On the Signed Complete Graphs with Maximum Index
aa r X i v : . [ m a t h . C O ] F e b On the Signed Complete Graphs withMaximum Index
N. Kafai a , F. Heydari a , N. Jafari Rad b , M. Maghasedi a a Department of Mathematics, Karaj Branch, Islamic Azad University, Karaj,Iran b Department of Mathematics, Shahed University, Tehran, Iran Abstract
Let Γ = ( K n , H − ) be a signed complete graph whose negativeedges induce a subgraph H . The index of Γ is the largest eigenvalueof its adjacency matrix. In this paper we study the index of Γ when H is a unicyclic graph. We show that among all signed complete graphsof order n > k and maximizes the index, the negative edges induce a triangle withall remaining vertices being pendant at the same vertex of the triangle. Keywords:
Signed complete graph, Unicyclic graph, Index.
MSC:
Let G be a simple graph with the vertex set V ( G ) and the edge set E ( G ).The order of G is defined as | V ( G ) | . The degree of a vertex v in G is denotedby deg G ( v ) . A vertex of degree one is called a pendant vertex . We denotethe set of all neighbors of v in G by N G ( v ). Let K n be the complete graph of order n and K ,k denote the star graph of order k + 1. A tree containingexactly two non-pendant vertices is called a double-star . A double-star with E-mail addresses : [email protected] (N. Kafai), [email protected] (F. Heydari),[email protected] (N.J. Rad), [email protected] (M. Maghasedi). s + 1 , t + 1 , , . . . ,
1) is denoted by D s,t . By C k we denote acycle of length k . A unicyclic graph is a connected graph containing exactlyone cycle. A cactus is a connected graph in which any two cycles have atmost one common vertex.A signed graph Γ is an ordered pair (
G, σ ), where G is a simple graph(called the underlying graph ), and σ : E ( G ) −→ {− , + } is a mapping definedon the edge set of G (called the signature ). If all edges of a signed graph ( G, σ )are positive (resp. negative), then we denote it by ( G, +) (resp. ( G, − )). Fora subset X ⊆ V (Γ), the subgraph induced by X is denoted by Γ[ X ]. Let A ( G ) = ( a ij ) be the adjacency matrix of G . The adjacency matrix of asigned graph Γ = ( G, σ ) is defined as a square matrix A (Γ) = ( a σij ), where a σij = σ ( v i v j ) a ij . The characteristic polynomial of a matrix A is denoted by ϕ ( A, λ ). The characteristic polynomial of A (Γ) is called the characteristicpolynomial of the signed graph Γ and denoted by ϕ (Γ , λ ). Also, the spectrumof A (Γ) is called the spectrum of Γ and the largest eigenvalue is often calledthe index . The spectrum of signed graphs has been studied by many authors,for instance see [1, 3, 8, 9].Let Γ = ( G, σ ) be a signed graph and v ∈ V (Γ). We obtain a new graphΓ ′ from Γ if we change the signs of all edges incident with v . We call v a switching vertex . A switching of a signed graph Γ is a graph that can beobtained by applying finitely many switching operations. We call two graphsΓ and Γ ′ switching equivalent if Γ ′ is a switching of Γ and we write Γ ∼ Γ ′ .The adjacency matrices of two switching equivalent signed graphs Γ and Γ ′ are similar and hence they have the same spectrum, see [10].In [7], the connected signed graphs of fixed order, size, and number ofnegative edges with maximum index have been studied. It was conjecturedin [7] that if Γ is a signed complete graph of order n with k negative edges, k < n − K ,k . In [2], the authors proved that this conjecture holds for signedcomplete graphs whose negative edges form a tree. Recently, Ghorbani et al.[6] proved the conjecture. Let ( K n , H − ) denote a signed complete graph oforder n whose negative edges induce a subgraph H . They introduced a familyof graphs H m,n for positive integers n and m with m ≤ ⌊ n ⌋ and proved thatamong the signed complete graphs with n vertices and m negative edges,( K n , H − ) has the maximum index if and only if H is isomorphic to a H n,m .In this paper we focus on the signed complete graphs whose negativeedges induce a unicyclic graph. We show that among all signed completegraphs of order n > and maximizes the index, the negative edges induce a triangle with allremaining vertices being pendant at the same vertex of the triangle. Thisresult with a result of [2] on trees lead to a conjecture on signed completegraphs whose negative edges induce a cactus graph. The spectral theory of signed graphs has more varieties than unsignedgraphs. But an important tool works in a similar way for signed graphs,which is a consequence of [5, Theorem 1 . . Theorem 1 (Interlacing Theorem for signed graphs) Let Γ be a signed graphwith n vertices and eigenvalues λ ≥ · · · ≥ λ n , and let Γ ′ be an inducedsubgraph of Γ with m vertices. If the eigenvalues of Γ ′ are µ ≥ · · · ≥ µ m ,then λ n − m + i ≤ µ i ≤ λ i for i = 1 , . . . , m. The following result will be useful in the sequel.
