Mixed graphs with cut vertices having exactly two positive eigenvalues

Xiaocong He, Lihua Feng***Corresponding author.
Email addresses<EMAIL_ADDRESS>(X.C. He<EMAIL_ADDRESS>(L.H. Feng)

School of Mathematics and Statistics, HNP-LAMA, Central South University,
Changsha, Hunan 410083, P. R. China

Abstract: A mixed graph is obtained by orienting some edges of a simple graph.
The positive inertia index of a mixed graph is defined as the number of
positive eigenvalues of its Hermitian adjacency matrix, including
multiplicities. This matrix was introduced by Liu and Li, independently by Guo
and Mohar, in the study of graph energy. Recently, Yuan et al. characterized
the mixed graphs with exactly one positive eigenvalue. In this paper, we study
the positive inertia indices of mixed graphs and characterize the mixed graphs
with cut vertices having positive inertia index 2.

Keywords: Mixed graph; Cut vertices; Hermitian adjacency matrix; Positive
inertia index

##  1 Introduction

We will start with introducing some background that will lead to our main
results. Some significant previously established facts will also be presented.

###  1.1 Background

Let $G$ be a simple graph with vertex set
$V(G)=\\{v_{1},v_{2},\ldots,v_{n}\\}$ and edge set $E(G)$. A mixed graph
$\widetilde{G}$ is obtained by orienting some edges of $G$. Accordingly, $G$
is referred to as the underlying graph of $\widetilde{G}$. Consequently, a
mixed graph $\widetilde{G}$ is an ordered triple $(V(G),E_{0},E_{1})$, where
$E_{0}$ is the undirected edge set and $E_{1}$ is the directed edge (arc) set.
It is obvious that $\widetilde{G}$ is undirected if $E_{0}=E(G)$ while
$\widetilde{G}$ is an oriented graph if $E_{0}=\emptyset$. Thus, mixed graphs
are the generalizations of simple graphs and oriented graphs.

The $Hermitian$ $adjacency$ $matrix$ $H(\widetilde{G})=(h_{st})_{n\times n}$
of $\widetilde{G}$ was proposed, independently, by Liu and Li [28], and Guo
and Mohar [13]. It is defined as

$h_{st}=\left\\{\begin{array}[]{ll}1,&\hbox{if $v_{s}v_{t}$ is an undirected
edge;}\\\\[2.84526pt] i,&\hbox{if $\overrightarrow{v_{s}v_{t}}$ is an
arc;}\\\\[2.84526pt] -i,&\hbox{if $\overrightarrow{v_{t}v_{s}}$ is an
arc;}\\\\[2.84526pt] 0,&\hbox{otherwise,}\end{array}\right.$

where $i=\sqrt{-1}$. Note that $H(\widetilde{G})$ is Hermitian, that is,
$H(\widetilde{G})^{*}=H(\widetilde{G})$, where $H(\widetilde{G})^{*}$ denotes
the conjugate transpose of $H(\widetilde{G})$. Thus the eigenvalues of
$H(\widetilde{G})$ are real, called the $H$-eigenvalues of $\widetilde{G}$.
The $H$-rank of $\widetilde{G}$, denoted by $rk(\widetilde{G})$, is the rank
of $H(\widetilde{G})$, and the number of positive, negative and zero
eigenvalues of $H(\widetilde{G})$ are defined as $positive$ $inertia$ $index$,
$negative$ $inertia$ $index$ and $nullity$ of $\widetilde{G}$, denoted by
$p(\widetilde{G})$, $n(\widetilde{G})$ and $\eta(\widetilde{G})$,
respectively. Obviously,
$p(\widetilde{G})+n(\widetilde{G})=rk(\widetilde{G})$. The inertia index of
$\widetilde{G}$ is the triple ${\rm
In}(\widetilde{G})=(p(\widetilde{G}),n(\widetilde{G}),\eta(\widetilde{G}))$.
Clearly, the Hermitian adjacency matrix of a mixed graph is a natural
generalization of the adjacency matrix of a simple graph. Therefore the mixed
graph attracts more and more researchers’ attention; see [13, 21, 24, 25, 28,
35, 36, 52, 51, 50, 59, 60].

In [13, 35], the authors considered the so-called four-way switching operation
keeping the $H$-spectrum of a mixed graph unchanged. We describe the
definition in the matrix-theoretic language below.

######  Definition 1.

[35] Let $\widetilde{G_{1}}$ and $\widetilde{G_{2}}$ be two mixed graphs. A
four-way switching (FWS for short) is the operation of changing
$\widetilde{G_{1}}$ into $\widetilde{G_{2}}$, if there exists a diagonal
matrix $Q$ with $Q_{vv}\in\\{\pm 1,\pm i\\}$ such that
$H(\widetilde{G_{2}})=Q^{-1}H(\widetilde{G_{1}})Q$.

In fact, it is often difficult to check whether a mixed graph
$\widetilde{G_{2}}$ can be obtained from a mixed graph $\widetilde{G_{1}}$ by
a four-way switching. Because of this, Mohar [35] gave some special cases of
four-way switching called two-way switching, which can be described as a
special similarity transformation. There are really two different cases as
follows.

  * (1)

Directed two-way switching. Reversing all edges in an edge cut consisting of
directed edges only.

  * (2)

Mixed two-way switching. If there is an edge cut which contains undirected
edges (possibly empty) and directed edges (possibly empty) in one direction
only, replacing the each directed edge in the cut by a single undirected edge,
and replacing each former undirected edge by a single directed edge in the
direction opposite to the direction of former directed edges.

The converse ${\widetilde{G}}^{\top}$ of a mixed graph $\widetilde{G}$ is
obtained from $\widetilde{G}$ by reversing all directed edges in
$\widetilde{G}$. They are cospectral since
$H(\widetilde{G}^{\top})=H(\widetilde{G})^{\top}$. Two mixed graphs
$\widetilde{G_{1}}$ and $\widetilde{G_{2}}$ are said to be switching
equivalent if $\widetilde{G_{1}}$ can be changed into $\widetilde{G_{2}}$ by a
FWS and taking a possible converse.

The circle group, which is denoted by $\mathbb{T}=\\{z\in\mathbb{C}:|z|=1\\}$,
is a subgroup of the multiplicative group of all nonzero complex numbers
$\mathbb{C}^{\times}$. If $v_{s}v_{t}\in E(G)$, we denote by $e_{v_{s}v_{t}}$
the oriented edge from $v_{s}$ to $v_{t}$. Let $\overrightarrow{E}$ be the set
of oriented edges of $G$, i.e.,
$\overrightarrow{E}=\\{e_{v_{s}v_{t}},e_{v_{t}v_{s}}:v_{s}v_{t}\in E(G)\\}$. A
complex unit gain graph $\Phi=(G,\mathbb{T},\varphi)$ is a graph with a gain
function $\varphi:\overrightarrow{E}\rightarrow\mathbb{T}$, such that
$\varphi(e_{v_{s}v_{t}})=\varphi(e_{v_{t}v_{s}})^{-1}=\overline{\varphi(e_{v_{t}v_{s}})}$
for any $v_{s}v_{t}\in E(G)$. Clearly, the complex unit gain graph is a
natural generalization of a simple graph. So, the complex unit gain graph
attracts much plenty of researchers’ attention; see [18, 24, 30, 31, 29, 34,
43, 55, 58, 61].

We usually write $G^{\varphi}$ for a complex unit gain graph
$\Phi=(G,\mathbb{T},\varphi)$. The adjacency matrix of $G^{\varphi}$ is the
Hermitian matrix $A(G^{\varphi})=(a_{st}^{\varphi})_{n\times n}$, where

$a_{st}^{\varphi}=\left\\{\begin{array}[]{ll}\varphi(e_{v_{s}v_{t}}),&\hbox{if
$v_{s}v_{t}\in E(G)$;}\\\ 0,&\hbox{otherwise.}\end{array}\right.$

The $rank$ of $G^{\varphi}$, denoted by $rk(G^{\varphi})$, is the rank of
$A(G^{\varphi})$. The $positive$ $inertia$ $index$ $p(G^{\varphi})$, the
$negative$ $inertia$ $index$ $n(G^{\varphi})$ and the $nullity$
$\eta(G^{\varphi})$ of $G^{\varphi}$ are defined to be the number of positive
eigenvalues, negative eigenvalues and zero eigenvalues of $A(G^{\varphi})$,
respectively. Obviously, $p(G^{\varphi})+n(G^{\varphi})=rk(G^{\varphi})$.

If $\varphi(\overrightarrow{E})\subseteq\\{1,\pm i\\}\subseteq\mathbb{T}$,
then for any $v_{s}v_{t}\in E(G)$, we define

$h_{st}=\left\\{\begin{array}[]{ll}\varphi(e_{v_{s}v_{t}}),&\hbox{if
$v_{s}v_{t}\in E(G)$;}\\\ 0,&\hbox{otherwise.}\end{array}\right.$

Then $(h_{st})_{n\times n}$ is the Hermitian adjacency matrix of a mixed graph
$\widetilde{G}$ and we have $H(\widetilde{G})=(h_{st})_{n\times
n}=A(G^{\varphi})$. Hence, the complex unit gain graphs are the
generalizations of mixed graphs. We can view a mixed graph as a special case
of a complex unit gain graph.

Two complex unit gain graphs $G^{\varphi}$ and $G^{\varphi^{\prime}}$ are
switching equivalent if there is a mapping $\theta:V(G)\rightarrow\mathbb{T}$
such that $\varphi^{\prime}(uv)=\theta(u)^{-1}\varphi(uv)\theta(v)$. In this
case, the mapping $\varphi^{\prime}$ can be written as $\varphi\theta$. It
leads to that

$A(G^{\varphi^{\prime}})=diag(\theta(v_{1}),\theta(v_{2}),\ldots,\theta(v_{n}))^{-1}A(G^{\varphi})diag(\theta(v_{1}),\theta(v_{2}),\ldots,\theta(v_{n})).$

Therefore, $G^{\varphi}$ and $G^{\varphi^{\prime}}$ share the same spectrum.
It is clear that the switching equivalence is an equivalence relation.

We write $C_{n}$, $S_{n}$, $K_{n}$ for the cycle, star and complete graph of
order $n$, respectively. A mixed graph is called a mixed cycle, mixed star,
mixed complete graph, etc., if its underlying graph is a cycle, star, complete
graph, etc., respectively. A cut vertex in $G$ is a vertex whose removal
increases the number of connected components of $G$. For $v\in V$, $v$ is
called a $pendant$ $vertex$ if its degree $d_{G}(v)=1$. Let
$K_{n_{1},n_{2},\ldots,n_{k}}$ denote the complete $k$-partite graph with
$V_{1},V_{2},\ldots,V_{k}$ being the partition class such that $|V_{j}|=n_{j}$
for $1\leq j\leq k$. We follow [2] for notations and terminologies not defined
here.

A subgraph of $\widetilde{G}$ (resp. $G^{\varphi}$) is a subgraph of $G$ in
which each edge preserves the original direction (resp. gain fuction) in
$\widetilde{G}$ (resp. $G^{\varphi}$). Let $v\in V(G)$, we write
$\widetilde{G}-v$ (resp. $G^{\varphi}-v$) for the induced subgraph obtained
from $\widetilde{G}$ (resp. $G^{\varphi}$) by deleting the vertex $v$ and all
edges incident with $v$. For an induced subgraph $\widetilde{R}$ (resp.
$R^{\varphi}$) of $\widetilde{G}$ (resp. $G^{\varphi}$), denote by
$\widetilde{G}-\widetilde{R}$ (resp. $G^{\varphi}-R^{\varphi}$), the subgraph
obtained from $\widetilde{G}$ (resp. $G^{\varphi}$) by deleting all vertices
of $\widetilde{R}$ (resp. $R^{\varphi}$) and all incident edges. For a vertex
$v$ of $V(G)\backslash V(R)$, we write $\widetilde{R}+v$ (resp.
$R^{\varphi}+v$) to denote the induced subgraph of $\widetilde{G}$ (resp.
$G^{\varphi}$) with vertex set $V(R)\bigcup\\{v\\}$. A vertex of
$\widetilde{G}$ (resp. $G^{\varphi}$) is called a pendant vertex if it is of
degree one in $G$. A vertex of $\widetilde{G}$ (resp. $G^{\varphi}$) is called
a cut vertex if it is a cut vertex of $G$.

The spectral-based graph invariants are widely investigated in the literature,
such as inertia index [9, 12, 15, 23, 39, 37, 42, 52, 44, 49, 56, 57, 61] of
simple graphs, signed graphs, mixed graphs and complex unit gain graphs,
nullity [3, 7, 6, 8, 10, 14, 38, 41, 45, 56, 57, 64] of simple graphs and
signed graphs, rank [1, 11, 46, 47, 48, 54] of simple graphs, signed graphs
and complex unit gain graphs, the $H$-rank [4, 5, 50] of mixed graphs and
skew-rank [20, 22, 32, 33, 26, 27, 40, 53, 62] of oriented graphs. The
spectral parameter “inertia index” can determine the structure of a graph to
some extent and has attracted much attention recently. Smith [44]
characterized all (undirected) graphs with positive inertia index 1. Fan et
al. [9] determined sharp bounds for the positive and negative inertia index of
a graph. Oboudi [42] completely characterized the (undirected) graphs with
exactly two non-negative eigenvalues. Yu et al. [56] determined the signed
graphs with positive inertia index 1 and the signed graphs containing pendant
vertices with positive inertia index 2. Wang et al. [49] extended the above
work to the signed graphs containing cut vertices with positive inertia index
2. Gregory et al. [15] investigated the subadditivity of the inertia indices
and gave some properties of Hermitian rank that can determine the biclique
decomposition number. Gregory et al. [12] studied the inertia indices of a
partial join of two graphs and provided certain connections between biclique
decompositions of partial joins of graphs and the inertia indices. Geng et al.
[16] studied characterizations of graphs with given inertia index achieving
the maximum diameter.

Recently, much attention has been paid to the inertia index of mixed graphs.
Wissing and van Dam [52] characterized all mixed graphs with negative inertia
index 1. He et al. [17] studied the negative and positive inertia indices of
mixed unicyclic graphs and gave the upper and lower bounds of the negative and
positive inertia indices for mixed graphs. Zheng et al. [63] gave some
eigenvalue inequalities between two mixed graphs by means of inertia indices.
Wei et al. [51] investigated relations between the inertia indices of a mixed
graph and those of its underlying graph. Yuan et al. [60] provided a
characterization of mixed graphs with positive inertia index 1 and studied the
determination of some mixed graphs by their $H$-spectra.

Motivated by the results in [13, 28, 35, 51, 52, 49, 50, 56, 57, 60], in this
paper, we consider the mixed graphs with connected underlying graph and
characterize all mixed graphs with cut vertices which have positive inertia
index 2. For more general setting, we leave it for further research.

