On the Estrada index of unicyclic and bicyclic signed graphs

A signed graph $\Gamma$ is an ordered pair $(G,\sigma)$ in which $G$ is an underlying graph and $\sigma$ is a function from the edge set $E(G$) to $\{-1,1\}$, which is called a signing. For a signed graph $\Gamma=(G,\sigma)$ and its subgraph $H\subset G$, we use the notation $(H,\sigma)$ for writing the signed subgraph of $\Gamma=(G,\sigma)$, where $\sigma$ is the restriction of the mapping $\sigma:E(G)\rightarrow\{-1,1\}$ to the edge set $E(H)$. The adjacency matrix $A_{\Gamma}=\left(a_{ij}\right)$ of a signed graph $\Gamma=(G,\sigma)$ naturally arose from the unsigned graph by putting $-1$ or $1$, whenever the corresponding edge is either negative or positive, respectively. The characteristic polynomial, denoted by $\varphi(\Gamma,x)=\operatorname{det}\left(xI-A_{\Gamma}\right)$, is called the characteristic polynomial of the signed graph $\Gamma=(G,\sigma)$. For brevity, the spectrum of the adjacency matrix $A_{\Gamma}$ is called the spectrum of the signed graph $(G,\sigma)$. Let the signed graph $\Gamma$ of order $n$ has distinct eigenvalues $\mu_{1}(\Gamma),\mu_{2}(\Gamma),\ldots,\mu_{k}(\Gamma)$ (we drop $\Gamma$ where the signed graph is understood) and let their respective multiplicities be $m_{1},m_{2},\ldots,m_{k}$. The adjacency spectrum of $\Gamma$ is written as $Spec(\Gamma)=\{\mu^{(m_{1})}_{1}(\Gamma),\mu^{(m_{2})}_{2}(\Gamma),\ldots,\mu^{(m_{k})}_{k}(\Gamma)\}$. From the definition, it follows that the matrix $A_{\Gamma}$ is a real symmetric and hence the eigenvalues $\mu_{1}((G,\sigma))\geq\mu_{2}((G,\sigma))\geq\dots\geq\mu_{n}((G,\sigma))$ of the signed graph $(G,\sigma)$ are all real numbers. The largest eigenvalue $\mu_{1}(\Gamma)$ is also known as the index of the signed graph $\Gamma$.
The concept of signature switching is necessary when dealing with signed graphs. Let $Z$ be a subset of the vertex set $V(\Gamma)$. The switched signed graph $\Gamma^{Z}$ is obtained from $\Gamma$ by reversing the signs of the edges in the cut $[Z,V(\Gamma)\backslash Z]$. Clearly, we see that the signed graphs $\Gamma$ and $\Gamma^{Z}$ are switching equivalent. The switching equivalence is an equivalence relation that preserves the eigenvalues. The switching class of $\Gamma$ is denoted by $[\Gamma]$. The sign of a cycle is the product of the signs of its edges. A signed cycle $C_{n\sigma}$ on $n$ vertices is positive (or negative) if it contains an even (or odd) number of negative edges, respectively. A signed graph is said to be balanced if all of its signed cycles are positive, otherwise, it is unbalanced. That is, a signed graph is said to be balanced if it switches to the signed graph with all positive signature. Otherwise, it is said to be unbalanced. By $\sigma\sim+,$ we say that the signature $\sigma$ is equivalent to the all-positive signature, and the corresponding signed graph is equivalent to its underlying graph. In general, the signature is determined by the set of positive cycles. Hence, all trees are switching equivalent to the all-positive signature. Equivalently, we can say that the edge signs of bridges are irrelevant. Finally, the signature of the balanced signed graphs is switching equivalent to the all-positive one. In our drawings of signed graphs, we represent the negative edges with dashed lines and the positive edges with solid lines. A connected signed graph is said to be unicyclic if it has the same number of vertices and edges. If the number of edges is one more than the number of vertices, then it is said to be bicyclic. For definitions and notations of graphs, we refer to [21].
The signed graph $\Gamma$ is said to have the pairing property if $\mu$ is an eigenvalue whenever $-\mu$ is an eigenvalue of $\Gamma$ and both $\mu$ and $-\mu$ have the same multiplicities. The signed graph $\Gamma=(G,+)$ with all positive signature has the pairing property if and only if its underlying graph $G$ is bipartite. For any signature $\sigma$ it is not true.
The rest of the paper is organized as follows. In Section $2$, we extend the Estrada index to signed graphs. Furthermore, we characterize the signed unicyclic graphs with the maximum Estrada index. In Section $3$, we find the signed graphs in the set of all unbalanced unicyclic and bicyclic graphs having the pairing property with the maximal Estrada index respectively.

