R E S E A R C H
Open Access
On the sum of the two largest Laplacian
eigenvalues of unicyclic graphs
Yirong Zheng
1,2*, An Chang
2and Jianxi Li
2,3*Correspondence:
yrzheng@xmut.edu.cn 1School of Applied Mathematics,
Xiamen University of Technology, Xiamen, Fujian, P.R. China 2Center for Discrete Mathematics,
Fuzhou University, Fuzhou, Fujian, P.R. China
Full list of author information is available at the end of the article
Abstract
LetGbe a simple graph andS2(G) be the sum of the two largest Laplacian
eigenvalues ofG. Guanet al.(J. Inequal. Appl. 2014:242, 2014) determined the largest value forS2(T) among all trees of ordern. They also conjectured that among all
connected graphs of ordernwithm(n≤m≤2n– 3) edges,Gm,nis the unique graph which has maximal value ofS2(G), whereGm,nis a graph of ordernwithmedges which hasm–n+ 1 triangles with a common edge and 2n–m– 3 pendent edges incident with one end vertex of the common edge. In this paper, we confirm the conjecture withm=n.
MSC: 05C50; 15A48
Keywords: Laplacian eigenvalue; sum of eigenvalues; largest eigenvalue; unicyclic graph
1 Introduction
LetG= (V,E) be a simple connected graph with vertex setV(G) and edge setE(G). Its order is|V(G)|, denoted byn(G) (ornfor short), and its size is|E(G)|, denoted bym(G) (ormfor short). For a vertexv∈V(G), letN(v) be the set of all neighbors ofvinGand letd(v) =|N(v)|be the degree ofv. Particularly, denote by(G) the maximum degree ofG. A pendent vertex is a vertex with degree one. The diameter of a connected graphG, denoted byd(G), is the maximum distance among all pairs of vertices inG. LetSnandPn be the star and the path of ordern, respectively. LetSka,bbe the tree of ordernobtained from two starsSa+,Sb+by joining a path of lengthkbetween their central vertices (see Figure ). For all other notions and definitions, not given here, see, for example, [] or [] (for graph spectra).
LetA(G) andD(G) be the adjacency matrix and the diagonal matrix of vertex degrees of G, respectively. The matrixL(G) =D(G) –A(G) is calledLaplacianmatrix ofG. The Lapla-cianmatrix is an important topic in the theory of graph spectra. We use the notationIn for the identity matrix of ordernand denote byφ(G,x) =det(xIn–L(G)) theLaplacian characteristic polynomial ofG. It is well known thatL(G) is positive semidefinite sym-metric and singular. Denote its eigenvalues byμ(G)≥μ(G)≥ · · · ≥μn(G) (or simply μ≥μ≥ · · · ≥μn sometimes for convenience) which are always enumerated in non-increasing order and repeated according to their multiplicity. Note that each row sum of L(G) is and, therefore,μn(G) = . Fiedler [] showed that the second smallest eigenvalue μn–(G) ofL(G) is if and only ifGis disconnected. Thus the second smallest eigenvalue
Figure 1 Gn+1,n.Gm,nis a graph of ordernwithmedges which hasm–n+ 1
triangles with a common edge and 2n–m– 3 pendent edges incident with one end vertex of the common edge.
ofL(G) is popularly known as the algebraic connectivity ofG. The largest eigenvalueμ(G) ofL(G) is usually called theLaplacianspectral radius of the graphG. Recently, some of the research has been focused onμ,μorun–(see [–]).
LetSk(G) =i=ki=μi(G) be the sum of theklargestLaplacianeigenvalues ofG. Brouwer conjectured thatSk(G)≤m(G) +k+ fork= , , . . . ,n. The conjecture is still open, but some advances on the conjecture have been achieved (see [–]). Specially, fork= , Haemerset al. [] proved thatS(G)≤m(G) + for any graphG. When Gis a tree, Fritscheret al.[] improved this bound by givingS(T)≤m(T) + – n(T) , which im-plies that Haemers’ bound is always not attainable for trees. Therefore, it is interesting to determine which tree has maximal value ofS(T) among all trees of ordern. Guanet al.[] proved thatS(T)≤S(Tn–
,n–
) for any tree of ordern≥, and the equality
holds if and only ifT∼=Tn– ,n–
. For any graphGof ordernwithmedges, note that μ(G)≤n(G). Then Haemers’ bound is clearly not attainable when n(G) <m(G) + . For m(G) + ≤n(G), Guanet al.[] showed thatS(Gm,n) =m(Gm,n) + , whereGm,nis a graph of ordernsizemwhich hasm–n+ triangles with a common edge and n–m– pendent edges incident with one end vertex of the common edge (Gn+,n is illustrated in Figure ). This indicates that Haemers’ bound is always sharp for connected graphs (m≤n– ) other than trees. The following conjecture on the uniqueness of the extremal graph is also presented in [].