###  1.2 Main results

Before announcing the main theorem, we introduce one class mixed graph. Let
$\overrightarrow{C_{3}}(t_{1},t_{2},t_{3})$ be the mixed complete tripartite
graph shown in Figure. 1, where $|A|=t_{1},|B|=t_{2},|C|=t_{3}$. An odd
triangle is a mixed triangle containing odd number of directed edges. In
particular, $\overrightarrow{C_{3}}(1,1,1)$ is exactly an odd triangle.
Similarly, a mixed triangle containing even number of directed edges is called
an even triangle.

We now list our main results in the following.

######  Theorem 1.1.

Let $\widetilde{G}$ be a connected mixed graph with pendant vertices. Then
$p(\widetilde{G})=2$ if and only if $\widetilde{G}$ is obtained by adding some
undirected edges (possibly empty) and directed edges (possibly empty) with
arbitrary direction between the center of a mixed star and some vertices of a
mixed graph $\widetilde{F}$, where $\widetilde{F}$ is switching equivalent to
a complete multipartite graph or to some
$\overrightarrow{C_{3}}(t_{1},t_{2},t_{3})$.

A most interesting problem attracting us is to determine all mixed graphs
without pendant vertices whose positive inertia indices are 2. It seems
difficult for us to give a complete solution now. Note that a mixed graph with
a pendant vertex must have a cut vertex. In the following, we extend the
result of Theorem 1.1 to consider the mixed graphs with cut vertices which
have positive inertia index 2. In order to accomplish our goal, we should have
the aid of the following graph transformations and terminologies.

Let $\widetilde{G_{1}}$ and $\widetilde{G_{2}}$ be two mixed graphs. Let
$\widetilde{G_{1}}\bullet v\bullet\widetilde{G_{2}}$ be the mixed graph
obtained by identifying one vertex $v^{\prime}$ of $\widetilde{G_{1}}$ with
one vertex $v^{\prime\prime}$ of $\widetilde{G_{2}}$ (we denote the new vertex
by $v$).

Let $K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};p)$ be a simple graph obtained
from $K_{q_{1},\ldots,q_{r}}$ and $K_{n_{1},\ldots,n_{k}}$ by adding a new
vertex $v$ and adding edges such that $v$ is adjacent to all vertices of
$K_{q_{1},\ldots,q_{r}}$ and adjacent to all vertices of
$\bigcup_{j=1}^{p}V_{j}$, where $1\leq p\leq k$ and $V_{1},V_{2},\ldots,V_{k}$
are the partition classes of $V(K_{n_{1},n_{2},\ldots,n_{k}})$ such that
$|V_{j}|=n_{j}$ for $1\leq j\leq k$ (see Figure. 2). Similarly, let
$K(0;n_{1},\ldots,n_{k};p)$ be a simple graph obtained from
$K_{n_{1},\ldots,n_{k}}$ by adding a new vertex $v$ and adding edges such that
$v$ is adjacent to all vertices of $\bigcup_{j=1}^{p}V_{j}$, where $1\leq
p\leq k$. If $K_{q_{1},\ldots,q_{r}}$ and $K_{n_{1},\ldots,n_{k}}$ are
complete graphs $K_{r}$ and $K_{k}$, respectively, then we denote
$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};p)$ by $K(r;k;p)$. Similarly, if
$K_{n_{1},\ldots,n_{k}}$ is a complete graph $K_{k}$, then we denote
$K(0;n_{1},\ldots,n_{k};p)$ by $K(0;k;p)$.

For nonnegative integers $a,b,c$ and $d$ satisfying $1\leq a+b+c+d\leq k$, let

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a^{i},b^{-i},c^{1},d^{-1},k-a-b-
c-d)=(K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a+b+c+d),\mathbb{T},\varphi)$

be a complex unit gain graph with underlying graph
$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a+b+c+d)$ and gain function
$\varphi$:

$\left\\{\begin{array}[]{ll}\varphi(e_{vu})=\varphi(e_{uv})^{-1}=i,&\hbox{if
$u\in\bigcup_{j=1}^{a}V_{j}$;}\\\\[2.84526pt]
\varphi(e_{vu})=\varphi(e_{uv})^{-1}=-i,&\hbox{if
$u\in\bigcup_{j=a+1}^{a+b}V_{j}$;}\\\\[2.84526pt]
\varphi(e_{vu})=\varphi(e_{uv})^{-1}=1,&\hbox{if
$u\in\bigcup_{j=a+b+1}^{a+b+c}V_{j}$;}\\\\[2.84526pt]
\varphi(e_{vu})=\varphi(e_{uv})^{-1}=-1,&\hbox{if
$u\in\bigcup_{j=a+b+c+1}^{a+b+c+d}V_{j}$;}\\\\[2.84526pt]
\varphi(e_{uw})=\varphi(e_{wu})^{-1}=1,&\hbox{if $uw$ is any other edge of
$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a+b+c+d)${.}}\end{array}\right.$

Similarly, we can define
$K(0;n_{1},\ldots,n_{k};a^{i},b^{-i},c^{1},d^{-1},k-a-b-c-d)$. If
$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a+b+c+d)$ is $K(r;k;a+b+c+d)$, then
we denote

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a^{i},b^{-i},c^{1},d^{-1},k-a-b-c-d)$

by $K(r;k;a^{i},b^{-i},c^{1},d^{-1},k-a-b-c-d)$. Similarly, if
$K(0;n_{1},\ldots,n_{k};a+b+c+d)$ is $K(0;k;a+b+c+d)$, then we denote
$K(0;n_{1},\ldots,n_{k};a^{i},b^{-i},c^{1},d^{-1},k-a-b-c-d)$ by
$K(0;k;a^{i},b^{-i},c^{1},d^{-1},k-a-b-c-d)$.

Let $\widetilde{C_{n}}$ be a mixed cycle with vertex set
$\\{u_{1},u_{2},\ldots,u_{n}\\}$. The value of $\widetilde{C_{n}}$, denoted by
$h(\widetilde{C_{n}})$, is defined as $h_{12}h_{23}\ldots h_{n1}$. Clearly, if
in one direction the value of $\widetilde{C_{n}}$ is $h$, then its value is
$\overline{h}$ (the conjugate number of $h$) for the reverse direction. Thus,
we refer to the mixed cycle $\widetilde{C_{n}}$ with value 1 (resp. $-1$) as a
positive (resp. negative) mixed cycle regardless of the direction of
$\widetilde{C_{n}}$. For some orientation of $\widetilde{C_{n}}$, the
signature of $\widetilde{C_{n}}$, denoted by $\sigma(\widetilde{C_{n}})$, is
defined as the difference between the number of its forward and backward
directed edges in $\widetilde{C_{n}}$. Hence, a mixed cycle is positive (resp.
negative) if and only if its signature with respect to an arbitrary direction
is congruence to 0 (resp. 2) modulo 4. A mixed graph $\widetilde{G}$ is said
to be positive if all its mixed cycles are positive, otherwise it is called
non-positive.

######  Theorem 1.2.

Let $\widetilde{G}$ be a connected mixed graph with a cut vertex $v$ and
without pendant vertices. Then $p(\widetilde{G})=2$ if and only if
$\widetilde{G}$ is switching equivalent to one of the following mixed graphs:

  * (i)

$\widetilde{K}_{m_{1},m_{2},\ldots,m_{x}}\bullet
v\bullet\widetilde{K}_{m_{1}^{\prime},m_{2}^{\prime},\ldots,m^{\prime}_{y}}$,
where $\widetilde{K}_{m_{1},m_{2},\ldots,m_{x}}$ (resp.
$\widetilde{K}_{m_{1}^{\prime},m_{2}^{\prime},\ldots,m^{\prime}_{y}}$) is not
a mixed star, and $\widetilde{K}_{m_{1},m_{2},\ldots,m_{x}}$ (resp.
$\widetilde{K}_{m_{1}^{\prime},m_{2}^{\prime},\ldots,m^{\prime}_{y}}$) is
positive or switching equivalent to some
$\overrightarrow{C_{3}}(t_{1},t_{2},t_{3})$.

  * (ii)

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};p)$, where $r\geq 2,k\geq 2,p\geq 1$,
satisfying one of the following conditions:

    * (1)

$p=1$ and either $n_{1}=1$ and $k-p\geq 2$ or $n_{1}\geq 2$;

    * (2)

$p\geq 2$ and $k-p\leq 1$;

    * (3)

$p\geq 2,k-p\geq 2$ and $\frac{1}{r}+\frac{1}{p}+\frac{1}{k-p-1}\geq 1$.

  * (iii)

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a^{i},b^{-i},0^{1},0^{-1},s)$, where
$r\geq 2,k\geq 2,a\geq b\geq 1,s=k-a-b$, satisfying $a=b=1$ and $r=2$.

  * (iv)

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a^{i},0^{-i},c^{1},0^{-1},s)$, where
$r\geq 2,k\geq 2,a\geq c\geq 1,s=k-a-c$, satisfying one of the following
conditions:

    * (1)

$a=c=1$ and either $s=0$ or $s=1$;

    * (2)

$a=c=1,s=2$ and either $r=3$ or $r=4$;

    * (3)

$a=c=1$, $s=3$ and $r=3$;

    * (4)

$a=c=1$, $s\geq 2$ and $r=2$;

    * (5)

$a=c=2,s=0$ and $2\leq r\leq 4$;

    * (6)

$a=c=2,s=1$ and $r=2$;

    * (7)

$a=3,s=0$ and $r=c=2$;

    * (8)

$a=4,c=r=2$ and $s=0$;

    * (9)

$a\geq 2,c=1$ and $\frac{as-1}{a+s}\leq\frac{1}{r-1}$.

######  Example 1.3.

$\widetilde{G}=K(3,2;3,1;1^{i},1^{-i},0^{1},0^{-1},0)$ satisfies Theorem 1.2
(iii) and

$\widetilde{G^{\prime}}=K(q_{1},q_{2},q_{3};n_{1},n_{2},n_{3},n_{4};2^{i},0^{-i},1^{1},0^{-1},1)$

satisfies Theorem 1.2 (iv) (9) (see Figure. 3). It is easy to see that
$p(\widetilde{G})=p(\widetilde{G^{\prime}})=2$.

In the rest of this section we recall some known results. In Section 2 we give
the proof of Theorem 1.1. In Section 3 we first establish some technical
lemmas that help us characterize the mixed graphs with cut vertices having
exactly two positive eigenvalues and then present the proof of Theorem 1.2.

###  1.3 Preliminaries

In this subsection, we introduce some helpful lemmas and known results.

######  Lemma 1.4.

[19] (Sylvester’s law of inertia) If two Hermitian matrices $M$ and $N$ are
congruent, then they have the same positive (resp. negative) inertia indices.

######  Lemma 1.5.

[28] Let $\widetilde{G}$ be a mixed graph. Then the following conditions are
equivalent.

  * (1)

Every induced mixed cycle of $\widetilde{G}$ is positive.

  * (2)

$\widetilde{G}$ is positive.

  * (3)

$\widetilde{G}$ is switching equivalent to its underlying graph.

######  Lemma 1.6.

[51, 60] Let $v_{1}$ be a pendant vertex of mixed graph $\widetilde{G}$ with
neighbor $v_{2}$. Then

$p(\widetilde{G})=p(\widetilde{G}-v_{1}-v_{2})+1,\ \ \
n(\widetilde{G})=n(\widetilde{G}-v_{1}-v_{2})+1,\ \ \
\eta(\widetilde{G})=\eta(\widetilde{G}-v_{1}-v_{2}).$

######  Lemma 1.7.

[51] Let $\widetilde{G}$ be a mixed graph with $u\in V(G)$. Then

$\displaystyle p(\widetilde{G})-1\leq p(\widetilde{G}-u)\leq
p(\widetilde{G}),\ \ \ n(\widetilde{G})-1\leq n(\widetilde{G}-u)\leq
n(\widetilde{G}).$

######  Lemma 1.8.

[60] Let $v$ be a cut vertex of a mixed graph $\widetilde{G}$, and let
$\widetilde{G_{1}},\widetilde{G_{2}},\ldots,\widetilde{G_{l}}$ be all the
components of $\widetilde{G}-v$.

  * (1)

If there exists a component $\widetilde{G_{i}}$ satisfying
$rk(\widetilde{G_{i}}+v)=rk(\widetilde{G_{i}})+2$, then

${\rm In}(\widetilde{G})={\rm In}(\widetilde{G}-v)+(1,1,-1)=\sum_{j=1}^{l}{\rm
In}(\widetilde{G_{j}})+(1,1,-1).$

  * (2)

If there exists a component $\widetilde{G_{i}}$ satisfying
$rk(\widetilde{G_{i}}+v)=rk(\widetilde{G_{i}})$, then

${\rm In}(\widetilde{G})={\rm In}(\widetilde{G}_{i})+{\rm
In}(\widetilde{G}-\widetilde{G}_{i}).$

In particular, if $rk(\widetilde{G_{i}}+v)=rk(\widetilde{G_{i}})$ for each
$i=1,2,\ldots,l$, then

${\rm In}(\widetilde{G})={\rm In}(\widetilde{G}-v)+(0,0,1)=\sum_{j=1}^{l}{\rm
In}(\widetilde{G_{j}})+(0,0,1).$

######  Lemma 1.9.

[50] Let $\widetilde{C_{n}}$ be a mixed cycle of order $n$. Then

$\eta(\widetilde{C_{n}})=\left\\{\begin{array}[]{ll}1,&\hbox{if $n$ is odd,
$\sigma(\widetilde{C_{n}})$ is odd;}\\\\[2.84526pt] 0,&\hbox{if $n$ is odd,
$\sigma(\widetilde{C_{n}})$ is even;}\\\\[2.84526pt] 0,&\hbox{if $n$ is even,
$\sigma(\widetilde{C_{n}})$ is odd;}\\\\[2.84526pt] 0,&\hbox{if $n$ is even,
$\sigma(\widetilde{C_{n}})$ is even, $n+\sigma(\widetilde{C_{n}})\equiv
2(mod4)$;}\\\\[2.84526pt] 2,&\hbox{if $n$ is even, $\sigma(\widetilde{C_{n}})$
is even, $n+\sigma(\widetilde{C_{n}})\equiv 0(mod4)${.}}\end{array}\right.$

######  Lemma 1.10.

[60] Let $\widetilde{G}$ be a mixed graph. Then $p(\widetilde{G})=1$ if and
only if the subgraph induced by all non-isolated vertices of $\widetilde{G}$
is switching equivalent either to a complete multipartite graph or to some
$\overrightarrow{C_{3}}(t_{1},t_{2},t_{3})$.

##  2 Proof for Theorem 1.1

Proof of Theorem 1.1: The sufficiency can be verified by Lemma 1.6, and we
only prove necessity.