The Estrada index, a graph-spectrum-based structural descriptor, of a graph is defined as the trace of the adjacency matrix exponential and was first proposed by Estrada in $2000$. Pena et al. [19] recommended calling it the Estrada index, which has since become widely adopted. This index can be used to measure a range of things, including the degree of protein folding [6, 7, 8], the subgraph centrality and bipartivity of complex networks [9, 10]. Because of the graph Estrada index’s exceptional use, various Estrada indices based on the eigenvalues of other graph matrices were investigated. Estrada index-based invariant concerning the Laplacian matrix, signless Laplacian matrix, distance matrix, distance Laplacian matrix and distance signless Laplacian matrix have been studied, see [11].
In social networks, the balance (stability) [12, 13] of a signed network can be quantified by

$$k=\frac{tr(e^{A_{(G,\sigma)}})}{tr(e^{A_{(G,+)}})},$$

(2.1)

where $tr(X)$ denotes the trace of the matrix $X$. Motivated by Equation $(2.1)$, we define the Estrada index for the signed graph in full analogy with the Estrada index for a graph, as

$$EE(\Gamma)=EE((G,\sigma))=\sum_{i=1}^{n}e^{\mu_{i}},$$

(2.2)

where $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{n}$ are the eigenvalues of the signed graph $\Gamma$. The Seidel matrix $S_{G}$ of a simple graph $G$ with $n$ vertices and having the adjacency matrix $A_{G}$ is defined as $S_{G}=J-I-2A_{G}$. Obviously, the Seidel matrix is the adjacency matrix of some signed graph $\Gamma=(K_{n},\sigma)$, where $K_{n}$ is the complete graph on $n$ vertices. Therefore, Eq. $(2.2)$ is the extension of the Siedel Estrada index [1].
For non-negative integer $k$, let $M_{k}(\Gamma)=\sum_{i=1}^{n}\mu_{i}^{k}$ denote the $k$-th spectral moment of $\Gamma$. From the Taylor expansion of $\mathrm{e}^{\mu_{i}}$, $EE(\Gamma)$ in $(2.2)$ can be rewritten as

$$EE(\Gamma)=\sum_{k=0}^{\infty}\frac{M_{k}(\Gamma)}{k!}.$$

(2.3)

It is well-known that $M_{k}(\Gamma)$ is equal to the difference of the number of positive and negative closed walks of length $k$ in $\Gamma$. Mathematically, we have

$$M_{k}(\Gamma)=w^{+}(k)-w^{-}(k),$$

(2.4)

where $w^{+}(k)$ and $w^{-}(k)$ are, respectively, the number of positive and negative closed walks of length $k$ in $\Gamma$. In particular, we have

$M_{0}(\Gamma)=n$, $~{}M_{1}(\Gamma)=0$, $~{}M_{2}(\Gamma)=2m$ and $~{}M_{3}(\Gamma)=6(t^{+}-t^{-})$,

where $n$ is the number of vertices, $m$ is the number of edges, $t^{+}$ is the number of positive triangles and $t^{-}$ is the number of negative triangles in the signed graph $\Gamma$.

