Extremal results for Kr+1-free unbalanced signed graphs

Zhuang Xiong1, Yaoping Hou1
1College of Mathematics and Statistics, Hunan Normal University, small Changsha, Hunan 410081, China

Abstract

This paper investigates the Turan-like problem for Kr+1-free (r2) unbalanced signed graphs, where Kr+1 is the set of unbalanced signed complete graphs with r+1 vertices. The maximum number of edges and the maximum index for Kr+1-free unbalanced signed graphs are given. Moreover, the extremal Kr+1-free unbalanced signed graphs with the maximum index are characterized.

Keywords: Signed graph, Extremal graph, Clique, Index

1. Introduction

In this paper, write G˙=(G,σ) for a signed graph with underlying graph G and sign function σ on the edge set of G. Denote by V(G˙), E(G˙), e(G˙), and e(G˙) the vertex set, the edge set, the number of edges, and the number of negative edges of G˙, respectively. If two vertices u,vV(G˙) are joined by an edge, let the quantity auv be 1 or 1 depending on whether uv is a positive or a negative edge. The n×n adjacency matrix A (=A(G˙)) of G˙ is then defined by A=(auv). The eigenvalues of G˙ is that of A(G˙), which are denoted by λ1(G˙)λ2(G˙)λn(G˙). In particular, the largest eigenvalue of G˙ is called the index. The spectral radius of G˙ is the largest absolute value of the eigenvalues of G˙, denoted by ρ(G˙). These two coincide when the absolute values of the eigenvalues of G˙ do not exceed its index.

This paper aims to investigate the Turán-like problem for Kr+1-free (r2) unbalanced signed graphs, where Kr+1 is the set of all the r+1-vertex unbalanced signed complete graphs. Before presenting new theorems, we provide an introductory discussion.

Given a graph F, a graph G is called F-free, if it contains no F as a subgraph. A prime problem of extremal graph theory is to determine the maximum number of edges in an n-vertex F-free graph, known as Turán problem. Early in 1907, Mantel [10] proved that if G is an n-vertex triangle-free graph, then e(G)n2/4, with equality holding if and only if G=Kn2,n2, that is, G is an n-vertex complete bipartite graph with two almost equally size partitions. The study of extremal graph theory as a subject in its own right was formally initiated by Turán in 1940. Let r2 be an integer and Kr+1 be a complete graph on r+1 vertices. In his seminal papers [14,15], the maximum number of edges of a Kr+1-free graph was determined, and the unique extremal graph was also characterized. Turán’s graph, denoted by Tr(n), is the complete r-partite graph on n vertices which is the result of partitioning n vertices into r almost equally sized partitions (n/r,n/r) and taking all edges connecting two different partition classes (note that if nr then Tr(n)=Kn). Denote the number of edges in Turán’s graph by tr(n)=e(Tr(n)). Using these notations, Turán’s Theorem can be stated as follows.

Theorem 1.1. Let G be a graph on n vertices. If G is Kr+1-free (r2) then e(G)tr(n). Furthermore, equality holds if and only if G=Tr(n).

Turán’s Theorem is a fundamental theorem in extremal graph theory, providing insights into the structure of extremal graphs. In addition to Turán’s Theorem, extremal graph theory also encompasses many other important results and problems, such as the shortest path problems [1], graph decomposition problems [8], and so on. Research into these problems not only helps in understanding the relationship between the global and local structures of graphs, but also plays a crucial role in solving combinatorial optimization problems, network design, and information transmission. The readers can refer to [3] for a comprehensive overview.

In the past two decades, the spectral version of Turán problem is paid much attention by many researchers. Below we only mention the result raised by Nikiforov in 2007[11], which will be used for study later in this article. The readers can refer to an outstanding paper [9] to understand the current research status of such problems.

Theorem 1.2 ([11, Theorem 1]). Let G be a graph on n vertices. If G is Kr+1-free (r2) then λ1(G)λ1(Tr(n)). Furthermore, equality holds if and only if G=Tr(n).