“Necessity”. If $\widetilde{G}$ is a mixed graph with pendant vertices,
without loss of generality, let $v_{1}$ be a pendant vertex of $\widetilde{G}$
with neighbor $v_{2}$. Then by Lemma 1.6, we have
$p(\widetilde{G})=p(\widetilde{G}-v_{1}-v_{2})+1$. The condition
$p(\widetilde{G})=2$ implies $p(\widetilde{G}-v_{1}-v_{2})=1$. Thus, by Lemma
1.10, the subgraph, say $\widetilde{F}$, induced by all non-isolated vertices
of $\widetilde{G}-v_{1}-v_{2}$ is switching equivalent either to a complete
multipartite graph or to some $\overrightarrow{C_{3}}(t_{1},t_{2},t_{3})$.
Hence, $p(\widetilde{F})=1$. It is easy to see that $\widetilde{G}$ is
obtained by adding some undirected edges (possibly empty) and directed edges
(possibly empty) with arbitrary direction between the center of a mixed star
and some vertices of $\widetilde{F}$. This completes the proof. ∎

##  3 Proof for Theorem 1.2

###  3.1 Technical lemmas

In this subsection we present a few technical lemmas aiming to provide some
fundamental characterizations of mixed graphs with cut vertices and without
pendant vertices having exactly two positive eigenvalues.

Two vertices $u,w\in V(\widetilde{G})$ are called twins if $\widetilde{G}$ is
switching equivalent to $\widetilde{G^{\prime}}$ such that the rows of
$H(\widetilde{G^{\prime}})$ indexed by $u$ and $w$ are same. The relation of
being twins is an equivalence relation on vertex set of $\widetilde{G}$.
Denote by $[u]$ the equivalence class containing $u$. The twin reduction graph
of $\widetilde{G}$, written as $T_{\widetilde{G}}$, is a mixed graph whose
vertices are the equivalence classes and $[u][w]\in E_{0}(T_{\widetilde{G}})$
if and only if $uw\in E_{0}(\widetilde{G})$ and $\overrightarrow{[u][w]}\in
E_{1}(T_{\widetilde{G}})$ if and only if $\overrightarrow{uw}\in
E_{1}(\widetilde{G})$.

######  Lemma 3.1.

[35] Let $\widetilde{G}$ and $\widetilde{G^{\prime}}$ be mixed graphs with the
same underlying graph. Then they are switching equivalent if and only if
$T_{\widetilde{G}}$ and $T_{\widetilde{G^{\prime}}}$ are switching equivalent.

######  Lemma 3.2.

[50] Adding or removing twins preserves the $H$-rank of a mixed graph, and
also preserves the positive and negative inertia index of a mixed graph.

######  Lemma 3.3.

$rk(\overrightarrow{C_{3}}(t_{1},t_{2},t_{3}))=2$.

###### Proof.

It is easy to see that
$T_{\overrightarrow{C_{3}}(t_{1},t_{2},t_{3})}=\overrightarrow{C_{3}}(1,1,1)$
and $rk(\overrightarrow{C_{3}}(1,1,1))=2$. By Lemma 3.2, we have

$rk(\overrightarrow{C_{3}}(t_{1},t_{2},t_{3}))=rk(\overrightarrow{C_{3}}(1,1,1))=2.$

∎

######  Lemma 3.4.

[50] Let $\widetilde{G}$ be a connected mixed graph. Then
$rk(\widetilde{G})=3$ if and only if $T_{\widetilde{G}}$ is an even triangle.

######  Lemma 3.5.

Let $\widetilde{G}$ be a positive complete $k$-partite mixed graph with one
partition class containing exactly one vertex. If $k\geq 3$ and $v\in
V(\widetilde{G})$ is in one partition class containing exactly one vertex,
then $rk(\widetilde{G})=rk(\widetilde{G}-v)+1$.

###### Proof.

By Lemma 1.5, we have $\widetilde{G}$ is switching equivalent to its
underlying graph. Then the vertices in the same partition class of
$\widetilde{G}$ are twins. Thus, by Lemma 3.1, we have $T_{\widetilde{G}}$ is
switching equivalent to $K_{k}$. Using Lemma 3.2, we have
$rk(\widetilde{G})=rk(T_{\widetilde{G}})=rk(K_{k})=k$. Similarly, we have
$rk(\widetilde{G}-v)=rk(K_{k-1})=k-1$. Hence,
$rk(\widetilde{G})=rk(\widetilde{G}-v)+1$. ∎

######  Lemma 3.6.

Let $\widetilde{M_{1}}$ be a mixed graph with a vertex $v^{\prime}$ and
$\widetilde{M_{2}}$ be another mixed graph with a vertex $v^{\prime\prime}$,
and let $\widetilde{G}=\widetilde{M_{1}}\bullet v\bullet\widetilde{M_{2}}$.
Then
$p(\widetilde{M_{1}}-v^{\prime})+p(\widetilde{M_{2}}-v^{\prime\prime})\leq
p(\widetilde{G})\leq p(\widetilde{M_{1}})+p(\widetilde{M_{2}})$.

###### Proof.

By Lemma 1.7, we obtain

$\displaystyle p(\widetilde{M_{1}}-v^{\prime})\leq p(\widetilde{M_{1}})\leq
p(\widetilde{M_{1}}-v^{\prime})+1.$

Similarly,

$\displaystyle p(\widetilde{M_{2}}-v^{\prime\prime})\leq
p(\widetilde{M_{2}})\leq p(\widetilde{M_{2}}-v^{\prime\prime})+1.$

Lemma 1.7 applying to the induced subgraph
$\widetilde{G}-v=(\widetilde{M_{1}}-v^{\prime})\bigcup(\widetilde{M_{2}}-v^{\prime\prime})$
of $\widetilde{G}$ yields that

$\displaystyle
p(\widetilde{M_{1}}-v^{\prime})+p(\widetilde{M_{2}}-v^{\prime\prime})\leq
p(\widetilde{G})\leq
p(\widetilde{M_{1}}-v^{\prime})+p(\widetilde{M_{2}}-v^{\prime\prime})+1.$

If $p(\widetilde{M_{1}})=p(\widetilde{M_{1}}-v^{\prime})+1$, then

$\displaystyle p(\widetilde{G})\leq
p(\widetilde{M_{1}}-v^{\prime})+p(\widetilde{M_{2}}-v^{\prime\prime})+1\leq
p(\widetilde{M_{1}})+p(\widetilde{M_{2}}).$

If $p(\widetilde{M_{2}})=p(\widetilde{M_{2}}-v^{\prime\prime})+1$, we also
have

$\displaystyle p(\widetilde{G})\leq
p(\widetilde{M_{1}}-v^{\prime})+p(\widetilde{M_{2}}-v^{\prime\prime})+1\leq
p(\widetilde{M_{1}})+p(\widetilde{M_{2}}).$

Suppose now $p(\widetilde{M_{1}})=p(\widetilde{M_{1}}-v^{\prime})$ and
$p(\widetilde{M_{2}})=p(\widetilde{M_{2}}-v^{\prime\prime})$ and consider the
Hermitian adjacency matrix $H(\widetilde{G})$ of $\widetilde{G}$:

$\displaystyle
H(\widetilde{G})={\left(\begin{array}[]{ccc}H(\widetilde{M_{1}}-v^{\prime})&\alpha&0\\\\[3.0pt]
\alpha^{*}&0&\beta^{*}\\\\[3.0pt]
0&\beta&H(\widetilde{M_{2}}-v^{\prime\prime})\end{array}\right)}.$

The condition $p(\widetilde{M_{1}})=p(\widetilde{M_{1}}-v^{\prime})$ implies
that $rk(\widetilde{M_{1}})\leq rk(\widetilde{M_{1}}-v^{\prime})+1$, thus the
equation $H(\widetilde{M_{1}}-v^{\prime})X=\alpha$ has a solution, say
$X_{0}$. Similarly, suppose $Y_{0}$ is a solution of the equation
$H(\widetilde{M_{2}}-v^{\prime\prime})Y=\beta$ and let

$\displaystyle S={\left(\begin{array}[]{ccc}I&-X_{0}&0\\\\[3.0pt]
0&1&0\\\\[3.0pt] 0&-Y_{0}&I\end{array}\right)}.$

Then

$\displaystyle
S^{*}H(\widetilde{G})S={\left(\begin{array}[]{ccc}H(\widetilde{M_{1}}-v^{\prime})&0&0\\\\[3.0pt]
0&-\alpha^{*}X_{0}-\beta^{*}Y_{0}&0\\\\[3.0pt]
0&0&H(\widetilde{M_{2}}-v^{\prime\prime})\end{array}\right)}.$

From $p(\widetilde{M_{1}})=p(\widetilde{M_{1}}-v^{\prime})$ and
$p(\widetilde{M_{2}})=p(\widetilde{M_{2}}-v^{\prime\prime})$ it follows that
$-\alpha^{*}X_{0}\leq 0$ and $-\beta^{*}Y_{0}\leq 0$. Thus by Lemma 1.4,
$p(\widetilde{G})=p(S^{*}H(\widetilde{G})S)=p(\widetilde{M_{1}}-v^{\prime})+p(\widetilde{M_{2}}-v^{\prime\prime})=p(\widetilde{M_{1}})+p(\widetilde{M_{2}})$.
This completes the proof. ∎

######  Lemma 3.7.

Let $\widetilde{K}_{m_{1},m_{2},\ldots,m_{x}}$ and
$\widetilde{K}_{m^{\prime}_{1},m^{\prime}_{2},\ldots,m^{\prime}_{y}}$ be mixed
complete multipartite graphs without pendant vertices. Let $v^{\prime}$ (resp.
$v^{\prime\prime}$) be a vertex of $\widetilde{K}_{m_{1},m_{2},\ldots,m_{x}}$
(resp. $\widetilde{K}_{m^{\prime}_{1},m^{\prime}_{2},\ldots,m^{\prime}_{y}}$).
If $p(\widetilde{K}_{m_{1},m_{2},\ldots,m_{x}})=1$ and
$p(\widetilde{K}_{m^{\prime}_{1},m^{\prime}_{2},\ldots,m^{\prime}_{y}})=1$,
then $p(\widetilde{K}_{m_{1},m_{2},\ldots,m_{x}}\bullet
v\bullet\widetilde{K}_{m^{\prime}_{1},m^{\prime}_{2},\ldots,m^{\prime}_{y}})=2$.

###### Proof.

Since $\widetilde{K}_{m_{1},m_{2},\ldots,m_{x}}$ and
$\widetilde{K}_{m^{\prime}_{1},m^{\prime}_{2},\ldots,m^{\prime}_{y}}$ are
mixed complete multipartite graphs without pendant vertices, we have
$K_{m_{1},m_{2},\ldots,m_{x}}$ and
$K_{m^{\prime}_{1},m^{\prime}_{2},\ldots,m^{\prime}_{y}}$ are not mixed stars.
Thus, $\widetilde{K}_{m_{1},m_{2},\ldots,m_{x}}-v^{\prime}$ (resp.
$\widetilde{K}_{m^{\prime}_{1},m^{\prime}_{2},\ldots,m^{\prime}_{y}}-v^{\prime\prime}$)
has at least one edge. It follows that
$p(\widetilde{K}_{m_{1},m_{2},\ldots,m_{x}}-v^{\prime})\geq 1$ and
$p(\widetilde{K}_{m^{\prime}_{1},m^{\prime}_{2},\ldots,m^{\prime}_{y}}-v^{\prime\prime})\geq
1$. By Lemma 1.7, we have
$p(\widetilde{K}_{m_{1},m_{2},\ldots,m_{x}}-v^{\prime})=1$ and
$p(\widetilde{K}_{m^{\prime}_{1},m^{\prime}_{2},\ldots,m^{\prime}_{y}}-v^{\prime\prime})=1$.
In view of Lemma 3.6, we have

$\displaystyle 2=$ $\displaystyle
p(\widetilde{K}_{m_{1},m_{2},\ldots,m_{x}}-v^{\prime})+p(\widetilde{K}_{m^{\prime}_{1},m^{\prime}_{2},\ldots,m^{\prime}_{y}}-v^{\prime\prime})$
$\displaystyle\leq$ $\displaystyle
p(\widetilde{K}_{m_{1},m_{2},\ldots,m_{x}}\bullet
v\bullet\widetilde{K}_{m^{\prime}_{1},m^{\prime}_{2},\ldots,m^{\prime}_{y}})$
$\displaystyle\leq$ $\displaystyle
p(\widetilde{K}_{m_{1},m_{2},\ldots,m_{x}})+p(\widetilde{K}_{m^{\prime}_{1},m^{\prime}_{2},\ldots,m^{\prime}_{y}})$
$\displaystyle=$ $\displaystyle 2.$

This completes the proof. ∎

Let ${\bf J}_{n\times m}$ and ${\bf 0}_{n\times m}$ be respectively the all-
one and the all-zero $n\times m$ matrices. Let ${\bf 1}_{n}={\bf J}_{n\times
1}$. We often simply write ${\bf J},{\bf 0}$ and ${\bf 1}$, respectively, if
the order of these matrices is clear from the context.

######  Lemma 3.8.

For $r\geq 2,k\geq 2,a\geq b,a\geq 1$, and $s=k-a-b$, the positive inertia
index of $\widetilde{G}=K(r;k;a^{i},b^{-i},0^{1},0^{-1},s)$ is 2 if and only
if one of the following conditions holds:

  * (1)

$a=1,b=0$;

  * (2)

$a=b=1$ and $r=2$;

  * (3)

$a\geq 2,b=0$ and $s\leq 1$;

  * (4)

$a\geq 2,b=0,s\geq 2$ and $\frac{1}{r}+\frac{1}{a}+\frac{1}{s-1}\geq 1$.