Let $\Gamma_{1}$ and $\Gamma_{2}$ be two signed graphs. If $M_{k}\left(\Gamma_{1}\right)\geq M_{k}\left(\Gamma_{2}\right)$ holds for any positive integer $k$, then, by Eq. $(2.3)$ we get $EE\left(\Gamma_{1}\right)\geq EE\left(\Gamma_{2}\right)$. Further, if the strict inequality $M_{k}\left(\Gamma_{1}\right)>M_{k}\left(\Gamma_{2}\right)$ holds for at least one integer $k$, then $EE\left(\Gamma_{1}\right)>EE\left(\Gamma_{2}\right)$. It is easy to see that if $\Gamma$ has $q$ connected components $\Gamma_{1},\Gamma_{2},\ldots,\Gamma_{q}$, then $EE(\Gamma)=$ $\sum_{i=1}^{q}EE\left(\Gamma_{i}\right).$ So, we shall investigate the Estrada index of connected signed graph from now on. One classical problem of graph spectra is to identify the extremal graphs with respect to the Estrada index in some given class of graphs, for example, see [4, 5, 25]. For a signed tree, all signatures are equivalent. The following result shows that among all signed trees on $n$ vertices, the signed path $P_{n}$ has a minimum and the signed star $S_{n}$ has the maximum Estrada index.

Remark 1.1 Let $\Gamma=(G,+)$ be a connected signed graph of order $n$ with all positive signature and $e$ a positive edge. The signed graph $\Gamma^{\prime}=\Gamma+e$ is obtained from $\Gamma$ by adding the edge $e$. It is easy to see that any self-returning walk of length $k$ of $\Gamma$ is also a self-returning walk of length $k$ of $\Gamma^{\prime}$. Thus, $M_{k}\left(\Gamma^{\prime}\right)\geq M_{k}\left(\Gamma\right)$ and $EE(\Gamma^{\prime})\geq EE(\Gamma).$ But in general adding any signed edge between two non-adjacent vertices of the signed graph $\Gamma=(G,\sigma)$ may not increase the Estrada index. Consider the signed graphs $\Gamma_{i}$, $i=1,~{}2,\ldots,6$ as shown in Figure $1$. Their spectrum is given by $Spec(\Gamma_{1})=\{2,-2,0^{2}\}$, $Spec(\Gamma_{2})=\{\frac{-1+\sqrt{17}}{2},1,0,\frac{-1-\sqrt{17}}{2}\}$, $Spec(\Gamma_{3})=\{2,1,-1,-2\}$, $Spec(\Gamma_{4})=\{\sqrt{5},1,-1,-\sqrt{5}\}$, $Spec(\Gamma_{5})=\{2.17,0.31,-1,-1.48\}$ and $Spec(\Gamma_{6})=\{2,1,-1,-2\}$ respectively. The signed graph $\Gamma_{2}$ is obtained from $\Gamma_{1}$ by adding negative edge between two non-adjacent vertices and clearly $EE(\Gamma_{2})=8.55<9.52=EE(\Gamma_{1}).$ The signed graph $\Gamma_{4}$ is obtained from $\Gamma_{3}$ by adding negative edge between two non-adjacent vertices and $EE(\Gamma_{4})=12.54>10.61=EE(\Gamma_{3}).$ Furthermore, The signed graph $\Gamma_{6}$ is obtained from $\Gamma_{5}$ by adding positive edge and $EE(\Gamma_{5})=10.71>10.61=EE(\Gamma_{6}).$ Therefore, edge addition (deletion) technique cannot be used to compare the Estrada index in signed graphs.

Proof. Let $G$ be a graph on $n$ vertices. Put $\Gamma_{1}=(G,+)$ and $\Gamma_{2}=(G,-)$. Then, by Eq. $(2.2)$, we have

	$\displaystyle EE(\Gamma_{1})-EE(\Gamma_{2})$	$\displaystyle=\sum_{i=1}^{n}e^{\mu_{i}(\Gamma_{1})}-\sum_{i=1}^{n}e^{\mu_{i}(\Gamma_{2})}$		(2.5)
		$\displaystyle=\sum_{i=1}^{n}(e^{\mu_{i}(\Gamma_{1})}-e^{\mu_{i}(\Gamma_{2})}).$

The signed graph $\Gamma_{2}$ can be obtained from the signed graph $\Gamma_{1}$ by negating each edge. Therefore $Spec(\Gamma_{1})=-Spec(\Gamma_{2})$. Thus, by rearrangement of Eq. $(2.5)$ and using Taylor’s expansion, we have

	$\displaystyle EE(\Gamma_{1})-EE(\Gamma_{2})$	$\displaystyle=\sum_{i=1}^{n}(e^{\mu_{i}(\Gamma_{1})}-e^{-\mu_{i}(\Gamma_{1})})$		(2.6)
		$\displaystyle=2\sum_{k=0}^{\infty}\frac{M_{2k+1}(\Gamma_{1})}{(2k+1)!}.$

The signed graph $\Gamma_{1}$ has all positive signature and therefore by Eq. $(2.4)$, $M_{2k+1}(\Gamma_{1})\geq 0$. If $\Gamma_{1}$ has an odd cycle of size $l$, then $M_{l}(\Gamma_{1})>0$. Hence the result follows.  