Conjecture .[] Among all connected graphs of order n with m edges(n≤m≤n– ), Gm,nis the unique graph with maximal value of S(G),that is,S(Gm,n) =m(Gm,n) + .
In this paper, we confirm Conjecture . withm=n.
2 Preliminaries
In this section, we present some lemmas which will be useful in the subsequent sections. Forμ(G), the following results are well known.
Lemma . Let G be a connected graph of order n,di=d(vi)and mi=
vj∈N(vi)dj/di.Then () []μ(G)≤n(G)with equality if and only if the complement ofGis disconnected; () []μ(G)≤max{di+mi|vi∈V(G)}.
Lemma .[] Let G be a connected graph of order n≥with m edges.Then
μ(G) <max
(G),m–n–
+ .
Lemma .[] Let T be a tree of order n with d(T)≥.Thenμ(T) <n– ..
Proof For any treeT of ordernwithd(T)≥, it follows thatn≥ and(T)≤n– . That is,(T) + ≤n– andm–n– + =n– + ≤n– . Then the result follows from
Lemma ..
The following theorem from matrix theory plays a key role in our proofs. We denote the eigenvalues of a symmetric matrixMof ordernbyλ(M)≥λ(M)≥ · · · ≥λn(M).
Theorem .[] Let A and B be two real symmetric matrices of size n.Then, for any ≤k≤n,
k
i=
λi(A+B)≤ k
i= λi(A) +
k
i= λi(B).
From Theorem ., the following lemma is immediate.
Lemma . Let G, . . . ,Grbe some edge disjoint graphs.Then,for any k,
Sk(G∪ · · · ∪Gr)≤ r
i= Sk(Gi).
The following lemma can be found in [] and is well known as the interlacing theorem ofLaplacianeigenvalues.
Lemma .[] Let G be a graph of order n and let Gbe the graph obtained from G by inserting a new edge into G.Then the Laplacian eigenvalues of G and Ginterlace,that is,
μ
G≥μ(G)≥ · · ·μn
G≥μn(G) = .
Lemma .[] Let A be a symmetric matrix of order n with eigenvaluesλ≥λ≥ · · · ≥λn and B be a principal submatrix of A of order m with eigenvaluesμ≥μ≥ · · · ≥μm.Then the eigenvalues of B interlace the eigenvalues of A,that is,λi≥μi≥λn–m+ifor i= , . . . ,m. Specially,for v∈V(G),let Lv(G)be the principal submatrix of L(G)formed by deleting the row and column corresponding to vertex v,then the eigenvalues of Lv(G)interlace the eigenvalues of L(G).
Lemma . For any tree T of order n, if there exists an edge e ∈ E(T) such that
min{e(T),e(T)} ≥ and min{d(T),d(T)} ≥ ormax{d(T),d(T)} ≥,then S(T) < n(T) + ,where T,Tare the two components of T–e.
Proof LetT–e=T∪T. By Lemma ., it suffices to show thatS(T∪T) <n(T) – . If μ(T) =μ(T) andμ(T) =μ(T) (orμ(T) =μ(T) andμ(T) =μ(T)), then the result follows sinceS(T∪T) =S(Ti) <m(Ti) + ≤m(T) =n(T) – fori= , . In what follows, we assume thatS(T∪T) =μ(T) +μ(T). Ifmin{d(T),d(T)} ≥, then Lemma . implies thatS(T∪T) < (n(T) – .) + (n(T) – .) =n(T) – . Ifmax{d(T),d(T)} ≥, sayd(T)≥, then Lemmas . and . imply thatS(T∪T) <n(T) + (n(T) – ) =
n(T) – . This completes the proof.
