Seidel-Estrada index

Abstract

Let G be a simple graph with n vertices and $(0,1)$-adjacency matrix A. As usual, $S(G)=J-2A-I$ denotes the Seidel matrix of the graph G. Suppose $\theta_{1}, \theta_{2},\ldots, \theta_{n}$ and $\lambda_{1}, \lambda_{2},\ldots, \lambda_{n}$ are the eigenvalues of the adjacency matrix and the Seidel matrix of G, respectively. The Estrada index of the graph G is defined as $\sum_{i=1}^{n} e^{\theta_{i}}$. We define and investigate the Seidel-Estrada index, $SEE=SEE(G)=\sum_{i=1}^{n} e^{\lambda_{i}}$. In this paper the basic properties of the Seidel-Estrada index are investigated. Moreover, some lower and upper bounds for the Seidel-Estrada index in terms of the number of vertices are obtained. In addition, some relations between $SEE$ and the Seidel energy $E_{s}(G)$ are presented.

1 Introduction

Throughout this paper, let G be a simple graph with vertex set $V=\{v_{1}, v_{2}, \ldots, v_{n}\}$. The adjacency matrix $A(G)=[a_{ij}]$ of G is a binary matrix of order n such that $a_{ij}=1$ if the vertex $v_{i}$ is adjacent to the vertex $v_{j}$, and 0 otherwise. The Seidel matrix $S(G)=[s_{ij}]$ is equal to $\mathbb{J}_{n}-2A(G)-\mathbb{I}_{n}$, where the symbol $\mathbb{J}_{n}$ denotes the square matrix of order n all of whose entries are equal to 1. Since $A(G)$ and $S(G)$ are real symmetric matrices, their eigenvalues must be real. The eigenvalues of G are referred to as the eigenvalues of $A(G)$, denoted by $\theta_{1}(A(G)), \theta_{2}(A(G)), \ldots, \theta_{n}(A(G))$ and similarly, $\lambda_{1}(S(G))\geq\lambda_{2}(S(G))\geq\cdots\geq\lambda_{n}(S(G))$, the Seidel eigenvalues of G. For simplicity, we write $\lambda_{i}$ instead of $\lambda_{i}(S(G))$. The sequence of n Seidel eigenvalues is called the Seidel spectrum of G (for short $\mbox{S-spec}(G)$). We now present an example of pairs of graphs on n vertices with the same Seidel spectrum such that one of them is a connected graph and the other one is not.

Example 1

Here we address two examples from non-isomorphic graphs which are co-spectral:

$$\begin{aligned} (\mathrm{i})& \quad\mbox{S-spec}(K_{p, q})=\mbox{S-spec}( \overline{K}_{n}) \quad\mbox{if } p+q=n, \\ (\mathrm{ii})&\quad \mbox{S-spec}(K_{n/2}\cup K_{n/2})= \mbox{S-spec}(K_{n}) \quad(n\mbox{ is even}) . \end{aligned}$$

In our recent studies on Seidel eigenvalues it has been shown that a lower and upper bound exists for the sum of powers of the absolute eigenvalues of the Seidel matrix, suggesting a common core architecture similar to the cases of adjacency and signless Laplacian matrix [1]. The reader can find more information related to the eigenvalues of the adjacency matrix and the spectrum of G in [2]. The Estrada index of a graph G is defined as

$$\begin{aligned} EE(G)=\sum_{i=1}^{n} e^{\theta_{i}(A(G))}. \end{aligned}$$

(1)

This graph-spectrum-based structural descriptor was first proposed by Estrada in 2000; see [3–5]. Already, de la Peña et al. [6] proposed to call it the Estrada index, a name that in the meantime has been commonly accepted. Several kinds of Estrada indices were discussed in [7–13] and the references therein. For the recent work of the mathematical properties on the Estrada and signless Laplacian Estrada indices, see [6, 14]. In this review, we summarize some indirect evidence to support the concept of a Seidel matrix. Similarly, we define the Seidel-Estrada index for the graph G in full analogy with equation (1) as

$$\begin{aligned} SEE(G)=\sum_{i=1}^{n} e^{\lambda_{i}}. \end{aligned}$$

(2)

