The Majorization Theorems of Single-Cone Trees and Single-Cone Unicyclic Graphs

Abstract

A single-cone tree (unicyclic graph) is the join of a complete graph $K_1$ and a tree (unicyclic graph). Suppose $\pi =(d_1, d_2, \ldots , d_n)$ and $\pi ^{\,\prime }=(d_1^{\,\prime }, d_2^{\,\prime }, \ldots , d_n^{\,\prime })$ are two non-increasing degree sequences. We say $\pi$ is majorizated by $\pi ^{\,\prime }$, denoted by $\pi \lhd \pi ^{\,\prime }$, if and only if $\pi \ne \pi ^{\,\prime }$, $\sum \nolimits _{i=1}^{n} d_i=\sum \nolimits _{i=1}^{n} d_i^{\,^{\,\prime }}$, and $\sum \nolimits _{i=1}^j d_i\le \sum \nolimits _{i=1}^j d_i^{\,^{\,\prime }}$ for all $j=1, 2, \ldots , n-1$. We use $J_{\pi }$ to denote the class of single-cone trees (unicyclic graphs) with degree sequence $\pi$. Suppose that $\pi$ and $\pi ^{\,\prime }$ are two different non-increasing degree sequences of single-cone trees (unicyclic graphs). Let $\rho$ and $\rho ^{\,\prime }$ be the largest spectral radius of the graphs in $J_{\pi }$ and $J_{\pi ^{\,\prime }}$, respectively, $\mu$ and $\mu ^{\,\prime }$ be the largest signless Laplacian spectral radius of the graphs in $J_{\pi }$ and $J_{\pi ^{\,\prime }}$, respectively. In this paper, we prove that if $\pi \lhd \pi ^{\,\prime }$, then $\rho <\rho ^{\,\prime }$ and $\mu <\mu ^{\,\prime }$.

1 Introduction

Let $G=(V, E)$ be a simple undirected graph with vertex set $V= \{v_1, v_2, \ldots , v_n\}$ and edge set $E= \{e_1, e_2, \ldots , e_m\}$, i.e., $|V|=n$ and $|E|=m$. If $m=n+c-1$, then G is called a c-cyclic graph. Particularly, if $c=0, 1, 2, 3$, then G is called a tree, a unicyclic graph, a bicyclic graph, a tricyclic graph, respectively. For $v\in V(G)$, N(v) denotes the neighborhood of v in G and $d(v)=|N(v)|$ denotes the degree of vertex v. If $d_i=d_G(v_i)$ for $1\le i\le n$, then we call the sequence $\pi =(d_1, d_2, \ldots , d_n)$ the degree sequence of G. Throughout this paper, we enumerate the degrees in non-increasing order, i.e., $d_1\ge d_2\ge \cdots \ge d_n$. A non-increasing sequence $\pi =(d_1, d_2, \ldots , d_n)$ is called graphic if there exists a graph G having $\pi$ as its degree sequence. We use $C_{\pi }$ to denote the class of connected graphs with degree sequence $\pi$.

For a graph G, A(G) is its adjacency matrix and D(G) is the diagonal matrix of its degrees. The matrix $Q(G)=D(G)+A(G)$ is called the signless Laplacian matrix of G. The largest eigenvalue of A(G) (resp., Q(G)) is called the spectral radius (resp., signless Laplacian spectral radius) of G and denoted by $\rho (G)$ (resp., $\mu (G)$). If G is connected, then A(G) (resp., Q(G)) is irreducible and by the Perron–Frobenius theorem, $\rho (G)$ (resp., $\mu (G)$) has multiplicity one and there exists a unique positive unit eigenvector corresponding to $\rho (G)$ (resp., $\mu (G)$). In this paper, we use $f = \big (f(v_{1}),f(v_{2}), \ldots , f(v_{n})\big )^{T}$ to indicate the unique positive unit eigenvector corresponding to $\rho (G)$ (resp., $\mu (G)$), and call f the Perron vector of A(G) (resp., Q(G)). Furthermore, if $\rho (G)$ (resp., $\mu (G)$) is greatest in $C_{\pi }$, then G is called an extremal greatest graph of $C_{\pi }$ for $\rho (G)$ (resp., $\mu (G)$).

Suppose $\pi =(d_1, d_2, \ldots , d_n)$ and $\pi ^{\,\prime }=(d_1^{\,\prime }, d_2^{\,\prime }, \ldots , d_n^{\,\prime })$ are two non-increasing degree sequences. We say $\pi$ is majorizated by $\pi ^{\,\prime }$, denoted by $\pi \lhd \pi ^{\,\prime }$, if and only if $\pi \ne \pi ^{\,\prime }$, $\sum \nolimits _{i=1}^{n} d_i=\sum \nolimits _{i=1}^{n} d_i^{\,\prime }$, and $\sum \nolimits _{i=1}^j d_i\le \sum \nolimits _{i=1}^j d_i^{\,\prime }$ for all $j=1, 2, \ldots , n-1$.

In 2008, Bıyıkoğlu and Leydold connected the majorization of degree sequences with ordering graphs by their spectral radius, and they obtained the majorization theorem of trees as follows.