Figure 2 Two trees of ordern.
Proof Without loss of generality, we assume thatvvvvvvis a path of length inT. We now consider the following three cases.
Case .min{d(v),d(v)} ≥.
LetT,Tbe the two components ofT–vv. Then the result follows from Lemma .. Case .d(v) =d(v) = .
Note thatd(T) = . ThenT is isomorphic toSa,b (see Figure ), wherea,b≥ anda+ b+ =n. By an elementary calculation, we haveφ(Sa,b ,λ) =λ(λ– )n–f(λ), wheref
(λ) = λ– (n+ )λ+ (n+ab– )λ– (n+ ab– )λ+n. Denoted byλ
≥λ≥λ≥λ> are the four roots off(λ) = . Then we haveλ+λ+λ+λ=n+ since
i=n
i=μi= m. Note thatSa,b containsPas a subgraph. Then Lemma . implies thatλ≥μ(P) = . Thus,S(T) =λ+λ=n+ – (λ+λ) <n+ –λ<n– , as desired.
Case .d(v)≥ andd(v) = (ord(v)≥ andd(v) = ).
Without loss of generality, we assume thatd(v)≥ andd(v) = . Ifd(v)≥, letT– vv=T∪T; if there isP=uvv attached tov, letT –vv=T∪T, whereu,v= vi(i= , , . . . , ). Then the result follows from Lemma .. We now assume thatd(v) = d(v) = and all the neighbors of v except forv andvare pendent vertices. That is, T is isomorphic toHa,b, whereHa,b (see Figure ) is the tree of ordernobtained from vvvvvvby attachingaandb– pendent vertices tovandv, respectively, where a,b≥ anda+b+ =n. Note that the matrix ·In–L(Ha,b) hasaandbdifferent identical rows. Then the multiplicity of eigenvalue is at leastn– . Letλ≥λ≥λ≥λ≥λ≥ λ>λ= be the other seven eigenvalues. Thenλ+λ+λ+λ+λ+λ+λ=n+ since
i=n
i=μi= m. Fora,b≥,Ha,bcontainsH,as a subgraph. Then by Lemma . we have λ≥μ(H,) = . andλ≥μ(H,) = .. Therefore,S(T) =λ+λ<n+ – (λ+ λ) <n+ , as required. Ifa= , then by Lemmas . and . we haveμ(H,b)≤(n– ) +n–n– andμ(H,b)≤μ(Lv(H,b)) = .. That is,S(T) =S(H,b) =μ(H,b)+μ(H,b) <n+, as
required. Similarly, ifb= , then by Lemmas . and . we haveμ(Ha,)≤(n–)+n–n–and μ(Ha,)≤μ(Lv(Ha,)) = .. It follows thatS(T) =S(Ha,) =μ(Ha,) +μ(Ha,) <n+ ,
as required.
From the discussion above, the proof is completed.
Lemma . Let T be a tree of order n with d(T)≥.Then S(T) <n(T) + .
Proof We now consider the following two cases. Case .d(T)≥.
Letvvvvvvvvbe a path of length inTandT–vv=T∪T. Then the result follows from Lemma ..
Case .d(T) = .
Figure 3 An example of firefly graph.
T∪T (orT –vv=T∪T); ifd(v)≥ (ord(v)≥), letT –vv=T ∪T (or T–vv=T∪T). In each of the above cases, by Lemma ., we haveS(T) <n(T) + .
This completes the proof.
A firefly graphFs,t,n–s–t–(s≥,t≥ andn– s– t– ≥) is a graph withnvertices that consists ofstriangles,tpendent paths of length andn– s– t– pendent edges, sharing a common vertex. An example of a firefly graphF,,is illustrated in Figure . ClearlyF,,n–∼=SnandF,,n–∼=Gn,n.
Lemma . [] The second largest Laplacian eigenvalue of F,t,n–t– satisfies μ(F,t,n–t–) = .
Note that fort≥, the complement ofF,t,n–t–is connected. Hence Lemma . im-plies thatμ(F,t,n–t–) <n(F,t,n–t–). This together with Lemma . implies the follow-ing lemma.
Lemma . For t≥,S(F,t,n–t–) <n(F,t,n–t–) + .