For details on the theory of the Estrada index and several lower and upper bounds, see [6, 11, 13, 15]. Ayyaswamy et al. [14] gave a lower bound for a signless Laplacian of the graph using the numbers of vertices and edges. A conference matrix is a square matrix C of order n with zero diagonal and ±1 off the diagonal, such that $CC^{T} = (n-1)I$. If C is symmetric, then C is the Seidel matrix of a graph and this graph is called a conference graph; see [1, 2]. The aim of this paper is to find the upper and lower bounds for the Seidel-Estrada index of the graph G. The rest of the paper is organized as follows: In Section 2, we give some definitions and obtain some upper and lower bounds for the Seidel-Estrada index. In Section 3, we present a relation between the Seidel-Estrada index and the Seidel energy of a graph G, and we prove several results on the Seidel-Estrada index.

2 Estimates of the Seidel-Estrada index

Here we give some new lower and upper bounds on Seidel-Estrada index. For convenience, we give some notation and properties which will be used in the following proofs of our results. Let $S_{k}=S_{k}(G)=\sum_{i=1}^{n}(\lambda_{i})^{k}$, and $S^{k}(G)=\sum_{i=1}^{n}{|\lambda_{i}|}^{k}$. From the Taylor expansion of $e^{x}$, it is easy to see that the Seidel-Estrada index and $S_{k}(G)$ of G are related by

$$\begin{aligned} SEE(G)=\sum_{k=0}^{\infty}\frac{S_{k}(G)}{k!}. \end{aligned}$$

(3)

It is easy to see that any graph G of order $n\geq2$ has $SEE(G)>n$. (If equality holds, then $\lambda_{1}=\lambda_{2}=\cdots=\lambda_{n}=0$. By Lemma 2.1(ii), we can get a contradiction.)

Lemma 2.1

[1]

For any graph G with n vertices, we have

$$\begin{aligned} (\mathrm{i})&\quad S_{1}(G)=\sum_{i=1}^{n} \lambda_{i}=\operatorname{trace} \bigl(S(G) \bigr)=0, \\ (\mathrm{ii})&\quad S^{2}(G)=S_{2}(G) =\sum _{i=1}^{n} \lambda_{i}^{2} = \operatorname{trace} \bigl(S^{2}(G) \bigr)=(n-1)^{2}+(n-1)=n(n-1), \\ (\mathrm{iii})&\quad S_{3}(G)=\sum_{i=1}^{n} \lambda_{i}^{3} \leq S^{3}(G) \leq (n-1)^{3}+(n-1), \\ (\mathrm{iv})&\quad S^{3}(G)=\sum_{i=1}^{n} | \lambda_{i}|^{3} \geq n\sqrt {(n-1)^{3}}, \\ (\mathrm{v})&\quad S_{k}(G) \leq S^{k}(G)=\sum _{i=1}^{n} |\lambda_{i}|^{k} \leq (n-1)^{k}+(n-1),\quad k=3, 4,\ldots, \\ (\mathrm{vi})&\quad S^{k}(G)=\sum_{i=1}^{n} | \lambda_{i}|^{k} \geq n\sqrt{(n-1)^{k}},\quad k=3, 4, \ldots. \end{aligned}$$

Lemma 2.2

[16]

Let B be an $n\times n$ symmetric matrix with eigenvalues $\lambda_{1}\geq\lambda_{2}\geq\cdots \geq\lambda_{n}$ and let $B_{k}$ be its leading $k\times k$ submatrix of B. Then, for $i=1, 2,\ldots, k$,

$$ \lambda_{n-i+1}(B) \leq\lambda_{k-i+1}(B_{k}) \leq\lambda_{k-i+1}(B) , $$

(4)

where $\lambda_{i}(B)$ is the ith greatest eigenvalue of B.

Lemma 2.3

Let G be a graph of order $n\geq2$. Then $\lambda_{1}\geq1$.

