Some Results on Spectrum and Energy of Graphs with Loops

Abstract

Let $G_S$ be a graph with loops obtained from a graph G of order n and loops at $S \subseteq V(G).$ In this paper, we establish a neccesary and sufficient condition on the bipartititeness of a connected graph G and the spectrum ${\textrm{Spec}}(G_S)$ and ${\textrm{Spec}}(G_{V(G)\backslash S})$. We also prove that for every $S\subseteq V(G),$ ${\mathcal E}(G_S) \ge {\mathcal E}(G)$ when G is bipartite. Moreover, we provide an identification of the spectrum of complete graphs $K_n$ and complete bipartite graphs $K_{m,n}$ with loops. We characterize any graphs with loops of order n whose eigenvalues are all positive or non-negative, and also any graphs with a few distinct eigenvalues. Finally, we provide some bounds related to $G_S$.

1 Introduction

Let G be a simple graph and V(G) be the set of vertices and E(G) be the set of edges. We call G a graph of order n and size m if $|V(G)|=n$ and $|E(G)|=m$. Denote by $\overline{G}$ the complement of G. In this paper, the path of order n is denoted by $P_n$ whereas the complete graph of order n is denoted by $K_n$ and the complete bipartite graph with parts M and N with sizes m and n,  is denoted as $K_{m,n}$. Let $V(G)=\{v_1,\ldots , v_n\}$. For $i=1,2,\ldots ,n,$ we denote $d_i$ the degree of $v_i$ and $\Delta (G)=\max _{1 \le i \le n} d_i(G).$ In addition, G is called a (a, b)-semi-regular if $d_i=a$ or b for $i=1,\ldots , n.$

The adjacency matrix of G, denoted by $A(G)=(a_{ij}),$ whose entry is $a_{ij}=1$ if $v_i$ and $v_j$ are adjacent and $a_{ij}=0$ otherwise. The characteristic polynomial and eigenvalues of G are the characteristic polynomial and eigenvalues of A(G), respectively. As A(G) is a real symmetric matrix, all eigenvalues of G are real numbers and thus can be ordered as $\lambda _1(G) \ge \lambda _2(G)\ge \cdots \ge \lambda _n(G)$ with $\lambda _1(G)$ and $\lambda _n(G)$ being the largest and the smallest eigenvalue of G,  respectively. See [2, 3, 19] for more details. All eigenvalues of G with each respective algebraic multiplicity give the spectrum of G,  denoted by ${\textrm{Spec}}(G)= \begin{pmatrix} \lambda _1 &{} \lambda _2 &{} \cdots &{} \lambda _n \\ a_1 &{} a_2 &{} \cdots &{} a_n \end{pmatrix},$ where $a_i$ is the algebraic multiplicity of $\lambda _i$. The energy of G is defined as

$$\begin{aligned} {\mathcal E}(G)&=\sum ^n_{i=1} |\lambda _i(G)|, \quad (\text {cf. [6]}). \end{aligned}$$

In the early 1970 s, graph theory had been found to have an important application in the study of calculation of electron and molecules energy [16, 17]. This has initiated the emerging of the concept of energy of simple graphs in 1978 by Gutman [6] which greatly advances the research study of graphs and the energy of graphs from 1995 till now [5, 6, 9,10,11,12, 15, 18, 20].

The graph obtained from G by attaching a self-loop at each of the vertex in $S\subseteq V(G)$, is called the self-loop graph of G at S, denoted by $G_S$. Generalizing from the definition of A(G), $A(G_S)=J_S+A(G)$ where $(J_S)_{i,j}=1$ if $i=j$ and $v_i\in S$, and $(J_S)_{i,j}=0$ otherwise. Similar to G, the eigenvalues of $G_S$ are the eigenvalues of $A(G_S).$ It can be verified that all properties of eigenvalues of A(G) are attained by $A(G_S)$. The energy of $G_S$ of order n with $|S|=\sigma$ is defined as

$$\begin{aligned} {\mathcal E}(G_S)&=\sum ^n_{i=1} \left| \lambda _i(G_S)-\frac{\sigma }{n}\right| , \quad (\text {cf. [13]}). \end{aligned}$$

In some occasion, we also denote G as the self-loop graph and $G_0$ is the ordinary graph obtained from G by removing all its self-loops. When $\sigma =n,$ we write $G_S$ as $\widehat{G}.$

Self-loop graphs have been shown to play a significant role in the mathematical study of heteroconjugated molecules [7, 8, 16]. Recently in 2022, Gutman et al. have introduced the concept of energy of self-loops graphs for the first time in [13]. The study of energy of self-loop graphs is still very new with results appeared in only two papers [13, 14]. In [13], the following results were proved:

Theorem 1. Let G be a bipartite graph of order n, with vertex set V. Let S be a subset of V. Then, ${\mathcal E}(G_S)={\mathcal E}(G_{V\backslash S})$.

Theorem 2. Let $G_S$ be a self-loop graph of order n, with m edges, and $|S|=\sigma$. Let $\lambda _1\ge \lambda _2 \ge \cdots \ge \lambda _n$ be its eigenvalues. Then $\sum ^n_{i=1} \lambda _i^2 = 2\,m+\sigma .$

Theorem 3. Let $G_S$ be a self-loop graph of order n, with m edges, and $|S|=\sigma$. Then

$$\begin{aligned} {\mathcal E}(G_S)\le \sqrt{n\left( 2m+\sigma -\frac{\sigma ^2}{n}\right) }. \end{aligned}$$

This paper consists of four sections of main results. Section 2 first completely determines the ${\textrm{Spec}}((K_n)_S)$ and ${\textrm{Spec}}((K_{m,n})_S)$ for all $n,m \ge 1$ using Theorem 2. The results on ${\textrm{Spec}}((K_n)_S)$ are then used to completely characterize those $G_S$ with only positive or non-negative eigenvalues as well as with few distinct eigenvalues. In Sect. 3, a necessary and sufficient result on each $\lambda _i(G_{V\backslash S}) = 1-\lambda _i(G_S)$ in relation with the respective $\lambda _i(G_S)$ for every $i=1,\ldots ,n$ when G is bipartite, is obtained and this gives a simplified proof of Theorem 1. We also show that ${\mathcal E}(G_S)\ge {\mathcal E}(G)$ when G is bipartite and a conjecture is given. In Sect. 4, an alternative proof of Theorem 3 is given using Cauchy-Schwarz inequality. This approach leads to a result on the semi-regular graphs with self-loops. An upper bound for $\lambda _1(G_S)$ analogous to a classical bound for $\lambda _1(G)$ in terms of $\Delta (G)$ is obtained. The existence of a certain semi-regular graphs is shown when the upper bound for $\lambda _1(G_S)$ is attained.