Theorem 1.1

[1] Let $\pi$ and $\pi ^{\,\prime }$ be two different non-increasing degree sequences of trees with $\pi \lhd \pi ^{\,\prime }$. Suppose T and $T^{\,\prime }$ are the trees with the greatest spectral radius in $C_{\pi }$ and $C_{\pi ^{\,\prime }}$, respectively. Then, $\rho (T)<\rho (T^{\,\prime })$.

Almost at the same time, Zhang [15] proved the majorization theorem for the Laplacian spectral radius of trees. In the sequel, similar problems have been studied extensively. Liu et al. [9] and Zhang [16] proved the majorization theorems for the spectral radius and signless Laplacian spectral radius of unicyclic graphs, respectively. Jiang et al. [6] and Huang et al. [5] proved the majorization theorems for the spectral radius and signless Laplacian spectral radius of bicyclic graphs, respectively, and Jiang et al. [6] provided a counterexample to show that the majorization theorem cannot hold for tricyclic graphs. Liu and Liu et al. [7, 9, 12, 13] proved the majorization theorems for the spectral radius and signless Laplacian spectral radius of c-cyclic graphs with additional restrictions, respectively. Recently, Liu et al. [10] proved the majorization theorems for the spectral radius and signless Laplacian spectral radius of pseudographs. For more results, one may refer to [8, 11].

Let $G_1$ and $G_2$ be graphs with disjoint vertex sets, and $G_1\vee G_2$ denote the join of $G_1$ and $G_2$. If G is connected, then $K_1\vee G$ is called a single-cone graph. In particular, if G is a tree (unicyclic graph) of order $n-1$, then we call $K_1\vee G$ a single-cone tree (unicyclic graph) of order n. For a non-increasing graphic sequence $\pi =(d_1, d_2, \ldots , d_n)$, let

$$\begin{aligned} J_{\pi }=\{G: \,G\,\, \text{ is } \text{ a } \text{ single-cone } \text{ graph } \text{ with } \text{ degree } \text{ sequence }\,\, \pi \}. \end{aligned}$$

If $G\in J_{\pi }$ and $\rho (G)\ge \rho (G^{\,\prime })$ (resp., $\mu (G)\ge \mu (G^{\,\prime })$) for any other $G^{\,\prime }\in J_{\pi }$, then we call G has the greatest spectral radius (resp., signless Laplacian spectral radius) in $J_{\pi }$.

In this paper, we give the majorization theorems for the spectral radius and signless Laplacian spectral radius of single-cone trees and single-cone unicyclic graphs, and the main results can be stated as follows:

Theorem 1.2

Let $\pi$ and $\pi ^{\,\prime }$ be two different non-increasing degree sequences of single-cone trees with $\pi \lhd \pi ^{\,\prime }$. Suppose G and $G^{\,\prime }$ are the single-cone trees with the greatest spectral radius (resp., signless Laplacian spectral radius) in $J_{\pi }$ and $J_{\pi ^{\,\prime }}$, respectively. Then, $\rho (G)<\rho (G^{\,\prime })$ (resp., $\mu (G)<\mu (G^{\,\prime })$).

Theorem 1.3

Let $\pi$ and $\pi ^{\,\prime }$ be two different non-increasing degree sequences of single-cone unicyclic graphs with $\pi \lhd \pi ^{\,\prime }$. Suppose G and $G^{\,\prime }$ are the single-cone unicyclic graphs with the greatest spectral radius (resp., signless Laplacian spectral radius) in $J_{\pi }$ and $J_{\pi ^{\,\prime }}$, respectively. Then, $\rho (G)<\rho (G^{\,\prime })$ (resp., $\mu (G)<\mu (G^{\,\prime })$).

The rest of the paper is organized as follows. In Sect. 2, we recall some basic notions and lemmas used further, and prove a new lemma. In Sect. 3, we give the proof of Theorem 1.2. In Sect. 4, we give the proof of Theorem 1.3.

2 Preliminaries

Given a unit n-vector $g=(g_1, g_2, \ldots , g_n)^T \in \mathbf{R}^n$, g can be considered as a function defined on V(G), that is, each vertex $v_i$ is mapped to $g_i =g(v_i)$. The Rayleigh quotients of the adjacency matrix A(G) and signless Laplacian matrix Q(G) are defined, respectively, as follows:

$$\begin{aligned} {\mathcal {R}}_{A(G)}(g)=\sum _{uv\in E}2g(u)g(v)\,\,\text { and }\,\,{\mathcal {R}}_{Q(G)}(g)=\sum _{uv\in E}(g(u)+g(v))^2. \end{aligned}$$

It follows from the Rayleigh–Ritz theorem that

Lemma 2.1

[3, 4] Let S denote the set of unit vectors on V. Then,

$$\begin{aligned} \rho (G)= & {} \max \limits _{g\in S} {\mathcal {R}}_{A(G)}(g)=2\max \limits _{g\in S} \sum \limits _{uv\in E} g(u)g(v),\\ \mu (G)= & {} \max \limits _{g\in S} {\mathcal {R}}_{Q(G)}(g)=\max \limits _{g\in S} \sum \limits _{uv\in E} (g(u)+g(v))^2. \end{aligned}$$

Moreover, if ${\mathcal {R}}_{A(G)}(g)=\rho (G)$ (resp., ${\mathcal {R}}_{Q(G)}(g)=\mu (G)$) for a positive vector $g\in S$, then g is an eigenvector corresponding to $\rho (G)$ (resp., $\mu (G)$).