###### Proof.

The Hermitian adjacency matrix of
$\widetilde{G}=K(r;k;a^{i},b^{-i},0^{1},0^{-1},s)$ can be written as

$\displaystyle
H(\widetilde{G})={\left(\begin{array}[]{ccccc}H(K_{r})&\mathbf{1}&{\bf 0}&{\bf
0}&{\bf 0}\\\\[3.0pt]
\mathbf{1}^{\top}&0&i\mathbf{1}^{\top}&-i\mathbf{1}^{\top}&{\bf 0}\\\\[3.0pt]
{\bf 0}&-i\mathbf{1}&H(K_{a})&{\bf J}&{\bf J}\\\\[3.0pt] {\bf
0}&i\mathbf{1}&{\bf J}&H(K_{b})&{\bf J}\\\\[3.0pt] {\bf 0}&{\bf 0}&{\bf
J}&{\bf J}&H(K_{s})\end{array}\right)}.$

Case 1. $a=1$ and $b=0$. If $s=1$, then $\widetilde{G}$ has a pendant vertex,
by Theorem 1.6, it is easy to see that $p(\widetilde{G})=2$. If $s\geq 2$,
then

$\displaystyle
H(\widetilde{G})={\left(\begin{array}[]{cccc}H(K_{r})&\mathbf{1}&{\bf 0}&{\bf
0}\\\\[3.0pt] \mathbf{1}^{\top}&0&i&{\bf 0}\\\\[3.0pt] {\bf
0}&-i&0&\mathbf{1}^{\top}\\\\[3.0pt] {\bf 0}&{\bf
0}&\mathbf{1}&H(K_{s})\end{array}\right)}.$

Let

$\displaystyle
C={\left(\begin{array}[]{cccc}I_{r}&-\frac{1}{r-1}\mathbf{1}&{\bf 0}&{\bf
0}\\\\[3.0pt] {\bf 0}&1&0&{\bf 0}\\\\[3.0pt] {\bf 0}&0&1&{\bf 0}\\\\[3.0pt]
{\bf 0}&{\bf 0}&-\frac{1}{s-1}\mathbf{1}&I_{s}\end{array}\right)}.$

Then

$\displaystyle
C^{*}H(\widetilde{G})C={\left(\begin{array}[]{cccc}H(K_{r})&{\bf 0}&{\bf
0}&{\bf 0}\\\\[3.0pt] {\bf 0}&-\frac{r}{r-1}&i&{\bf 0}\\\\[3.0pt] {\bf
0}&-i&-\frac{s}{s-1}&{\bf 0}\\\\[3.0pt] {\bf 0}&{\bf 0}&{\bf
0}&H(K_{s})\end{array}\right)}.$

It is easy to see that the eigenvalues of

$\displaystyle{\left(\begin{array}[]{cc}-\frac{r}{r-1}&i\\\\[3.0pt]
-i&-\frac{s}{s-1}\end{array}\right)}$

are both negative, therefore

$p(C^{*}H(\widetilde{G})C)=p(H(K_{r}))+p(H(K_{s}))=2.$

Thus in this case, the positive inertia index of
$K(r;k;a^{i},b^{-i},0^{1},0^{-1},s)$ must be 2.

Case 2. $a=b=1$. If $s=0$, then

$\displaystyle
H(\widetilde{G})={\left(\begin{array}[]{cccc}H(K_{r})&\mathbf{1}&{\bf 0}&{\bf
0}\\\\[3.0pt] \mathbf{1}^{\top}&0&i&-i\\\\[3.0pt] {\bf 0}&-i&0&1\\\\[3.0pt]
{\bf 0}&i&1&0\end{array}\right)}.$

Let

$\displaystyle
C_{1}={\left(\begin{array}[]{cccc}I_{r}&-\frac{1}{r-1}\mathbf{1}&{\bf 0}&{\bf
0}\\\\[3.0pt] {\bf 0}&1&0&0\\\\[3.0pt] {\bf 0}&-i&1&0\\\\[3.0pt] {\bf
0}&i&0&1\end{array}\right)}.$

Then

$\displaystyle
C_{1}^{*}H(\widetilde{G})C_{1}={\left(\begin{array}[]{cccc}H(K_{r})&{\bf
0}&{\bf 0}&{\bf 0}\\\\[3.0pt] {\bf 0}&2-\frac{r}{r-1}&0&0\\\\[3.0pt] {\bf
0}&0&0&1\\\\[3.0pt] {\bf 0}&0&1&0\end{array}\right)}.$

Thus $p(\widetilde{G})=2$ if and only if $r=2$.

If $s=1$, then

$\displaystyle
H(\widetilde{G})={\left(\begin{array}[]{ccccc}H(K_{r})&\mathbf{1}&{\bf 0}&{\bf
0}&{\bf 0}\\\\[3.0pt] \mathbf{1}^{\top}&0&i&-i&0\\\\[3.0pt] {\bf
0}&-i&0&1&1\\\\[3.0pt] {\bf 0}&i&1&0&1\\\\[3.0pt] {\bf
0}&0&1&1&0\end{array}\right)}.$

Let

$\displaystyle
C_{2}={\left(\begin{array}[]{ccccc}I_{r}&-\frac{1}{r-1}\mathbf{1}&0&0&0\\\\[3.0pt]
0&1&0&0&0\\\\[3.0pt] 0&0&1&0&0\\\\[3.0pt] 0&0&-1&1&0\\\\[3.0pt]
0&-i&-1&0&1\end{array}\right)}.$

Then

$\displaystyle
C_{2}^{*}H(\widetilde{G})C_{2}={\left(\begin{array}[]{ccccc}H(K_{r})&{\bf
0}&{\bf 0}&{\bf 0}&{\bf 0}\\\\[3.0pt] {\bf 0}&-\frac{r}{r-1}&2i&0&0\\\\[3.0pt]
{\bf 0}&-2i&-2&0&0\\\\[3.0pt] {\bf 0}&0&0&0&1\\\\[3.0pt] {\bf
0}&0&0&1&0\end{array}\right)}.$

Thus $p(\widetilde{G})=2$ if and only if

$\displaystyle{\left(\begin{array}[]{cc}-\frac{r}{r-1}&2i\\\\[3.0pt]
-2i&-2\end{array}\right)}.$

has no positive eigenvalue, if and only if $r=2$.

If $s\geq 2$, let

$\displaystyle
C_{3}={\left(\begin{array}[]{ccccc}I_{r}&-\frac{1}{r-1}\mathbf{1}&{\bf 0}&{\bf
0}&{\bf 0}\\\\[3.0pt] {\bf 0}&1&0&0&{\bf 0}\\\\[3.0pt] {\bf 0}&0&1&0&{\bf
0}\\\\[3.0pt] {\bf 0}&0&0&1&{\bf 0}\\\\[3.0pt] {\bf 0}&{\bf
0}&-\frac{1}{s-1}\mathbf{1}&-\frac{1}{s-1}\mathbf{1}&I_{s}\end{array}\right)}.$

Then

$\displaystyle
C_{3}^{*}H(\widetilde{G})C_{3}={\left(\begin{array}[]{ccccc}H(K_{r})&{\bf
0}&{\bf 0}&{\bf 0}&{\bf 0}\\\\[3.0pt] {\bf 0}&-\frac{r}{r-1}&i&-i&{\bf
0}\\\\[3.0pt] {\bf 0}&-i&-\frac{s}{s-1}&-\frac{1}{s-1}&{\bf 0}\\\\[3.0pt] {\bf
0}&i&-\frac{1}{s-1}&-\frac{s}{s-1}&{\bf 0}\\\\[3.0pt] {\bf 0}&{\bf 0}&{\bf
0}&{\bf 0}&H(K_{s})\end{array}\right)}.$

If $r>2$, since the determinant of

$\displaystyle{\left(\begin{array}[]{ccc}-\frac{r}{r-1}&i&-i\\\\[3.0pt]
-i&-\frac{s}{s-1}&-\frac{1}{s-1}\\\\[3.0pt]
i&-\frac{1}{s-1}&-\frac{s}{s-1}\end{array}\right)}$

is positive, the positive inertia index of this matrix is at least 1, and thus
$p(\widetilde{G})\geq 3$. If $r=2$, it is easy to see that the above matrix
has 0 as an eigenvalue and the other two eigenvalues are negative, and thus
$p(\widetilde{G})=2$. Consequently, when $a=b=1$, $p(\widetilde{G})=2$ if and
only if $r=2$.

Case 3. $a\geq 2$. If $k=a=2$, then by Lemma 3.6,

$2=p(K_{r})+p(K_{k})\leq p(\widetilde{G})\leq
p(K_{r+1})+p(\widetilde{K_{k+1}})=2,$

and thus $p(\widetilde{G})=2$. Suppose $k\geq 3$. Let $U_{n}$ be the strictly
upper triangular matrix of order $n$ whose $(i,j)$-entries are 1 for all
$1\leq i<j\leq n$. Let
$D=diag(\frac{-1}{a-1},\frac{-1}{a},\cdots,\frac{-1}{k-2})$. Set

$\displaystyle C_{4}={\left(\begin{array}[]{ccc}I_{r}&{\bf 0}&{\bf
0}\\\\[3.0pt] {\bf 0}&1&{\bf 0}\\\\[3.0pt] {\bf 0}&{\bf
0}&P\end{array}\right)},$

where $P$ is the following matrix

$\displaystyle P={\left(\begin{array}[]{cccc}I_{a}&{\bf J}D\\\\[3.0pt] {\bf
0}&I_{k-a}+U_{k-a}D\end{array}\right)}.$

Then $C_{4}^{*}H(\widetilde{G})C_{4}$ is the following matrix:

$\displaystyle{\left(\begin{array}[]{cccccccccc}H(K_{r})&\mathbf{1}&{\bf
0}&{\bf 0}&{\bf 0}&\cdots&{\bf 0}&{\bf 0}&\cdots&{\bf 0}\\\\[3.0pt]
\mathbf{1}^{\top}&0&i\mathbf{1}^{\top}&\frac{(1-2a)i}{a-1}&\frac{(1-2a)i}{a}&\cdots&\frac{(1-2a)i}{a+b-2}&\frac{(b-a)i}{a+b-1}&\cdots&\frac{(b-a)i}{k-2}\\\\[3.0pt]
{\bf 0}&-i\mathbf{1}&H(K_{a})&0&0&\cdots&0&0&\cdots&0\\\\[3.0pt] {\bf
0}&\frac{(2a-1)i}{a-1}&0&\frac{-a}{a-1}&0&\cdots&0&0&\cdots&0\\\\[3.0pt] {\bf
0}&\frac{(2a-1)i}{a}&0&0&\frac{-(a+1)}{a}&\cdots&0&0&\cdots&0\\\\[3.0pt]
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\cdots&\vdots\\\\[3.0pt]
{\bf
0}&\frac{(2a-1)i}{a+b-2}&0&0&0&\cdots&\frac{-(a+b-1)}{a+b-2}&0&\cdots&0\\\\[3.0pt]
{\bf
0}&\frac{(a-b)i}{a+b-1}&0&0&0&\cdots&0&\frac{-(a+b)}{a+b-1}&\cdots&0\\\\[3.0pt]
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\cdots&\vdots\\\\[3.0pt]
{\bf
0}&\frac{(a-b)i}{k-2}&0&0&0&\cdots&0&0&\cdots&\frac{-(k-1)}{k-2}\end{array}\right)}.$

Let

$\displaystyle
C_{5}={\left(\begin{array}[]{cccccccccc}I_{r}&-\frac{1}{r-1}\mathbf{1}&{\bf
0}&{\bf 0}&{\bf 0}&\ldots&{\bf 0}&{\bf 0}&\ldots&{\bf 0}\\\\[3.0pt] {\bf
0}&1&0&0&0&\ldots&0&0&\ldots&0\\\\[3.0pt] {\bf
0}&\frac{i}{a-1}\mathbf{1}&I_{a}&0&0&\ldots&0&0&\ldots&0\\\\[3.0pt] {\bf
0}&\frac{(2a-1)i}{a}&0&1&0&\ldots&0&0&\ldots&0\\\\[3.0pt] {\bf
0}&\frac{(2a-1)i}{a+1}&0&0&1&\ldots&0&0&\ldots&0\\\\[3.0pt]
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\ldots&\vdots\\\\[3.0pt]
{\bf 0}&\frac{(2a-1)i}{a+b-1}&0&0&0&\ldots&1&0&\ldots&0\\\\[3.0pt] {\bf
0}&\frac{(a-b)i}{a+b}&0&0&0&\ldots&0&1&\ldots&0\\\\[3.0pt]
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\ldots&\vdots\\\\[3.0pt]
{\bf 0}&\frac{(a-b)i}{k-1}&0&0&0&\ldots&0&0&\ldots&1\end{array}\right)}.$

Then $C_{5}^{*}C_{4}^{*}H(\widetilde{G})C_{4}C_{5}$ is the following matrix:

$\displaystyle{\left(\begin{array}[]{cccccccccc}H(K_{r})&{\bf 0}&{\bf 0}&{\bf
0}&{\bf 0}&\ldots&{\bf 0}&{\bf 0}&\ldots&{\bf 0}\\\\[3.0pt] {\bf
0}&\rho&0&0&0&\ldots&0&0&\ldots&0\\\\[3.0pt] {\bf
0}&0&H(K_{a})&0&0&\ldots&0&0&\ldots&0\\\\[3.0pt] {\bf
0}&0&0&\frac{-a}{a-1}&0&\ldots&0&0&\ldots&0\\\\[3.0pt] {\bf
0}&0&0&0&\frac{-(a+1)}{a}&\ldots&0&0&\ldots&0\\\\[3.0pt]
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\ldots&\vdots\\\\[3.0pt]
{\bf 0}&0&0&0&0&\ldots&\frac{-(a+b-1)}{a+b-2}&0&\ldots&0\\\\[3.0pt] {\bf
0}&0&0&0&0&\ldots&0&\frac{-(a+b)}{a+b-1}&\ldots&0\\\\[3.0pt]
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\ldots&\vdots\\\\[3.0pt]
{\bf 0}&0&0&0&0&\ldots&0&0&\ldots&\frac{-(k-1)}{k-2}\end{array}\right)},$

where

$\displaystyle\rho$
$\displaystyle=(2a-1)^{2}\left(\frac{1}{a-1}-\frac{1}{a+b-1}\right)+(a-b)^{2}\left(\frac{1}{a+b-1}-\frac{1}{k-1}\right)-\frac{r}{r-1}-\frac{a}{a-1}$
$\displaystyle=\frac{b(2a-1)^{2}(k-1)+s(a-b)^{2}(a-1)}{(a-1)(a+b-1)(k-1)}-\frac{r}{r-1}-\frac{a}{a-1}.$

For convenience, set
$\phi=\frac{b(2a-1)^{2}(k-1)+s(a-b)^{2}(a-1)}{(a-1)(a+b-1)(k-1)}$. Clearly,
$p(\widetilde{G})=2$ if and only if $\rho\leq 0$, if and only if

$\phi\leq\frac{r}{r-1}+\frac{a}{a-1}.$

Clearly, $\phi\geq\frac{b(2a-1)^{2}}{(a-1)(a+b-1)}$ . If $b\geq 2$, since
$2a-1\geq a+b-1$ we have

$\phi\geq\frac{2(2a-1)}{a-1}>4\geq\frac{r}{r-1}+\frac{a}{a-1},$

and thus $p(\widetilde{G})>2$. If $b=1$, then

$\phi\geq\frac{(2a-1)^{2}}{a(a-1)}>4\geq\frac{r}{r-1}+\frac{a}{a-1},$

leading to $p(\widetilde{G})>2$. If $b=0$, then
$\phi=\frac{a^{2}s}{(a-1)(a+s-1)}$ . If $s=0$, then $\phi=0$. If $s=1$, then
$\phi=\frac{a}{a-1}$. In both cases, we have
$\phi\leq\frac{r}{r-1}+\frac{a}{a-1}$ and thus $p(\widetilde{G})=2$. If $s\geq
2$, the condition $\phi\leq\frac{r}{r-1}+\frac{a}{a-1}$ is equivalent to

$\frac{1}{r}+\frac{1}{a}+\frac{1}{s-1}\geq 1.$

This completes the proof. ∎

######  Corollary 3.9.

For $r\geq 2,k\geq 2$ and $p\geq 1$, the positive inertia index of
$\widetilde{G}=K(r;k;p)$ is 2 if and only if one of the following conditions
holds:

  * (1)

$p=1$;

  * (2)

$p\geq 2$ and $k-p\leq 1$;

  * (3)

$p\geq 2,k-p\geq 2$ and $\frac{1}{r}+\frac{1}{p}+\frac{1}{k-p-1}\geq 1$.