Note that Turán’s graph is both the extremal graph of number of edges and of spectral radius. Actually, the Turán problem and the spectral Turán problem are closely related. See [12] for more discussions on this topic. Different from aforementioned studies, we focus on signed graphs, and ask what are the maximum number of edges and the maximum spectral radius of an F-free signed graph of order n, where F is a set of signed graphs. That problem were initially proposed by Wang, Hou, and Li in their recent research [16], in which they referred to such problem as the Turán-like problem in the context of signed graphs.

Denote by Kr+1 (r2) the set of all unbalanced signed complete graphs on r+1 vertices. Let G˙y1,,yr be a Kr+1-free unbalanced signed graph obtained from a signed clique Kr on vertex set X={v1,,vr} with one negative edge v1v2 and an all-positive clique Knr on vertex set Y={vr+1,,vn} by adding positive edges between each vertex of Y and all but one vertex of X. The superscript yi (1ir) denotes the number of vertices not adjacent to vi. It is obvious that the number of vertices in Y not adjacent to vi (3ir) is not greater than r2, that is, y3++yrr2. In addition, y1++yn=nr. Note that, due to the unbalance, if r=2, then y1,y21, and if r3, then y1,,yr0. Here we draw the signed graphs when r=2,3 to illustrate this structure, as shown in Fig. 1, noting that thin solid lines (resp., thin dashed lines) represent positive edges (resp., negative edges), thick solid lines between two vertex sets represent the connection of all possible positive edges, and the solid circle represents all-positive clique.

An important feature of signed graphs is the concept of switching the signature. For a signed graph G˙ and UV(G˙), the operation that changes the signs of all edges between U and V(G˙)U is called a switching operation. We say that the signed graphs G˙ and G˙~ obtained by a switching operation from G˙ is switching equivalent. It is important to observe that switching equivalent signed graphs have similar adjacency matrices and so have the same eigenvalues (see [22]).

Now we present some interesting results about the Turán-like problem.

Theorem 1.3 ([16, Theorem 1.2]). If G˙ is a connected K3-free unbalanced signed graph of order n, then e(G˙)n(n1)2(n2), with equality holding if and only if G˙ is switching equivalent to G˙y1,y2 (see Figure 1), where y1+y2=n2 and y1,y21.

Figure 1 The signed graphs G˙y1,y2 and G˙y1,y2,y3
Theorem 1.4 ([16, Theorem 1.3]). If G˙ is a connected K3-free unbalanced signed graph of order n, then ρ(G˙)12(n28+n4), with equality holding if and only if G˙ is switching equivalent to G˙n3,1.

Remark 1.5. In Theorems 1.3 and 1.4, the condition of connectivity can be omitted. In fact, for the former theorem, if the signed extremal graph is disconnected, we can add an edge between two connected components to obtain a K3-free unbalanced signed graph, leading to a contradiction. For the latter theorem, we know that for a K3-free unbalanced signed graph with the maximum spectral radius, its spectral radius is equal to its index (cf. [16, p. 62]). Therefore, according to Proposition 3.2 and Lemma 3.3 , which we will prove later, if the signed spectral extremal graph is disconnected, we can add a positive edge between two connected components, getting a K3-free unbalanced signed graph with larger index, a contradiction.

In this context, for some special F, we reframe the Turán-like problem in signed graphs to connect it with the spectral Turán problem. Let Kr+1+ (r2) be the set of balanced signed complete graphs on r+1 vertices. Note that, if F=Kr+1+, then the all-negative signed complete graph is the unique (up to switching equivalence) F-free unbalanced signed graph attaining the maximum spectral radius (see [5, Theorem 3.1]). This is the trivial case. Thus, we only detect among Kr+1+-free balanced signed graphs those having the maximum spectral radius. Moreover, balanced signed graphs are switching equivalent to all-positive signed graphs. We can only look for the desired signed graphs in all-positive signed graphs and this problem becomes the spectral Turán problem for Kr+1-free graphs, when considering the spectral radius.