Lemma 1 [2, Lemma 3]
Let
Γ = ( K n , K − ,k ) be a signed complete graph.Then ϕ (Γ , λ ) = ( λ + 1) n − (cid:16) λ + (3 − n ) λ + (3 − n ) λ + 4 k ( n − k −
1) + 1 − n (cid:17) . Also, we need to introduce an additional notation. Assume that A is asymmetric real matrix of order n and { X , . . . , X m } is a partition of [ n ] = { , . . . , n } . Let { X , . . . , X m } partition the rows and columns of A, as follows, A , · · · A ,m ... ... A m, · · · A m,m , where A i,j denotes the submatrix of A formed by rows in X i and the columnsin X j . Then the m × m matrix B = ( b ij ) is called the quotient matrix relatedto that partition, where b ij denotes the average row sum of A i,j . If the rowsum of each A i,j is constant, then the partition is called equitable , see [4].If X = ( x , . . . , x n ) T is an eigenvector corresponding to the eigenvalue λ of a signed graph Γ = ( G, σ ), then we assume that the component x v v . So the following is the eigenvalue equation for v : λx v = X u ∈ N G ( v ) σ ( uv ) x u . The next lemma is one of the most used tools in the identifications ofgraphs with maximum index.
Lemma 2 [7, Lemma 5.1(i)]
Let u , v and w be distinct vertices of a signedgraph Γ and let X = ( x , . . . , x n ) T be an eigenvector corresponding to theindex λ (Γ) . Let Γ ′ be obtained by reversing the sign of the positive edge uv and the negative edge uw . If x u ≥ , x v ≤ x w or x u ≤ , x v ≥ x w , then λ (Γ) ≤ λ (Γ ′ ) . If at least one inequality is strict, then λ (Γ) < λ (Γ ′ ) . If u , v and w are the vertices given in Lemma 2, then R ( u, v, w ) refers tothe relocation described in the lemma. ( K n , U − ) with maximum index One classical problem of graph spectra is to identify the maximal graphs withrespect to the index in a given class of graphs. In this section, we determine( K n , U − ) with maximum index, where U is a unicyclic subgraph of K n oforder k . We begin with the following lemma. Lemma 3
Let
Γ = ( K n , Q − ) be a signed complete graph with k ≥ negativeedges, where Q is the graph depicted in Fig. 1. Then ϕ (Γ , λ ) = ( λ + 1) n − (cid:16) λ + (5 − n ) λ + (10 − n ) λ + (12 k − n + 4 ku − λ +(24 k − n + 8 ku − λ + 127 n − k − ku − (cid:17) , where u = n − k .Proof . First assume that k < n . By switching Γ at vertex v , one candeduce that Γ = ( K n , Q − ) is switching equivalent to ( K n , D − n − k, ). Thusthe characteristic polynomials of Γ = ( K n , Q − ) and ( K n , D − n − k, ) are thesame. Hence by [2, Theorem 4] and [2, Remark 5], the result holds. Now,4ssume that k = n . Again by switching Γ at vertex v , we conclude thatΓ = ( K n , Q − ) is switching equivalent to ( K n , K − , ). Therefore, by Lemma 1, ϕ (Γ , λ ) = ( λ + 1) n − (cid:16) λ + (3 − n ) λ + (3 − n ) λ + 11 n − (cid:17) =( λ +1) n − (cid:16) λ +(5 − n ) λ +(10 − n ) λ +(6 n − λ +(20 n − λ +11 n − (cid:17) . Hence the proof is complete. (cid:3) k − v v v v Q s tv v v v k = s + t + 4 Q ( s, t )Figure 1: The unicyclic graphs Q , Q ( s, t ).Let Γ = ( K n , σ ) be a signed complete graph. Let ⊓ = { X , . . . , X p , X p +1 , . . . ,X p + q } be a partition of V (Γ) such that all edges between X i and X j have thesame sign for each i, j , Γ[ X i ] = ( K n i , +) for i = 1 , . . . , p , and Γ[ X i ] = ( K n i , − )for i = p + 1 , . . . , p + q , where | X i | = n i for i = 1 , . . . , p + q . Obviously, ⊓ isan equitable partition of V (Γ). Moreover, we have the next theorem. Theorem 2