###### Proof.

Using mixed two-way switching of mixed graphs we have $\widetilde{G}=K(r;k;p)$
is switching equivalent to the mixed graph
$\widetilde{G}=K(r;k;p^{i},0^{-i},0^{1},0^{-1},k-p)$. Thus, the result follows
from Lemma 3.8. ∎

######  Lemma 3.10.

For $r\geq 2,k\geq 2,a\geq c,a\geq 1$ and $s=k-a-c$, the positive inertia
index of $\widetilde{G}=K(r;k;a^{i},0^{-i},c^{1},0^{-1},s)$ is 2 if and only
if one of the following conditions holds:

  * (1)

$a=1,c=0$;

  * (2)

$a=c=1$ and either $s=0$ or $s=1$;

  * (3)

$a=c=1,s=2$ and either $r=3$ or $r=4$;

  * (4)

$a=c=1$, $s=3$ and $r=3$;

  * (5)

$a=c=1$, $s\geq 2$ and $r=2$;

  * (6)

$a=c=2,s=0$ and $2\leq r\leq 4$;

  * (7)

$a=c=2,s=1$ and $r=2$;

  * (8)

$a=3,s=0$ and $r=c=2$;

  * (9)

$a=4,c=r=2$ and $s=0$;

  * (10)

$a\geq 2,c=1$ and $\frac{as-1}{a+s}\leq\frac{1}{r-1}$;

  * (11)

$a\geq 2,c=0$ and either $s=0$ or $s=1$;

  * (12)

$a\geq 2,c=0,s\geq 2$ and $\frac{1}{r}+\frac{1}{a}+\frac{1}{s-1}\geq 1$.

###### Proof.

The Hermitian adjacency matrix of
$\widetilde{G}=K(r;k;a^{i},0^{-i},c^{1},0^{-1},s)$ can be written as

$\displaystyle
H(\widetilde{G})={\left(\begin{array}[]{ccccc}H(K_{r})&\mathbf{1}&{\bf 0}&{\bf
0}&{\bf 0}\\\\[3.0pt]
\mathbf{1}^{\top}&0&i\mathbf{1}^{\top}&\mathbf{1}^{\top}&{\bf 0}\\\\[3.0pt]
{\bf 0}&-i\mathbf{1}&H(K_{a})&{\bf J}&{\bf J}\\\\[3.0pt] {\bf
0}&\mathbf{1}&{\bf J}&H(K_{c})&{\bf J}\\\\[3.0pt] {\bf 0}&{\bf 0}&{\bf J}&{\bf
J}&H(K_{s})\end{array}\right)}.$

Case 1. $a=1$ and $c=0$. The result follows from Lemma 3.8 (1). Thus in this
case, the positive inertia index of $K(r;k;a^{i},0^{-i},c^{1},0^{-1},s)$ must
be 2.

Case 2. $a=c=1$. If $s=0$, then

$\displaystyle
H(\widetilde{G})={\left(\begin{array}[]{cccc}H(K_{r})&\mathbf{1}&{\bf 0}&{\bf
0}\\\\[3.0pt] \mathbf{1}^{\top}&0&i&1\\\\[3.0pt] {\bf 0}&-i&0&1\\\\[3.0pt]
{\bf 0}&1&1&0\end{array}\right)}.$

Let

$\displaystyle
T_{1}={\left(\begin{array}[]{cccc}I_{r}&-\frac{1}{r-1}\mathbf{1}&{\bf 0}&{\bf
0}\\\\[3.0pt] {\bf 0}&1&0&0\\\\[3.0pt] {\bf 0}&-1&1&0\\\\[3.0pt] {\bf
0}&i&0&1\end{array}\right)}.$

Then

$\displaystyle
T_{1}^{*}H(\widetilde{G})T_{1}={\left(\begin{array}[]{cccc}H(K_{r})&{\bf
0}&{\bf 0}&{\bf 0}\\\\[3.0pt] {\bf 0}&-\frac{r}{r-1}&0&0\\\\[3.0pt] {\bf
0}&0&0&1\\\\[3.0pt] {\bf 0}&0&1&0\end{array}\right)}.$

Thus $p(\widetilde{G})=2$ since $r\geq 2$. When $s=1$, then

$\displaystyle
H(\widetilde{G})={\left(\begin{array}[]{ccccc}H(K_{r})&\mathbf{1}&{\bf 0}&{\bf
0}&{\bf 0}\\\\[3.0pt] \mathbf{1}^{\top}&0&i&1&0\\\\[3.0pt] {\bf
0}&-i&0&1&1\\\\[3.0pt] {\bf 0}&1&1&0&1\\\\[3.0pt] {\bf
0}&0&1&1&0\end{array}\right)}.$

Let

$\displaystyle
T_{2}={\left(\begin{array}[]{ccccc}I_{r}&-\frac{1}{r-1}\mathbf{1}&{\bf 0}&{\bf
0}&{\bf 0}\\\\[3.0pt] {\bf 0}&1&0&0&0\\\\[3.0pt] {\bf 0}&0&1&0&0\\\\[3.0pt]
{\bf 0}&0&-1&1&0\\\\[3.0pt] {\bf 0}&-1&-1&0&1\end{array}\right)}.$

Then

$\displaystyle
T_{2}^{*}H(\widetilde{G})T_{2}={\left(\begin{array}[]{ccccc}H(K_{r})&{\bf
0}&{\bf 0}&{\bf 0}&{\bf 0}\\\\[3.0pt] {\bf
0}&-\frac{r}{r-1}&i-1&0&0\\\\[3.0pt] {\bf 0}&-i-1&-2&0&0\\\\[3.0pt] {\bf
0}&0&0&0&1\\\\[3.0pt] {\bf 0}&0&0&1&0\end{array}\right)}.$

Thus $p(\widetilde{G})=2$ if and only if

$\displaystyle{\left(\begin{array}[]{cc}-\frac{r}{r-1}&i-1\\\\[3.0pt]
-i-1&-2\end{array}\right)}$

has no positive eigenvalue, if and only if $r\geq 2$. When $s\geq 2$, let

$\displaystyle
T_{3}={\left(\begin{array}[]{ccccc}I_{r}&-\frac{1}{r-1}\mathbf{1}&{\bf 0}&{\bf
0}&{\bf 0}\\\\[3.0pt] {\bf 0}&1&0&0&{\bf 0}\\\\[3.0pt] {\bf 0}&0&1&0&{\bf
0}\\\\[3.0pt] {\bf 0}&0&0&1&{\bf 0}\\\\[3.0pt] {\bf 0}&{\bf
0}&-\frac{1}{s-1}\mathbf{1}&-\frac{1}{s-1}\mathbf{1}&I_{s}\end{array}\right)}.$

Then

$\displaystyle
T_{3}^{*}H(\widetilde{G})T_{3}={\left(\begin{array}[]{ccccc}H(K_{r})&{\bf
0}&{\bf 0}&{\bf 0}&{\bf 0}\\\\[3.0pt] {\bf 0}&-\frac{r}{r-1}&i&1&{\bf
0}\\\\[3.0pt] {\bf 0}&-i&-\frac{s}{s-1}&-\frac{1}{s-1}&{\bf 0}\\\\[3.0pt] {\bf
0}&1&-\frac{1}{s-1}&-\frac{s}{s-1}&{\bf 0}\\\\[3.0pt] {\bf 0}&{\bf 0}&{\bf
0}&{\bf 0}&H(K_{s})\end{array}\right)}.$

If $r\geq 3$, since the determinant of

$\displaystyle{\left(\begin{array}[]{ccc}-\frac{r}{r-1}&i&1\\\\[3.0pt]
-i&-\frac{s}{s-1}&-\frac{1}{s-1}\\\\[3.0pt]
1&-\frac{1}{s-1}&-\frac{s}{s-1}\end{array}\right)}$ (3.4)

is $\frac{r(s-1)-2s}{(r-1)(s-1)}$. If $s\geq 4$, then
$\frac{r(s-1)-2s}{(r-1)(s-1)}\geq\frac{s-3}{(r-1)(s-1)}>0$. Thus the positive
inertia index of matrix (3.1) is at least 1, and thus $p(\widetilde{G})\geq
3$. If $s=3$ and $r=3$, matrix (3.1) has eigenvalues 0, $-\frac{3}{2}$ and
$-3$ and thus $p(\widetilde{G})=2$. If $s=3$ and $r>3$, then
$\frac{r(s-1)-2s}{(r-1)(s-1)}=\frac{r-3}{r-1}>0$. Thus the positive inertia
index of matrix (3.1) is at least 1, and thus $p(\widetilde{G})\geq 3$. If
$s=2$ and $r>4$, then $\frac{r(s-1)-2s}{(r-1)(s-1)}=\frac{r-4}{r-1}>0$. Thus
the positive inertia index of matrix (3.1) is at least 1, and thus
$p(\widetilde{G})\geq 3$. If $s=2$ and $r=4$, matrix (3.1) has eigenvalues 0,
$\frac{-8+\sqrt{7}}{3}$ and $\frac{-8-\sqrt{7}}{3}$. If $s=2$ and $r=3$,
matrix (3.1) has eigenvalues $-3.5878$, $-1.8363$ and $-0.0759$. In both
cases, we have the positive inertia index of matrix (3.1) is 0, and thus
$p(\widetilde{G})=2$.

If $r=2$, the characteristic polynomial of matrix (3.1) is
$f(\lambda)=\lambda^{3}+\frac{2(2s-1)}{s-1}\lambda^{2}+\frac{3(s+1)}{(s-1)}\lambda+\frac{2}{s-1}$.
We have
$f^{\prime}(\lambda)=3\lambda^{2}+\frac{4(2s-1)}{s-1}\lambda+\frac{3(s+1)}{(s-1)}$.
It is easy to see that $f^{\prime}(\lambda)>0$ for $\lambda\geq 0$. Thus
$f(\lambda)$ is strictly increasing in $[0,+\infty)$. Since
$f(0)=\frac{2}{s-1}>0$, we have matrix (3.1) has no positive eigenvalue, and
thus $p(\widetilde{G})=2$.

Case 3. $a\geq 2$. If $k=a=2$, then by Lemma 3.6,

$2=p(K_{r})+p(K_{k})\leq p(\widetilde{G})\leq
p(K_{r+1})+p(\widetilde{K_{k+1}})=2,$

and thus $p(\widetilde{G})=2$. Suppose $k\geq 3$. Similarly, as in the proof
of Case 3 of Lemma 3.8, set

$\displaystyle T_{4}={\left(\begin{array}[]{ccc}I_{r}&{\bf 0}&{\bf
0}\\\\[3.0pt] {\bf 0}&1&{\bf 0}\\\\[3.0pt] {\bf 0}&{\bf
0}&P\end{array}\right)}.$

Then $T_{4}^{*}H(\widetilde{G})T_{4}$ is the following matrix:

$\displaystyle{\left(\begin{array}[]{cccccccccc}H(K_{r})&\mathbf{1}&{\bf
0}&{\bf 0}&{\bf 0}&\cdots&{\bf 0}&{\bf 0}&\cdots&{\bf 0}\\\\[3.0pt]
\mathbf{1}^{\top}&0&i\mathbf{1}^{\top}&\frac{(a-1)-ai}{a-1}&\frac{(a-1)-ai}{a}&\cdots&\frac{(a-1)-ai}{a+c-2}&\frac{-c-ai}{a+c-1}&\cdots&\frac{-c-ai}{k-2}\\\\[3.0pt]
{\bf 0}&-i\mathbf{1}&H(K_{a})&0&0&\cdots&0&0&\cdots&0\\\\[3.0pt] {\bf
0}&\frac{(a-1)+ai}{a-1}&0&\frac{-a}{a-1}&0&\cdots&0&0&\cdots&0\\\\[3.0pt] {\bf
0}&\frac{(a-1)+ai}{a}&0&0&\frac{-(a+1)}{a}&\cdots&0&0&\cdots&0\\\\[3.0pt]
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\cdots&\vdots\\\\[3.0pt]
{\bf
0}&\frac{(a-1)+ai}{a+c-2}&0&0&0&\cdots&\frac{-(a+c-1)}{a+c-2}&0&\cdots&0\\\\[3.0pt]
{\bf
0}&\frac{-c+ai}{a+c-1}&0&0&0&\cdots&0&\frac{-(a+c)}{a+c-1}&\cdots&0\\\\[3.0pt]
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\cdots&\vdots\\\\[3.0pt]
{\bf
0}&\frac{-c+ai}{k-2}&0&0&0&\cdots&0&0&\cdots&\frac{-(k-1)}{k-2}\end{array}\right)}.$

Let

$\displaystyle
T_{5}={\left(\begin{array}[]{cccccccccc}I_{r}&-\frac{1}{r-1}\mathbf{1}&{\bf
0}&{\bf 0}&{\bf 0}&\cdots&{\bf 0}&{\bf 0}&\cdots&{\bf 0}\\\\[3.0pt] {\bf
0}&1&0&0&0&\cdots&0&0&\cdots&0\\\\[3.0pt] {\bf
0}&\frac{i}{a-1}\mathbf{1}&I_{a}&0&0&\cdots&0&0&\cdots&0\\\\[3.0pt] {\bf
0}&\frac{(a-1)+ai}{a}&0&1&0&\cdots&0&0&\cdots&0\\\\[3.0pt] {\bf
0}&\frac{(a-1)+ai}{a+1}&0&0&1&\cdots&0&0&\cdots&0\\\\[3.0pt]
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\cdots&\vdots\\\\[3.0pt]
{\bf 0}&\frac{(a-1)+ai}{a+c-1}&0&0&0&\cdots&1&0&\cdots&0\\\\[3.0pt] {\bf
0}&\frac{-c+ai}{a+c}&0&0&0&\cdots&0&1&\cdots&0\\\\[3.0pt]
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\cdots&\vdots\\\\[3.0pt]
{\bf 0}&\frac{-c+ai}{k-1}&0&0&0&\cdots&0&0&\cdots&1\end{array}\right)}.$

Then $T_{5}^{*}T_{4}^{*}H(\widetilde{G})T_{4}T_{5}$ is the following matrix:

$\displaystyle{\left(\begin{array}[]{cccccccccc}H(K_{r})&{\bf 0}&{\bf 0}&{\bf
0}&{\bf 0}&\ldots&{\bf 0}&{\bf 0}&\ldots&{\bf 0}\\\\[3.0pt] {\bf
0}&\xi&0&0&0&\ldots&0&0&\ldots&0\\\\[3.0pt] {\bf
0}&0&H(K_{a})&0&0&\ldots&0&0&\ldots&0\\\\[3.0pt] {\bf
0}&0&0&\frac{-a}{a-1}&0&\ldots&0&0&\ldots&0\\\\[3.0pt] {\bf
0}&0&0&0&\frac{-(a+1)}{a}&\ldots&0&0&\ldots&0\\\\[3.0pt]
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\ldots&\vdots\\\\[3.0pt]
{\bf 0}&0&0&0&0&\ldots&\frac{-(a+c-1)}{a+c-2}&0&\ldots&0\\\\[3.0pt] {\bf
0}&0&0&0&0&\ldots&0&\frac{-(a+c)}{a+c-1}&\ldots&0\\\\[3.0pt]
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\ldots&\vdots\\\\[3.0pt]
{\bf 0}&0&0&0&0&\ldots&0&0&\ldots&\frac{-(k-1)}{k-2}\end{array}\right)},$

where

$\displaystyle\xi$
$\displaystyle=[(a-1)^{2}+a^{2}]\left(\frac{1}{a-1}-\frac{1}{a+c-1}\right)+(a^{2}+c^{2})\left(\frac{1}{a+c-1}-\frac{1}{k-1}\right)-\frac{r}{r-1}-\frac{a}{a-1}$
$\displaystyle=\frac{c[(a-1)^{2}+a^{2}](k-1)+s(a-1)(a^{2}+c^{2})}{(a-1)(a+c-1)(k-1)}-\frac{r}{r-1}-\frac{a}{a-1}.$

For convenience, set
$\gamma=\frac{c[(a-1)^{2}+a^{2}](k-1)+s(a-1)(a^{2}+c^{2})}{(a-1)(a+c-1)(k-1)}$.
Clearly, $p(\widetilde{G})=2$ if and only if $\xi\leq 0$, if and only if

$\gamma\leq\frac{r}{r-1}+\frac{a}{a-1}.$

If $c\geq 4$, we have

$\gamma\geq\frac{c[(a-1)^{2}+a^{2}]}{(a-1)(a+c-1)}\geq\frac{4[(a-1)^{2}+a^{2}]}{(a-1)(2a-1)}=\frac{4(2a^{2}-2a+1)}{2a^{2}-3a+1}>4\geq\frac{r}{r-1}+\frac{a}{a-1},$

and thus $p(\widetilde{G})>2$.

If $c=3$ and $a\geq 4$, then we have

$\gamma\geq\frac{c[(a-1)^{2}+a^{2}]}{(a-1)(a+c-1)}=\frac{3[(a-1)^{2}+a^{2}]}{(a-1)(a+2)}>4\geq\frac{r}{r-1}+\frac{a}{a-1},$

and thus $p(\widetilde{G})>2$.

If $c=a=3$, then $\gamma=\frac{39}{10}+\frac{18s}{5(5+s)}$, where $k=a+c+s$.
If $\gamma\leq\frac{r}{r-1}+\frac{a}{a-1}=\frac{r}{r-1}+\frac{3}{2}$, then we
have $\frac{12}{5}+\frac{18s}{5(5+s)}\leq\frac{r}{r-1}$. Hence, we obtain
$2<\frac{r}{r-1}$, i.e. $r<2$, a contradiction. Thus, $p(\widetilde{G})>2$.

If $c=2$ and $a\geq 4$, we now consider three cases. If $r\geq 3$, then
$\widetilde{R_{1}}$ (see Fig. 4) is an induced subgraph of $\widetilde{G}$. If
$s\geq 1$, then $\widetilde{R_{2}}$ (see Figure. 4) is an induced subgraph of
$\widetilde{G}$. By direct calculation, we have
$p(\widetilde{R_{1}})=p(\widetilde{R_{2}})=3$. Hence, $p(\widetilde{G})=2$
implies $r=2$ and $s=0$. Let
$\gamma=\frac{c[(a-1)^{2}+a^{2}]}{(a-1)(a+c-1)}=\frac{2(2a^{2}-2a+1)}{a^{2}-1}\leq\frac{r}{r-1}+\frac{a}{a-1}=2+\frac{a}{a-1}$.
We have $a=4$, $r=2$ and $s=0$. Hence, in this case, $p(\widetilde{G})=2$ if
and only if $a=4,r=2$ and $s=0$.

If $c=2$ and $a=3$, then $\gamma=\frac{13}{4}+\frac{13s}{4(4+s)}$, where
$k=a+c+s$. If
$\gamma\leq\frac{r}{r-1}+\frac{a}{a-1}=\frac{r}{r-1}+\frac{3}{2}$, then we
have $(3+4s)r\leq 7+5s$. Note that $r\geq 2$, we obtain $2(3+4s)\leq 7+5s$.
Thus, we have $s=0$. In addition, $3r\leq 7$ i.e. $r=2$. Thus, in this case,
$p(\widetilde{G})=2$ if and only if $r=2$ and $s=0$.

If $c=2$ and $a=2$, then $\gamma=\frac{10(3+s)+8s}{3(3+s)}$, where $k=a+c+s$.
If $\gamma\leq\frac{r}{r-1}+\frac{a}{a-1}=\frac{r}{r-1}+2$, then we have
$(3+9s)r\leq 12+12s$. Note that $r\geq 2$, we obtain $2(3+9s)\leq 12+12s$.
Thus, we have $s=0$ or 1. If $s=0$, we have $2\leq r\leq 4$. If $s=1$, we
obtain $r=2$. Thus, in this case, $p(\widetilde{G})=2$ if and only if (i)
$s=0$ and $2\leq r\leq 4$ or (ii) $s=1$ and $r=2$.

If $c=1$, then
$\gamma=\frac{[(a-1)^{2}+a^{2}](k-1)+s(a-1)(a^{2}+1)}{a(a-1)(k-1)}=\frac{[(a-1)^{2}+a^{2}](a+s)+s(a-1)(a^{2}+1)}{a(a-1)(a+s)}$,
where $k=a+c+s$. The condition $\gamma\leq\frac{r}{r-1}+\frac{a}{a-1}$ is
equivalent to $\frac{as-1}{a+s}\leq\frac{1}{r-1}$.

If $c=0$, then $\gamma=\frac{a^{2}s}{(a-1)(k-1)}=\frac{a^{2}s}{(a-1)(a+s-1)}$,
where $k=a+c+s$. If $s=0$, then $\gamma=0$. If $s=1$, then
$\gamma=\frac{a}{a-1}$. In both cases, we have
$\gamma\leq\frac{r}{r-1}+\frac{a}{a-1}$ and thus $p(\widetilde{G})=2$. If
$s\geq 2$, the condition $\gamma\leq\frac{r}{r-1}+\frac{a}{a-1}$ is equivalent
to

$\frac{1}{r}+\frac{1}{a}+\frac{1}{s-1}\geq 1.$

This completes the proof. ∎

######  Lemma 3.11.

Let $\widetilde{F_{1}}$ be a connected mixed graph with
$p(\widetilde{F_{1}})=1$. Let $\widetilde{F_{2}}$ be a mixed graph obtained
from $\widetilde{F_{1}}$ by adding a new vertex $v$ and adding some mixed
edges between $v$ and some vertices of $\widetilde{F_{1}}$. If
$rk(\widetilde{F_{2}})=rk(\widetilde{F_{1}})+1$ and $p(\widetilde{F_{2}})=2$,
then $\widetilde{F_{1}}$ is switching equivalent to some complete $k$-partite
graph $K_{n_{1}^{\prime},n_{2}^{\prime},\ldots,n_{k}^{\prime}}$ and
$\widetilde{F_{2}}$ is switching equivalent to some complex unit gain graph
$K(0;n_{1}^{\prime\prime},n_{2}^{\prime\prime},\ldots,n_{k}^{\prime\prime};a_{*}^{i},b_{*}^{-i},c_{*}^{1},d_{*}^{-1},k-a_{*}-b_{*}-c_{*}-d_{*})$.

###### Proof.

Since $p(\widetilde{F_{1}})=1$, by Lemma 1.10, we have $\widetilde{F_{1}}$ is
switching equivalent to a complete multipartite graph or some
$\overrightarrow{C_{3}}(t_{1},t_{2},t_{3})$. In view of Lemma 3.3, we have
$rk(\overrightarrow{C_{3}}(t_{1},t_{2},t_{3}))=2$. If $\widetilde{F_{1}}$ is
switching equivalent to some $\overrightarrow{C_{3}}(t_{1},t_{2},t_{3})$, then
combining $rk(\overrightarrow{C_{3}}(t_{1},t_{2},t_{3}))=2$ and
$rk(\widetilde{F_{2}})=rk(\widetilde{F_{1}})+1$, we have
$rk(\widetilde{F_{2}})=3$. By Lemma 3.4, we have $T_{\widetilde{F_{2}}}$ is an
even triangle. Since $\widetilde{F_{1}}$ is a subgraph of $\widetilde{F_{2}}$,
hence $\widetilde{F_{2}}$ contains an odd triangle. By Lemma 1.9, odd triangle
is not equivalent to even triangle. Thus, we obtain $T_{\widetilde{F_{2}}}$ is
not an even triangle, a contradiction. Hence, we have $\widetilde{F_{1}}$ is
switching equivalent to a complete multipartite graph.

Without loss of generality, assume that $\widetilde{F_{1}}$ is switching
equivalent to $K_{n_{1}^{\prime},n_{2}^{\prime},\ldots,n_{k}^{\prime}}$. Then
$\widetilde{F_{2}}$ is switching equivalent to a complex unit gain graph
$F_{2}^{\varphi}$ with
$F_{2}^{\varphi}-v=K_{n_{1}^{\prime},n_{2}^{\prime},\ldots,n_{k}^{\prime}}$,
where $\varphi(\overrightarrow{E(F_{2})})\subseteq\\{\pm 1,\pm i\\}$. Now we
consider $F_{2}^{\varphi}$. Let $V_{1},V_{2},\ldots,V_{k}$ be the partition
classes of $V(K_{n_{1}^{\prime},n_{2}^{\prime},\ldots,n_{k}^{\prime}})$ such
that $|V_{j}|=n_{j}^{\prime}$ for $j=1,2,\ldots,k$. Since
$rk(\widetilde{F_{2}})=rk(\widetilde{F_{1}})+1$, we have $v$ is not an
isolated vertex. In the following, we will prove the following two facts.

######  Fact 1.

Let $1\leq j\leq k$. If there exists one edge connecting $v$ and a vertex of
$V_{j}$ in $F_{2}$, then $v$ is adjacent to all vertices of $V_{j}$ in
$F_{2}$.

Proof of Fact 1 Suppose on the contrary, without loss of generality, let
${u,w}\in V_{j}$ such that $vu\in E(F_{2})$ and $vw\notin E(F_{2})$. Without
loss of generality, choose arbitrary vertex $v_{t}\in V_{t}$ for $1\leq t\leq
k,t\neq j$. The Hermitian adjacency matrix of the complex unit gain graph
induced by vertex subset

$\\{v,u,w\\}\bigcup\Big{(}\bigcup_{1\leq t\leq k,t\neq j}\\{v_{t}\\}\Big{)}$

of $F_{2}^{\varphi}$ can be written as

$\displaystyle N={\left(\begin{array}[]{cccc}0&0&\epsilon&\eta^{*}\\\\[3.0pt]
0&0&0&\mathbf{1}^{\top}\\\\[3.0pt]
\overline{\epsilon}&0&0&\mathbf{1}^{\top}\\\\[3.0pt]
\eta&\mathbf{1}&\mathbf{1}&\mathbf{J}_{k-1}-\mathbf{I}_{k-1}\end{array}\right)},$

where $\epsilon\in\\{\pm 1,\pm i\\}$. Set

$\displaystyle Q_{1}={\left(\begin{array}[]{cccc}1&0&0&\mathbf{0}\\\\[3.0pt]
0&1&0&\mathbf{0}\\\\[3.0pt] 0&-1&1&\mathbf{0}\\\\[3.0pt]
\mathbf{0}&\mathbf{0}&\mathbf{0}&\mathbf{I}_{k-1}\end{array}\right)},$

then $Q_{1}^{*}NQ_{1}$ is the following matrix:

$\displaystyle{\left(\begin{array}[]{cccc}0&-\epsilon&\epsilon&\eta^{*}\\\\[3.0pt]
-\overline{\epsilon}&0&0&\mathbf{0}\\\\[3.0pt]
\overline{\epsilon}&0&0&\mathbf{1}^{\top}\\\\[3.0pt]
\eta&\mathbf{0}&\mathbf{1}&\mathbf{J}_{k-1}-\mathbf{I}_{k-1}\end{array}\right)}.$

Note that the equation

$\displaystyle\left(\begin{array}[]{cc}0&\mathbf{1}^{\top}\\\\[3.0pt]
\mathbf{1}&\mathbf{J}_{k-1}-\mathbf{I}_{k-1}\end{array}\right)X=\left(\begin{array}[]{c}\overline{\epsilon}\\\
\eta\\\ \end{array}\right)$

has a solution, say $\left(\begin{array}[]{c}\mu\\\ \vartheta\\\
\end{array}\right).$ Let

$\displaystyle Q_{2}={\left(\begin{array}[]{cccc}1&0&0&\mathbf{0}\\\\[3.0pt]
0&1&0&\mathbf{0}\\\\[3.0pt] -\mu&0&1&\mathbf{0}\\\\[3.0pt]
-\vartheta&\mathbf{0}&\mathbf{0}&\mathbf{I}_{k-1}\end{array}\right)}.$

Then $Q_{2}^{*}Q_{1}^{*}NQ_{1}Q_{2}$ is the following matrix:

$\displaystyle{\left(\begin{array}[]{cccc}-\epsilon\mu-\eta^{*}\vartheta&-\epsilon&0&\mathbf{0}\\\\[3.0pt]
-\overline{\epsilon}&0&0&\mathbf{0}\\\\[3.0pt]
0&0&0&\mathbf{1}^{\top}\\\\[3.0pt]
\mathbf{0}&\mathbf{0}&\mathbf{1}&\mathbf{J}_{k-1}-\mathbf{I}_{k-1}\end{array}\right)}.$

Thus

$\displaystyle rank(N)$ $\displaystyle=rank(Q_{2}^{*}Q_{1}^{*}NQ_{1}Q_{2})$
$\displaystyle=rank{\left(\begin{array}[]{cc}-\epsilon\mu-\eta^{*}\vartheta&-\epsilon\\\\[3.0pt]
-\overline{\epsilon}&0\end{array}\right)}+rank{\left(\begin{array}[]{cc}0&\mathbf{1}^{\top}\\\\[3.0pt]
\mathbf{1}&\mathbf{J}_{k-1}-\mathbf{I}_{k-1}\end{array}\right)}$
$\displaystyle=2+k$ $\displaystyle=2+rk(\widetilde{F}_{1}).$

Since $rk(\widetilde{F}_{2})=rk(F_{2}^{\varphi})\geq rank(N)$, we have
$rk(\widetilde{F}_{2})\geq 2+rk(\widetilde{F}_{1})$, a contradiction.∎

######  Fact 2.

$\forall 1\leq j\leq k$, $A(F_{2}^{\varphi})_{vu}=A(F_{2}^{\varphi})_{vw}$ for
all $\\{u,w\\}\subseteq V_{j}$.