2 Some Characterization of Self-loop Graphs by Its Eigenvalues

In this section, we aim to provide some characterization of self-loop graphs with positive and non-negative eigenvalues, as well as those with few distinct eigenvalues. Before that, an identification of ${\textrm{Spec}}((K_n)_S)$ and ${\textrm{Spec}}((K_{m,n})_S)$ are needed for arbitrary S. The results we obtained are a generalisation of the classical spectrum result of ${\textrm{Spec}}(K_n)$ and ${\textrm{Spec}}(K_{m,n})$ when $\sigma =0.$

The self-loop spectrum characterization for both $K_n$ and $K_{m,n}$ are technical and require careful examinations of several cases. As a rule of thumb, consider G as $K_n$ or $K_{m,n},$ then the steps to determine ${\textrm{Spec}}(G_S)$ are as follows.

Step 1. Determination of the multiplicity of eigenvalue 0.

Determine the rank of $A(G_S)$ through each of its submatrices. By the Rank-Nullity Theorem, this implies the nullity of $A(G_S),$ the multiplicity $m_0$ of the eigenvalue 0.

Step 2. Determination of multiplicity of eigenvalue 1 (for $(K_n)_S$) or $-1$ (for $(K_{m,n})_S$).

Repeat Step 1 for the matrix $A(G_S)-I_n$ or $A(G_S)+I_n$ to obtain the multiplicity $m_1$ of the eigenvalue 1 or $-1,$ respectively.

Step 3. Determination of the remaining eigenvalues and its multiplicities.

Find the remaining $n-m_0-m_1$ many eigenvalues by appropriate methods. If there are only two eigenvalues $\lambda _1(G_S),\lambda _2(G_S)$ left, then they can be obtained by solving the simultaneous equations in Lemma 2.1 below.

Lemma 2.1

[13] Let $G_S$ be a self-loop graph of order n and m edges. Let $\lambda _1(G_S),\ldots ,\lambda _n(G_S)$ be its eigenvalues. Then,

• (i) $\displaystyle \sum ^n_{i=1} \lambda _i(G_S)=\sigma ,$

• (ii) $\displaystyle \sum ^n_{i=1} \lambda ^2_i (G_S)=2m+\sigma .$

For $(K_{m,n})_S,$ there are occasions where we need to determine three eigenvalues $\lambda _i((K_{m,n})_S),$ $i=1,2,3.$ In particular, an additional formula of $\sum _i \lambda ^3_i((K_{m,n})_S)$ will be introduced, see Lemma 2.3.

Without loss of generality, let $S=\{1,2,\ldots , \sigma \}.$ Let $J_{k\times \ell }$ denote the $k \times \ell$ matrix whose all entries are 1 and $j_k$ be the $k \times 1$ vector with all entries 1.

2.1 Identification of ${\textrm{Spec}}((K_n)_S)$

In this subsection, ${\textrm{Spec}}((K_n)_S)$ for arbitrary S are identified according to $\sigma .$

Theorem 2.2

Let $(K_n)_S$ be the self-loop graph of $K_n$ and $|S|=\sigma .$ Then, ${\textrm{Spec}}((K_n)_S)$ are determined by the following three cases:

Case 1. $\sigma =0.$ Then, $A((K_n)_S)=A(K_n).$ This is the classical case where

$$\begin{aligned} {\textrm{Spec}}((K_n)_S) ={\textrm{Spec}}(K_n) = \begin{pmatrix} n-1 &{} -1 \\ 1 &{} n-1 \end{pmatrix}. \end{aligned}$$

Case 2. $0<\sigma <n.$ Then,

$$\begin{aligned} A((K_n)_S)= \left[ \begin{array}{c|c} J_{\sigma \times \sigma } &{} J_{\sigma \times (n-\sigma )} \\ \hline J_{(n-\sigma ) \times \sigma } &{} J_{(n-\sigma ) \times (n-\sigma )} - I_{n-\sigma } \end{array} \right] = \left[ \begin{array}{c} B \\ \hline C \end{array} \right] . \end{aligned}$$

Clearly, ${\textrm{rank}}(B)=1.$ On the other hand, $J_{(n-\sigma )\times (n-\sigma )} - I_{n-\sigma }$ is the adjacency matrix of $K_{n-\sigma },$ which follows from Case 1 there is no zero eigenvalue, thus it is invertible. Thus, all rows in C are linearly independent. We claim that all rows of C are linearly independent to $j_n^T$, a row of B. Let $\alpha _i$ be the i-th row of the adjacency matrix of $K_{n-\sigma }.$ Then, $\{\alpha _1,\ldots , \alpha _{n-\sigma }\}$ is a basis for ${\mathbb R}^{n-\sigma }.$ Clearly, $j^T_{n-\sigma }=\sum ^{n-\sigma }_{i=1}\frac{1}{n-\sigma -1}\alpha _i$ and this implies that no rows of B is a linear combination of rows of C. Thus,

$$\begin{aligned} {\textrm{rank}}(A(K_n)_S) = n-\sigma +1, \quad {\textrm{null}}(A(K_n)_S)= \sigma -1. \end{aligned}$$

This completes Step 1.

Now, we consider

$$\begin{aligned} A(K_n)_S + I_n = \left[ \begin{array}{c|c} J_{\sigma \times \sigma }+ I_\sigma &{} J_{\sigma \times (n-\sigma )} \\ \hline J_{(n-\sigma ) \times \sigma } &{} J_{(n-\sigma ) \times (n-\sigma )} \end{array} \right] = \left[ \begin{array}{c} B \\ \hline C \end{array} \right] . \end{aligned}$$

We have ${\textrm{rank}}(C)=1.$ Let $B'=J_{\sigma \times \sigma } + I_\sigma .$ Then,

$$\begin{aligned} B'=J_{\sigma \times \sigma }+ I_\sigma = 2I_\sigma + A(K_\sigma ). \end{aligned}$$

The eigenvalues of $B'$ are $\sigma +1$ with multiplicity 1 and 1 with multiplicity $\sigma -1$. Hence, $B'$ are invertible. By a similar argument as in Step 1, all rows of B are linearly independent to $j_n^T$, a row of C. Thus, we have

$$\begin{aligned} {\textrm{rank}}(A(K_n)_S + I_n)=\sigma +1, \quad {\textrm{null}}(A(K_n)_S +I_n) = n-\sigma -1. \end{aligned}$$

This completes Step 2.