Let $f=(f(v_1), f(v_2), \ldots , f(v_n))^T$ be the Perron vector of A(G) (resp., Q(G)). Then, $\rho (G)f(v)=\sum \nolimits _{uv\in E} f(u)$ (resp., $\mu (G)f(v)$$=d(v)f(v)+\sum _{uv\in E} f(u)$) for $v\in V(G)$. We will refer to such an equation as the eigenvalue equation of $\rho (G)$ (resp., $\mu (G)$). Let $G-u$ denote the graph that arises from G by deleting the vertex $u\in V(G)$ and all the edges incident with u, and $G-uv$ denote the graph that arises from G by deleting the edge $uv\in E(G)$. Similarly, $G+uv$ is the graph that arises from G by adding an edge $uv\notin E(G)$, where $u, v\in V(G)$.

In order to complete the proof of the theorems, we also need the following definition and lemmas.

Lemma 2.2

[4, 14] Let u, v be two vertices of the connected graph G. Suppose $v_1, v_2, \ldots ,$$v_s\in N_G(v)\backslash N_G(u)~(1\le s\le d_G(v))$, and f is the Perron vector of A(G) (resp., Q(G)). Let $G^{\,'}$ be the graph obtained from G by deleting the edges $vv_i$ and adding the edges $uv_i~(1\le i\le s)$. If $f(u)\ge f(v)$, then $\rho (G^{\,'})>\rho (G)$ (resp., $\mu (G^{\,'})>\mu (G)$).

Lemma 2.3

[1, 15] Let G be a connected graph of order n such that $v_1v_3$, $v_2v_4\in E(G)$, $v_1v_2$, $v_3v_4\notin E(G)$. Let $G^{\,'}=G-v_1v_3-v_2v_4+v_1v_2+v_3v_4$. Suppose f is the Perron vector of A(G) (resp., Q(G)); if $f(v_1)\ge f(v_4)$ and $f(v_2)\ge f(v_3)$, then $\rho (G^{\,'})\ge \rho (G)$ (resp., $\mu (G^{\,'})\ge \mu (G)$), where the equalities hold if and only if $f(v_1)=f(v_4)$ and $f(v_2)=f(v_3)$.

Lemma 2.4

[8] Let G be a connected graph and f be the Perron vector of A(G) (resp., Q(G)). Let $G\,'$ be a connected graph obtained from G by deleting $t \, (\ge 1)$ edges and adding another t new edges such that $G\not \cong G\,'$. Suppose that there exists a vertex $v\in V(G)$ such that $N_{G}(v)\subset N_{G\,'}(v)$ or $N_{G\,'}(v)\subset N_{G}(v)$. If ${\mathcal {R}}_{A(G\,')}(f)\ge {\mathcal {R}}_{A(G)}(f)$ (resp., ${\mathcal {R}}_{Q(G\,')}(f)\ge {\mathcal {R}}_{Q(G)}(f)$), then $\rho (G\,')>\rho (G)$ (resp., $\mu (G\,')>\mu (G)$).

Lemma 2.5

[2, 9] Let $\pi =(d_1, d_2, \ldots , d_n)$ be a sequence with $d_1\ge d_2\ge \cdots \ge d_n>0$. Then, $\pi$ is graphic if and only if

$$\begin{aligned} \sum \limits _{i=1}^{n} d_i \text{ is } \text{ even } \text{ and } \sum \limits _{i=1}^{k} d_i\le k(k-1)+\sum \limits _{i=k+1}^{n} \min \{d_i, k\} \quad \text{ for } \text{ all } \quad k=1, 2, \ldots , n-1. \end{aligned}$$

Lemma 2.6

[12] Let $\pi =(d_1, d_2, \ldots , d_n)$ and $\pi ^{\,\prime }=(d_1^{\,\prime }, d_2^{\,\prime }, \ldots , d_n^{\,\prime })$ be two non-increasing degree sequences with $\pi \lhd \pi ^{\,\prime }$. Then, $d_n\ge d_n^{\,\prime }$.

Lemma 2.7

[2, 15] Let $\pi$ and $\pi ^{\,\prime }$ be two non-increasing graphic degree sequences. If $\pi \lhd \pi ^{\,\prime }$, then there exists a series non-increasing graphic degree sequences $\pi _1, \ldots , \pi _k$ such that $(\pi =)$$\pi _0\lhd \pi _1\lhd \cdots \lhd \pi _k \lhd \pi _{k+1}(=\pi ^{\,\prime })$, and $\pi _i$ and $\pi _{i+1}$ differ only at two positions, where the differences are 1 for $0\le i\le k$.