There exist exactly two switching classes on the signings of an odd unicyclic graph (whose unique cycle has odd girth). The cycle with all positive signature and the cycle with all negative signature. The following result is an immediate consequence of Theorem 2.2.

Let $G$ be a bipartite unicyclic graph of girth $l$ and order $n$ and let $\Gamma_{1}$ be any balanced signed graph on $G$ and $\Gamma_{2}$ be any unbalanced one. It is easy to see that $M_{2k+1}(\Gamma_{1})=0=M_{2k+1}(\Gamma_{2})$ for each $k\geq 0$, $M_{2k}(\Gamma_{1})=M_{2k}(\Gamma_{2})$ for $2k\leq l-2$ and $M_{2k}(\Gamma_{1})>M_{2k}(\Gamma_{2})$ for $2k\geq l$. In particular, $M_{l}(\Gamma_{1})=M_{l}(\Gamma_{2})+4l.$ Thus, by Eq.s $(2.3)$ and $(2.4)$, we have the following lemma.

The following result is directly obtained from Corollary 2.3, Lemma 2.4 and Lemma 2.5.

Next, we show that the Estrada index of the balanced cycle $C_{n+}$ is almost equal to the Estrada index of the unbalanced cycle ${C}_{n-}$.

Proof. The Estrada index of the $n$-vertex signed cycle $C_{n+}$ can be approximated [15] as

$$EE\left(C_{n+}\right)\approx nJ_{0},$$

(2.7)

where $J_{0}=\sum_{r\geqslant 0}\frac{1}{(r!)^{2}}=2.27958530\ldots.$
The eigenvalues of the unbalanced cycle $C_{n-}$ are given by $2\cos\frac{(2r+1)\pi}{n},\quad r=0,1,2,\ldots,n-1$. Therefore

$$EE\left(C_{n-}\right)=\sum_{r=0}^{n-1}\mathrm{e}^{2\cos((2r+1)\pi/n)}.$$

The angles $(2r+1)\pi/n$, for $r=0,1,2,\ldots,n-1,$ uniformly cover the semi-closed interval $[0,2\pi)$. Now, using the property of definite integrals as a sum, we have

$$EE\left(C_{n-}\right)=n\left(\frac{1}{n}\sum_{r=0}^{n-1}\mathrm{e}^{2\cos((2r+1)\pi/n)}\right)\approx n\left(\frac{1}{2\pi}\int_{0}^{2\pi}\mathrm{e}^{2\cos x}\mathrm{~{}d}x\right).$$

As $\mathrm{e}^{2\cos x}$ is an even function, therefore

$$\int_{0}^{2\pi}\mathrm{e}^{2\cos x}\mathrm{~{}d}x=2\int_{0}^{\pi}\mathrm{e}^{2\cos x}\mathrm{~{}d}x=\pi J_{0},$$

where $J_{0}$ is a special value of the function encountered in the theory of Bessel function and can be seen in [14] as

$$J_{0}=\sum_{r=0}^{\infty}\frac{1}{(r!)^{2}}=2.27958530\ldots.$$

In view of this,

$$EE\left(C_{n-}\right)\approx nJ_{0}=2.27958530n.$$

(2.8)

Eqs. $(2.7)$ and $(2.8)$ give

$$EE\left(C_{n-}\right)\approx nJ_{0}\approx EE\left(C_{n+}\right).$$

(2.9)

Note that $M_{k}(C_{n+})=M_{k}(C_{n-})$ for $k\leq n-1$ and $M_{k}(C_{n+})\geq M_{k}(C_{n-})$ for $k\geq n$. In particular, $M_{n}(C_{n+})=M_{n}(C_{n-})+4n.$ Thus, by Eq.s $(2.3)$ and $(2.4)$, we get

$$EE(C_{n+})-EE(C_{n-})=\frac{4n}{n!}+\sum_{k=n+1}^{\infty}\frac{M_{k}(C_{n+})-M_{k}(C_{n-})}{k!}.$$