Let
Γ = ( K n , σ ) be a signed complete graph. Let ⊓ = { X , . . . , X p ,X p +1 , . . . , X p + q } be a partition of V (Γ) with the above properties and B be thequotient matrix of A (Γ) related to ⊓ . If m = P pi =1 n i and m = P p + qi = p +1 n i ,then ϕ (Γ , λ ) = ( λ + 1) m − p ( λ − m − q ϕ ( B, λ ) . Proof . Suppose that X = { v , . . . , v n } and X i = { v n + ··· + n i − +1 , . . . , v n + ··· + n i } ,for i = 2 , . . . , p + q . Then we have, λI − A ( K n , σ ) = (cid:20) λI − A ( K n , +) ∗∗ ∗ (cid:21) .
5e apply finitely many elementary row and column operations on thematrix λI − A ( K n , σ ). First, subtract the n th row from all the upper rows.This leads to the following matrix, λ + 1 − λ − . . . ... λ + 1 − λ − − λ ∗∗ ∗ . Now, adding the first n − n th column, we obtain thefollowing matrix, λ + 1 0 . . . ... λ + 1 0 − λ + 1 − n ∗∗ ∗ . Thus the following holds: ϕ (Γ , λ ) = ( λ + 1) n − det (cid:20) λ + 1 − n ∗∗ ∗ (cid:21) . By repeating the same procedure, one can deduce that ϕ (Γ , λ ) = ( λ + 1) m − p ( λ − m − q ϕ ( B, λ ) , as desired. (cid:3) Lemma 4
Let
Γ = ( K n , Q ( s, t ) − ) , s, t ≥ , be a signed complete graph with k negative edges, where Q ( s, t ) is the graph depicted in Fig. 1. Then ϕ (Γ , λ ) = ( λ +1) n − (cid:16) λ +(7 − n ) λ +(21 − n ) λ +(12 k − n +4 ku +8 st − λ +(48 k − n + 16 ku + 32 st − λ + (113 n − k − ku − st ( u − − λ +(250 n − k − ku − st ( u +1) − λ +127 n − k − ku +24 st (2 u − − (cid:17) , where u = n − k . roof . Assume that k < n . Suppose that X = { v } , X = { v } , X = { v } , X = { v } , X = N Q ( s,t ) ( v ) \ { v , v } , X = N Q ( s,t ) ( v ) \ { v , v } , and X = V ( K n ) \ V ( Q ( s, t )), see Fig. 1. Let B be the quotient matrix of A (Γ) relatedto { X , . . . , X } . It is not hard to see that the characteristic polynomial of B is ϕ ( B, λ ) = λ + (7 − n ) λ + (21 − n ) λ + (12 k − n + 4 ku + 8 st − λ +(48 k − n + 16 ku + 32 st − λ + (113 n − k − ku − st ( u − − λ +(250 n − k − ku − st ( u +1) − λ +127 n − k − ku +24 st (2 u − − , where u = n − k . By Theorem 2, we have ϕ (Γ , λ ) = ( λ + 1) n − ϕ ( B, λ ) . If k = n , let X , . . . , X be as above and B ′ be the quotient matrix of A (Γ)related to { X , . . . , X } . Then ϕ (Γ , λ ) = ( λ + 1) n − ϕ ( B ′ , λ ) . So one candeduce that ϕ (Γ , λ ) = ( λ + 1) n − (cid:16) λ + (6 − n ) λ + (15 − n ) λ + (2 n + 8 st − λ +(26 n + 24 st − λ + (31 n − st − λ + 11 n − st − (cid:17) , and hence ϕ (Γ , λ ) = ( λ + 1) n − (cid:16) λ + (7 − n ) λ + (21 − n ) λ + ( − n + 8 st − λ +(28 n +32 st − λ +(57 n +16 st − λ +(42 n − st − λ +11 n − st − (cid:17) . This completes the proof. (cid:3)
Corollary 1
Let
Γ = ( K n , Q ( s, t ) − ) , s, t ≥ , and Γ ′ = ( K n , Q − ) be twosigned complete graphs with k negative edges, where Q ( s, t ) and Q are thegraphs depicted in Fig. 1. Then λ (Γ) < λ (Γ ′ ) . Proof . Let λ = λ (Γ). By Lemmas 3 and 4, we deduce that ϕ (Γ ′ , λ ) − ϕ (Γ , λ ) = − st ( λ + 1) n − (cid:0) λ + 4 λ − n − k − λ − n − k + 1) λ +3(2 n − k − (cid:1) . Setting λ = λ , we have ϕ (Γ ′ , λ ) = − st ( λ +1) n − (cid:0) λ +4 λ − n − k − λ − n − k +1) λ +3(2 n − k − (cid:1) . − v v v Figure 2: The unicyclic graph U .Since Γ has ( K n − , +) as an induced subgraph, by Theorem 1, we concludethat n − k + 1 < n − ≤ λ . Therefore, ϕ (Γ ′ , λ (Γ)) < λ (Γ) < λ (Γ ′ ) . (cid:3) By a proof similar to the proof of Lemma 4, we have the following lemma.