3 Main result
A unicyclic graph is a connected graph whose number of edgesmis equal to the number of verticesn. It is easy to see that each unicyclic graph can be obtained by attaching rooted trees to the vertices of a cycleCkfor somek. Thus ifR, . . . ,Rkarekrooted trees (of orders n, . . . ,nk, say), then we adopt the notationU(R, . . . ,Rk) to denote the unicyclic graphG (of ordern=n+· · ·+nk) obtained by attaching the rooted treeRito the vertexviof a cycleCk=vv· · ·vkv(i.e., by identifying the root ofRiwith the vertexvifori= , . . . ,k). Denote bye(vi) the maximum distance betweenviand any vertex ofRi. In the special case whenRiis a rooted starK,ai with the center of the star as its root (that is,e(vi) = ), we
will simplify the notation by replacingRiby the numberai.
The following lemma is immediate from Lemmas ., . and ..
Lemma . For any unicyclic graph G of order n with m edges,if there exists an edge e∈ E(G)such that G–e is a tree with d(G–e)≥,then S(G) <m(G) + .
Letuvwu be a triangle andTa,b,c be the graph obtained by attachinga,b,c pendent vertices tou,v,w, respectively, wherea+b+c=n– anda≥b≥c≥. Note thatTn–,,∼= Gn,n. Denote byQa,bthe graph obtained by attachingaandbpendent vertices to two non-adjacent vertices of a quadrangle, respectively, wherea+b=n– anda≥b≥.Ta,b,c andQa,bare illustrated in Figure .
Figure 4 Two unicyclic graphs of ordern.Ta,b,cis a unicyclic graph of ordernobtained by attachinga,b,c
pendent vertices to three vertices of a triangleuvw, respectively, andQa,bis a unicyclic graph of ordern
obtained by attachingaandbpendent vertices to two non-adjacent vertices of a quadrangle, respectively.
Proof Note that the matrix ·In–L(Ta,b,c) has a, band cdifferent identical rows. So the multiplicity of eigenvalue is at leastn– . Letλ≥λ≥λ≥λ≥λ>λ= be the other six eigenvalues ofL(Ta,b,c). Then we haveλ+λ+λ+λ+λ=n+ since
i=n
i=μi= m(G). Forc≥,Ta,b,ccontainsT,,as a subgraph. Then by Lemma . we haveλ≥μ(T,,) = .. Therefore,S(Ta,b,c) =λ+λ=n+–(λ+λ+λ) <n+–λ≤ n+ =m(Ta,b,c) + , as required. Forc= andb≥,Ta,b,ccontainsT,, as a subgraph. Then by Lemma . we haveλ≥μ(T,,) = .. That is,S(Ta,b,) =n+ – (λ+λ+λ) < n+ –λ≤n+ =m(Ta,b,c) + , as required. Ifc= andb= , then by Lemmas . and ., we haveμ(Ta,,)≤n– + n– andμ(Ta,,)≤μ(Lu(Ta,,)) = .. It follows that S(Ta,,) =μ(Ta,,) +μ(Ta,,) <m(Ta,,) + , as required. Ifc= andb= , then by Lem-mas . and ., we haveμ(Ta,,)≤n– +n– andμ(Ta,,)≤μ(Lu(Ta,,)) = .. That is,S(Ta,,) =μ(Ta,,) +μ(Ta,,) <m(Ta,,) + forn≥ (that is,a≥), as required. A direct calculation shows thatS(T,,) <m(T,,) + (orS(T,,) <m(T,,) + ). Sim-ilarly, ifc= andb= , then by Lemmas . and ., we haveμ(Ta,,)≤n– +n– and μ(Ta,,)≤μ(Lu(Ta,,)) = .. That is,S(Ta,,) =μ(Ta,,) +μ(Ta,,) <m(Ta,,) + , for n≥ (that is,a≥), as required. A direct calculation shows thatS(T,,) <m(T,,) + (orS(T,,) <m(T,,) + orS(T,,) <m(T,,) + ). Next, we assumec= . By an ele-mentary calculation, we haveφ(Ta,b,,λ) =λ(λ– )n–f(λ), wheref(λ) =λ– (n+ )λ+ (n+ab+ )λ– (n+ ab+ )λ+ n. Letx≥x≥x≥xbe the roots off(λ) = . Then
x+x+x+x=n+ , (.)
xx+xx+xx+xx+xx+xx= n+ab+ , (.)
xxx+xxx+xxx+xxx= n+ ab+ . (.)