For Step 3, note that $n-(\sigma -1)-(n-\sigma -1)=2.$ By Lemma 2.1, we have

$$\begin{aligned} \lambda _1 ((K_n)_S) + \lambda _2 ((K_n)_S)&= n-1, \\ \lambda ^2_1 ((K_n)_S) + \lambda ^2_2 ((K_n)_S)&= n^2-2n+2\sigma +1. \end{aligned}$$

Solving this, we obtain

$$\begin{aligned} \lambda _{1,2}((K_n)_S)= \frac{(n-1)\pm \sqrt{(n-1)^2+4\sigma }}{2}. \end{aligned}$$

As a conclusion, we have

$$\begin{aligned} {\textrm{Spec}}((K_n)_S)= \begin{pmatrix} \frac{(n-1)+ \sqrt{(n-1)^2+4\sigma }}{2} &{} 0 &{} -1 &{} \frac{(n-1)- \sqrt{(n-1)^2+4\sigma }}{2} \\ 1 &{} \sigma -1 &{} n-\sigma -1 &{} 1 \end{pmatrix}. \end{aligned}$$

(2.1)

Case 3. $\sigma =n.$ Let $\widehat{K_n}=(K_n)_S.$ Then, $A(\widehat{K_n}) = A(K_n) + I_n.$ Thus, we have

$$\begin{aligned} {\textrm{Spec}}(\widehat{K_n})= \begin{pmatrix} n &{} 0 \\ 1 &{} n-1 \end{pmatrix}. \end{aligned}$$

(2.2)

2.2 Identification of ${\textrm{Spec}}((K_{m,n})_S)$

In this subsection, ${\textrm{Spec}}((K_{m,n})_S)$ for arbitrary S are identified according to $\sigma .$ Before that, we first show a crucial lemma.

Lemma 2.3

Let $G=K_{m,n}$ with $S=S_M \cup S_N \subseteq V(G)$ with $S_M \subseteq M$ and $S_N \subseteq N.$ Let $|S_M|=\sigma _M,$ $|S_N|=\sigma _N,$ and $|S|=\sigma =\sigma _M+\sigma _N$. If $\lambda _1(G_S), \ldots ,\lambda _n(G_S)$ are the eigenvalues of $G_S,$ then

$$\begin{aligned} \sum ^n_{i=1} \lambda ^3_i(G_S) = 3(m \sigma _N + n\sigma _M)+\sigma . \end{aligned}$$

Proof

By [2, Lemma 2.5 & Result 2 h], $\sum ^n_{i=1} \lambda ^3_i(G_S)$ is the number of closed walks of length 3. Generally, a closed walk of length 3 in $G_S$ is either a triple self-looping or three walks that involve two edges and a loop. Without loss of generality, we write vertices that have a loop by $\mathring{v},$ vertices without loop by $\bar{v}$, and $\ell$ as a loop. A direct counting of total number of closed walks of length 3 in $G_S$ is as follows.

Case 1. Starting from a $v \in M$ that is incident with $u \in N$ via an edge e.

• (1) For $\mathring{v},$ there are only two closed walks of length 3: $\mathring{v} \ell \mathring{v} e u e \mathring{v}$ and $\mathring{v} e u e \mathring{v} \ell \mathring{v}.$ This gives a total of $2n\sigma _M$ closed walks of length 3.

• (2) For $\bar{v},$ it has to be incident with $\mathring{u} \in N$. There is only one such walk: $\bar{v} e \mathring{u} \ell \mathring{u} e \bar{v}.$ This gives a total of $m\sigma _N$ closed walks of length 3.

Thus, totally there are $2n\sigma _M +m\sigma _N$ closed walks of length 3.

Case 2. Starting from $u \in N$ that is incident with $\mathring{v}$ or $\bar{v} \in M$ via an edge e. Similar to Case 1, there is a total of $2m\sigma _N +n\sigma _M$ closed walks of length 3.

Case 3. Triple self-looping at $\mathring{v} \in M$ or $\mathring{u} \in N.$ There are $\sigma$ triple self-loopings.

Thus, the total number of closed walks of length 3 is $3(m\sigma _N+n\sigma _M)+\sigma .$ $\square$

We are now ready to characterize the spectrum of $G_S.$

Theorem 2.4

Let $(K_{m,n})_S$ be the self-loop graph of $K_{m,n}$ for $S \subseteq V(K_{m,n})$ with $|S|=\sigma .$ Assume that, if $0<\sigma \le m,$ all loops are within M,  and if $m<\sigma \le m+n,$ there are m loops in M and $\sigma -m$ loops in N,  respectively. Then, the spectrum ${\textrm{Spec}}((K_{m,n})_S)$ are determined by the following five cases:

For convenience, we let $G=K_{m,n}$. Recall that the adjacency matrix of G is given by

$$\begin{aligned} A(G) = \left[ \begin{array}{c|c} {\textbf {0}}_{m} &{} J_{m \times n} \\ \hline J_{n \times m} &{} {\textbf {0}}_{n} \end{array} \right] . \end{aligned}$$

(2.3)

Case 1. $\sigma =0$. It is clear that $A(G_S)=A(G)$ and so

$$\begin{aligned} {\textrm{Spec}}(G_S)={\textrm{Spec}}(K_{m,n})= \begin{pmatrix} \sqrt{mn} &{} 0 &{} -\sqrt{mn} \\ 1 &{} m+n-2 &{} 1 \end{pmatrix}. \end{aligned}$$

Case 2. $0<\sigma < m$. The adjacency matrix of $G_S$ is

$$\begin{aligned} A(G_S)= \left[ \begin{array}{c|c} J_S &{} J_{m \times n} \\ \hline J_{n \times m} &{} {\textbf {0}}_{n} \end{array} \right] = \left[ \begin{array}{ccc} I_\sigma &{} {\textbf {0}}_{\sigma \times (m-\sigma )} &{} J_{\sigma \times n}\\ \hline {\textbf {0}}_{(m-\sigma ) \times \sigma }&{} {\textbf {0}}_{m-\sigma } &{} J_{(m-\sigma ) \times n} \\ \hline J_{n \times \sigma } &{} J_{n \times (m-\sigma )} &{} {\textbf {0}}_{n} \\ \end{array} \right] = \left[ \begin{array}{c} B\\ \hline C\\ \hline D\\ \end{array} \right] . \end{aligned}$$

We proceed with Step 1. It is straightforward to observe that both submatrices C and D have rank 1 due to repeated rows. One verifies that ${\textrm{rank}}(A(G_S))= \sigma +2$ and we obtain the eigenvalue 0 with multiplicity ${\textrm{null}}(A)=m+n-\sigma -2.$

For Step 2, consider the matrix $A(G_S) - I_{m+n},$ which can be viewed in three parts as well:

$$\begin{aligned} A(G_S)-I_{m+n}= \left[ \begin{array}{ccc} {\textbf {0}}_\sigma &{} {\textbf {0}}_{\sigma \times (m-\sigma )} &{} J_{\sigma \times n}\\ \hline {\textbf {0}}_{(m-\sigma ) \times \sigma }&{} -I_{m-\sigma } &{} J_{(m-\sigma ) \times n} \\ \hline J_{n \times \sigma } &{} J_{n \times (m-\sigma )} &{} -I_{n} \\ \end{array} \right] = \left[ \begin{array}{c} B\\ \hline C\\ \hline D\\ \end{array} \right] . \end{aligned}$$