Definition 2.8

[6, 10] Let $\pi =(d_1, d_2, \ldots , d_n)$ and $\pi ^{\,\prime }=(d_1^{\,\prime }, d_2^{\,\prime }, \ldots , d_n^{\,\prime })$ be two different non-increasing degree sequences. We say $\pi$ is star majorizated by $\pi ^{\,\prime }$, denoted by $\pi \lhd ^* \pi ^{\,\prime }$, if and only if $\pi \lhd \pi ^{\,\prime }$ and only two components of $\pi$ and $\pi ^{\,\prime }$ are different by 1, that is, $d_i=d_i^{\, \prime }$ for $i\ne k, l$, $1\le k<l\le n$ and $d_k^{\, \prime }=d_k+1$, $d_l^{\, \prime }=d_l-1$.

Lemma 2.9

Suppose $\pi =(d_1, d_2, \ldots , d_n)$ is a non-increasing degree sequence. If G is an extremal greatest single-cone graph for $\rho (G)$ (resp., $\mu (G)$) in $J_{\pi }$ with Perron vector f, then there exists an ordering of the vertices of G such that $d(v_i)=d_i$ for $1\le i\le n$ and $f(v_1)\ge f(v_2)\ge \cdots \ge f(v_n)$.

Proof

Since G is a single-cone graph, there exists a vertex $v_1$ such that $d_G(v_1)=n-1$ and $G-v_1$ is a connected graph. Create an ordering of the vertices of G beginning with $v_1$ and appending other vertices after it. We use the notation $v_i\prec v_j$ to indicate that the vertex $v_i$ precedes the vertex $v_j$ in the ordering of vertices. Clearly, $v_1\prec v_i$ for $i=2, 3, \ldots , n$. The order of other vertices is defined as follows: if $d_G(v_i)>d_G(v_j)$, or $d_G(v_i)=d_G(v_j)$ and $f(v_i)\ge f(v_j)$, then $v_i\prec v_j$. It is easy to see that this ordering satisfies $d_G(v_1)\ge d_G(v_2)\ge \cdots \ge d_G(v_n)$. We will prove that $f(v_1)\ge f(v_2)\ge \cdots \ge f(v_n)$.

Firstly, we claim that $f(v_1)\ge f(v_i)$ for $2\le i\le n$. Otherwise, we suppose that there exists some vertex $v_i$ such that $f(v_1)<f(v_i)$. If $d_G(v_i)=n-1$, then $N_G(v_1){\setminus } \{v_i\}=N_G(v_i){\setminus } \{v_1\}$. By the eigenvalue equation of $\rho (G)$ (resp., $\mu (G)$), we have $f(v_1)=f(v_i)$, contradicting $f(v_1)<f(v_i)$. If $d_G(v_i)<n-1$, then $N_G(v_1){\setminus } (N_G(v_i)\cup \{v_i\})\ne \varnothing$. Let

$$\begin{aligned} G^{\,\prime }=G-\sum _{u\in N_G(v_1){\setminus } (N_G(v_i)\cup \{v_i\})}v_1u+\sum _{u\in N_G(v_1){\setminus } (N_G(v_i)\cup \{v_i\})}v_iu. \end{aligned}$$

Then, $d_{G^{\,\prime }}(v_i)=d_{G}(v_1)$, $d_{G^{\,\prime }}(v_1)=d_G(v_i)$ and $d_{G^{\,\prime }}(v)=d_G(v)$ for $v\in V(G){\setminus } \{v_1, v_i\}$. Noting that $G-v_1$ is a connected graph and the neighbors of $v_i$ in $G-v_1$ are adjacent to $v_1$ in $G^{\,\prime }-v_i$, we have $G^{\,\prime }-v_i$ which is a connected graph. This implies that $G^{\,\prime }\in J_{\pi }$. By Lemma 2.2, we have $\rho (G^{\,\prime })>\rho (G)$ ( resp., $\mu (G^{\,\prime })>\mu (G)$), a contradiction because G is an extremal greatest single-cone graph for $\rho (G)$ (resp., $\mu (G)$) in $J_{\pi }$. Therefore, $f(v_1)\ge f(v_i)$ for $2\le i\le n$.

Secondly, we show that $v_i\prec v_j$ implies $f(v_i)\ge f(v_j)$ for all $v_i, v_j\in V(G){\setminus } \{v_1\}$. Otherwise, we suppose that there exist two vertices such that $v_i\prec v_j$ but $f(v_j)>f(v_i)$. Then, $d(v_i)\ge d(v_j)$. Noting that $v_i, v_j\in V(G){\setminus } \{v_1\}$, there exists a shortest path $P_{ij}$ from $v_i$ to $v_j$ in $G-v_1$. If $d_G(v_i)>d_G(v_j)$, let $k=d_G(v_i)-d_G(v_j)$, $v_l\in V(P_{ij})$ and $v_lv_i\in E(G)$. Then, there exist k vertices $u_1, \ldots , u_k\in N_G(v_i){\setminus } (N_G(v_j)\cup \{v_l\})$. Let

$$\begin{aligned} G^{\,\prime }=G-\sum _{s=1}^k v_iu_s+\sum _{s=1}^k v_ju_s. \end{aligned}$$

Then, $G^{\,\prime }\in J_{\pi }$. By Lemma 2.2, we have $\rho (G^{\,\prime })>\rho (G)$ (resp., $\mu (G^{\,\prime })>\mu (G)$), a contradiction because G is an extremal greatest single-cone graph for $\rho (G)$ (resp., $\mu (G)$) in $J_{\pi }$. If $d_G(v_i)=d_G(v_j)$, noting that $v_i\prec v_j$, we have $f(v_i)\ge f(v_j)$, contradicting $f(v_j)> f(v_i)$.

