(}\hat{Z}^{2}\big{)}+a_{\alpha}^{2}{\rm
Tr}\big{(}\hat{Z}^{*}\big{)}^{2}\right),$

$\displaystyle F_{1}=-\frac{f_{1}^{2}}{2\tau^{2}c_{1}}{\rm
Tr}\Big{(}\Gamma_{1}-\sum_{\alpha=2}^{t}\widetilde{T}_{{\rm
d},\alpha}\Big{)}^{2}-\sum_{\alpha=2}^{t}\frac{f_{\alpha}^{2}}{2\tau^{2}c_{\alpha}}{\rm
Tr}\big{(}\widetilde{T}_{{\rm d},\alpha}^{2}\big{)}-\frac{1}{\tau}{\rm
Tr}\big{(}Q_{t+1}A_{t+1}^{*}A_{t+1}Q_{t+1}^{*}\big{)}$ (3.27)
$\displaystyle+\frac{N^{-\frac{1}{2}}}{\tau}\left(\sum_{\alpha=2}^{t}{\rm
Tr}\Big{(}\big{(}\overline{a}_{\alpha}\hat{Z}+a_{\alpha}\hat{Z}^{*}\big{)}\widetilde{T}_{{\rm
d},\alpha}\Big{)}+{\rm
Tr}\Big{(}\big{(}\overline{a}_{1}\hat{Z}+a_{1}\hat{Z}^{*}\big{)}\big{(}\Gamma_{1}-\sum_{\alpha=2}^{t}\widetilde{T}_{{\rm
d},\alpha}\big{)}\Big{)}\right)$
$\displaystyle-\sum_{\alpha=2}^{t}\frac{f_{\alpha}}{\tau}{\rm
Tr}\big{(}T_{{\rm u},\alpha}T_{{\rm
u},\alpha}^{*}\big{)}-\frac{f_{1}^{2}}{\tau^{2}c_{1}}{\rm
Tr}\Big{(}G_{1}-\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}T_{{\rm
u},\alpha}\Big{)}\Big{(}G_{1}-\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}T_{{\rm u},\alpha}\Big{)}^{*}$
$\displaystyle+\frac{1}{\tau}\sum_{1\leq i<j\leq
n}\Big{|}a_{1}(G_{1})_{i,j}+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}(a_{\alpha}-a_{1})t_{i,j}^{(\alpha)}\Big{|}^{2}$
$\displaystyle-\frac{P_{1}}{2}{\rm
Tr}(YY^{*})^{2}+N^{-\frac{1}{2}}\left(\overline{P_{0}}{\rm
Tr}\big{(}Y^{*}Y\hat{Z}\big{)}+P_{0}{\rm
Tr}\big{(}YY^{*}\hat{Z}^{*}\big{)}\right),$

and

$\displaystyle
F_{2}=N^{-\frac{3}{2}}+\|Y\|^{6}+N^{-1}\|Y\|^{2}+N^{-\frac{1}{2}}\|Y\|^{4}+\big{(}\Omega_{\rm
error}+N^{-\frac{1}{2}}\big{)}^{3}.$

For Taylor expansion of $g(Y,U,T)$ defined in (2.10), we still have (3.17),
(3.18) and (3.19), while as $z_{0}=0$,

$\det\begin{bmatrix}\sqrt{\gamma_{N}}Z&-Y^{*}\\\
Y&\sqrt{\gamma_{N}}Z^{*}\end{bmatrix}=\big{(}\det(YY^{*})\big{)}^{n}+O\Big{(}\sum_{\alpha,\beta}\|Y\|^{\alpha}N^{-\frac{\beta}{2}}\Big{)},$

where $\gamma_{N}=\frac{N}{N-n}$, the finite non-negative integer pair
$(\alpha,\beta)$ is such that

$\frac{\alpha}{4}+\frac{\beta}{2}\geq\frac{n^{2}}{2}+\frac{1}{4}.$ (3.28)

The remaining part is devoted to the last determinant in (2.10).