Proof

Since $n\geq2$, therefore $\overline{K}_{2}$ or ${K}_{2}$ must be an induced subgraph of G. Since $\lambda_{1}(K_{2})=\lambda_{1}(\overline{K}_{2})=1$, by Lemma 2.2, we get the required result. □

Theorem 2.4

Let G be a simple graph with $n\geq2$ and $\operatorname{det} S(G)\neq0$. Then the Seidel-Estrada index of G is bounded by

$$\begin{aligned} \sqrt{n(3n-2)}< SEE(G) < n-1+e^{\sqrt{n(n-1)}} . \end{aligned}$$

(5)

Proof

(a) To prove this theorem, we apply a technique similar to the proof of Theorem 1 in [6]. At first we prove that the left inequality of (5):

From (2), we get

$$\begin{aligned} SEE^{2}(G)=\sum_{i=1}^{n} e^{2\lambda_{i}}+2\sum_{i< j} e^{\lambda _{i}}e^{\lambda_{j}}. \end{aligned}$$

(6)

In view by the inequality between the geometric and arithmetic mean, we get

$$\begin{aligned} 2\sum_{i< j} e^{\lambda_{i}}e^{\lambda_{j}} \geq& n(n-1) \biggl(\prod_{i< j}e^{\lambda_{i}}e^{\lambda_{j}} \biggr)^{\frac{2}{n(n-1)}}=n(n-1) \Biggl[ \Biggl(\prod_{i=1}^{n} e^{\lambda_{i}} \Biggr)^{n-1} \Biggr]^{\frac{2}{n(n-1)}} \\ =&n(n-1) \bigl(e^{S_{1}(G)} \bigr)^{\frac{2}{n}}= n(n-1). \end{aligned}$$

(7)

By using the power series expansion, and Lemma 2.1, we get

$$\begin{aligned} \sum_{i=1}^{n} e^{2\lambda_{i}} =&\sum _{i=1}^{n} \sum_{k=0}^{\infty} \frac {(2\lambda_{i})^{k}}{k!}=\sum_{i=1}^{n} \frac{(2\lambda_{i})^{0}}{0!}+\sum_{i=1}^{n} \frac{(2\lambda_{i})^{1}}{1!}+\sum_{i=1}^{n} \frac{(2\lambda _{i})^{2}}{2!}+\sum_{i=1}^{n}\sum _{k\geq 3}\frac{(2\lambda_{i})^{k}}{k!} \\ =&S_{0}+2S_{1}+2S_{2}+\sum _{i=1}^{n}\sum_{k\geq3} \frac{(2\lambda_{i})^{k}}{k!} =n+0+2n(n-1)+\sum_{i=1}^{n} \sum_{k\geq3}\frac{(2\lambda_{i})^{k}}{k!} . \end{aligned}$$

Since $\sum_{k\geq3}\frac{(2\lambda_{i})^{k}}{k!}\geq8\sum_{k\geq3}\frac {(\lambda_{i})^{k}}{k!}$, we shall use a multiplier $\gamma\in[0, 8]$, so as to arrive at

$$\begin{aligned} \sum_{i=1}^{n} e^{2\lambda_{i}} \geq& n+2n(n-1)+\gamma\sum_{i=1}^{n} \sum_{k\geq3}\frac{(\lambda_{i})^{k}}{k!} \\ =&n+2n(n-1)-\gamma n-\frac{1}{2}\gamma n(n-1)+\gamma\sum _{i=1}^{n}\sum_{k\geq0} \frac{(\lambda_{i})^{k}}{k!} \\ =&n+2n(n-1)-\gamma n -\frac{1}{2}\gamma n(n-1)+\gamma SEE(G). \end{aligned}$$

(8)

By substituting (7) and (8) back into (6) and solving for $SEE(G)$, we obtain

$$ SEE(G)\geq\frac{\gamma}{2} +\sqrt{\frac{\gamma^{2}}{4}+n^{2}(3-\gamma /2)-n(2+\gamma/2)} . $$

(9)

Now, we consider a function

$$\begin{aligned} f(x)=\frac{x}{2} +\sqrt{\frac{x^{2}}{4}+n^{2} \biggl(3- \frac{x}{2} \biggr)-n \biggl(2+\frac {x}{2} \biggr)}. \end{aligned}$$

(10)