Combining the above arguments, we have $f(v_1)\ge f(v_2)\ge \cdots \ge f(v_n)$. $\square$

3 The proof of Theorem 1.2

The proof of Theorem 1.2

Since $\pi \lhd \pi ^{\,\prime }$, it follows from Lemma 2.7 and Definition 2.8 that there exists a series non-increasing graphic degree sequences $\pi _1, \ldots , \pi _k$ such that $(\pi =)$$\pi _0\lhd ^* \pi _1\lhd ^* \cdots \lhd ^* \pi _k \lhd ^* \pi _{k+1}(=\pi ^{\,\prime })$. Let $\pi _i=(d_1^{\,(i)}, d_2^{\,(i)}, \ldots , d_n^{\,(i)})$ for $0\le i\le k$. Clearly, $d_1=d_1^{\,(1)}=\cdots =d_1^{\,(k)}=d_1^{\,\prime }=n-1$. By Lemma 2.6, we have $d_n\ge d_n^{\, (1)}\ge \cdots \ge d_n^{\,(k)}\ge d_n^{\,\prime }$. Noting that $\pi$ and $\pi '$ are two different non-increasing degree sequences of single-cone trees, we have $d_n=d_n^{\,\prime }=2$. This implies that $d_n=d_n^{\,(1)}=\cdots =d_n^{\, (k)}=d_n^{\,\prime }=2$.

Since $\pi \lhd ^* \pi _1$, without loss of generality, we suppose that $d_k+1=d_k^{\,(1)}$, $d_l-1=d_l^{\,(1)}$, $d_j=d_j^{\,(1)}$ for $j\notin \{ k,l\}$, and $1<k<l<n$. Let f be the Perron vector of A(G) (resp., Q(G)). Then, Lemma 2.9 implies that there exists an ordering of the vertices of G such that $d(v_i)=d_i$ for $1\le i\le n$ and $f(v_1)\ge f(v_2)\ge \cdots \ge f(v_n)$. Particularly, $f(v_k)\ge f(v_l)$.

Assume that $G-v_1$ is a tree and $P_{kl}$ is a shortest path from $v_k$ to $v_l$ in $G-v_1$. Noting that $d_l=d_l^{\,(1)}+1>2$, there must exist some $w\in N_{G-v_1}(v_l){\setminus }{N_{G-v_1}(v_k)}$ such that $w\notin V(P_{kl})$. Let $G_1=G-v_lw+v_kw$. Then, $G_1-v_1$ is a tree, $d_{G_1}(v_k)=d_G(v_k)+1$, $d_{G_1}(v_l)=d_G(v_l)-1$, and $d_{G_1}(v)=d_G(v)$ for $v\in V(G){\setminus } \{v_k, v_l\}$. This implies that $G_1$ is a single-cone tree and $G_1\in J_{\pi _1}$. Noting that $f(v_k)\ge f(v_l)$, by Lemma 2.2, we have $\rho (G)<\rho (G_1)$ (resp., $\mu (G)<\mu (G_1)$). Let $G_1^*$ be the single-cone tree with the greatest spectral radius (resp., signless Laplacian spectral radius) in $J_{\pi _1}$. Then, $\rho (G)<\rho (G_1)\le \rho (G_1^*)$ (resp., $\mu (G)<\mu (G_1)\le \mu (G_1^*)$).

By a similar reasoning as the above, we can obtain that $\pi _i$ is a non-increasing degree sequence of a single-cone tree for each $2\le i\le k$. Let $G_i^*$ be a single-cone tree with the greatest spectral radius (resp., signless Laplacian spectral radius) in $J_{\pi _i}$. Then, we have $\rho (G)<\rho (G_1^*)<\cdots<\rho (G_k^*)<\rho (G^{\,\prime })$ (resp., $\mu (G)<\mu (G_1^*)<\cdots<\mu (G_k^*)<\mu (G^{\,\prime }))$. $\square$

4 The proof of Theorem 1.3

Lemma 4.1

Let $\pi =(d_1, d_2, \ldots , d_n)$ be a non-increasing degree sequence of a single-cone unicyclic graph, and G be an extremal greatest single-cone unicyclic graph for $\rho (G)$ (resp., $\mu (G)$) in $J_{\pi }$. Suppose $\pi ^{\,\prime }=(d_1^{\,\prime }, d_2^{\,\prime }, \ldots , d_n^{\,\prime })$ ($d_n^{\,\prime }\ge 2$) is a non-increasing graphic degree sequence such that $\pi \lhd ^* \pi ^{\,\prime }$. Then, there exists a single-cone unicyclic graph $G^{\,\prime }\in J_{\pi ^{\,\prime }}$ such that $\rho (G)<\rho (G^{\,\prime })$ (resp., $\mu (G)<\mu (G^{\,\prime })$).