Recently, several researches have explored related issues (see [7,20,19,17]), and here we recall several results that will be helpful for study later. Chen and Yuan, in [7], investigated the Turán-like problem for K4-free signed graphs, and their results are as follows:

Theorem 1.6 ([7, Theorem 1.5]). If G˙ is a K4-free unbalanced signed graph of order n (n7), then e(G˙)n(n1)2(n3).

Remark 1.7. Note that all the K4-free unbalanced signed graphs attaining above upper bound are determined in their paper. Here, we do not list them and observe that the condition n7 can be removed from Theorem 1.6 by consulting the tables of signed graphs with order at most 6 (cf. [6]).

Theorem 1.8 ([7], Theorem 1.6). If G˙ is a K4-free unbalanced signed graph of order n, then ρ(G˙)n2, with equality holding if and only if G˙ is switching equivalent to G˙n3,0,0.

Now we are in a position to present our results. The following theorem concerns the maximum of the number of edges.

Theorem 1.9. If G˙ is a Kr+1-free (2rn1) unbalanced signed graph of order n, then e(G˙)n(n1)2(nr). Furthermore, if e(G˙)=1, then the equality holds if and only if G˙=G˙y1,,yr, where y1++yr=nr, y3++yrr2, and y1,y21 when r=2, y1,,yr0 when r3.

Below we show that the Kr+1-free unbalanced signed graph with maximum index is switching equivalent to G˙nr,0,,0.

Theorem 1.10. If G˙ is a Kr+1-free (3rn1) unbalanced signed graph of order n, then λ1(G˙)λ1(G˙nr,0,,0), with equality holding if and only if G˙ is switching equivalent to G˙nr,0,,0, where the number of 0 in the superscript is r1.

The proposition below adds some numerical estimates to Theorem 1.10.

Proposition 1.11. The index λ1(G˙nr,0,,0) equals the largest root of the polynomial f(x)=x3+(3n)x2+(3nr)x+(n+4)r(r2+n+7), and satisfies n2λ1(G˙nr,0,,0)<n1.

In particular, λ1(G˙nr,0,,0)=n2 if and only if r=3.

The remainder of this article is organized as follows: in Section 2, we present some fundamental properties and conclusions that will be used in the sequel. In Section 3 , we provide the proofs of Theorems 1.9,1.10 and Proposition 1.11. In the concluding remarks, we analyze our results and provide some comments.

2. Preliminaries

In this section we introduce some results which will be useful in the sequel. The lemma below concerns equitable partitions. Consider a partition P={V1,,Vm} of the set V={1,,n}. The characteristic matrix χP of P is the n×m matrix whose columns are the characteristic vectors of V1,,Vm. Consider a symmetric matrix M of order n, with rows and columns are partitioned according to P. The partition of M is equitable if each submatrix Mi,j formed by the rows of Vi and the columns of Vj has constant row sums qi,j. The m×m matrix Q=(qi,j)1i,jm is called the quotient matrix of M with respect to the equitable partition P.

Lemma 2.1 ([4, p.30]) The matrix M has the following two kinds of eigenvectors and eigenvalues:

(i) The eigenvectors in the column space of χP; the corresponding eigenvalues coincide with the eigenvalues of Q,

(ii) The eigenvectors orthogonal to the columns of χP; the corresponding eigenvalues of M remain unchanged if some scalar multiple of the all-one block J is added to block Mi,j for each i,j{1,,m}.

Next we present a celebrated result of signed graphs, which is an important method for determining the switching equivalence of two signed graphs with the same underlying graph.

Lemma 2.2 ([21], Proposition 3.2]). Two signed graphs on the same underlying graph are switching equivalent if and only if they have the same list of balanced cycles.