We have $f'(x)<0$ for $x\geq0$. Thus $f(x)$ is a monotonically decreasing function for $x>0$. Consequently, the best lower bound for $SEE(G)$ is attained $\gamma=0$. Setting $\gamma=0$ in (9), we arrive at the first half of Theorem 2.4:

$$ SEE(G)\geq\sqrt{n(3n-2)} . $$

Now, we have to prove that the lower bound is strict. For this purpose, we assume that the left equality holds in (5). Then we have

$$e^{\lambda_{i}+\lambda_{j}}=e^{\lambda_{k}+\lambda_{\ell}} , \quad\mbox{for any } i, j, k, \ell\in\{1, 2, \ldots, n\}, $$

that is,

$$e^{\lambda_{1}+\lambda_{2}}=e^{\lambda_{1}+\lambda_{3}}=\cdots=e^{\lambda _{1}+\lambda_{n}}=e^{\lambda_{2}+\lambda_{3}} , $$

and hence

$$\lambda_{1}=\lambda_{2}=\cdots=\lambda_{n} . $$

By Lemma 2.3 and the trace of $S(G)$, we can get a contradiction. Thus the left equality in (5) is strict.

(b) Let us prove now the right inequality.

Since $f(x)=e^{x}$ monotonically increases in the interval $(-\infty, \infty)$, we starting with equation (2), we get

$$\begin{aligned} SEE(G) =&n+\sum_{i=1}^{n} \sum_{k=1}^{\infty}\frac{(\lambda_{i})^{k}}{k!} \leq n+\sum _{i=1}^{n}\sum_{k\geq1} \frac{(|\lambda_{i}|)^{k}}{k!} \\ =&n+\sum_{k\geq1}\sum_{i=1}^{n} \frac{[(\lambda_{i})^{2}]^{\frac{k}{2}}}{k!} \\ \leq&n+\sum_{k\geq1}\frac{1}{k!} \bigl(S_{2}(G) \bigr)^{\frac{k}{2}}=n+\sum _{k\geq 1}\frac{(\sqrt{n(n-1)})^{k}}{k!} \\ =&n-1+\sum_{k=0}^{\infty}\frac{(\sqrt{n(n-1)})^{k}}{k!}=n-1+e^{\sqrt {n(n-1)}}. \end{aligned}$$

(11)

Suppose that the right equality holds in (5). Then the equality holds in (11). Thus we have $\lambda_{i}=|\lambda_{i}|$, $i=1, 2,\ldots, n$. Since $\lambda_{1}\geq1$ and by Lemma 2.1(i), again we get a contradiction. Hence the right inequality in (5) is strict. □

Theorem 2.5

Let G be a conference graph. Then the Seidel-Estrada index of G is equal to

$$\begin{aligned} SEE(G)=n ch(\sqrt{n-1}), \end{aligned}$$

(12)

where $ch(x)$ is the hyperbolic cosine of x defined as follows:

$$ch(x)=\frac{e^{x}+e^{-x}}{2} . $$

Proof

Since G is a conference graph, the Seidel matrix of a graph is symmetric and $SS^{T}=(n-1)I$, thus each Seidel eigenvalue equals $\lambda_{i}=\sqrt{n-1}$ or $\lambda_{i}=-\sqrt{n-1}$. Let the number of positive eigenvalues of Seidel matrix $S(G)$ be $n_{+}$. Hence, $\lambda _{i}=\pm\sqrt{n-1}$ and $S_{1}=\sum_{i=1}^{n}\lambda_{i}=\sum_{i=1}^{n_{+}} \sqrt{n-1}+\sum_{i=n_{+}+1}^{n}-\sqrt{n-1}=0$, then $n_{+}=\frac{n}{2}$. Therefore

$$\begin{aligned} SEE(G)&=\sum_{i=1}^{n}e^{\lambda_{i}}= \sum_{i=1}^{n_{+}}e^{\lambda_{i}}+\sum _{i=n_{+}+1}^{n}e^{\lambda_{i}}= \sum _{i=1}^{n_{+}}e^{\sqrt{n-1}}+\sum _{i=n_{+}+1}^{n}e^{-\sqrt{n-1}} \\ &=\sum_{i=1}^{n_{+}} \bigl(e^{\sqrt{n-1}}+e^{-\sqrt{n-1}} \bigr) =2\sum_{i=1}^{\frac{n}{2}} ch( \sqrt{n-1})=n ch(\sqrt{n-1}). \end{aligned}$$