(2.10)

The eigenvalues of $C_{n+}$ and $C_{n-}$ are, respectively, given by $2\cos\frac{2(r-1)\pi}{n}$ and $2\cos\frac{(2r+1)\pi}{n},\quad r=0,1,2,\ldots,n-1.$ Therefore

$\displaystyle\sum_{k=n+1}^{\infty}\frac{M_{k}(C_{n+})-M_{k}(C_{n-})}{k!}$

$\displaystyle=\sum_{k=n+1}^{\infty}\frac{\sum_{r=0}^{n-1}2^{k}(\cos^{k}\frac{2(r-1)\pi}{n}-\cos^{k}\frac{(2r+1)\pi}{n})}{k!}.$

(2.11)

The maximum value of the function $\cos^{k}\frac{2(r-1)\pi}{n}-\cos^{k}\frac{(2r+1)\pi}{n}$ is $2$. Thus, by Eq. $(2.11)$, we have

	$\displaystyle\sum_{k=n+1}^{\infty}\frac{M_{k}(C_{n+})-M_{k}(C_{n-})}{k!}$	$\displaystyle\leq\sum_{k=n+1}^{\infty}\frac{\sum_{r=0}^{n-1}2^{k+1}}{k!}$		(2.12)
		$\displaystyle=\sum_{k=n+1}^{\infty}\frac{2^{k+1}n}{k!}$
		$\displaystyle=\frac{2^{n+2}n}{(n+1)!}+\frac{2^{n+3}n}{(n+2)!}+\frac{2^{n+4}n}{(n+3)!}+\cdots$
		$\displaystyle\leq\frac{2^{n+2}n}{n!}\left(\frac{1}{(n+1)}+\frac{2}{(n+1)^{2}}+\frac{2^{2}}{(n+1)^{3}}+\cdots\right).$

The series $\frac{1}{(n+1)}+\frac{2}{(n+1)^{2}}+\frac{2^{2}}{(n+1)^{3}}+\dots$ is an infinite geometric progression with common ratio $\frac{2}{(n+1)}$. By inequality $(2.12)$, we obtain

$\displaystyle\sum_{k=n+1}^{\infty}\frac{M_{k}(C_{n+})-M_{k}(C_{n-})}{k!}$

$\displaystyle\leq\frac{2^{n+2}n}{n!(n-1)}.$

(2.13)

Eq.s $(2.10)$ and $(2.13)$ imply that

$\displaystyle EE(C_{n+})-EE(C_{n-})$

$\displaystyle\leq\frac{2^{n+2}n+4n(n-1)}{n!(n-1)},$

(2.14)

where the term $\frac{2^{n+2}n+4n(n-1)}{n!(n-1)}\sim\frac{2^{n+2}}{n!}$ tends to zero as the girth $n$ becomes large enough. Moreover, the accuracy of the Eq. $(2.9)$ can be seen from the data given in Table 1. As seen from this data, except for the first few values of $n$ $(n\leq 9)$, the accuracy is more than sufficient.
Table 1. Approximate and exact values of the Estrada index of the $n$-vertex signed cycles $\left(C_{n+}\right)$ and $\left(C_{n-}\right).$

$n$	$EE\left(C_{n+}\right)$	$nJ_{0}$	$EE\left(C_{n-}\right)$
3	$8.1248150$	$6.8387561$	$5.571899$
4	$9.5243914$	$9.1183414$	$8.7127342$
5	$11.4961863$	$11.3979268$	$11.2993665$
6	$13.6967139$	$13.6775122$	$13.658309$
7	$15.9602421$	$15.9570975$	$15.9533523$
8	$18.2371256$	$18.2366829$	$18.2368574$
9	$20.5163225$	$20.5162683$	$20.5163962$
10	$22.7958591$	$22.7958536$	$22.7958491$
11	$25.0754389$	$25.0754390$	$25.0754200$
12	$27.3550237$	$27.3550243$	$27.3550195$
13	$29.6346089$	$29.6346097$	$29.6345864$
14	$31.9141942$	$31.9141951$	$31.9141892$
15	$34.1937795$	$34.1937804$	$34.1937780$

Hence the result follows.  