The balanced clique number of a signed graph G˙, denoted by ωb(G˙), is the maximum order of a balanced clique in G˙. The following lemma gives an upper bound on the index of a signed graph in terms of its order and balanced clique number.

Lemma 2.3 ([18], Proposition 5]). Let G˙ be a signed graph of order n. Then λ1(G˙)n(11ωb(G˙)).

We end this section by following lemma which says that the index of a signed graph is not greater than that of its underlying graph.

Lemma 2.4 ([5], Theorem 2.1 and Proposition 2.5]). For a non-empty signed graph G˙ of order n, λ1(G˙)λ1(G). Furthermore, λ1(G˙)n1, with equality if and only if G˙ is balanced and complete.

3. Proofs

This section is devoted to the proofs of Theorems 1.9, 1.10, and Proposition 1.11. For notations and concepts of signed graphs undefined here, we refer the reader to [21]. For introductory material on the matrix theory of signed graphs see the survey of Zaslavsky [22] and its references. In particular, let G˙ be a signed graph, and X and Y be disjoint sets of vertices of G˙. We write:

V(G˙)={v1,,vn} for the set of vertices of G˙;

G˙[X] for the signed graph induced by X, and e(X) for e(G˙[X]);

e(X,Y) for the number of edges joining vertices in X to vertices in Y;

NG˙(v) for the set of neighbors of a vertex v in G˙, and dG˙(v) for |NG˙(v)|;

u+v, uv, and uv for that u is adjacent to v by a positive edge, a negative edge, and non-edge, respectively;

G˙G˙ for that G˙ is switching equivalent to G˙.

Proof of Theorem 1.9. From Theorem 1.3, we know that our assertion holds for r=2. We will prove our result by induction on r. Assume that the assertion is true for all rt1 (2t1n2), and we prove it for t. Let G˙ be a Kt+1-free unbalanced signed graph with maximum possible number of edges. Note that G˙y1,,yt is Kt+1-free, and so e(G˙)e(G˙y1,,yt).

First we claim that G˙ contains an unbalanced signed clique of order t. Otherwise, by induction hypothesis we know e(G˙)n(n1)2(nt+1)<e(G˙y1,,yt), a contradiction.

Let X be the vertex set of an unbalanced signed complete graph with order t and let Y be its complement. Since each vertex in Y can have at most t1 neighbours in X, the number of edges between X and Y is at most (t1)(nt). We see that e(G˙)=e(X)+e(Y)+e(X,Y)(t2)+(nt2)+(t1)(nt)=n(n1)2(nt).

If e(G˙)=1, then the equality holds if and only if the unbalanced t-vertex clique contains the negative edge, G˙[Y] is an all-positive clique, each vertex of Y is adjacent to t1 vertices of X by positive edges, and the number of vertices adjacent to both v1 and v2 is not great than t2.

Thus, we have proven that the conclusion holds when r=t. Based on the induction hypothesis, we complete the proof of Theorem 1.9. ◻

The approach for proving Theorem 1.9 is inspired by the proof of Turán’s Theorem, with the key difference being that we use induction on r here, whereas his proof employs induction on n.

Proof of Proposition 1.11. We give a vertex partition as V1={v1}, V2={v2}, V3={v3,,vr}, and V4={vr+1,,vn}. Then the adjacency matrix of G˙nr,0,,0 and its quotient matrix

where 0 and j represent the zero vector and the all-ones vector of appropriate dimensions, respectively, and I and J denote the identity matrix and all-ones matrix of appropriate orders, respectively. By Lemma 2.1, the eigenvalues of Q1 are that of A and the other eigenvalues of A remain if we add some scalar multiple of j or J from the blocks equal to 1, j, J, and JI. Then A and Q1 become

The eigenvalues of matrix A+ except the eigenvalues of Q1+ are 1 with multiplicity n4. Then the eigenvalues of G˙nr,0,,0 are the eigenvalues of Q1 and 1 with multiplicity n4. Therefore, λ1(G˙nr,0,,0)=λ1(Q1). By direct calculation, the characteristic polynomial of the matrix Q1 is g(x)=(x+1)(x3+(3n)x2+(3nr)x+(n+4)r(r2+n+7))=(x+1)f(x).