 □

3 Relation between Seidel-Estrada index and Seidel energy

Let G be a simple graph of order n, and its Seidel eigenvalues will be denoted by $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq \lambda_{n}$. The Seidel energy $E_{s}(G)$ of graph G is defined by $E_{s}(G)=E_{s}=\sum_{i=1}^{n} |\lambda_{i}|$ [1]. Since

$$\sum^{n}_{i=1} \lambda_{i}=0, $$

we have

$$ E_{s}(G)=2 \sum^{n_{+}}_{i=1} \lambda_{i}=-2 \sum^{n}_{i=n_{+}+1} \lambda_{i} . $$

(13)

In this section, we investigate the relation between the Seidel-Estrada index and the Seidel energy.

Theorem 3.1

The Seidel-Estrada index $SEE(G)$ and the Seidel energy $E_{s}(G)$ satisfy the following inequality:

$$ \frac{e}{2} E_{s}(G) +(n-n_{+})e^{-\frac{E_{s}(G)}{2 (n -n_{+})}} \leq SEE(G)\leq n-1+e^{E_{s}(G)} $$

(14)

with left equality holding if and only if $G\cong K_{n}$.

Proof

(a) At first, we prove the left inequality of (14).

For $G\cong K_{n}$, $SEE(G)=(n-1) e+e^{-n+1}$ and hence the left equality holds in (14). Otherwise, we have to prove that the lower bound is strict for $G\ncong K_{n}$. We have $e^{x}\geq e x$ with equality holding if and only if $x=1$. By the arithmetic-geometric mean inequality, we get

$$\sum_{i=n_{+}+1}^{n} e^{\lambda_{i}}\geq (n-n_{+}) \Biggl(\prod^{n}_{i=n_{+}+1} e^{\lambda_{i}} \Biggr)^{\frac{1}{n-n_{+}}}=(n-n_{+}) \bigl(e^{\sum ^{n}_{i=n_{+}+1} \lambda_{i}} \bigr)^{\frac{1}{n-n_{+}}}=(n-n_{+}) e^{-\frac {E_{s}(G)}{2 (n-n_{+})}} . $$

Using the above result, we have

$$\begin{aligned} SEE(G) =&\sum_{i=1}^{n} e^{\lambda_{i}} = \sum_{\lambda_{i} > 0} e^{\lambda _{i}}+\sum _{\lambda_{i} \leq0} e^{\lambda_{i}} \\ \geq&\sum_{i=1}^{n_{+}} e \lambda_{i} +(n-n_{+}) e^{-\frac{E_{s}(G)}{2 (n-n_{+})}} \\ =&\frac{e}{2} E_{s}(G) +(n-n_{+})e^{-\frac{E_{s}(G)}{2 (n -n_{+})}}. \end{aligned}$$

Suppose that the left equality holds in (14) for $G\ncong K_{n}$. Then we must have

$$\lambda_{1}=\lambda_{2}=\cdots=\lambda_{n_{+}}=1 \quad\mbox{and}\quad \lambda _{n_{+}+1}=\lambda_{n_{+}+2}=\cdots= \lambda_{n} . $$

Since $G\ncong K_{n}$, we see that $K_{1, 2}$ is an induced subgraph G or $K_{2}\cup K_{2}$ is an induced subgraph of G. We have $\lambda_{1}(G)\geq\lambda_{1}(K_{1, 2})=2$ and $\lambda _{1}(G)\geq\lambda_{1}(K_{2}\cup K_{2})=3$. In both cases, we get a contradiction.

(b) Upper bound:

Starting with equation (2), we get

$$\begin{aligned} SEE(G) =&n+\sum_{i=1}^{n} \sum_{k\geq1} \frac{(\lambda _{i})^{k}}{k!} \leq n+\sum_{i=1}^{n} \sum _{k\geq1} \frac{|\lambda_{i}|^{k}}{k!} \\ \leq& n+\sum_{k\geq1}\frac{1}{k!} \Biggl(\sum _{i=1}^{n} |\lambda_{i}| \Biggr)^{k} =n-1+\sum_{k=0}^{\infty} \frac{(E_{s})^{k}}{k!}=n-1+e^{E_{s}(G)} . \end{aligned}$$