Lemma 5
Let
Γ = ( K n , U − ) be a signed complete graph with k ≥ negativeedges, where U is the graph depicted in Fig. 2. Then ϕ (Γ , λ ) = ( λ +1) n − ( λ − (cid:16) λ +(6 − n ) λ +(16 − n ) λ +(4 k − n +4 ku +18) λ +28 k − n + 12 ku + 7 (cid:17) , where u = n − k . Now, we can state the main result of this paper.
Theorem 3
Among all signed complete graphs of order n > whose nega-tive edges induce a unicyclic graph of order k and maximizes the index, thenegative edges induce a triangle with all remaining vertices being pendant atthe same vertex of the triangle.Proof . Suppose that k negative edges induce a unicyclic graph U and max-imizes the index, where k ≤ n . Let Γ = ( K n , U − ) and λ = λ (Γ). Clearly,( K n − k +1 , +) is an induced subgraph of Γ. Hence by Theorem 1, we deducethat n − k ≤ λ . Suppose U consists of the cycle, say C , of length g and a cer-tain number of trees attached at vertices of C . Let V ( U ) = { v , . . . , v k } and8 ( C ) = { v , . . . , v g } . Let X = ( x , . . . , x n ) T be an eigenvector correspondingto λ .We claim that there is an integer i , 1 ≤ i ≤ g , such that x i = 0. To seethis, by contrary assume that x i = 0, for i = 1 , . . . , g . Let v p and v q be twoconsecutive vertices of C , that is v p v q ∈ E ( C ), and suppose that two vertices v j ( g < j ≤ k ) and v p are adjacent in U . If x j = 0, then by Lemma 2, therelocation R ( v j , v q , v p ) would contradict the maximality of λ . By the sameargument, one can prove that x i = 0 for i = 1 , . . . , k . Thus k < n . Let x t be an entry of X having the largest absolute value. We may assume that x t > − X instead of X ). By the eigenvalue equationfor v t , we have P ni = k +1 ,i = t x i = λ x t which implies that λ ≤ n − k −
1, acontradiction. The claim is proved.Suppose that g >
3. Without loss of generality, assume that x >
0. If x g − ≤ x , then the possibility of R ( v , v g − , v ) contradicts the maximality of λ . So x < x g − . Also, x g < x since otherwise, the relocation R ( v , v , v g )leads to a contradiction. Now, if x ≥ x g ≥ R ( v , v g , v ) (resp. R ( v g , v , v g − )) contradicts the maximality of λ . Hence, x , x g <
0. Thus for g ≥
5, if x g − ≥ x (resp. x ≥ x g − ), then by R ( v , v g − , v ) (resp. R ( v g , v , v g − )), we get a contradiction. Therefore, g =4. From the previous statement, we have x > x , x <
0. Also, x > R ( v , v , v ) concludes a contradiction. Sup-pose that v v p , v v q ∈ E ( U ), where p, q >
4. If x p ≥ x q ≥ R ( v p , v , v ) (resp. R ( v q , v , v )) contradicts the maximality of λ . Thus x p , x q <
0. So if x ≥ x (resp. x ≥ x ), then by R ( v p , v , v )(resp. R ( v q , v , v )), we find a contradiction. It follows that deg U ( v ) = 2or deg U ( v ) = 2. Similarly, if v v p , v v q ∈ E ( U ) and p, q >
4, then we havea contradiction and hence deg U ( v ) = 2 or deg U ( v ) = 2. Without loss ofgenerality, assume that deg U ( v ) = deg U ( v ) = 2.Now, we prove that the trees attached to vertices v , v in U , if any, arestars. For proving this, first assume that v v p , v p v q ∈ E ( U ), where p, q > x p ≥
0, then the relocation R ( v p , v , v ) gives a contradiction and hence x p <