One checks that ${\textrm{rank}}(B)=1,$ ${\textrm{rank}}(C)=m-\sigma ,$ and ${\textrm{rank}}(D)=n.$ Thus, ${\textrm{rank}}(A(G_S) - I_{m+n})=m+n -\sigma +1.$ So, $G_S$ has the eigenvalue 1 with multiplicity ${\textrm{null}}(A(G_S)-I_{m+n})=\sigma -1.$

For Step 3, there are $m+n-{\textrm{null}}(A(G_S))-{\textrm{null}}(A(G_S)-I_{m+n})=3$ eigenvalues $\lambda _i=\lambda _i(G_S),$ $i=1,2,3,$ yet to be determined. Now, by Lemma 2.1 and Lemma 2.3, we obtain

$$\begin{aligned} \lambda _1 + \lambda _2 + \lambda _3&= 1, \\ \lambda ^2_1 + \lambda ^2_2 + \lambda ^2_3&= 2mn+1, \\ \lambda ^3_1 + \lambda ^3_2 + \lambda ^3_3&= 3n\sigma +1. \end{aligned}$$

By a fundamental property of a cubic polynomial of $\lambda _i,$ $i=1,2,3,$ we write

$$\begin{aligned} p(\lambda ) =(\lambda - \lambda _1)(\lambda - \lambda _2)(\lambda - \lambda _3) =\lambda ^3+a\lambda ^2+b\lambda +c. \end{aligned}$$

Solving the coefficients yield

$$\begin{aligned} a&=-1, \quad b = \frac{1}{2}\left( \left( \sum ^3_{i=1}\lambda _i\right) ^2 - \sum ^3_{i=1}\lambda _i^2\right) = -mn, \\ c&= \frac{(\sum ^3_{i=1}\lambda _i)^3 - (\sum ^3_{i=1}\lambda _i^3) - 3(\sum ^3_{i=1}\lambda _i)(\sum _{i\ne j}\lambda _i\lambda _j) }{3} = n(m-\sigma ). \nonumber \end{aligned}$$

Thus, $\lambda _1,\lambda _2,\lambda _3$ are exactly the roots of $p(\lambda )=\lambda ^3-\lambda ^2-mn\lambda + n(m-\sigma ).$ Note that $p(0)>0,$ $p(1)=-n\sigma <0,$ $\lim _{\lambda \rightarrow -\infty } p(\lambda )= -\infty ,$ and $\lim _{\lambda \rightarrow +\infty }p(\lambda ) = +\infty .$ Thus, by Intermediate Value Theorem, the three roots of $p(\lambda )$ are in the intervals $(-\infty ,0), (0,1),$ and $(1,+\infty ),$ respectively. Hence, $\lambda _1,\lambda _2,$ and $\lambda _3$ are all distinct with multiplicity 1.

As a conclusion, when $0<\sigma <m,$ we have

$$\begin{aligned} {\textrm{Spec}}(G_S) = \begin{pmatrix} 1 &{} 0 &{} \lambda _1 &{} \lambda _2 &{} \lambda _3\\ \sigma -1 &{} m+n-\sigma -2 &{} 1 &{} 1 &{} 1 \end{pmatrix} \end{aligned}$$

where $\lambda _1,\lambda _2,\lambda _3$ are the roots of $p(\lambda )=\lambda ^3-\lambda ^2-mn\lambda +n(m-\sigma ).$

Case 3. $\sigma = m$. From (), we have

$$\begin{aligned} A(G_S)= \left[ \begin{array}{c|c} I_{m} &{} J_{m \times n}\\ \hline J_{n \times m} &{} {\textbf {0}}_{n} \end{array} \right] . \end{aligned}$$

Clearly, we have ${\textrm{rank}}(A(G_S))=m+1$ and ${\textrm{null}}(A(G_S))=n-1.$ Hence, $G_S$ has the eigenvalue 0 with multiplicity $n-1.$ Step 1 is complete.

Next, we consider

$$\begin{aligned} A(G_S)-I_{m+n} = \left[ \begin{array}{c|c} {\textbf {0}}_{m} &{} J_{m \times n}\\ \hline J_{n \times m} &{} -I_{n} \end{array} \right] . \end{aligned}$$

Similarly, ${\textrm{rank}}(A(G_S)-I_{m+n})=n+1$ and ${\textrm{null}}(A(G_S)-I_{m+n})=m-1.$ Thus, $G_S$ has the eigenvalue 1 with multiplicity $m-1.$ Step 2 is now complete.

For Step 3, there are only $(m+n)-(n-1)-(m-1)=2$ eigenvalues $\lambda _i=\lambda _i(G_S),$ $i=1,2,$ left to be determined. By Lemma 2.1, we have

$$\begin{aligned} \lambda _1+\lambda _2&= 1 \\ \lambda ^2_1+\lambda ^2_2&= 2mn+1. \end{aligned}$$

So, we have

$$\begin{aligned} \lambda _{1,2} = \frac{1 \pm \sqrt{1+4mn}}{2}. \end{aligned}$$

As a conclusion, when $\sigma =m,$ we have

$$\begin{aligned} {\textrm{Spec}}(G_S) = \begin{pmatrix} \frac{1+\sqrt{1+4mn}}{2}&{} 1 &{} 0 &{} \frac{1-\sqrt{1+4mn}}{2} \\ 1 &{} m-1 &{} n-1 &{} 1 \end{pmatrix}. \end{aligned}$$

Case 4. $m<\sigma <m+n$. In this case, the adjacency matrix is

$$\begin{aligned} A(G_S)= \left[ \begin{array}{ccc} I_m &{} J_{m \times (\sigma -m)} &{} J_{m \times (m+n-\sigma )}\\ \hline J_{(\sigma -m) \times m}&{} I_{(\sigma -m)} &{} {\textbf {0}}_{(\sigma -m) \times (m+n-\sigma )} \\ \hline J_{(m+n-\sigma )\times m} &{} {\textbf {0}}_{(m+n-\sigma ) \times (\sigma -m)} &{} {\textbf {0}}_{(m+n-\sigma )} \\ \end{array} \right] = \left[ \begin{array}{c} B\\ \hline C\\ \hline D\\ \end{array} \right] . \end{aligned}$$