Proof

Since $\pi \lhd ^* \pi ^{\,\prime }$, without loss of generality, we suppose that $d_k+1=d_k^{\,\prime }$, $d_l-1=d_l^{\,\prime }$, and $d_i=d_i^{\,\prime }$ for $i\ne k,l$. Since $\pi$ is a non-increasing degree sequence of a single-cone unicyclic graph, then $d_1=d_1^{\,\prime }=n-1$, $d_i\ge 2$ for $1\le i\le n$, and $1<k<l\le n$. Assume that $G-v_1$ is a unicyclic graph. Let $P_{kl}$ be a shortest path from $v_k$ to $v_l$ in $G-v_1$, $u\in N_{G-v_1}(v_l)\cap V(P_{kl})$ and f be the Perron vector of A(G) (resp., Q(G)). By Lemma 2.9, there exists an ordering of the vertices of G such that $d(v_i)=d_i$ for $1\le i\le n$ and $f(v_1)\ge f(v_2)\ge \cdots \ge f(v_n)$. Particularly, $f(v_k)\ge f(v_l)$.

Case 1:

$N_{G-v_1}(v_l){\setminus } (N_{G-v_1}(v_k)\cup \{u\})\ne \emptyset$. Assume $w\in N_{G-v_1}(v_l){\setminus } (N_{G-v_1}(v_k)\cup \{u\})$. Let $G^{\,\prime }=G-v_lw+v_kw$. Then, $G^{\,\prime }-v_1$ is a unicyclic graph, $d_{G^{\,\prime }}(v_k)=d_G(v_k)+1$, $d_{G^{\,\prime }}(v_l)=d_G(v_l)-1$, and $d_{G^{\,\prime }}(v)=d_G(v)$ for $v\in V(G){\setminus } \{v_k, v_l\}$. This implies that $G^{\,\prime }$ is a single-cone unicyclic graph and $G^{\,\prime }\in J_{\pi '}$. Noting that $f(v_k)\ge f(v_l)$, by Lemma 2.2, we have $\rho (G)<\rho (G^{\,\prime })$ (resp., $\mu (G)<\mu (G^{\,\prime })$).

Case 2:

$N_{G-v_1}(v_l){\setminus } (N_{G-v_1}(v_k)\cup \{u\})=\emptyset$. Noting that G is a single-cone unicyclic graph, we have $d_l\le 3$, and so $d_l^{\,\prime }\le 2$. Since $d_l^{\,\prime }\ge d_n^{\,\prime } \ge 2$, it follows that $d_l^{\,\prime }=2$, $d_l=3$, $d_i=d_i^{\,\prime }=2$ for $l+1\le i\le n$. Let $w\in N_{G-v_1}(v_l){\setminus } \{u\}$. Then, $w\in N_{G-v_1}(v_k)$. This implies that $|V(P_{kl})|\le 3$.

Subcase 2.1:

$|V(P_{kl})|=2$. In this case, $C_3=v_kwv_lv_k$ is the unique cycle of $G-v_1$. We claim that $l\ge 5$. Otherwise, we suppose $l\le 4$. Noting that $\pi ^{\,\prime }$ is a non-increasing degree sequence and $d_l^{\,\prime }=2$, we have $d_i^{\,\prime }=2$ for $4\le i\le n$. By $\pi \lhd \pi ^{\,\prime }$, we have $\sum \nolimits _{i=1}^{n} d_i^{\,\prime }=\sum \nolimits _{i=1}^{n} d_i=2(2n-2)$. It follows that

$$\begin{aligned} \sum \limits _{i=1}^{3} d_i^{\,\prime }=\sum \limits _{i=1}^{n} d_i^{\,\prime }-\sum \limits _{i=4}^{n} d_i^{\,\prime }=2n+2>2n=3(3-1)+\sum \limits _{i=4}^{n} \min \{d_i^{\,\prime }, 3\}, \end{aligned}$$

a contradiction to Lemma 2.5. Therefore, $l\ge 5$. This implies that there must exist vertices a, b such that $a\notin \{v_1, v_k, v_l, w\}$, $d_G(a)\ge 3$, $d_G(b)=2$, and $ab\in E(G)$.

If $av_k\notin E(G)$, noting that $d_G(v_l)=3$ and $d_G(b)=2$, we have $f(v_l)\ge f(b)$, $av_l\notin E(G)$, and $bv_k\notin E(G)$. We claim that $f(v_k)\ge f(a)$. Otherwise, we suppose $f(a)>f(v_k)$. Let $G^*=G-ab-v_lv_k+av_l+bv_k$. By Lemma 2.3, we have $\rho (G)<\rho (G^*)$ (resp., $\mu (G)<\mu (G^*)$). It is easy to see that $G^*\in J_{\pi }$, which is a contradiction because G has the greatest spectral radius (resp., signless Laplacian spectral radius) in $J_{\pi }$. Therefore, $f(v_k)\ge f(a)$. Noting that $d_i=d_i^{\,\prime }=2$ for $l+1\le i\le n$, we have $a\prec v_l$. It follows that $f(a)\ge f(v_l)$. Let $G^{\,\prime }=G-v_kv_l-ab+v_ka+v_kb$. Then, $d_{G^{\,\prime }}(v_k)=d_G(v_k)+1$, $d_{G^{\,\prime }}(v_l)=d_G(v_l)-1$, and $d_{G^{\,\prime }}(v)=d_G(v)$ for $v\in V(G){\setminus } \{v_k, v_l\}$. It is not difficult to see that $G^{\,\prime }-v_1$ is a unicyclic graph, $G^{\,\prime }\in J_{\pi '}$,