Proof of Fact 2 Let $1\leq j\leq k$. If there are no edges between $v$ and
$V_{j}$, then it is easy to see that
$A(F_{2}^{\varphi})_{vu}=A(F_{2}^{\varphi})_{vw}=0$ for all
$\\{u,w\\}\subseteq V_{j}$. If there are edges between $v$ and $V_{j}$, then
by $\mathbf{Fact\ 1}$, we have $vu$ and $vw$ are edges of $F_{2}$ for all
$\\{u,w\\}\subseteq V_{j}$. Suppose $A(F_{2}^{\varphi})_{vu}\neq
A(F_{2}^{\varphi})_{vw}$ for some $\\{u,w\\}\subseteq V_{j}$. Without loss of
generality, choose arbitrary vertex $v_{t}\in V_{t}$ for $1\leq t\leq k$ and
$t\neq j$. The Hermitian adjacency matrix of the complex unit gain graph
induced by vertex subset

$\\{v,u,w\\}\bigcup(\bigcup_{1\leq t\leq k,t\neq j}\\{v_{t}\\})$

of $F_{2}^{\varphi}$ can be written as

$\displaystyle
M={\left(\begin{array}[]{cccc}0&A(F_{2}^{\varphi})_{vu}&A(F_{2}^{\varphi})_{vw}&\zeta^{*}\\\\[3.0pt]
\overline{A(F_{2}^{\varphi})_{vu}}&0&0&\mathbf{1}^{\top}\\\\[3.0pt]
\overline{A(F_{2}^{\varphi})_{vw}}&0&0&\mathbf{1}^{\top}\\\\[3.0pt]
\zeta&\mathbf{1}&\mathbf{1}&\mathbf{J}_{k-1}-\mathbf{I}_{k-1}\end{array}\right)}.$

Set

$\displaystyle Q_{3}={\left(\begin{array}[]{cccc}1&0&0&\mathbf{0}\\\\[3.0pt]
0&1&0&\mathbf{0}\\\\[3.0pt] 0&-1&1&\mathbf{0}\\\\[3.0pt]
\mathbf{0}&\mathbf{0}&\mathbf{0}&\mathbf{I}_{k-1}\end{array}\right)},$

then $Q_{3}^{*}MQ_{3}$ is the following matrix:

$\displaystyle{\left(\begin{array}[]{cccc}0&A(F_{2}^{\varphi})_{vu}-A(F_{2}^{\varphi})_{vw}&A(F_{2}^{\varphi})_{vw}&\zeta^{*}\\\\[3.0pt]
\overline{A(F_{2}^{\varphi})_{vu}}-\overline{A(F_{2}^{\varphi})_{vw}}&0&0&\mathbf{0}\\\\[3.0pt]
\overline{A(F_{2}^{\varphi})_{vw}}&0&0&\mathbf{1}^{\top}\\\\[3.0pt]
\zeta&\mathbf{0}&\mathbf{1}&\mathbf{J}_{k-1}-\mathbf{I}_{k-1}\end{array}\right)}.$

Note that the equation

$\displaystyle\left(\begin{array}[]{cc}0&\mathbf{1}^{\top}\\\\[3.0pt]
\mathbf{1}&\mathbf{J}_{k-1}-\mathbf{I}_{k-1}\end{array}\right)X=\left(\begin{array}[]{c}\overline{A(F_{2}^{\varphi})_{vw}}\\\
\zeta\\\ \end{array}\right)$

has a solution, say $\left(\begin{array}[]{c}\nu\\\ \delta\\\
\end{array}\right).$ Let

$\displaystyle Q_{4}={\left(\begin{array}[]{cccc}1&0&0&\mathbf{0}\\\\[3.0pt]
0&1&0&\mathbf{0}\\\\[3.0pt] -\nu&0&1&\mathbf{0}\\\\[3.0pt]
-\delta&\mathbf{0}&\mathbf{0}&\mathbf{I}_{k-1}\end{array}\right)}.$

Then $Q_{4}^{*}Q_{3}^{*}NQ_{3}Q_{4}$ is the following matrix:

$\displaystyle{\left(\begin{array}[]{cccc}-\nu
A(F_{2}^{\varphi})_{vw}-\zeta^{*}\delta&A(F_{2}^{\varphi})_{vu}-A(F_{2}^{\varphi})_{vw}&0&\mathbf{0}\\\\[3.0pt]
\overline{A(F_{2}^{\varphi})_{vu}}-\overline{A(F_{2}^{\varphi})_{vw}}&0&0&\mathbf{0}\\\\[3.0pt]
0&0&0&\mathbf{1}^{\top}\\\\[3.0pt]
\mathbf{0}&\mathbf{0}&\mathbf{1}&\mathbf{J}_{k-1}-\mathbf{I}_{k-1}\end{array}\right)}.$

Since $A(F_{2}^{\varphi})_{vu}\neq A(F_{2}^{\varphi})_{vw}$, we obtain

$\displaystyle rank(M)$ $\displaystyle=rank(Q_{4}^{*}Q_{3}^{*}MQ_{3}Q_{4})$
$\displaystyle=rank{\left(\begin{array}[]{cc}-\nu
A(F_{2}^{\varphi})_{vw}-\zeta^{*}\delta&A(F_{2}^{\varphi})_{vu}-A(F_{2}^{\varphi})_{vw}\\\\[3.0pt]
\overline{A(F_{2}^{\varphi})_{vu}}-\overline{A(F_{2}^{\varphi})_{vw}}&0\end{array}\right)}+rank{\left(\begin{array}[]{cc}0&\mathbf{1}^{\top}\\\\[3.0pt]
\mathbf{1}&\mathbf{J}_{k-1}-\mathbf{I}_{k-1}\end{array}\right)}$
$\displaystyle=2+k$ $\displaystyle=2+rk(\widetilde{F_{1}}).$

Since $rk(\widetilde{F_{2}})=rk(F_{2}^{\varphi})\geq rk(M)$, we have
$rk(\widetilde{F_{2}})\geq 2+rk(\widetilde{F_{1}})$, a contradiction. This
completes the proof of Fact 2.∎

Combining Facts 1 and 2, we have $\widetilde{F_{2}}$ is switching equivalent
to some complex unit gain graph

$K(0;n_{1}^{\prime\prime},n_{2}^{\prime\prime},\ldots,n_{k}^{\prime\prime};a_{*}^{i},b_{*}^{-i},c_{*}^{1},d_{*}^{-1},k-a_{*}-b_{*}-c_{*}-d_{*}).$

∎

###  3.2 Proof for Theorem 1.2

Now, we are ready to prove the second main result of this paper.

Proof of Theorem 1.2 “Sufficiency”. If $\widetilde{G}$ is a mixed graph
switching equivalent to a mixed graph satisfying (i), then by Lemmas 1.10 and
3.7, the result is naturally obtained. If $\widetilde{G}$ is a mixed graph
switching equivalent to a mixed graph satisfying (ii), then by Lemma 3.1,
$T_{\widetilde{G}}$ is switching equivalent to $K(r;k;p)$ satisfying one of
the three conditions of Corollary 3.9. By Corollary 3.9, we have
$p(K(r;k;p))=2$. In view of Lemmas 3.1 and 3.2, we have
$p(\widetilde{G})=p(T_{\widetilde{G}})=p(K(r;k;p))=2$. If $\widetilde{G}$ is a
mixed graph switching equivalent to a mixed graph satisfying (iii), then by
Lemma 3.1, $T_{\widetilde{G}}$ is switching equivalent to
$K(2;k;1^{i},1^{-i},0^{1},0^{-1},s)$ satisfying the condition (2) of Lemma
3.8. By Lemma 3.8, we have $p(K(2;k;1^{i},1^{-i},0^{1},0^{-1},s)=2$. In view
of Lemmas 3.1 and 3.2, we have
$p(\widetilde{G})=p(T_{\widetilde{G}})=p(K(r;k;a^{i},b^{-i},0^{1},0^{-1},s)=2$.
If $\widetilde{G}$ is a mixed graph switching equivalent to a mixed graph
satisfying (iv), then by Lemma 3.1, $T_{\widetilde{G}}$ is switching
equivalent to $K(r;k;a^{i},0^{-i},c^{1},0^{-1},s)$ satisfying one of the
conditions (2)-(10) of Lemma 3.10. By Lemma 3.10, we have
$p(K(r;k;a^{i},0^{-i},c^{1},0^{-1},s))=2$. In view of Lemmas 3.1 and 3.2, we
have
$p(\widetilde{G})=p(T_{\widetilde{G}})=p(K(r;k;a^{i},0^{-i},c^{1},0^{-1},s))=2$.

“Necessity”. We proceed by establishing some claims.

######  Claim 1.

$\widetilde{G}-v$ has exactly two components.

Proof of Claim 1. Suppose that
$\widetilde{G}-v=\widetilde{G_{1}}\cup\widetilde{G_{2}}\cup\ldots\cup\widetilde{G_{t}}$
is the disjoint union of different components of $\widetilde{G}-v$. Since $v$
is a cut vertex of $\widetilde{G}$, we have $t\geq 2$. Since $\widetilde{G}$
has no pendant vertices, we have $\widetilde{G_{i}}$ has at least one edge for
all $i=1,2,\ldots,t$. Hence, $p(\widetilde{G_{i}})\geq 1$ for all
$i=1,2,\ldots,t$. From $2=p(\widetilde{G})\geq
p(\widetilde{G}-v)=p(\widetilde{G}_{1})+\ldots+p(\widetilde{G}_{t})$ it
follows that $t=2$. Hence, $\widetilde{G}-v$ has exactly two components
$\widetilde{G_{1}}$ and $\widetilde{G_{2}}$. ∎

######  Claim 2.

$p(\widetilde{G_{i}})=1$ for each $i=1,2$.

Proof of Claim 2. Noting that $p(\widetilde{G_{1}})\geq 1$ and
$p(\widetilde{G_{2}})\geq 1$, then from

$\displaystyle 2=p(\widetilde{G})\geq
p(\widetilde{G_{1}})+p(\widetilde{G_{2}})\geq 2$

we have $p(\widetilde{G_{i}})=1$ for each $i=1,2$. ∎

By Claim 2 and Lemma 1.10, we have $\widetilde{G_{1}}$ (resp.
$\widetilde{G_{2}}$) is switching equivalent to a complete multipartite graph
or some $\overrightarrow{C_{3}}(t_{1},t_{2},t_{3})$. Suppose, without loss of
generality, that $p(\widetilde{G_{1}}+v)\leq p(\widetilde{G_{2}}+v)$.

######  Claim 3.

$p(\widetilde{G_{1}}+v)=1$.

Proof of Claim 3. We distinguish the following three cases.

######  Case 1.

There exists $\widetilde{G_{i}}$ such that
$rk(\widetilde{G_{i}}+v)=rk(\widetilde{G_{i}})+2$.

By Lemma 1.8 (1), we have
$p(\widetilde{G})=p(\widetilde{G}-v)+1=p(\widetilde{G_{1}})+p(\widetilde{G_{2}})+1=3$,
a contradiction.

######  Case 2.

There exists $\widetilde{G_{i}}$ such that
$rk(\widetilde{G_{i}}+v)=rk(\widetilde{G_{i}})$.

######  Subcase 1.

$rk(\widetilde{G_{1}}+v)=rk(\widetilde{G_{1}})$.

By Lemma 1.8 (2), we have
$p(\widetilde{G})=p(\widetilde{G_{1}})+p(\widetilde{G}-\widetilde{G_{1}})=p(\widetilde{G_{1}})+p(\widetilde{G_{2}}+v)$.
Hence, $1\leq p(\widetilde{G_{1}}+v)\leq p(\widetilde{G_{2}}+v)=1$. It follows
that $p(\widetilde{G_{1}}+v)=p(\widetilde{G_{2}}+v)=1$.

######  Subcase 2.

$rk(\widetilde{G_{2}}+v)=rk(\widetilde{G_{2}})$.

By Lemma 1.8 (2),
$p(\widetilde{G})=p(\widetilde{G_{2}})+p(\widetilde{G}-\widetilde{G_{2}})=p(\widetilde{G_{2}})+p(\widetilde{G_{1}}+v)$.
Hence, $p(\widetilde{G_{1}}+v)=1$. Furthermore, from
$rk(\widetilde{G_{2}}+v)=rk(\widetilde{G}_{2})$, we have
$p(\widetilde{G_{2}}+v)=p(\widetilde{G_{2}})=1$.

######  Case 3.

$rk(\widetilde{G_{i}}+v)=rk(\widetilde{G_{i}})+1$ for each $i=1,2$.

Consider the Hermitian adjacency matrix of $\widetilde{G}$:

$\displaystyle
H(\widetilde{G})={\left(\begin{array}[]{ccc}H(\widetilde{G_{1}})&\tau&0\\\\[3.0pt]
\tau^{*}&0&\psi^{*}\\\\[3.0pt]
0&\psi&H(\widetilde{G_{2}})\end{array}\right)}.$

The condition $rk(\widetilde{G_{1}}+v)=rk(\widetilde{G_{1}})+1$ implies that
the equation $H(\widetilde{G_{1}})X=\tau$ has a solution, say $X_{1}$.
Similarly, suppose $Y_{1}$ is a solution of the equation
$H(\widetilde{G_{2}})Y=\psi$ and let

$\displaystyle W={\left(\begin{array}[]{ccc}I&-X_{1}&0\\\\[3.0pt]
0&1&0\\\\[3.0pt] 0&-Y_{1}&I\end{array}\right)}.$

Then

$\displaystyle
W^{*}H(\widetilde{G})W={\left(\begin{array}[]{ccc}H(\widetilde{G_{1}})&0&0\\\\[3.0pt]
0&-\tau^{*}X_{1}-\psi^{*}Y_{1}&0\\\\[3.0pt]
0&0&H(\widetilde{G_{2}})\end{array}\right)}.$

From $2=p(\widetilde{G})=p(\widetilde{G_{1}})+p(\widetilde{G_{2}})$ and Lemma
1.4, we have $-\tau^{*}X_{1}-\psi^{*}Y_{1}\leq 0$. Note that
$-\tau^{*}X_{1}\neq 0$ and $-\psi^{*}Y_{1}\neq 0$. Hence, at least one of
$-\tau^{*}X_{1}$ and $-\psi^{*}Y_{1}$ is less than 0. If $-\psi^{*}Y_{1}>0$,
then $-\tau^{*}X_{1}<0$. Hence, we have
$p(\widetilde{G_{1}}+v)=p(\widetilde{G_{1}})=1$. If $-\psi^{*}Y_{1}<0$, we
have $p(\widetilde{G_{2}}+v)=p(\widetilde{G_{2}})=1$. Hence, by the condition
of $p(\widetilde{G_{1}}+v)\leq p(\widetilde{G_{2}}+v)$, we have
$p(\widetilde{G_{1}}+v)=1$. ∎

Combining the proof of Cases $1\verb|--|3$ in Claim 3, we have the following
observation.