Recalling $\hat{L}_{1}$ and $\hat{L}_{2}$ defined in (2.3) and (2.5), since
$0$ is not an eigenvalue of $A_{t+1}$ and all $a_{\alpha}\neq 0$,
$\widehat{L}_{1}$ is invertible when $N$ is sufficiently large. Together with
the decomposition

$\widehat{L}_{2}=\left[\begin{smallmatrix}\mathbb{I}_{2n}\otimes T_{1}^{*}\\\
\vdots\\\ \mathbb{I}_{2n}\otimes T_{t}^{*}\\\
\left[\begin{smallmatrix}\mathbb{I}_{n}\otimes A_{t+1}Q_{t+1}^{*}&\\\
&\mathbb{I}_{n}\otimes
A_{t+1}^{*}Q_{t+1}^{*}\end{smallmatrix}\right]\end{smallmatrix}\right]\left[\begin{smallmatrix}\left[\begin{smallmatrix}a_{1}\mathbb{I}_{n}&\\\
&\overline{a}_{1}\mathbb{I}_{n}\end{smallmatrix}\right]\otimes
T_{1}&,\cdots,&\left[\begin{smallmatrix}a_{t}\mathbb{I}_{n}&\\\
&\overline{a}_{t}\mathbb{I}_{n}\end{smallmatrix}\right]\otimes
T_{t},&\mathbb{I}_{2n}\otimes Q_{t+1}\end{smallmatrix}\right],$

change the order of the above matrix product and we obtain

$\displaystyle\det\\!\bigg{(}\widehat{L}_{1}+\sqrt{\gamma_{N}}\widehat{L}_{2}\bigg{)}=\big{|}\det\big{(}A_{t+1}\big{)}\big{|}^{2n}\Big{(}\prod_{\alpha=1}^{t}\det\big{(}A_{\alpha}\big{)}\Big{)}^{n}\big{(}1+O(\|Y\|+N^{-\frac{1}{4}})\big{)}$
(3.29)
$\displaystyle\times\det\bigg{(}\mathbb{I}_{2n^{2}}+\sqrt{\gamma_{N}}\begin{bmatrix}Q_{1,1}&Q_{1,2}\\\
Q_{2,1}&Q_{2,2}\end{bmatrix}+O\big{(}\|Q_{t+1}\|^{2}\big{)}\bigg{)},$

where

$\begin{bmatrix}Q_{1,1}&Q_{1,2}\\\
Q_{2,1}&Q_{2,2}\end{bmatrix}:=\sum_{\alpha=1}^{t}\left(A_{\alpha}^{-1}\left[\begin{smallmatrix}a_{\alpha}\mathbb{I}_{n}&\\\
&\overline{a}_{\alpha}\mathbb{I}_{n}\end{smallmatrix}\right]\right)\otimes(T_{\alpha}T_{\alpha}^{*}).$

Obviously, by (2.2), setting
$Z_{\alpha}=\sqrt{\gamma_{N}}(Z-a_{\alpha}\mathbb{I}_{n})$ we have

$\displaystyle
Q_{1,1}=\sum_{\alpha=1}^{t}a_{\alpha}\Big{(}Z_{\alpha}+Y^{*}\big{(}Z_{\alpha}^{*}\big{)}^{-1}Y\Big{)}^{-1}\otimes(T_{\alpha}T_{\alpha}^{*}),\quad
Q_{2,2}=\sum_{\alpha=1}^{t}\overline{a}_{\alpha}\big{(}Z_{\alpha}^{*}+YZ_{\alpha}^{-1}Y^{*}\big{)}^{-1}\otimes(T_{\alpha}T_{\alpha}^{*}),$
$\displaystyle
Q_{1,2}=\sum_{\alpha=1}^{t}\overline{a}_{\alpha}Z_{\alpha}^{-1}Y^{*}\big{(}Z_{\alpha}^{*}+YZ_{\alpha}^{-1}Y^{*}\big{)}^{-1}\otimes(T_{\alpha}T_{\alpha}^{*}),$
$\displaystyle
Q_{2,1}=-\sum_{\alpha=1}^{t}a_{\alpha}\big{(}Z_{\alpha}^{*}\big{)}^{-1}Y\Big{(}Z_{\alpha}+Y^{*}\big{(}Z_{\alpha}^{*}\big{)}^{-1}Y\Big{)}^{-1}\otimes(T_{\alpha}T_{\alpha}^{*}).$

In view of (3.20) and (3.21), we have

$\mathbb{I}_{n^{2}}+\sqrt{\gamma_{N}}Q_{1,1}=O(\Delta),$ (3.30)

where

$\Delta=N^{-\frac{1}{2}}+\Omega_{\rm error}+\|Y\|^{2}.$
$\mathbb{I}_{n^{2}}+\sqrt{\gamma_{N}}Q_{2,2}=O(\Delta).$ (3.31)