(15)

Suppose now that the right equality holds in (14). Then all the above inequalities must be equalities. From (15), we have $|\lambda_{i}|=\lambda_{i}$, for all i. By the trace of $S(G)$, we have $\lambda_{1}=\lambda_{2}=\cdots=\lambda _{n}=0$, a contradiction by Lemma 2.3. This completes the proof of the theorem. □

Remark 3.2

From equation (15) and Lemma 2.1, we get

$$\begin{aligned} SEE(G) \leq{}& n+\sum_{i=1}^{n} \sum _{k\geq1} \frac{|\lambda_{i}|^{k}}{k!} = n+E_{s}(G) + \sum_{i=1}^{n} \sum _{k\geq2} \frac{|\lambda_{i}|^{k}}{k!} \\ ={}& n+E_{s}(G) +\sum_{i=1}^{n} \sum_{k\geq2} \frac{[(\lambda_{i})^{2}]^{\frac{k}{2}}}{k!} \leq n+E_{s}(G) + \sum_{k\geq2} \frac{1}{k!} \Biggl[\sum_{i=1}^{n}(\lambda _{i})^{2} \Biggr]^{\frac{k}{2}} \\ ={}&n+E_{s}(G) -1-\sqrt{n(n-1)}+ \sum_{k \geq0} \frac{(\sqrt{n(n-1)})^{k}}{k!} . \end{aligned}$$

Hence

$$\begin{aligned} SEE(G)-E_{s}(G) \leq n-1-\sqrt{n(n-1)}+e^{\sqrt{n(n-1)}}. \end{aligned}$$

(16)

Equality does not hold because if the equality is to occur, then we have $|\lambda_{i}|=\lambda_{i}$, for all i. Hence by Lemma 2.1(i), again we get a contradiction. We also have

$$SEE(G)-E_{s}(G)< \sqrt{n-1}(\sqrt{n-1}-\sqrt{n})+e^{\sqrt{n(n-1)}} $$

and $SEE(G)< n-1+e^{E_{s}}$; we also give an inequality between the $SEE(G)$ and $E_{s}(G)$.

Theorem 3.3

Let G be a simple graph with n vertices. Then

$$\begin{aligned} e^{|\lambda_{1}|}+e^{|\lambda_{2}|}+\cdots+e^{|\lambda_{n}|} \geq 1+e^{\frac{2E_{s}}{n}}+(n-2)e^{\frac{E_{s}}{n}} \end{aligned}$$

(17)

with equality holding if and only if $|\lambda_{1}|=\frac{2E_{s}}{n} , |\lambda_{2}|=|\lambda_{3}|=\cdots=|\lambda_{n-1}|=\frac{E_{s}}{n} , |\lambda_{n}|=0$.

Proof

We have the Seidel eigenvalues $\lambda_{1},\lambda_{2},\ldots, \lambda_{n}$ with $|\lambda_{1}|>0$, $|\lambda_{n}|\geq0$. Then, by the arithmetic-geometric mean inequality, we get

$$\begin{aligned} e^{|\lambda_{1}|}+e^{|\lambda_{2}|}+\cdots+e^{|\lambda_{n}|} \geq& e^{|\lambda _{1}|}+e^{|\lambda_{n}|}+(n-2) \Biggl(\prod _{i=2}^{n-1}e^{|\lambda_{i}|} \Biggr)^{\frac {1}{n-2}} \\ =& e^{|\lambda_{1}|}+e^{|\lambda_{n}|}+(n-2) \bigl(e^{E_{s}-|\lambda_{1}|-|\lambda _{n}|} \bigr)^{\frac{1}{n-2}} \end{aligned}$$

(18)

as $E_{s}=\sum_{i=1}^{n}|\lambda_{i}|$. Now, we consider the function

$$f(x,y)=e^{x}+e^{y}+(n-2)e^{\frac{E_{s}-x-y}{n-2}}, \quad\mbox{for }x>0, y \geq0. $$