For Step 1, it is clear that ${\textrm{rank}}(D)=1$ and both B and C have full rank. Thus, ${\textrm{rank}}(A(G_S))=\sigma +1$ and $G_S$ has the eigenvalue 0 with multiplicity $m+n-\sigma -1.$ For Step 2, consider the matrix

$$\begin{aligned} A(G_S)-I_{m+n}= \left[ \begin{array}{ccc} {\textbf {0}}_m &{} J_{m \times (\sigma -m)} &{} J_{m \times (m+n-\sigma )}\\ \hline J_{(\sigma -m) \times m}&{} {\textbf {0}}_{(\sigma -m)} &{} {\textbf {0}}_{(\sigma -m) \times (m+n-\sigma )} \\ \hline J_{(m+n-\sigma )\times m} &{} {\textbf {0}}_{(m+n-\sigma ) \times (\sigma -m)} &{} -I_{(m+n-\sigma )} \\ \end{array} \right] = \left[ \begin{array}{c} B\\ \hline C\\ \hline D\\ \end{array} \right] . \end{aligned}$$

We have ${\textrm{rank}}(B)={\textrm{rank}}(C)=1$ and D has full rank. Thus, ${\textrm{rank}}(A(G_S))=m+n-\sigma +2$ and $G_S$ has the eigenvalue 1 with multiplicity $\sigma -2.$

For Step 3, there are $(m+n)-(m+n-\sigma -1)-(\sigma -2)=3$ eigenvalues $\lambda _i=\lambda _i(G_S),$ $i=1,2,3,$ left to be determined. Proceed similarly as in Case 2, we apply Lemma 2.1 and Lemma 2.3 to get

$$\begin{aligned} \lambda _1 + \lambda _2 + \lambda _3&= 2, \\ \lambda ^2_1 + \lambda ^2_2 + \lambda ^2_3&= 2mn+2, \\ \lambda ^3_1 + \lambda ^3_2 + \lambda ^3_3&= 3m(\sigma + n-m) +2. \end{aligned}$$

The corresponding cubic polynomial is $p(\lambda )=\lambda ^3-2\lambda ^2+(1-mn)\lambda +m(m+n-\sigma ).$ Using the method as in Case 2, these eigenvalues are distinct, each has multiplicity 1. As a conclusion, when $m<\sigma <m+n,$ we have

$$\begin{aligned} {\textrm{Spec}}(G_S) = \begin{pmatrix} 1 &{} 0 &{} \lambda _1 &{} \lambda _2 &{} \lambda _3\\ \sigma -2 &{} m+n-\sigma -1 &{} 1 &{} 1 &{} 1 \end{pmatrix}, \end{aligned}$$

where $\lambda _1,\lambda _2,\lambda _3$ are the roots of $p(\lambda )=\lambda ^3-2\lambda ^2+(1-mn)\lambda +m(m+n-\sigma ).$

Case 5. $\sigma =m+n$. In this case, the adjacency matrix takes the form $A(G_S)= A(G)+I_{m+n}.$ Thus, its spectrum can be obtained by shifting from Case 1:

$$\begin{aligned} {\textrm{Spec}}(G_S)= \begin{pmatrix} 1+\sqrt{mn} &{} 1 &{} 1- \sqrt{mn} \\ 1 &{} m+n-2 &{} 1 \end{pmatrix}. \end{aligned}$$

Now, we are ready to prove the main results of this section. Recall that if the eigenvalues are in non-increasing order, then we have the Courant-Weyl Inequality:

Theorem 2.5

[3, Theorem 1.3.15] Let A and B be $n\times n$ real symmetric matrix. Then

$$\begin{aligned}&\lambda _i(A+B)\le \lambda _j(A)+ \lambda _{i-j+1}(B) \quad \text { for } n\ge i\ge j\ge 1,\\&\lambda _i(A+B)\ge \lambda _j(A)+ \lambda _{i-j+n}(B) \quad \text { for } 1\le i\le j\le n. \end{aligned}$$

By choosing $i=j=n$ in the second inequality above, we obtain the inequality $\lambda _{\min }(A+B)\ge \lambda _{\min }(A)+ \lambda _{\min }(B)$ which is essential in the proof of our next theorem.

Theorem 2.6

Let G be a self-loop graph of order n and has eigenvalues $\lambda _1(G) \ge \lambda _2(G)\ge \cdots \ge \lambda _n(G)$.

• (i) If $\lambda _i(G)>0$, for $i=1,2,\ldots ,n$, then G is disjoint unions of n $\widehat{K_1}$.

• (ii) If $\lambda _i(G)\ge 0$, for $i=1,2,\ldots ,n$, then every connected component of G is either $K_1$ or $\widehat{K_r}$ for some $r\in \{1,2,\ldots ,n\}$.

Proof

• (i) Let H be a connected component of G. We shall show that $H_0$ is a complete graph. Suppose on the contrary that $H_0$ is not a complete graph. Thus, $H_0$ contains $P_3$ as a vertex-induced subgraph. Note that $\lambda _{\min }(P_3)=-\sqrt{2}.$ So, by Interlacing Theorem (cf. [3, Cor 1.3.12]), we obtain $\lambda _{\min }(H_0)\le -\sqrt{2}$. Note that $J_S=A(H)-A(H_0)$. By the Courant-Weyl Inequalities, we have $\lambda _{\min }(H_0)\ge \lambda _{\min }(H)+\lambda _{\min }(-J_S)$. This implies that

  $$\begin{aligned} \lambda _{\min }(G) \le \lambda _{\min }(H)\le \lambda _{\min }(H_0)-\lambda _{\min }(-J_S)\le -\sqrt{2}+1, \end{aligned}$$

  this is a contradiction because G has only positive eigenvalues. Therefore, $H_0$ is a complete graph. By Theorem 2.2, $H=\widehat{K_1}$ and we are done.

• (ii) With a slight modification of the proof in Part (i), we can deduce that if H is a connected component of G, then $H_0$ is a complete graph. Now, by Theorem 2.2, $H=K_1$ or $H=\widehat{K_r}$ for some $r\in \{1,2,\ldots ,n\}$. The proof is complete. $\square$

In the following, we characterize self-loop graphs of order n with a few distinct eigenvalues.