Also, we have

$\displaystyle\sqrt{\gamma_{N}}Q_{1,2}$
$\displaystyle=\sum_{\alpha=1}^{t}\frac{\overline{a}_{\alpha}}{-a_{\alpha}}\frac{Y^{*}}{\overline{-a_{\alpha}}}\otimes\Big{(}\frac{\tau
c_{\alpha}}{f_{\alpha}}\mathbb{I}_{n}\Big{)}+O(\Delta)$ (3.32)
$\displaystyle=\tau\overline{P}_{0}(Y^{*}\otimes\mathbb{I}_{n})+O(\Delta),$

and

$\sqrt{\gamma_{N}}Q_{2,1}=-\tau P_{0}(Y\otimes\mathbb{I}_{n})+O(\Delta).$
(3.33)

Combing (3.29), (3.30), (3.31), (3.32) and (3.33) we have

$\displaystyle\det\\!\bigg{(}\widehat{L}_{1}+\sqrt{\gamma_{N}}\widehat{L}_{2}\bigg{)}=\big{|}\det\big{(}A_{t+1}\big{)}\big{|}^{2n}\Big{(}\prod_{\alpha=1}^{t}f_{\alpha}^{n^{2}}+O\big{(}N^{-\frac{1}{2}}+\|Y\|\big{)}\Big{)}$
$\displaystyle\times\left(\tau^{2n^{2}}|P_{0}|^{2n^{2}}\big{(}\det(YY^{*})\big{)}^{n}+O\Big{(}\sum_{\alpha,\beta}\|Y\|^{\alpha}\big{(}N^{-\frac{1}{2}}+\Omega_{\rm
error}\big{)}^{\beta}\Big{)}\right),$

where the finite non-negative integer pair $(\alpha,\beta)$ is such that
(3.28).

For the final proof, computation of $e^{NF_{0}}D_{N,n}$ can be obtained
directly by setting $r_{0}=0$ and $z_{0}=0$ in (2.45).

By repeating almost the same procedure we can get the desired result. Last but
not least, in case $z_{0}=0$, the change of integration variables is

$Y\rightarrow
N^{-\frac{1}{4}}Y,\quad\big{(}G_{1},\Gamma_{1},\widetilde{T}_{{\rm
d},\alpha},T_{{\rm u},\alpha}\big{)}\rightarrow
N^{-\frac{1}{2}}\big{(}G_{1},\Gamma_{1},\widetilde{T}_{{\rm d},\alpha},T_{{\rm
u},\alpha}\big{)},\quad\alpha=2,\cdots,t.$

When comparing (3.27) with (2.32), we discover that $\Gamma_{1}$ takes place
of $(Q_{0}Q_{0}^{*})^{({\rm d})}$, while $G_{1}$ takes place of
$(Q_{0}Q_{0}^{*})^{({\rm off},{\rm u})}$, other terms remain unchanged. This
is caused by the difference of typical sizes of $\Gamma_{1}$ and $G_{1}$
between $z_{0}=0$ and $z_{0}\not=0$. So when dealing with the term (3.25), we
first need to integrate out $\\{T_{{\rm u},\alpha}|\alpha=2,\cdots,t\\}$.
Applying Proposition A.3 gets to the fact that the result does not depend on
$G_{1}$. When dealing with the terms (3.26), again we need to integrate out
$\\{\widetilde{T}_{{\rm d},\alpha}|\alpha=2,\cdots,t\\}$. We then get a
similar result in form with (2.43).

Acknowledgements

This work was supported by the National Natural Science Foundation of China #
12371157 and #12090012.

##  Appendix A Properties on matrices and integrals

The following well-known property for tensor product (see e.g. [20, eqn(6)]).

######  Proposition A.1.

For a $p\times q$ matrix A and an $m\times n$ matrix B, there exist
permutation matrices $\mathbb{I}_{p,m}$, which only depend on $p,m$ and
satisfy $\mathbb{I}_{p,m}^{-1}=\mathbb{I}_{m,p}=\mathbb{I}_{p,m}^{t}$, such
that

$\mathbb{I}_{p,m}\left(A\otimes B\right)\mathbb{I}_{q,n}^{-1}=B\otimes A.$
(A.1)

An integral over non-positive definite matrices will be useful, for proof see
[25, Proposition A.2].

######  Proposition A.2.

For a non-positive definite Hermitian matrix
$H_{n}=\left[h_{i,j}\right]_{i,j=1}^{n}$, given an integer $r_{0}\geq n$, then

$\int_{H_{n}\leq 0}\left(\det(H_{n})\right)^{r_{0}-n}\prod_{i<j}^{n}{\rm
d}h_{i,j}=\pi^{\frac{n(n-1)}{2}}((r_{0}-1)!)^{-n}\prod_{k=1}^{n}(r_{0}-k)!\prod_{j=1}^{n}(h_{j,j})^{r_{0}-1}.$
(A.2)

We also need a Gaussian integral over several strictly upper triangular
matrices.