Thus λ1(G˙nr,0,,0) is the largest root of f(x)=x3+(3n)x2+(3nr)x+(n+4)r(r2+n+7). By simple calculations f(n2)=(r3)20, and so the equality holds if and only if r=3. So we have λ1(G˙nr,0,,0)n2, and λ1(G˙nr,0,,0)<n1 from Lemma 2.4. ◻

To simplify the proof of Theorem 1.10, we shall prove several auxiliary results. First, note that according to Perron-Frobenius theory, there exists a strictly positive eigenvector corresponding to the index of a simple connected graph G. Furthermore, if we add some edges in G, the index of the resulting graph is larger than that of G. These are incorrect for signed graphs, but we have following results.

Lemma 3.1 ([13, Lemma 1]). Let G˙ be a signed graph with n vertices. Then there exists a signed graph G˙ switching equivalent to G˙ such that λ1(G˙) has a non-negative eigenvector.

Let x=(x1,x2,,xn) be an eigenvector corresponding to the index λ1(G˙) of a signed graph G˙. The entry xi corresponds to the vertex vi of G˙. So the eigenvalue equation for vi reads as follows λ1(G˙)xi=vjNG˙(vi)σ(vivj)xj. The following proposition presents a crucial method investigating the maximum index in signed graphs, which can be proven based on [13, Theorem 3] or [2, Proposition 2.1]. For the sake of completeness, we provide a detailed proof here.

Proposition 3.2. Let G˙ be a signed graph with a non-negative unit eigenvector x=(x1,x2,,xn) corresponding to the index λ1(G˙). If we perform one of the following perturbations in G˙:

(i) Adding some positive edges,

(ii) Removing some negative edges,

(iii) Reversing the signs of some negative edges, resulting in a new signed graph G˙, then λ1(G˙)λ1(G˙). The equality holds if and only if the entries of x corresponding to the endpoints of these edges are all zeros.

And if we perform one of the following perturbations in G˙:

(iv) Rotating the positive edge vivj to the non-edge position vivk, where xjxk,

(v) Reversing the sign of the positive edge vivj and the negative edge vivk, where xjxk, resulting in a new signed graph G˙, then λ1(G˙)λ1(G˙). The equality holds if and only if xi=0 and xj=xk.

Proof. For (i), we denote by E1E(G˙) the set of added positive edges. By Rayleigh Principle, we have λ1(G˙)λ1(G˙)=maxy=1yA(G˙)yxA(G˙)xxA(G˙)xxA(G˙)x=2vivjE1xixj0.

If λ1(G˙)=λ1(G˙), then all the equalities hold and so x is an eigenvector of A(G˙) corresponding to the eigenvalue λ1(G˙). Take one positive edge from E1, say vkvl. We will show xl=0. Assume that the added positive edges with one endpoint vk are vkvl,vkvk1,vkvk2,,vkvks. According to the following eigenvalue equations, λ1(G˙)xk=vhNG˙(vk)σ(vhvk)xh,λ1(G˙)xk=vhNG˙(vk)σ(vhvk)xh+j=1sxkj+xl, we obtain xl=xk1==xkj=0. By similar analysis as above, the entries of x corresponding to the endpoints of added positive edges are all zeros.

The proof for (ii) is similar to that for (i).

Note that changing negative edges to positive is equivalent to deleting the negative edges and then adding positive edges, so (iii) follows easily from (i) and (ii).