######  Observation 1.

If $rk(\widetilde{G_{i}}+v)=rk(\widetilde{G_{i}})+1$ for each $i=1,2$, and
$-\psi^{*}Y_{1}>0$, where $\psi$ and $Y_{1}$ are defined in Case 3 of Claim 3,
then $p(\widetilde{G_{2}}+v)=2$. Otherwise, $p(\widetilde{G_{2}}+v)=1$.

If $rk(\widetilde{G_{i}}+v)=rk(\widetilde{G_{i}})$ for each $i=1,2$, then
combining Claim 3 and Observation 1, we have
$p(\widetilde{G_{1}}+v)=p(\widetilde{G_{2}}+v)=1$. In view of Lemma 1.10, we
obtain $\widetilde{G_{1}}+v$ (resp. $\widetilde{G_{2}}+v$) is switching
equivalent to complete multipartite graph or some
$\overrightarrow{C_{3}}(t_{1},t_{2},t_{3})$. Furthermore, if
$\widetilde{G_{1}}+v$ (resp. $\widetilde{G_{2}}+v$) is switching equivalent to
complete multipartite graph, then $v$ is in the partition class containing at
least two vertices of $V(G_{1}+v)$ (resp. $V(G_{2}+v)$).

If $rk(\widetilde{G_{i}}+v)=rk(\widetilde{G_{i}})$ and
$rk(\widetilde{G_{j}}+v)=rk(\widetilde{G_{j}})+1$ for $i\neq j$. Without loss
of generality, assume $rk(\widetilde{G_{1}}+v)=rk(\widetilde{G_{1}})$ and
$rk(\widetilde{G_{2}}+v)=rk(\widetilde{G_{2}})+1$. By Claim 3 and Observation
1, we have $p(\widetilde{G_{1}}+v)=p(\widetilde{G_{2}}+v)=1$. In view of Lemma
1.10, we obtain $\widetilde{G_{1}}+v$ is switching equivalent to complete
multipartite graph or some $\overrightarrow{C_{3}}(t_{1},t_{2},t_{3})$.
Furthermore, if $\widetilde{G_{1}}+v$ is switching equivalent to complete
multipartite graph, then by $rk(\widetilde{G_{1}}+v)=rk(\widetilde{G_{1}})$,
we have $v$ is in the partition class containing at least two vertices of
$V(G_{1}+v)$. Combining $rk(\widetilde{G_{2}}+v)=rk(\widetilde{G_{2}})+1$ and
Lemma 1.10, we obtain $\widetilde{G_{2}}+v$ is switching equivalent to some
complete $k^{\prime}$-partite graph satisfying $k^{\prime}\geq 3$ and $v$ is
in the partition class containing exactly one vertex of $V(G_{2}+v)$.

If $rk(\widetilde{G_{i}}+v)=rk(\widetilde{G_{i}})+1$ for each $i=1,2$, and
$-\psi^{*}Y_{1}<0$, then by Claim 3 and Observation 1, we have
$p(\widetilde{G_{1}}+v)=p(\widetilde{G_{2}}+v)=1$. Combining Lemma 1.10 and
$rk(\widetilde{G_{i}}+v)=rk(\widetilde{G_{i}})+1$ for each $i=1,2$, we have
$\widetilde{G_{1}}+v$ (resp. $\widetilde{G_{2}}+v$) is switching equivalent to
some complete $k_{1}$-partite (resp. $k_{2}$-partite) graph satisfying
$k_{1}\geq 3$ (resp. $k_{2}\geq 3$) and $v$ is in the partition class
containing exactly one vertex of $V(G_{1}+v)$ (resp. $V(G_{2}+v)$).

If $rk(\widetilde{G_{i}}+v)=rk(\widetilde{G_{i}})+1$ for each $i=1,2$, and
$-\psi^{*}Y_{1}>0$, then by Claim 3 and Observation 1, we have
$p(\widetilde{G_{1}}+v)=1$ and $p(\widetilde{G_{2}}+v)=2$. Combining Lemma
1.10 and $rk(\widetilde{G_{1}}+v)=rk(\widetilde{G_{1}})+1$, we have
$\widetilde{G_{1}}+v$ is switching equivalent to some complete $(r+1)$-partite
graph $K_{q_{1},q_{2},\ldots,q_{r},1}$ satisfying $r\geq 2$ and $v$ is in the
partition class containing exactly one vertex. In view of Lemmas 1.10 and
3.11, we obtain $\widetilde{G_{2}}$ is switching equivalent to some complete
$k$-partite graph $K_{n^{\prime}_{1},n^{\prime}_{2},\ldots,n^{\prime}_{k}}$
and $\widetilde{G_{2}}+v$ is switching equivalent to some complex unit gain
graph
$K(0;n_{1}^{\prime\prime},\ldots,n_{k}^{\prime\prime};{a_{*}}^{i},{b_{*}}^{-i},{c_{*}}^{1},{d_{*}}^{-1},k-{a_{*}}-{b_{*}}-{c_{*}}-{d_{*}})$.
Thus $\widetilde{G}$ is switching equivalent to some complex unit gain graph
$K(q_{1},\ldots,q_{r};n_{1}^{\prime\prime\prime},\ldots,n_{k}^{\prime\prime\prime};a^{i},b^{-i},c^{1},d^{-1},k-a-
b-c-d)$. In the following, we will prove at least two of $a,b,c,d$ are equal
to 0. Otherwise, we have $abc\neq 0$, $abd\neq 0$, $acd\neq 0$ or $bcd\neq 0$.

Let $H$ denote the Hermitian adjacency matrix of
$K(q_{1},\ldots,q_{r};n_{1}^{\prime\prime\prime},\ldots,n_{k}^{\prime\prime\prime};a^{i},b^{-i},c^{1},d^{-1},k-a-
b-c-d)$. Suppose $abc\neq 0$ and note that $r\geq 2$, then

$\displaystyle H_{1}={\left(\begin{array}[]{cccccc}0&1&1&0&0&0\\\\[3.0pt]
1&0&1&0&0&0\\\\[3.0pt] 1&1&0&i&-i&1\\\\[3.0pt] 0&0&-i&0&1&1\\\\[3.0pt]
0&0&i&1&0&1\\\\[3.0pt] 0&0&1&1&1&0\\\\[3.0pt] \end{array}\right)}$

is a submatrix of $H$. By a direct calculation, we have $p(H_{1})=3$. Thus,
$p(\widetilde{G})=p(H)\geq p(H_{1})=3$, a contradiction. Similarly, we also
have $abd=0,acd=0$ and $bcd=0$. Thus, we have at least two of $a,b,c,d$ are
equal to 0.

If $d=0$, then
$K(q_{1},\ldots,q_{r};n_{1}^{\prime\prime\prime},\ldots,n_{k}^{\prime\prime\prime};a^{i},b^{-i},c^{1},0^{-1},k-a-
b-c)$ is a mixed graph. Hence, we have $\widetilde{G}$ is switching equivalent
to mixed graph
$K(q_{1},\ldots,q_{r};n_{1}^{\prime\prime\prime},\ldots,n_{k}^{\prime\prime\prime};a^{i},b^{-i},c^{1},0^{-1},k-a-
b-c)$.

If $d>0$, we proceed by considering four cases.

Case 1. $a,b,c$ are all equal to 0.

In this case, there exists a diagonal matrix $D_{1}$ with $(D_{1})_{uu}=-1$ if
$u\in V(G_{1}+v)$. Otherwise, $(D_{1})_{uu}=1$. It is easy to see that
$D_{1}^{-1}HD_{1}$ is the adjacency matrix of some simple graph
$K(q_{1},\ldots,q_{r};n_{1}^{\prime\prime\prime},\ldots,n_{k}^{\prime\prime\prime};d)$.
Thus $\widetilde{G}$ is switching equivalent to
$K(q_{1},\ldots,q_{r};n_{1}^{\prime\prime\prime},\ldots,n_{k}^{\prime\prime\prime};d)$.

Case 2. $a>0$ and $b=c=0$.

In this case, there exists a diagonal matrix $D_{2}$ with $(D_{2})_{uu}=i$ if
$u\in V(G_{1}+v)$. Otherwise, $(D_{2})_{uu}=1$. It is easy to see that
$D_{2}^{-1}HD_{2}$ is the Hermitian adjacency matrix of some mixed graph

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};d^{i},0^{-i},a^{1},0^{-1},k-a-d).$

Thus $\widetilde{G}$ is switching equivalent to some mixed graph

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};d^{i},0^{-i},a^{1},0^{-1},k-a-d).$

Case 3. $b>0$ and $a=c=0$.

In this case, there exists a diagonal matrix $D_{3}$ with $(D_{3})_{uu}=-i$ if
$u\in V(G_{1}+v)$. Otherwise, $(D_{3})_{uu}=1$. It is easy to see that
$D_{3}^{-1}HD_{3}$ is the Hermitian adjacency matrix of some mixed graph

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};0^{i},d^{-i},b^{1},0^{-1},k-b-d).$

Thus $\widetilde{G}$ is switching equivalent to some mixed graph

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};0^{i},d^{-i},b^{1},0^{-1},k-b-d).$

Case 4. $c>0$ and $a=b=0$.

In this case, there exists a diagonal matrix $D_{4}$ with $(D_{4})_{uu}=-i$ if
$u\in V(G_{1}+v)$. Otherwise, $(D_{4})_{uu}=1$. It is easy to see that
$D_{4}^{-1}HD_{4}$ is the Hermitian adjacency matrix of some mixed graph

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};c^{i},d^{-i},0^{1},0^{-1},k-c-d).$

Thus $\widetilde{G}$ is switching equivalent to some mixed graph

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};c^{i},d^{-i},0^{1},0^{-1},k-c-d).$

Since simple graphs are mixed graphs, according to the above analysis, without
loss of generality, assume that $\widetilde{G}$ is switching equivalent to
some mixed graph
$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a^{i},b^{-i},c^{1},0^{-1},k-a-b-c)$,
where $a,b,c$ are nonnegative integers such that at least one of them is equal
to 0 and $1\leq a+b+c\leq k$.

Note that $[{v},V(G_{2})]$ is an edge cut of
$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a^{i},b^{-i},c^{1},0^{-1},k-a-b-c)$.
We can perform a sequence of two-way switching and operations of taking the
converse so that $\widetilde{G}$ is switching equivalent to one of the
following mixed graphs:

  * (i)

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};p)$;

  * (ii)

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a^{i},b^{-i},0^{1},0^{-1},k-a-b)$,
where $a\geq b\geq 1$;

  * (iii)

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a^{i},0^{-i},c^{1},0^{-1},k-a-c)$,
where $a\geq c\geq 1$.

If $\widetilde{G}$ is switching equivalent to
$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};p)$, then according to Lemmas 3.1,
3.2 and Corollary 3.9, $T_{\widetilde{G}}$ is switching equivalent to
$K(r;k;p)$, where $r\geq 2,k\geq 2,p\geq 1$, satisfying one of the following
conditions:

  * (1)

$p=1$;

  * (2)

$p\geq 2$ and $k-p\leq 1$;

  * (3)

$p\geq 2,k-p\geq 2$ and $\frac{1}{r}+\frac{1}{p}+\frac{1}{k-p-1}\geq 1$.

By the definition of twin reduction graph and $G$ without pendant vertices, it
is easy to see that $\widetilde{G}$ is switching equivalent to
$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};p)$, where $r\geq 2,k\geq 2,p\geq 1$,
satisfying one of the conditions stated in Theorem 1.2 “only if ” part (ii).

If $\widetilde{G}$ is switching equivalent to
$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a^{i},b^{-i},0^{1},0^{-1},k-a-b)$,
where $a\geq b\geq 1$, then according to Lemmas 3.1, 3.2 and 3.8,
$T_{\widetilde{G}}$ is switching equivalent to
$K(r;k;a^{i},b^{-i},0^{1},0^{-1},s)$, where $r\geq 2,k\geq 2,a\geq b\geq
1,s=k-a-b$, satisfying $a=b=1$ and $r=2$. By the definition of twin reduction
graph and $G$ without pendant vertices, it is easy to see that $\widetilde{G}$
is switching equivalent to

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a^{i},b^{-i},0^{1},0^{-1},s)$

where $r\geq 2,k\geq 2,a\geq b\geq 1,s=k-a-b,$ satisfying $a=b=1$ and $r=2$.

If $\widetilde{G}$ is switching equivalent to
$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a^{i},0^{-i},c^{1},0^{-1},k-a-c)$,
where $a\geq c\geq 1$, then according to Lemmas 3.1, 3.2 and 3.10,
$T_{\widetilde{G}}$ is switching equivalent to
$K(r;k;a^{i},0^{-i},c^{1},0^{-1},s)$, with $r\geq 2,k\geq 2,a\geq c\geq
1,s=k-a-c$, satisfying one of the following conditions:

  * (1)

$a=c=1$ and either $s=0$ or $s=1$;

  * (2)

$a=c=1,s=2$ and either $r=3$ or $r=4$;

  * (3)

$a=c=1$, $s=3$ and $r=3$;

  * (4)

$a=c=1$, $s\geq 2$ and $r=2$;

  * (5)

$a=c=2,s=0$ and $2\leq r\leq 4$;

  * (6)

$a=c=2,s=1$ and $r=2$;

  * (7)

$a=3,s=0$ and $r=c=2$;

  * (8)

$a=4,c=r=2$ and $s=0$;

  * (9)

$a\geq 2,c=1$ and $\frac{as-1}{a+s}\leq\frac{1}{r-1}$.

By the definition of twin reduction graph and $G$ without pendant vertices, it
is easy to see that $\widetilde{G}$ is switching equivalent to

$K(q_{1},\ldots,q_{r};n_{1},\ldots,n_{k};a^{i},0^{-i},c^{1},0^{-1},s)$

with $r\geq 2,k\geq 2,a\geq c\geq 1,s=k-a-c$, satisfying one of the conditions
stated in Theorem 1.2 “only if ” part (iv). This completes the proof. ∎

## Acknowledgement

This work was supported by NSFC (Grant Nos. 11871479, 12071484), Hunan
Provincial Natural Science Foundation (Grant Nos. 2018JJ2479, 2020JJ 4675).

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