######  Proposition A.3.

Let T, $\\{G_{\alpha}|\alpha=2,\cdots,t\\}$ be strictly upper triangular
matrices of size $n\times n$. Assume that $\sum_{\alpha=1}^{t}\frac{\tau
c_{\alpha}}{f_{\alpha}}=1$, with $f_{\alpha}=|z_{0}-a_{\alpha}|^{2}$ and
$P_{0}=\sum_{\alpha=1}^{t}\frac{c_{\alpha}(a_{\alpha}-z_{0})}{f_{\alpha}^{2}}$,
then

$\displaystyle\int\cdots\int\exp\Big{\\{}-\frac{f_{1}^{2}}{\tau^{2}c_{1}}{\rm
Tr}\big{(}T+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}G_{\alpha}\big{)}\big{(}T+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}G_{\alpha}\big{)}^{*}-\sum_{\alpha=2}^{t}\frac{f_{\alpha}}{\tau}{\rm
Tr}\big{(}G_{\alpha}G_{\alpha}^{*}\big{)}$ (A.3)
$\displaystyle+\frac{1}{\tau}\sum_{1\leq i<j\leq
n}\Big{|}-(a_{1}-z_{0})T_{i,j}+\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}}}(a_{\alpha}-a_{1})(G_{\alpha})_{i,j}\Big{|}^{2}\Big{\\}}\prod_{\alpha=2}^{t}{\rm
d}G_{\alpha}$
$\displaystyle=\tau^{\frac{n(n-1)}{2}(t-2)}\Big{(}\frac{f_{1}}{c_{1}}\prod_{\alpha=1}^{t}f_{\alpha}|P_{0}|^{2}\Big{)}^{-\frac{n(n-1)}{2}}\pi^{\frac{n(n-1)(t-1)}{2}}.$

###### Proof.

By letting $a_{\alpha}\rightarrow\sqrt{\tau}a_{\alpha}$ and
$z_{0}\rightarrow\sqrt{\tau}z_{0}$, without loss of generality, it suffices to
prove the conclusion with $\tau=1.$

Denote the exponential term on the left hand side of (A.3) as
$L(G_{\alpha},T)$. From basic computations, we can rewrite $L(G_{\alpha},T)$
as the polynomials of $(G_{\alpha})_{i,j}$ for $\alpha=2,\cdots,t,\ 1\leq
i<j\leq n$,

$\displaystyle L(G_{\alpha},T)=\sum_{1\leq i<j\leq
n}\Big{(}-(\vec{G}_{i,j}-T_{i,j}\vec{d}A^{-1})A(\vec{G}_{i,j}-T_{i,j}\vec{d}A^{-1})^{*}+|T_{i,j}|^{2}\big{(}\vec{d}A^{-1}\vec{d}^{*}-\frac{f_{1}^{2}}{c_{1}}+f_{1}\big{)}\Big{)},$

where

$\displaystyle\vec{G}_{i,j}=\big{(}(G_{2})_{i,j},\cdots,(G_{t})_{i,j}\big{)},\quad
A=[A_{\alpha,\beta}]_{\alpha,\beta=2}^{t},\quad\vec{d}=\big{(}d_{2},\cdots,d_{t}\big{)},$
$\displaystyle
A_{\alpha,\beta}=\Big{(}\frac{f_{1}^{2}}{c_{1}}-(a_{\alpha}-a_{1})\overline{a_{\beta}-a_{1}}\Big{)}\sqrt{\frac{c_{\alpha}c_{\beta}}{f_{\alpha}f_{\beta}}}+f_{\alpha}\delta_{\alpha,\beta},\quad\alpha,\beta=2,\cdots,t,$
$\displaystyle
d_{\alpha}=-\sqrt{\frac{c_{\alpha}}{f_{\alpha}}}\Big{(}(a_{1}-z_{0})\overline{a_{\alpha}-a_{1}}+\frac{f_{1}^{2}}{c_{1}}\Big{)},\quad\alpha=2,\cdots,t,$

Calculate the gaussian integrals according to row vectors $\vec{G}_{i,j}$ and
we obtain