For (iv), we have λ1(G˙)λ1(G˙)xA(G˙)xxA(G˙)x=2xi(xkxj)0. If λ1(G˙)=λ1(G˙), then all the equalities hold and so x is an eigenvector of A(G˙) corresponding to the eigenvalue λ1(G˙). In view of the following eigenvalue equations, λ1(G˙)xi=vhNG˙(vi)vjσ(vhvi)xh+xj,λ1(G˙)xi=vhNG˙(vi)vjσ(vhvi)xh+xk,λ1(G˙)xj=vhNG˙(vj)viσ(vhvj)xh+xi,λ1(G˙)xj=vhNG˙(vj)viσ(vhvj)xh, λ1(G˙)xk=vhNG˙(vk)σ(vhvk)xh,λ1(G˙)xk=vhNG˙(vk)σ(vhvk)xh+xi, we have xi=0 and xj=xk.

The proof for (v) is similar to that for (vi) and we omit it. The converse is clear. ◻

Lemma 3.3. Let G˙ be a signed graph with a unit eigenvector x=(x1,x2,,xn) corresponding to λ1(G˙). If λ1(G˙)>nk, then x has at most k2 zero components.

Proof. Without loss of generality, assume for a contradiction that x1=x2==xk1=0. Delete the corresponding vertices from G˙ to obtain a signed graph G˙. Then by Rayleigh Principle and Lemma 2.4, λ1(G˙)=(xk,,xn)A(G˙)(xk,,xn)λ1(G˙)λ1(Knk+1)=nk, a contradiction. ◻

Remark 3.4. Note from Proposition 1.11 that G˙nr,0,,0 is a Kr+1-free unbalanced signed graph with index λ1(G˙nr,0,,0)>n2, for r4. Combining this with Proposition 3.2 and Lemma 3.3, we know that if G˙ is a Kr+1-free unbalanced signed graph with maximum index, then it is connected. Furthermore, if r4, then we can find an eigenvector corresponding to λ1(G˙) with no zero components.

Proof of Theorem 1.10. From Theorem 1.8 we know that our assertion holds when r=3. Now by induction on r, assume the assertion is true for all rt1 (3t1n2) and prove it for t.

Let G˙ be a signed graph having the maximum index over all Kt+1-free unbalanced signed graphs. In view of Lemma 3.1 and Remark 3.4 we can find G˙~G˙ with a positive unit eigenvector x=(x1,,xn) corresponding to λ1(G˙~)>n2. Note from Lemma 2.2 that G˙~ is also a connected Kt+1-free unbalanced signed graph. We will show G˙~=G˙nt,0,,0 step by step.

First, since G˙~ is unbalanced, there exist at least one negative edge and at least one negative cycle. Take a negative cycle C=v1v2vlv1 of the shortest length from G˙~. We claim l=3, otherwise G˙~ is K3-free, and so λ1(G˙~)12(n28+n4)<n2 by Theorem 1.4, a contradiction. Thus, C is a negative triangle on vertices v1, v2, and v3.

Secondly, we say that all the negative edges of G˙~ are contained in C. Indeed, if there exists a negative edge not in C, by Proposition 3.2 we may delete it resulting in a Kt+1free unbalanced signed graph with larger index, a contradiction. Therefore, the number of negative edges of G˙~ is either 1 or 3.

We conclude that G˙~ contains only one negative edge. Actually, if not, then C is a negative triangle with three negative edges and by Proposition 3.2 we may reverse signs of two of those, resulting in a Kt+1-free unbalanced signed graph with larger index, which leads to a contradiction.

Next we claim that G˙~ contains an unbalanced t-vertex clique with one negative edge v1v2, written as Kt. If not, G˙~ is Kt-free and by induction hypothesis and Proposition 3.2 we know that λ1(G˙~)λ1(G˙nt+1,0,,0)<λ1(G˙nt,0,,0), a contradiction. Without loss of generality, suppose that X=V(Kt)={v1,,vt} and Y=V(G˙~)X, and further that x1x2 and x3xt. Let W1=NG˙~(v1)NG˙~(v2),W2=NG˙~(v2)NG˙~(v1), and W=NG˙~(v1)NG˙~(v2)X. We claim that W1=, otherwise there exists a vertex vk satisfying vk+v1 and vkv2, and by Proposition 3.2 we can rotate the positive edge vkv1 to the non-edge position vkv2, getting a Kt+1-free unbalanced signed graph with larger index than G˙~, a contradiction.