The main tool used to prove Lemma 2.5 is the construction of mappings which increases the $k$-th spectral moment for each $k$ and using Eq.$(2.3)$. For example consider the following result.

But in signed unicyclic graphs, in general, we cannot construct the mapping which increases the $k$-th spectral moment for each $k$. To defend this statement, consider the signed unicyclic graphs $\Gamma_{5}^{4-}$ and $\Gamma_{5}^{5-}$. Their spectra is given by $Spec(\Gamma_{5}^{5-})=\{\frac{1+\sqrt{5}}{2}^{2},\frac{1-\sqrt{5}}{2}^{2},-2\}$ and $Spec(\Gamma_{5}^{4-})=\{\sqrt{3},\sqrt{2},0,-\sqrt{2},-\sqrt{3}\}$, respectively. It is easy to see that not only $EE(\Gamma_{5}^{5-})=11.30>11.18=EE(\Gamma_{5}^{4-})$ but also $M_{4}(\Gamma_{5}^{5-})=30>26=M_{4}(\Gamma_{5}^{4-})$ and $M_{5}(\Gamma_{5}^{5-})=-10<0=M_{5}(\Gamma_{5}^{4-})$. Thus the problem of finding the unbalanced unicyclic graphs with extremal Estrada index is interesting.

In this section, we first characterize the unbalanced bipartite unicyclic graphs with the maximal Estrada index. Given two non-increasing real number sequences $\alpha=\{\alpha_{1},\alpha_{2},\alpha_{3},\ldots,\alpha_{n}\}$ and $\beta=\{\beta_{1},\beta_{2},\beta_{3},\ldots,\beta_{n}\}$, we say that $\alpha$ majorizes $\beta$, denoted by $\alpha\succeq\beta$, if $\sum_{j=1}^{t}\alpha_{j}\geq\sum_{j=1}^{t}\beta_{j}$ for each $t=1,\ldots,n$, with equality for $t=n.$ Also, if $\alpha\neq\beta$ then $\alpha\succ\beta$.

Let $\Gamma_{p}(n,m)$ denote the set of all signed graphs on $n$ vertices and $m$ edges with the pairing property. The following result will be useful in the sequel.

Proof. Let $\Gamma_{1}$, $\Gamma_{2}$ $\in\Gamma_{p}(n,m)$. Therefore, we have $\sum_{i=1}^{n}\mu_{i}^{2}(\Gamma_{1})=\sum_{i=1}^{n}\mu_{i}^{2}(\Gamma_{2})=2m$. The signed graphs $\Gamma_{1}$ and $\Gamma_{2}$ have the pairing property, so we get $M_{2k+1}(\Gamma_{1})=0=M_{2k+1}(\Gamma_{2})$ for each $k\geq 0$. Let $\mu_{1}(\Gamma_{2})\geq\mu_{2}(\Gamma_{2})\geq\mu_{3}(\Gamma_{2})\geq\dots\geq\mu_{2r}(\Gamma_{2})$, $r\geq 2$, be the non-zero eigenvalues of $\Gamma_{2}$. Thus, by Eq. $(2.3)$ and using the pairing property, we have

$$EE\left(\Gamma_{1}\right)=n-2r+2\sum_{k=0}^{\infty}\frac{g_{k}\left(\mu_{1}^{2}\left(\Gamma_{1}\right),\mu_{2}^{2}\left(\Gamma_{1}\right),0,\dots,0\right)}{(2k)!}$$

(3.15)

and

$$EE\left(\Gamma_{2}\right)=n-2r+2\sum_{k=0}^{\infty}\frac{g_{k}\left(\mu_{1}^{2}\left(\Gamma_{2}\right),\mu_{2}^{2}\left(\Gamma_{2}\right),\dots,\mu_{r}^{2}\left(\Gamma_{2}\right)\right)}{(2k)!},$$