0. If x q ≤
0, then the relocation R ( v q , v , v p ) leads to a contradiction.Thus x q >
0. Therefore, if x ≤ x p (resp. x p ≤ x ), then by R ( v q , v , v p )(resp. R ( v , v p , v )), we get a contradiction. This implies that v p is a pendantvertex of U . By repeating the same procedure one can easily prove that thetree which is attached to the vertex v in U , if any, is a star. So we concludethat U may be equal to the graphs Q or Q ( s, t ), depicted in Fig. 1. By9orollary 1, λ ( K n , Q ( s, t ) − ) < λ ( K n , Q − ) and hence U = Q ( s, t ).Next, suppose that g = 3. We first prove that at least two vertices ofthe cycle C have degree 2 in U . By contrary assume that v v p , v v q ∈ E ( U ),where p, q >
3. If x = x = 0, then R ( v p , v , v ) implies that x p = 0.Hence by R ( v , v p , v ), we deduce that x = 0, a contradiction. Withoutloss of generality, assume that x >
0. If x q ≤ x (resp. x q ≤ x ), thenthe relocation R ( v , v q , v ) (resp. R ( v , v q , v )) contradicts the maximalityof λ . Thus x q > x , x . Therefore, x > R ( v , v q , v )). If x ≤ x ,then R ( v q , v , v ) yields a contradiction. So x < x and hence x p < R ( v p , v , v )). Now, R ( v , v p , v ) gives the final contradiction. This completesthe assertion.We now prove that the tree attached to a vertex of C in U , if any, is astar. By contrary assume that v v p , v p v q ∈ E ( U ), where p, q >
3. Withoutloss of generality, assume that x q ≥ − X instead of X ). If x < x p , then the relocation R ( v q , v , v p ) contradicts the maximalityof λ . Hence x p ≤ x . Thus the relocations R ( v , v p , v ) and R ( v , v p , v ),respectively, imply that x ≤ x ≤
0. Therefore, by R ( v , v q , v ), wededuce that x = x = x q = 0. Finally, the relocation R ( v , v q , v ) yields x = 0 which is impossible, and we are done.So the candidates for maximizers are: Γ = ( K n , U − ), and Γ ′ = ( K n , Q − ),see Figs. 1 and 2. We just have to compare the indices. By Lemmas 3 and5, we find that ϕ (Γ , λ ) − ϕ (Γ ′ , λ ) = − λ +1) n − (cid:0) ( k − λ +2( n − λ +12 n − k − k ( n − k ) − (cid:1) . Let λ = λ (Γ ′ ). Since Γ ′ has ( K n − , +) as an induced subgraph, by Theorem1, we conclude that λ ≥ n −
3. Setting λ = λ , we have ϕ (Γ , λ ) = − λ + 1) n − (cid:0) ( k − λ + 2( n − λ + 12 n − k − k ( n − k ) − (cid:1) . First assume that k ≥
5. Thus ( k − λ + 2( n − λ + 12 n ≥ ( k − n − + 2( n − n −
3) + 12 n = kn + 26 n + 9 k − n − kn −
15. It is nothard to see that kn + 2 k + 26 n > n + 8 kn + 2 k + 20 which implies that kn +26 n +9 k − n − kn − > k +2 k ( n − k )+5. Hence ϕ (Γ , λ ) < λ (Γ ′ ) < λ (Γ). This is exactly what we need here. Note that if n = k = 5,by switching Γ ′ at vertex v , we have Γ ′ ∼ Γ and hence λ (Γ ′ ) = λ (Γ).Finally, assume that k = 4. Thus ϕ (Γ , λ ) = 8( λ + 1) n − (cid:0) λ − n − λ − n + 17 (cid:1) . λ − n − λ − n + 17 are n − ± √ n − n + 8. If n > n − − √ n − n + 8 < n − ≤ λ ≤ n − < n − √ n − n + 8,so ϕ (Γ , λ ) < λ (Γ ′ ) < λ (Γ). By a computer search, onecan see that the result holds for n = 6 ,
7, but λ ( K , U − ) < λ ( K , C − ) and λ ( K , U − ) < λ ( K , C − ). The proof is complete. (cid:3) The following is a direct consequence of Theorem 3 which confirms theKoledin-Stani´c conjecture for signed complete graphs whose negative edgesinduce a unicyclic graph.