We proceed with our proof and establish that x2<x3. Assume for a contradiction that x2x3. If there exists V1V(G˙~) such that G˙~[V1{v1,v3}] is all-positive clique Kt+1, then by W1= we have that each vertex in V1 is adjacent to v2 and so G˙~[V1{v1,v2}] is an unbalanced clique of order t+1, a contradiction. So there exist no V1V(G˙~) such that G˙~[V1{v1,v3}] is Kt+1, and by Proposition 3.2 we may reverse the signs of v1v2 and v1v3, resulting in a Kt+1-free unbalanced signed graph with larger index than G˙~, a contradiction. Then we know that x1x2<x3xt.

Note that each vertex in Y is adjacent to at most t1 vertices in X. Then we claim that W=. Otherwise, there exists a vertex viW not adjacent to a vertex vjX{v1,v2}, and then we can rotate the positive edge viv1 to the non-edge position vivj, resulting a signed graph with larger index than G˙~ by Proposition 3.2 and the fact xj>x1, a contradiction.

Summing up, G˙~ must be a subgraph of G˙nt,0,,0. Furthermore, combining Proposition 3.2 and the fact that the eigenvector x is positive, G˙~ actually is G˙nt,0,,0. Thus, we have proven that the conclusion holds when r=t. Based on the induction hypothesis, we complete the proof of Theorem 1.10. ◻

4. Concluding remarks

In this paper, we study the problems what is the maximum number of edges and what is the maximum index for Kr+1-free (r2) unbalanced signed graphs. The condition of unbalance is necessary, otherwise, up to switching equivalence, the all-positive complete graph Kn is the unique Kr+1-free signed graph with e(G˙)=n(n1)/2, and is the unique Kr+1-free signed graph with ρ(G˙)=n1. In this time, this problem is trivial.

Our result partly solve the following problem, which called as the spectral Turán-like problem for Kr+1-free signed graphs. To see this, we need introduce some notations and notions.

Problem 4.1. What is the maximum spectral radius among all Kr+1-free (2rn1) unbalanced signed graphs?

Combining Theorem 1.10 and Lemma 2.3, we can drive the following proposition. Note that the negation of G˙ (denoted by G˙) is obtained by reversing the sign of each edge in G˙. Clearly, the eigenvalues of G˙ are obtained by reversing the signs of the eigenvalues of G˙.

Proposition 4.2. If G˙ is a Kr+1-free (3rn/2) unbalanced signed graph of order n, then ρ(G˙)ρ(G˙nr,0,,0), with equality holding if and only if G˙G˙nr,0,,0.

Proof. The assertion holds for r=3 by Theorem 1.8. Now assume that 4rn/2 and G˙ has the maximum spectral radius among Kr+1-free unbalanced signed graphs. Note that G˙nr,0,,0 is Kr+1-free and so ρ(G˙)ρ(G˙nr,0,,0)>n2.

We claim that ρ(G˙)=λ1(G˙). If not, then ρ(G˙)=λn(G˙). Since G˙ contains no r+1-vertex balanced clique, then ωb(G˙)r. Using Lemma 2.3, we have n2<ρ(G˙)=λn(G˙)=λ1(G˙)n(11wb(G˙))r1rn, which contradicts to rn/2.

So in this time, the problem finding the maximum spectral radius among Kr+1-free (3rn/2) unbalanced signed graphs becomes finding the maximum index among Kr+1-free (3rn/2) unbalanced signed graphs. The latter is solved by Theorem 1.10. ◻

Proposition 4.2 solves Problem 4.1 under the restriction 3rn/2. The case of r>n/2 is left and seems more challenging for further study.

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