If
x+x=n+ , (.)
then, by (.), we have
x+x= . (.)
From (.)-(.) it follows that
(n+ )xx+ xx= n+ ab+ . (.)
By (.) and (.), we have
xx= . (.)
Combining (.) and (.), we have
x=x= . (.)
Then f() = –ab= , which implies that b= . Therefore, if b≥, thenS(Ta,b,) < m(Ta,b,) + . A direct calculation shows thatS(Tn–,,) =m(Tn–,,) + . This completes
the proof.
Lemma . For Qa,b,S(Qa,b) <m(Qa,b) + .
Proof By a direct calculation, we have φ(Qa,b,λ) =λ(λ– )n–f(λ), where f(λ) = (λ– )(λ– (n+ )λ+ (n+ab+ )λ– (n+ ab– )λ+ n). Letλ
≥λ≥λ≥λ≥λ> be the five roots off(λ) = . Then we haveλ+λ+λ+λ+λ=n+ since
i=n
i=μi= m(G). Forb≥,Qa,bcontainsQ,as a subgraph. Then by Lemma . we haveλ≥μ(Q,) = andλ≥μ(Q,) = .. Therefore,S(Qa,b) =λ+λ=n+ –λ–λ–λ<n+ . In what follows, we assumeb= . Sincef() = , we can rewriteφ(Qa,,λ) =λ(λ– )n–f(λ), where f(λ) = (λ– )f(λ). Letλ≥λ≥λ≥λ> be the four roots off(λ) = . Then we have λ+λ+λ+λ=n+ sincei=ni=μi= m(G). Fora≥,Qa,containsQ,as a subgraph. Then by Lemma . we haveλ≥μ(Q,) = . Thus,S(Qa,b) =λ+λ=n+ –λ–λ< n+ . Ifa= , thenS(Qa,b) =S(C) <m(C) + by a straight calculation. This completes
the proof.
Now, we come to the main results of this paper.
Theorem . For any unicyclic graph G, S(G)≤m(G) + with equality if and only if G∼=T(n– , , ).
Proof For any unicyclic graphG, we assume that Ck =vv· · ·vkv is the unique cycle in G(for somek) andGhas the form U(R, . . . ,Rk). Fork≥,G–vv is a tree with d(G–vv)≥. Then by Lemma . we haveS(G) <m(G) + . We now consider the fol-lowing two cases.
Case .k= .
LetC=vvvvvbe the unique cycle inG. If there existe(vi)≥, saye(v)≥, then G–vvis a tree withd(G–vv)≥; if there are two adjacent vertices inC=vvvvv, sayvandv, such thate(v)≥ ande(v)≥, thenG–vvis a tree withd(G–vv)≥. Then by Lemma . we haveS(G) <m(G) + . We now assumeG∼=Qa,b(see Figure ). Then the result follows from Lemma ..
Case .k= .
m(G) + . We now assumeG∼=F,t,n–t–(t≥) orG∼=Ta,b,c. Then the result follows from Lemma . or Lemma ..
This completes the proof.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
YZ carried out the proofs of the main results in the manuscript. AC and JL participated in the design of the study and drafted the manuscript. All the authors read and approved the final manuscript.
Author details
1School of Applied Mathematics, Xiamen University of Technology, Xiamen, Fujian, P.R. China.2Center for Discrete
Mathematics, Fuzhou University, Fuzhou, Fujian, P.R. China. 3School of Mathematics and Statistics, Minnan Normal University, Zhangzhou, Fujian, P.R. China.
Acknowledgements
The authors would like to thank the anonymous referees for their constructive corrections and valuable comments on this paper, which have considerably improved the presentation of this paper. This project is supported by NSF of China (Nos. 11471077, 11301440), China Postdoctoral Science Foundation (No. 2014M551831), NSF of Fujian (No. 2014J01020), the Foundation to the Educational Committee of Fujian (JA13240, JA15381, JB13155).
Received: 31 March 2015 Accepted: 23 August 2015
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