We have

$$\begin{aligned}& \frac{\partial f}{\partial x}=f_{x}=e^{x}-e^{\frac {E_{s}-x-y}{n-2}},\qquad \frac{\partial f}{\partial y}=f_{y}=e^{y}-e^{\frac {E_{s}-x-y}{n-2}} \\& f_{xx}=e^{x}+\frac{1}{n-2} e^{\frac {E_{s}-x-y}{n-2}},\qquad f_{yy}=e^{y}+\frac{1}{n-2} e^{\frac{E_{s}-x-y}{n-2}} \\& f_{xy}=f_{yx}=\frac{1}{n-2} e^{\frac{E_{s}-x-y}{n-2}} . \end{aligned}$$

To find the minimum of the function of $f(x,y)$, we get

$$\begin{aligned} f_{x}=f_{y}=0 \quad\Rightarrow\quad (n-1)x+y=E_{s},\qquad x+(n-1)y=E_{s} \quad\Rightarrow\quad x+y=\frac{2E_{s}}{n}. \end{aligned}$$

(19)

For $x+y=\frac{2E_{s}}{n}$, we have $f_{xx}>0$ and

$$\begin{aligned} f_{xx}f_{yy}-f^{2}_{xy} =& \biggl(e^{x}+\frac{1}{n-2}e^{\frac {E_{s}-x-y}{n-2}} \biggr) \biggl(e^{y}+\frac{1}{n-2}e^{\frac{E_{s}-x-y}{n-2}} \biggr)- \biggl( \frac {1}{n-2}e^{\frac{E_{s}-x-y}{n-2}} \biggr)^{2} \\ =&e^{x+y}+\frac{1}{n-2}e^{\frac{E_{s}-x-y}{n-2}} \bigl(e^{x}+e^{y} \bigr)=e^{\frac {2E_{s}}{n}}+\frac{1}{n-2}e^{\frac{E_{s}}{n}} \bigl(e^{x}+e^{\frac {2E_{s}}{n}-x} \bigr)>0. \end{aligned}$$

From the above, we conclude that $f(x,y)$ has a minimum value at $x+y=\frac{2E_{s}}{n}$ and the minimum value is $e^{y}+e^{\frac{2E_{s}}{n}-y}+(n-2)e^{\frac{E_{s}-\frac{2E_{s}}{n}}{n-2}}$. Now we can see easily that $g(y)=e^{y}+e^{\frac{2E_{s}}{n}-y}+(n-2)e^{\frac{E_{s}}{n}}$ is an increasing function for $y\geq0$. Thus

$$e^{|\lambda_{n}|}+e^{\frac{2E_{s}}{n}-|\lambda_{n}|}+(n-2)e^{\frac {E_{s}}{n}}\geq e^{0}+e^{\frac{2E_{s}}{n}-0}+(n-2)e^{\frac{E_{s}}{n}}=1+e^{\frac {2E_{s}}{n}}+(n-2)e^{\frac{E_{s}}{n}}. $$

Hence we get the required result in (17).

Now suppose that equality holds in (17). Then all inequalities in the above argument must be equalities. From equality in (18) and $E_{s} =\sum_{i=1}^{n} |\lambda_{i}|$, we get $|\lambda_{2}|=|\lambda_{3}|=\cdots=|\lambda _{n-1}|=\frac{E_{s}}{n}$ as $|\lambda_{1}|+|\lambda_{n}|=\frac{2E_{s}}{n}$. Thus, $|\lambda_{1}|=\frac{2E_{s}}{n} , |\lambda_{2}|=|\lambda_{3}|=\cdots =|\lambda_{n-1}|=\frac{E_{s}}{n} , |\lambda_{n}|=0$.

Conversely, one can easily see that equality holds in (17) for a Seidel matrix of graph by $|\lambda_{1}|=\frac{2E_{s}}{n} , |\lambda_{2}|=|\lambda_{3}|=\cdots=|\lambda_{n-1}|=\frac{E_{s}}{n} , |\lambda_{n}|=0$. □

4 Conclusion

In this paper, we investigate the Seidel matrix and Seidel eigenvalues. Moreover, we defined the Seidel-Estrada index and Seidel energy, and computed the upper and lower bounds for the Seidel-Estrada index. We obtained a relation between the Seidel-Estrada index and the Seidel energy of a graph G, and we proved several theorems on the Seidel-Estrada index. The reader can use these results to calculate the Seidel energy and the Seidel-Estrada index.