$\int e^{L(G_{\alpha},T)}\prod_{\alpha=2}^{t}{\rm
d}G_{\alpha}=\exp\Big{\\{}\big{(}\vec{d}A^{-1}\vec{d}^{*}-\frac{f_{1}^{2}}{c_{1}}+f_{1}\big{)}{\rm
Tr}(TT^{*})\Big{\\}}\big{(}\det(A)\big{)}^{-\frac{n(n-1)}{2}}\pi^{\frac{n(n-1)}{2}(t-1)},$

Later we will prove

$\det(A)=\frac{f_{1}}{c_{1}}\Big{(}\prod_{\alpha=1}^{t}f_{\alpha}\Big{)}|P_{0}|^{2}$
(A.4)

and

$\vec{d}A^{-1}\vec{d}^{*}-\frac{f_{1}^{2}}{c_{1}}+f_{1}=0,$ (A.5)

these two facts complete the proof.

To prove(A.4), first rewrite A as

$A={\rm
diag}\Big{(}\sqrt{\frac{c_{\alpha}}{f_{\alpha}}}\Big{)}_{\alpha=2}^{t}\Big{(}\frac{f_{1}^{2}}{c_{1}}\vec{1}^{t}\vec{1}+\widehat{A}\Big{)}{\rm
diag}\Big{(}\sqrt{\frac{c_{\beta}}{f_{\beta}}}\Big{)}_{\beta=2}^{t},$

where

$\vec{d}A^{-1}\vec{d}^{*}-\frac{f_{1}^{2}}{c_{1}}+f_{1}=\frac{f_{1}^{2}}{K_{1}c_{1}}\Big{(}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}}-1\Big{)}=0,$

Then

$\det(A)=\det\big{(}\widehat{A}\big{)}\big{(}1+\frac{f_{1}^{2}}{c_{1}}\vec{1}\widehat{A}^{-1}\vec{1}^{t}\big{)}\prod_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}}$
(A.6)

From basic computations, for $\alpha,\beta=2,\cdots,t$,

$\displaystyle\big{(}\widehat{A}^{-1}\big{)}_{\alpha,\beta}=\frac{c_{\alpha}}{f_{\alpha}^{2}}\delta_{\alpha,\beta}+\frac{c_{\alpha}c_{\beta}}{f_{\alpha}^{2}f_{\beta}^{2}}(a_{\alpha}-a_{1})\overline{a_{\beta}-a_{1}}\Big{(}1-\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}|a_{\alpha}-a_{1}|^{2}\Big{)}^{-1},$
(A.7)
$\displaystyle\det\big{(}\widehat{A}\big{)}=\prod_{\alpha=2}^{t}\frac{f_{\alpha}^{2}}{c_{\alpha}}\Big{(}1-\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}|a_{\alpha}-a_{1}|^{2}\Big{)}.$

Recall the definition of $P_{0}$ and $P_{1}$ in (1.7) and (1.8),

$\Big{|}\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}(a_{\alpha}-a_{1})\Big{|}^{2}=\Big{|}\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}\big{(}(a_{\alpha}-z_{0})-(a_{1}-z_{0})\big{)}\Big{|}^{2}=|P_{0}-(a_{1}-z_{0})P_{1}|^{2}.$

Also from $P_{00}(z_{0})=\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}}=1,$

$\displaystyle\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}|a_{\alpha}-a_{1}|^{2}$
$\displaystyle=\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}\big{|}(a_{\alpha}-z_{0})-(a_{1}-z_{0})\big{|}^{2}$
$\displaystyle=1-\overline{a_{1}-z_{0}}P_{0}-(a_{1}-z_{0})\overline{P}_{0}+|a_{1}-z_{0}|^{2}P_{1}.$

so from (A.7) and basic computations we obtain

$\big{(}1+\frac{f_{1}^{2}}{c_{1}}\vec{1}\widehat{A}^{-1}\vec{1}^{t}\big{)}\Big{(}1-\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}|a_{\alpha}-a_{1}|^{2}\Big{)}=\frac{f_{1}^{2}}{c_{1}}|P_{0}|^{2}$

Hence, Combining (A.6) and (A.7) we obtain (A.4).