Theorem 2.7

Let G be a self-loop graph of order n. Then,

• (i) G has exactly one eigenvalue if and only if $G=\overline{K}_n$ or every connected component of G is $\widehat{K_1}.$

• (ii) G has exactly two distinct eigenvalues if and only if all connected components of G are the same and each is $\widehat{K_r}$ for some r,  or each component of G is either $K_1$ or $\widehat{K_r}.$

Proof

• (i) Assume that G has $\sigma$ loops and $a=\lambda _1(G)=\cdots =\lambda _n(G).$ Then, $\sigma =\textrm{Tr}(A(G))=na.$ This implies that $a=\displaystyle \frac{\sigma }{n}.$ Since a is a root of a monic polynomial with integer coefficients, it follows that a is an integer. Since $0 \le \sigma \le n,$ we have $\sigma =0$ or $\sigma =n.$

  • (a) If $\sigma =0,$ then $a=0.$ This implies $A(G)={\textbf {0}}_n.$ Thus, $G=\overline{K}_n.$

  • (b) If $\sigma =n,$ then $a=1.$ So, $A(G)-I_n$ is the adjacency matrix of an ordinary graph with only eigenvalue 0. It follows that $A(G)-I_n={\textbf {0}}_n,$ that is, $A(G)=I_n$ and each component of G is $\widehat{K_1}.$

• (ii) Let $A=A(G).$ If G has two eigenvalues $\lambda _1$ and $\lambda _2,$ then the minimal polynomial of A has the form

  $$\begin{aligned} p(\lambda )=(\lambda -\lambda _1)(\lambda -\lambda _2). \end{aligned}$$

  So, we have

  $$\begin{aligned} A^2-(\lambda _2+\lambda _2)A+\lambda _1\lambda _2I_n={\textbf {0}}_n. \end{aligned}$$

  (2.4)

  Let H be a connected component of G. We show that $H_0$ is a complete graph. If $H_0$ is not a complete graph, then there is a vertex induced path of order 3 in $H_0$ such as $\widehat{P}_3:$ $v_rv_sv_k.$ Now, we have $(A^2)_{rk} > 0$ with $A_{rk}=0.$ This is a contradiction because the (r, k)-entry of (2.4) is positive on the left, but the (r, k)-entry on the right side is 0. Thus, $H_0$ is a complete graph. By Theorem 2.2, $H=K_r$ or $H=\widehat{K_r}$ for some r. If one of the components of G is $K_r$ (resp. $\widehat{K_r}$), then the other components of G should also be $K_r$ (resp. $\widehat{K_r}$) because otherwise G has more than two distinct eigenvalues. If $r=1,$ then G is union of finitely many $K_1$ or $\widehat{K_r}.$ The opposite direction is obvious and the proof is complete. $\square$

3 Bipartite Graphs and Eigenvalues of Self-loop Graphs

In this section, we provide a necessary and sufficient condition for a graph G being bipartite according to the eigenvalues of G and $G_{V(G)\backslash S}$ for arbitrary $S \subseteq V(G).$ First, we give an observation on similarity.

Lemma 3.1

Let G be a bipartite graph of part sizes m and n and $|S|=\sigma .$ Then, $J_S + A(G)$ is similar to the matrix $J_S-A(G).$

Proof

One can easily see that if

$$\begin{aligned} P= \begin{bmatrix} I_m &{} {\textbf {0}} \\ {\textbf {0}} &{} -I_n \end{bmatrix}, \end{aligned}$$

then, by (2.3), $P(J_S + A(G))P^{-1} = P(J_S + A(G))P = J_S - A(G).$ Thus, $J_S + A(G)$ is similar to $J_S - A(G).$ $\square$

Theorem 3.2

Let G be a bipartite graph of order n and $|S|=\sigma .$ Then, $1 - \lambda _n(G_S) \ge \cdots \ge 1 - \lambda _1(G_S)$ are the eigenvalues of $G_{V(G)\backslash S},$ where $\lambda _1(G_S)\ge \cdots \ge \lambda _n(G_S)$ are the eigenvalues of $G_S.$

Proof

Note that $A(G_S) = J_S + A(G)$ and $A(G_{V(G)\backslash S}) = J_{V(G) \backslash S} + A(G).$ Hence, we find

$$\begin{aligned} J_{V(G) \backslash S} - A(G) = I_n -(J_S + A(G)) = I_n - A(G_S). \end{aligned}$$

Thus, we have $1 -\lambda _i(G_S),$ $i=1, \ldots ,n,$ the eigenvalues of $J_{V(G)\backslash S}- A(G),$ which coincide with the eigenvalues of $A(G_{V(G)\backslash S})$ by similarity in the previous lemma. $\square$

Theorem 3.3

Let G be a connected graph of order n. Let $G_S$ be its self-loop graph with eigenvalues $\lambda _1(G_S) \ge \cdots \ge \lambda _n(G_S)$ and $S \subseteq V(G).$ Then, the eigenvalues of $G_{V(G)\backslash S}$ are $1-\lambda _n(G_S) \ge \cdots \ge 1-\lambda _1(G_S)$ if and only if G is bipartite.

Proof

By Theorem 3.2, it suffices to prove that if the spectrum of $G_{V(G)\backslash S}$ is $1-\lambda _n(G_S) \ge \cdots \ge 1-\lambda _1(G_S),$ then G is bipartite.

Since $1-\lambda _1(G_S)$ is the smallest eigenvalues of $G_{V(G) \backslash S},$ it follows that the largest eigenvalue of the matrix $I_n-A(G_{V(G)\backslash S})$ is $\lambda _1(G_S).$ Also note that $I_n-A(G_{V(G)\backslash S})=I_n-(J_{V(G)\backslash S}+A(G))=J_S-A(G)$. Thus, we have

$$\begin{aligned} \lambda _1(G_S)&=\max _{||x||=1} x^T(J_S + A(G))x \\&=\max _{||x||=1} x^T(I_n -A(G_{V(G) \backslash S})x\\&=\max _{||x||=1} x^T(J_S - A(G))x. \end{aligned}$$

Suppose that $\lambda _1(G_S)=z^T(J_S-A(G))z$ for some z with $\Vert z\Vert =1.$ Define

$$\begin{aligned} |z|^T = (|z_1|, |z_2|, \cdots , |z_n|). \end{aligned}$$

Then, we have

$$\begin{aligned} \lambda _1(G_S)=z^T(J_S - A(G))z \le |z|^T (J_S +A(G)) |z| \le \lambda _1(G_S). \end{aligned}$$

Thus, $z^T(J_S-A(G))z=|z|^T(J_S+A(G))|z|.$ This implies that $z^T(-A(G))z=|z|^T(A(G))|z|$, or equivalently

$$\begin{aligned} \sum _{1\le i,j\le n}(-a_{ij})z_iz_j=\sum _{1\le i,j\le n}a_{ij}|z_i||z_j|. \end{aligned}$$

(3.1)

Observe that for each $1\le i,j\le n$, we have

$$\begin{aligned} (-a_{ij})z_iz_j\le a_{ij}|z_i||z_j|. \end{aligned}$$

(3.2)

Since $\Vert |z|\Vert =1$ and $|z|^T(J_S+A(G))|z|=\lambda _1(G_S),$ we conclude that |z| is an eigenvector corresponding to $\lambda _1(G_S)$ for the matrix $J_S +A(G).$ Since G is connected, by Perron-Frobenius Theorem [4, §2], $\lambda _1(G_S)$ is a simple eigenvalue and there exists an eigenvector $\alpha$ for $\lambda _1(G_S)$ whose all entries are positive. Then |z| is a multiple of $\alpha .$ This implies that $z_i \ne 0$ for $i=1,\ldots , n.$ Note that if there exist some i and j with $1\le i,j\le n$ such that $a_{ij}=1$ and $z_iz_j>0,$ then (3.2) becomes a strict inequality. Thus, by taking the summation on both sides over i, j,  we obtain a strict inequality that contradicts (3.1). We shall show that this contradiction occurs if G is not bipartite.