$$\begin{aligned} {\mathcal {R}}_{A(G^{\,\prime })}(f)-{\mathcal {R}}_{A(G)}(f)= & {} 2\sum \limits _{xy\in E(G^{\,\prime })} f(x)f(y)-2\sum \limits _{xy\in E(G)} f(x)f(y)\\= & {} 2f(v_k)f(a)+2f(v_k)f(b)-2f(a)f(b)-2f(v_k)f(v_l)\\= & {} 2f(v_k)(f(a)-f(v_l))+2f(b)(f(v_k)-f(a))\ge 0, \end{aligned}$$

and

$$\begin{aligned} {\mathcal {R}}_{Q(G^{\,\prime })}(f)-{\mathcal {R}}_{Q(G)}(f)= & {} \sum \limits _{xy\in E(G^{\,\prime })} (f(x)+f(y))^2-\sum \limits _{xy\in E(G)} (f(x)+f(y))^2\\= & {} f^2(v_k)-f^2(v_l)+2f(v_k)(f(a)-f(v_l))\\&+2f(b)(f(v_k)-f(a))\ge 0. \end{aligned}$$

By Lemma 2.1, we have $\rho (G^{\,\prime })\ge {\mathcal {R}}_{A(G^{\,\prime })}(f)\ge {\mathcal {R}}_{A(G)}(f)=\rho (G)$ (resp., $\mu (G^{\,\prime })\ge {\mathcal {R}}_{Q(G^{\,\prime })}(f)\ge {\mathcal {R}}_{Q(G)}(f)=\mu (G)$). Noting that $N_{G^{\,\prime }}(v_l)\subset N_G(v_l)$, by Lemma 2.4, we have $\rho (G^{\,\prime })>\rho (G)$ (resp., $\mu (G^{\,\prime })>\mu (G)$).

If $av_k\in E(G)$, we can show $f(v_k)\ge f(a)$ similarly. Let $G^{\,\prime }=G-wv_l-ab+wa+v_kb$. Then, $d_{G^{\,\prime }}(v_k)=d_G(v_k)+1$, $d_{G^{\,\prime }}(v_l)=d_G(v_l)-1$, and $d_{G^{\,\prime }}(v)=d_G(v)$ for $v\in V(G){\setminus } \{v_k, v_l\}$. It is not difficult to see that $G^{\,\prime }-v_1$ is a unicyclic graph, $G^{\,\prime }\in J_{\pi '}$,

$$\begin{aligned} {\mathcal {R}}_{A(G^{\,\prime })}(f)-{\mathcal {R}}_{A(G)}(f)= & {} 2\sum \limits _{xy\in E(G^{\,\prime })} f(x)f(y)-2\sum \limits _{xy\in E(G)} f(x)f(y)\\= & {} 2f(w)f(a)+2f(v_k)f(b)-2f(a)f(b)-2f(w)f(v_l)\\= & {} 2f(w)(f(a)-f(v_l))+2f(b)(f(v_k)-f(a))\ge 0, \end{aligned}$$

and

$$\begin{aligned} {\mathcal {R}}_{Q(G^{\,\prime })}(f)-{\mathcal {R}}_{Q(G)}(f)= & {} \sum \limits _{xy\in E(G^{\,\prime })} (f(x)+f(y))^2-\sum \limits _{xy\in E(G)} (f(x)+f(y))^2\\= & {} f^2(v_k)-f^2(v_l)+2f(w)(f(a)-f(v_l))\\&+\,2f(b)(f(v_k)-f(a))\ge 0. \end{aligned}$$

By Lemma 2.1, we have $\rho (G^{\,\prime })\ge {\mathcal {R}}_{A(G^{\,\prime })}(f)\ge {\mathcal {R}}_{A(G)}(f)=\rho (G)$ (resp., $\mu (G^{\,\prime })\ge {\mathcal {R}}_{Q(G^{\,\prime })}(f)\ge {\mathcal {R}}_{Q(G)}(f)=\mu (G)$). Noting that $N_{G^{\,\prime }}(v_l)\subset N_G(v_l)$, by Lemma 2.4, we have $\rho (G^{\,\prime })>\rho (G)$ (resp., $\mu (G^{\,\prime })>\mu (G)$).

Subcase 2.2:

$|V(P_{kl})|=3$. In this case, $P_{kl}=v_kuv_l$ and $C_4=v_kwv_luv_k$ is the unique cycle of $G-v_1$. This implies that $d_G(v_k)\ge 3$, $d_G(w)\ge 3$, $d_G(u)\ge 3$. By $d_i=d^{\,\prime }_i=2$ for $l+1\le i\le n$, we have $w\prec v_l$ and $u\prec v_l$. It follows that $f(w)\ge f(v_l)$ and $f(u)\ge f(v_l)$.