Corollary 2
Let ( K n , U − ) be a signed complete graph whose negative edgesinduce a unicyclic graph U of order k < n . Then λ ( K n , U − ) < λ ( K n , K − ,k ) . Proof . Suppose that Γ = ( K n , K − ,k ) and Γ ′ = ( K n , U − ), where U is thegraph depicted in Fig. 2. Let λ = λ (Γ ′ ). Since Γ ′ has ( K n − , +) as aninduced subgraph, by Theorem 1, we conclude that λ ≥ n −
3. By Lemmas1 and 5, we deduce that ϕ (Γ , λ ) − ϕ (Γ ′ , λ ) = − λ + 1) n − (cid:0) ( k − λ − n − k + 1) λ +4 n − k − k ( n − k ) − (cid:1) . Setting λ = λ , we have ϕ (Γ , λ ) = − λ + 1) n − (cid:0) ( k − λ − n − k + 1) λ +4 n − k − k ( n − k ) − (cid:1) . If k = 3, then ϕ (Γ , λ ) = − λ + 1) n − (cid:0) λ − n + 4 (cid:1) . Thus ϕ (Γ , λ ) < λ (Γ ′ ) < λ (Γ) . Note that n ≥ k >
3. By a computer search, one can see that λ ( K , U − ) < λ ( K , C − ) < λ ( K , K − , ), so the result holds for n = 5.In what follows, we consider two cases. Case n > k −
1. We claim that ϕ (Γ , λ ) <
0. We have ( k − λ +4 n ≥ ( k − n − + 4 n = kn + 10 n + 9 k − n − kn −
9. Since λ ≤ n −
1, hence2( n − k +1) λ +3 k +2 k ( n − k )+1 ≤ n − k +1)( n − k +2 k ( n − k )+1 =2 n +7 k − k − kn −
1. It is easy to see that kn +2 k +10 n +2 k > n +4 kn +8which yields that kn + 10 n + 9 k − n − kn − > n + 7 k − k − kn − λ (Γ ′ ) < λ (Γ) . Case n ≤ k −
1. Then ( k − λ − n − k + 1) λ + 4 n ≥ ( k − n − − n − k + 1)( n −
3) + 4 n . It is not hard to check that kn + 2 k + 14 n > n +6 k +4 kn +4 which implies that ( k − n − − n − k +1)( n − n > k + 2 k ( n − k ) + 1. Thus ϕ (Γ , λ ) < λ (Γ ′ ) < λ (Γ) . Therefore, by Theorem 3, the proof is complete. (cid:3) k − t t Figure 3: The cactus graph G t .Let G t be the cactus graph with k edges and t cycles, depicted in Fig. 3( G is the star graph K ,k ). We close this paper with the following conjecture. Conjecture 1
Among all signed complete graphs of order n > whose neg-ative edges induce a cactus graph with k edges and t cycles ( k − t < n ), andmaximizes the index, negative edges induce the graph G t . We note that Conjecture 1 is true for t = 0 ,
1, according to Theorem 3and a result of [2].
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