Assume G is not bipartite, that is, G contains an odd cycle with vertices $v_1,v_2,\ldots ,$ $v_{2k+1}.$ Then, for $i=1,\ldots , 2k+1$, we have $a_{i,r_i}=1,$ where $r_i=i+1\mod (2k+1)$. Note that $z_\ell z_{r_{\ell }}$ cannot be negative for every $1\le \ell \le 2k+1$ because there are odd distinct vertices. So, there exists an $\ell$ with $1 \le \ell \le 2k+1$ such that $z_\ell z_{r_{\ell }}>0.$ Thus, G is bipartite. The proof is complete. $\square$

Remark 3.4

Let $S \subseteq V(G).$ Theorem 3.3 provides a way of determining the bipartiteness of a graph directly from the eigenvalues of its self-loop graphs $G_S$ and the eigenvalues of $G_{V(G)\backslash S}.$ Indeed, if we have $\lambda _1(G_S)$ and $\lambda _n(G_{V(G)\backslash S}),$ we can determine whether G is bipartite.

Another immediate consequence of Theorem 3.3 is the following corollary.

Corollary 3.5

[13, Theorem 3] Let G be a bipartite graph of order n with vertex set V(G). Let S be a nonempty subset of V(G). Then, ${\mathcal E}(G_S)={\mathcal E}(G_{V(G)\backslash S}).$

Before closing this section, we discuss a case that answers [13, Conjecture 2]. As a motivation, let us consider the case $G=K_{3,3}.$ Using the spectrum determined Sect. 2.2, the energy of $(K_{3,3})_S$ is obtained as follows.

Table 1 The energy of $(K_{3,3})_S$

From the table, we observe that when there are loops, $(K_{3,3})_S$ has at least the energy of the ordinary graph $K_{3,3}.$ Thus, it is natural to ask whether the same is true for all bipartite graphs. Indeed, we provide an affirmative answer to this question. Before proving this, let us recall an inequality in [1].

Theorem 3.6

[1] Let A be a Hermitian matrix in the following block form:

$$\begin{aligned} A= \begin{bmatrix} B &{} D \\ D^* &{} C \end{bmatrix}. \end{aligned}$$

Then, ${\mathcal E}(A)\ge 2 {\mathcal E}(D).$

Theorem 3.7

If G is a bipartite graph and $S \subseteq V(G),$ then, ${\mathcal E}(G_S) \ge {\mathcal E}(G).$

Proof

Let $A(G)=\begin{bmatrix} {\textbf {0}} &{} D \\ D^T &{} {\textbf {0}} \end{bmatrix}$ be its adjacency matrix. Note that

$$\begin{aligned} {\mathcal E}(G_S) = {\mathcal E}(A(G_S) - \frac{\sigma }{n} I_n), \end{aligned}$$

where $\sigma =|S|$ and $n=|V(G)|.$ By Theorem 3.6,

$$\begin{aligned} {\mathcal E}(G) =2{\mathcal E}(D) \le {\mathcal E}(A(G_S)-\frac{\sigma }{n}I_n) = {\mathcal E}(G_S). \end{aligned}$$

$\square$

Theorem 3.7 answers a special case (bipartite graphs) of [13, Conjecture 2]. Lastly, we propose the following conjecture.

Conjecture 1

For every simple graph G,  there exists $S \subseteq V(G)$ such that ${\mathcal E}(G_S) > {\mathcal E}(G).$

4 Upper Bound for the Energy of $G_S$

We first recall the following theorem.

Theorem 4.1

[5, Theorem 4] For any $(a_{ij})=A \in M_n({\mathbb C})$ with eigenvalues $\lambda _1,\ldots , \lambda _n,$

$$\begin{aligned} \sum ^n_{i=1} |\lambda _i| \le \sum ^n_{i=1} \Vert A_i\Vert , \end{aligned}$$

where $A_i$ denotes the i-th row of A and $\Vert A_i\Vert = \sqrt{a^2_{i1} + a^2_{i2} + \cdots + a^2_{in}}.$

For $A \in M_n({\mathbb C}),$ the energy of a complex matrix is defined to be

$$\begin{aligned} {\mathcal E}(A)= \sum ^n_{i}|\lambda _i(A)|, \end{aligned}$$

where $\lambda _1(A),\ldots , \lambda _n(A)$ are eigenvalues of A. Thus, ${\mathcal E}(A)\le \sum ^n_{i=1} \Vert A_i\Vert .$

As a consequence, we obtain an alternative proof of an upper bound for the energy of $G_S$ given by Gutman et al. and discuss the equality case.

Corollary 4.2

[13, Theorem 6] Let $G_S$ be a self-loop graph of order n and size m with $|S|=\sigma$. Then,

$$\begin{aligned} {\mathcal E}(G_S) \le \sqrt{n\left( 2m+ \sigma - \frac{\sigma ^2}{n}\right) }. \end{aligned}$$

(4.1)

Suppose the equality holds, then $G_S$ is (a, b)-semi-regular, where

$$\begin{aligned} a = \frac{2m}{n} + \frac{3\sigma }{n} - \frac{2\sigma ^2}{n^2} -1, \quad b = \frac{2m}{n} + \frac{\sigma }{n} -\frac{2\sigma ^2}{n^2}. \end{aligned}$$

Proof

Let $\displaystyle B=A(G_S) - \frac{\sigma }{n} I_n.$ By Theorem 4.1, we have

$$\begin{aligned} {\mathcal E}(G_S)= \sum ^n_{i=1} \left| \lambda _i(B) \right| =\sum ^n_{i=1} \left| \lambda _i(G_S) - \frac{\sigma }{n} \right| \le \sum ^n_{i=1} \Vert B_i\Vert . \end{aligned}$$

Hence, we have

$$\begin{aligned} \sum ^n_{i=1} \Vert B_i\Vert&= \sqrt{d_1 + \left( 1-\frac{\sigma }{n}\right) ^2} + \cdots + \sqrt{d_\sigma + \left( 1-\frac{\sigma }{n}\right) ^2} \\&\qquad \qquad + \sqrt{d_{\sigma +1} + \left( \frac{\sigma }{n}\right) ^2} + \cdots + \sqrt{d_n + \left( \frac{\sigma }{n}\right) ^2}. \nonumber \end{aligned}$$

By Cauchy-Schwarz inequality, we obtain

$$\begin{aligned} {\mathcal E}(G_S) \le \sqrt{n\left( 2m+ \sigma - \frac{\sigma ^2}{n}\right) }. \end{aligned}$$