To prove (A.5), first we have

$\big{(}A^{-1}\big{)}_{\alpha,\beta}=\sqrt{\frac{f_{\alpha}f_{\beta}}{c_{\alpha}c_{\beta}}}\frac{c_{\alpha}c_{\beta}}{f_{\alpha}^{2}f_{\beta}^{2}K_{1}}\big{(}\frac{f_{\alpha}^{2}}{c_{\alpha}}\delta_{\alpha,\beta}K_{1}+K_{\alpha,\beta}\big{)},$
(A.8)

with

$K_{1}=1+\frac{f_{1}^{2}}{c_{1}}G_{2}-G_{1}\big{(}1+\frac{f_{1}^{2}}{c_{1}}G_{2}\big{)}+\frac{f_{1}^{2}}{c_{1}}|G_{3}|^{2},$
(A.9)

and

$K_{\alpha,\beta}=(a_{\alpha}-a_{1})\overline{a_{\beta}-a_{1}}-\frac{f_{1}^{2}}{c_{1}}+\frac{f_{1}^{2}}{c_{1}}G_{1}-\frac{f_{1}^{2}}{c_{1}}\overline{a_{\beta}-a_{1}}G_{3}-\frac{f_{1}^{2}}{c_{1}}(a_{\alpha}-a_{1})\overline{G}_{3}+\frac{f_{1}^{2}}{c_{1}}(a_{\alpha}-a_{1})\overline{a_{\beta}-a_{1}}G_{2},$
(A.10)

where

$G_{1}=\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}|a_{\alpha}-a_{1}|^{2},\quad
G_{2}=\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}},\quad
G_{3}=\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}(a_{\alpha}-a_{1}).$

(A.8) can be seen by verifying the equality

$\sum_{\gamma=2}^{t}A_{\alpha,\gamma}\big{(}A^{-1}\big{)}_{\gamma,\beta}=\delta_{\alpha,\beta}$

directly. To compute
$\vec{d}A^{-1}\vec{d}^{*}=\sum_{\alpha,\beta=2}^{t}d_{\alpha}\big{(}A^{-1}\big{)}_{\alpha,\beta}d_{\beta}$,
we first expand the right hand side according to the terms of $K_{1}$ and
$K_{\alpha,\beta}$ in (A.9) and (A.10), then for each term we compute the sum
$\sum_{\alpha,\beta=2}^{t}$ to the form represented by $G_{1}$, $G_{2}$ and
$G_{3}$, for example, for the first term
$(a_{\alpha}-a_{1})\overline{a_{\beta}-a_{1}}$ in $K_{\alpha,\beta}$(cf
(A.10)), we have

$\displaystyle\sum_{\alpha,\beta=2}^{t}\frac{c_{\alpha}c_{\beta}}{f_{\alpha}^{2}f_{\beta}^{2}}(a_{\alpha}-a_{1})\overline{a_{\beta}-a_{1}}\big{(}(a_{1}-z_{0})\overline{a_{\alpha}-a_{1}}+\frac{f_{1}^{2}}{c_{1}}\big{)}\big{(}\overline{a_{1}-z_{0}}(a_{\beta}-a_{1})+\frac{f_{1}^{2}}{c_{1}}\big{)}$
$\displaystyle=\big{(}(a_{1}-z_{0})G_{1}+\frac{f_{1}^{2}}{c_{1}}G_{3}\big{)}\big{(}\overline{a_{1}-z_{0}}G_{1}+D\frac{f_{1}^{2}}{c_{1}}\overline{G}_{3}\big{)},$

Following this process, after basic calculations we obtain

$\displaystyle\vec{d}A^{-1}\vec{d}^{*}-\frac{f_{1}^{2}}{c_{1}}+f_{1}=\frac{f_{1}^{2}}{K_{1}c_{1}}\Big{(}\frac{c_{1}}{f_{1}}+f_{1}G_{2}-1+G_{1}+(a_{1}-z_{0})\overline{G}_{3}+\overline{a_{1}-z_{0}}G_{3}\Big{)}.$
(A.11)

Now rewrite $a_{\alpha}-a_{1}$ as $(a_{\alpha}-z_{0})-(a_{1}-z_{0})$, then we
have

$\displaystyle
G_{1}=\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}}-\overline{a_{1}-z_{0}}\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}(a_{\alpha}-z_{0})-(a_{1}-z_{0})\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}\overline{a_{\alpha}-z_{0}}+f_{1}\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}$
$\displaystyle
G_{2}=\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}},\quad
G_{3}=\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}(a_{\alpha}-z_{0})-(a_{1}-z_{0})\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}^{2}}.$

Substituting the above into (A.11), we get that

$\vec{d}A^{-1}\vec{d}^{*}-\frac{f_{1}^{2}}{c_{1}}+f_{1}=\frac{f_{1}^{2}}{K_{1}c_{1}}\Big{(}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}}-1\Big{)}=0,$

and finally we get (A.5). ∎

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