If $f(v_k)\ge f(w)$, let $G^{\,\prime }=G-wv_l-uv_l+v_kv_l+wu$. Then, $d_{G^{\,\prime }}(v_k)=d_G(v_k)+1$, $d_{G^{\,\prime }}(v_l)=d_G(v_l)-1$, $d_{G^{\,\prime }}(v)=d_G(v)$ for $v\in V(G){\setminus } \{v_k, v_l\}$. This implies that $G^{\,\prime }-v_1$ is a unicyclic graph, $G^{\,\prime }\in J_{\pi '}$,

$$\begin{aligned} {\mathcal {R}}_{A(G^{\,\prime })}(f)-{\mathcal {R}}_{A(G)}(f)= & {} 2\sum \limits _{xy\in E(G^{\,\prime })} f(x)f(y)-2\sum \limits _{xy\in E(G)} f(x)f(y)\\= & {} 2f(v_k)f(v_l)+2f(w)f(u)-2f(w)f(v_l)-2f(u)f(v_l)\\= & {} 2f(v_l)(f(v_k)-f(w))+2f(u)(f(w)-f(v_l))\ge 0, \end{aligned}$$

and

$$\begin{aligned} {\mathcal {R}}_{Q(G^{\,\prime })}(f)-{\mathcal {R}}_{Q(G)}(f)= & {} \sum \limits _{xy\in E(G^{\,\prime })} (f(x)+f(y))^2-\sum \limits _{xy\in E(G)} (f(x)+f(y))^2\\= & {} f^2(v_k)-f^2(v_l)+2f(v_l)(f(v_k)-f(w))\\&+\,2f(u)(f(w)-f(v_l))\ge 0. \end{aligned}$$

By Lemma 2.1, we have $\rho (G^{\,\prime })\ge {\mathcal {R}}_{A(G^{\,\prime })}(f)\ge {\mathcal {R}}_{A(G)}(f)=\rho (G)$ (resp., $\mu (G^{\,\prime })\ge {\mathcal {R}}_{Q(G^{\,\prime })}(f)\ge {\mathcal {R}}_{Q(G)}(f)=\mu (G)$. Noting that $N_G(v_k)\subset N_{G^{\,\prime }}(v_k)$, by Lemma 2.4, we have $\rho (G^{\,\prime })>\rho (G)$ (resp., $\mu (G^{\,\prime })>\mu (G)$).

If $f(w)>f(v_k)$, let $G^{\,\prime }=G-wv_l-uv_k+wu+v_kv_l$. It is easy to see that $G^{\,\prime }-v_1$ is a unicyclic graph and $d_{G^{\,\prime }}(v)=d_G(v)$ for $v\in V(G)$. This implies that $G^{\,\prime }\in J_{\pi }$. By Lemma 2.3, we have $\rho (G^{\,\prime })>\rho (G)$ (resp., $\mu (G^{\,\prime })>\mu (G)$), a contradiction because G has the greatest spectral radius (resp., signless Laplacian spectral radius) in $J_{\pi }$. $\square$

The proof of Theorem 1.3

Since $\pi \lhd \pi ^{\,\prime }$, it follows from Lemma 2.7 and Definition 2.8 that there exists a series non-increasing graphic degree sequences $\pi _1, \ldots , \pi _k$ such that $(\pi =)$$\pi _0\lhd ^* \pi _1\lhd ^* \cdots \lhd ^* \pi _k \lhd ^* \pi _{k+1}(=\pi ^{\,\prime })$. Let $\pi _i=(d_1^{\,(i)}, d_2^{\,(i)}, \ldots , d_n^{\,(i)})$ for $0\le i\le k$. By Lemma 2.6, we have $d_n\ge d_n^{\,(1)}\ge \cdots \ge d_n^{\,(k)}\ge d_n^{\,\prime }\ge 2$.

For $\pi$ and $\pi _1$, Lemma 4.1 implies that there exists a single-cone unicyclic graph $G_1\in J_{\pi _1}$ such that $\rho (G)<\rho (G_1)$ (resp., $\mu (G)<\mu (G_1)$). It follows that $\pi _1$ is a non-increasing degree sequence of a single-cone unicyclic graph. Let $G_1^*$ be a single-cone unicyclic graph with the greatest spectral radius (resp., signless Laplacian spectral radius) in $J_{\pi _1}$. Then, $\rho (G)<\rho (G_1)\le \rho (G_1^*)$ (resp., $\mu (G)<\mu (G_1)\le \mu (G_1^*)$).

By a similar reasoning as the above, we can obtain that $\pi _j$ is a non-increasing degree sequence of a single-cone unicyclic graph for each $2\le j\le k$. Let $G_j^*$ be a single-cone unicyclic graph with the greatest spectral radius (resp., signless Laplacian spectral radius) in $J_{\pi _j}$ for $2\le j\le k$. By Lemma 4.1, we have $\rho (G)<\rho (G_1^*)<\cdots<\rho (G_k^*)<\rho (G^{\,\prime })$ (resp., $\mu (G)<\mu (G_1^*)<\cdots<\mu (G_k^*)<\mu (G^{\,\prime })$). $\square$