If the equality holds, then $d_i+(1-\frac{\sigma }{n})^2=d_{i+1}+(1-\frac{\sigma }{n})^2$ for $1\le i \le \sigma -1$ and $d_j+(\frac{\sigma }{n})^2=d_{j+1} + (\frac{\sigma }{n})^2$ for $\sigma +1 \le i \le n-1.$ One deduces that $a=d_1=\cdots =d_\sigma$ and $b=d_{\sigma +1} = \cdots =d_{n}.$ Thus, the (a, b)-semi-regularity can be obtained by solving the simultaneous equations $2m=\sigma a + (n-\sigma )b$ and $a+(1-\frac{\sigma }{n})^2=b+\frac{\sigma ^2}{n^2}.$ This completes the proof. $\square$

5 Analogous Classical Upper Bound for $\lambda _1$ in Terms of Maximum Degree

It is well-known that if G is a graph, then $\lambda _1(G)\le \Delta (G)$. In the next theorem, we generalize it to graphs with self-loop graphs by showing that $\lambda _1(G_S) \le \Delta (G)+1$.

Theorem 5.1

Let $G_S$ be a self-loop graph of order n. Then $\lambda _1(G_S)\le \Delta (G)+1\le n.$ Moreover, $\lambda _1(G_S)=n$ if and only if $G_S=\widehat{K_n}.$

Proof

Since $A(G_S)$ is a non-negative matrix, by Perron-Frobenius Theorem, there exists an eigenvector x corresponding to eigenvalue $\lambda _1(G_S)$ such that $x^T= [x_1, x_2, \ldots , x_n]$ and $x_i \ge 0$ for $i=1,2,\ldots ,n.$

Assume that $\displaystyle x_r=\max _{1\le i\le n}x_i,$ for some $r\in \{1,2,\ldots ,n\}$. By considering the r-th entry of both sides of $A(G_S)x=\lambda _1(G_S)x$, we have

$$\begin{aligned} x_{i_1}+x_{i_2}+\cdots +x_{i_{d(v_r)}}+\gamma x_r=\lambda _1(G_S) x_r,\quad \gamma \in \{0,1\}. \end{aligned}$$

(5.1)

This implies that

$$\begin{aligned} \lambda _1(G_S) x_r\le (\Delta (G)+1)x_r. \end{aligned}$$

Hence, $\lambda _1(G_S)\le \Delta (G)+1$ as $x_r>0$. Since $\Delta (G)\le n-1,$ $\lambda _1(G_S) \le n.$

If $\lambda _1(G_S)=n$, then Eq. (5.1) becomes

$$\begin{aligned} x_{i_1}+x_{i_2}+\cdots +x_{i_{d(v_r)}}+\gamma x_r=n x_r,\quad \gamma \in \{0,1\}, \end{aligned}$$

with $d(v_r)=\Delta (G)=n-1$, $\gamma =1$, and $x_{i_k}=x_r$ for $k=1,2,\ldots ,n-1$. Hence, $A(G_S)=J_{n\times n}$ indicating $G_S=\widehat{K_n}$. Conversely, if $G_S=\widehat{K_n}$, then by (2.2), we have

$$\begin{aligned} \text {Spec}(G_S)= \begin{pmatrix} n &{} 0 \\ 1 &{} n-1 \end{pmatrix}, \end{aligned}$$

which gives $\lambda _1(G_S)=n$. $\square$

The next theorem provides a sharp lower bound for $\lambda _1(G)$ for any connected graph with loops.

Theorem 5.2

Let G be a connected graph of order n and size m. If $S\subseteq V(G)$ with $|S|=\sigma$, then

$$\begin{aligned} \lambda _1(G_S)\ge \frac{2m}{n}+\frac{\sigma }{n}. \end{aligned}$$

Moreover, if G is a $\left( k,k+1\right)$-semi-regular graph for some natural number k, 

$$\begin{aligned} d_G(v)={\left\{ \begin{array}{ll} k,\quad &{} \text {if }v\in S,\\ k+1,\quad &{}\text {if }v\in V(G)\backslash S, \end{array}\right. } \end{aligned}$$

where $d_G$ is the degree of vertices of G, then

$$\begin{aligned} \lambda _1(G_S)=\frac{2m}{n}+\frac{\sigma }{n}. \end{aligned}$$

Proof

Since $J_S+A(G)$ is a real symmetric matrix, by Rayleigh quotient, we obtain

$$\begin{aligned} \lambda _1(G_S)&=\max _{0\ne x\in \mathbb {R}^n} \frac{x^T(J_S+A(G))x}{x^Tx}\\&\ge \frac{j^T(J_S+A(G))j}{j^Tj}\\&=\frac{j^T(J_S)j}{j^Tj}+\frac{j^T(A(G))j}{j^Tj}=\frac{2m}{n}+\frac{\sigma }{n}. \end{aligned}$$

To show the equality, we consider a $\left( k,k+1\right)$-semi-regular graph G, $1\le k\le n$, such that

$$\begin{aligned} d_G(v)= {\left\{ \begin{array}{ll}k,\quad &{} \text {if}~ \, v\in S,\\ k+1,\quad &{}\text {if}~\, v\in V(G)\backslash S. \end{array}\right. } \end{aligned}$$

Then, $A(G_S)=J_S+A(G)$ gives $(J_S+A(G))j=(k+1)j$. This shows that j is an eigenvector corresponding to the eigenvalue $k+1$, and thus $k+1\in \text {Spec}(G_S)$. Notice that for G, 

$$\begin{aligned} 2m=\sum _{i=1}^n d_{G}(v_i)=\sigma k +(n-\sigma )(k+1) = -\sigma + nk + n. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \frac{2m}{n}+\frac{\sigma }{n}=k+1\in \text {Spec}(G_S), \end{aligned}$$

and j is an eigenvector corresponding to $\frac{2m}{n}+\frac{\sigma }{n}$.

To show that $\frac{2m}{n}+\frac{\sigma }{n}$ is the largest eigenvalue of graph $G_S$, we suppose on the contrary that $\frac{2m}{n}+\frac{\sigma }{n}=\lambda _i(G_S)$ for some $i\ge 2$. Since $A(G_S)$ is a non-negative matrix, by Perron-Frobenius Theorem, there exists an eigenvector x corresponding to eigenvalue $\lambda _1$ such that $x^T=[x_1,\ldots , x_n]$ and $x_i>0$ for $i=1,2,\ldots ,n.$ Since $\lambda _i\ne \lambda _1$, it follows that the eigenvector x and j are orthogonal, a contradiction. Hence, $\lambda _1(G_S)=\frac{2m}{n}+\frac{\sigma }{n}$. $\square$