(3.16)

where $g_{k}(x_{1},x_{2},x_{3},\ldots,x_{r})=x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+\dots+x_{r}^{k}$, $k$ is a positive integer and $\mu_{1}^{2}\left(\Gamma_{1}\right)+\mu_{2}^{2}\left(\Gamma_{1}\right)=m=\mu_{1}^{2}\left(\Gamma_{2}\right)+\mu_{2}^{2}\left(\Gamma_{2}\right)+\cdots+\mu_{r}^{2}\left(\Gamma_{2}\right)$. Now, Eq.s $(3.15)$ and $(3.16)$ imply that

$$EE\left(\Gamma_{1}\right)-EE\left(\Gamma_{2}\right)=2\sum_{k=2}^{\infty}\frac{[g_{k}\left(\mu_{1}^{2}\left(\Gamma_{1}\right),\mu_{2}^{2}\left(\Gamma_{1}\right),0,\ldots,0\right)-g_{k}\left(\mu_{1}^{2}\left(\Gamma_{2}\right),\mu_{2}^{2}\left(\Gamma_{2}\right),\dots,\mu_{r}^{2}\left(\Gamma_{2}\right)\right)]}{(2k)!}.$$

(3.17)

We know that the function $f(x)=x^{2k}$ is strictly convex for every positive integer $k.$
As $\mu_{1}\left(\Gamma_{1}\right)>\mu_{1}\left(\Gamma_{2}\right)$ and $\sum_{i=1}^{n}\mu_{i}^{2}(\Gamma_{1})=\sum_{i=1}^{n}\mu_{i}^{2}(\Gamma_{2})=2m$, therefore the vector
$\alpha=(\mu_{1}^{2}\left(\Gamma_{1}\right),\mu_{2}^{2}\left(\Gamma_{1}\right),0,\ldots,0)$ majorizes $\beta=(\mu_{1}^{2}\left(\Gamma_{2}\right),\mu_{2}^{2}\left(\Gamma_{2}\right),\cdots,\mu_{r}^{2}\left(\Gamma_{2}\right)),$ that is $\alpha\succ\beta$. Thus, by Lemma 3.1, we have

$$g_{k}\left(\mu_{1}^{2}\left(\Gamma_{1}\right),\mu_{2}^{2}\left(\Gamma_{1}\right),0,\ldots,0\right)>g_{k}\left(\mu_{1}^{2}\left(\Gamma_{2}\right),\mu_{2}^{2}\left(\Gamma_{2}\right),\dots,\mu_{r}^{2}\left(\Gamma_{2}\right)\right),$$

for $k\geq 2.$ Hence, by Eq. $(3.17)$, we get $EE\left(\Gamma_{1}\right)>EE\left(\Gamma_{2}\right)$ and the result follows.  

A signed unicyclic graph has the pairing property if and only if its underlying graph is bipartite because it has a unique cycle. Next, we characterize the unique unbalanced bipartite unicyclic graphs with the maximum Estrada index among all unbalanced bipartite unicyclic graphs.

Proof. Let $\Gamma\in\Gamma_{p}^{-}(n,n)$ be an unbalanced bipartite unicyclic graph on $n$ vertices. By applying Lemma 3.3, we get $\mu_{1}(\Gamma)\leq\mu_{1}(\Gamma_{n}^{4-})$ with equality if and only if $\Gamma$ is switching equivalent to $\Gamma_{n}^{4-}$. Now, by Lemma 3.5, the characteristic polynomial of $\Gamma_{n}^{4-}$ is given by

$\varphi(\Gamma_{n}^{4-},x)=x^{n-4}\{x^{4}-nx^{2}+2(n-2)\}.$

Clearly, the signed graph $\Gamma_{n}^{4-}$ has four non-zero eigenvalues. Let the signed graph $\Gamma$, where $\Gamma\in\Gamma_{p}^{-}(n,n)$, contains a cycle of length $l\geq 4$. Therefore, the unbalanced cycle $C_{l-}$ is an induced signed subgraph of $\Gamma$. The eigenvalues of $C_{l-}$ are given by $2\cos\frac{(2r+1)\pi}{l},\quad r=0,1,2,\ldots,l-1$. Since $l$ is a positive even integer and thus all the eigenvalues of $C_{l-}$ are non-zero. Hence the result follows by Theorem 3.2 and Lemma 3.4.