R E S E A R C H
Open Access
The extremal graphs with respect to their
nullity
Sa Rula
1*, An Chang
1and Yirong Zheng
1,2*Correspondence:
changsarula163@163.com 1Center for Discrete Mathematics,
Fuzhou University, Fuzhou, Fujian, P.R. China
Full list of author information is available at the end of the article
Abstract
The nullity of a graphG, denoted by
η(G), is the multiplicity of the eigenvalue zero of
its adjacency matrix. In this paper, we determine all graphs withη(G) =
n(G) – 2m(G) –c(G), wherec(G) =e(G) –n(G) +ω(G), and
n(G),e(G),m(G), andω(G) are the vertex number, edge number, matching number, and the number of
connected components ofG, respectively.MSC: 05C50
Keywords: nullity of graph; matching number
1 Introduction
All graphs considered in this paper are finite, undirected, and simple. LetGbe a graph of ordern(G) and sizee(G). The adjacency matrix ofG, denoted byA(G) = (aij)n×n, is the
n×nmatrix such thataij= if verticesviandvjare adjacent andaij= otherwise,i,j=
, . . . ,n. The rank of a graphG, denoted byr(G), is the rank of its adjacency matrixA(G). The multiplicity of the eigenvalue zero ofA(G) is called the nullity ofGand is denoted byη(G). It is obvious thatη(G) =n(G) –r(G). The graphGis called singular ifη(G) > and called nonsingular ifη(G) = . Collatz and Sinogowitz [] first raised the problem of characterizing all singular or nonsingular graphs. This problem is hard to be solved. The nullity is relevant to the rank of symmetric matrices described by graphs. On the other hand, the nullity has strong chemical background. A singular bipartite graph expresses the chemical instability of the molecule corresponding to the bipartite graph. Due to all these reasons, the nullity aroused the interest of many mathematicians and chemists. The topics on the nullity of graphs includes the computing nullity, the nullity distribution, bounds on nullity, characterization of graphs with certain nullity, and so on. For some particular classes of graphs, some preliminary results about nullity have been released [–].
It is known that ≤η(G)≤n(G) – ifGis a graph of ordern(G) containing at least one edge []. Cheng and Liu [] characterized all graphs with nullityn(G) – andn(G) – . Changet al.[, ] characterized all graphs of ranks and (i.e., with nullityn(G) – and n(G) – ). Along this way, Zhuet al.[] gave the characterization of the unicyclic graphs with nullityn(G) – andn(G) – . Liet al.[] characterized two kinds of bicyclic graphs withn(G) – , whereas the entire characterization of graphs with nullityn(G) – ,n(G) – , etc.is still unfinished.
A matching in a graphGis a set of pairwise nonadjacent edges. A maximum matching is one that contains as many edges ofGas possible. The matching number ofG, denoted bym(G), is the size of a maximum matching inG. Letc(G) =e(G) –n(G) +ω(G), where
ω(G) is the number of connected components ofG. Wang and Wong [] obtained the following bounds of the matching number.
Theorem .([]) Let G be a graph of rank r(G).Then
r(G) –c(G)
≤m(G)≤
r(G) + c(G)
.
Sinceη(G) =n(G) –r(G), this theorem can be equivalently rewritten in the following form in terms of the nullity ofG.
Theorem . Let G be a graph with nullityη(G).Then
n(G) – m(G) –c(G)≤η(G)≤n(G) – m(G) + c(G).
Recently, Songet al.[] characterized all graphs attaining the upper bound, that is, the graphs with nullityη(G) =n(G) – m(G) + c(G).
Theorem .([]) For any graph G,η(G) =n(G) – m(G) + c(G)if and only if the fol-lowing three conditions are all satisfied:
() All cycles ofGare pairwise vertex-disjoint. () The length of each cycle ofGis a multiple of.
() m(TG) =m(TG–WG),whereTGis obtained fromGby contracting each cycle into a vertex,andWGconsists of new vertices obtained by contracting each cycle ofG.
Motivated by the works mentioned, in this paper, we characterize all graphs Gwith nullityη(G) =n(G) – m(G) –c(G). The main result is as follows.
Theorem . For any graph G,η(G) =n(G) – m(G) –c(G)if and only if the following three conditions are all satisfied:
(a) Cycles ofG(if any)are pairwise vertex-disjoint. (b) Each cycle ofG(if any)is odd cycle.
(c) m(TG) =m(TG–WG).
2 Preliminaries
LetGbe a graph with vertex setV(G). We denote byG–Uthe graph obtained fromGby removing the vertices inUtogether with all incident edges, whereU⊆V(G). Especially, whenU={x}, we writeG–U simply asG–x. Sometimes we use the notationG–G
instead ofG–V(G) whenGis an induced subgraph ofG. IfGis an induced subgraph
ofGandxis a vertex not inG, we write the subgraph ofGinduced byV(G)∪ {x}simply
asG+x.
LetCnandPnbe the cycle and path onnvertices, respectively. LetB(l,x,k) be the graph
shown in Figure , which is obtained from two vertex-disjoint cyclesClandCkby
Figure 1 B(l,x,k) andθ(l,x,k).
x= , it means that we identifyv andvx). Letθ(l,x,k) be the graph shown in Figure ,
which is the union of three internally disjoint pathsPl+,Px+, andPk+of lengthsl,x,k,
respectively, with common end vertices, wherel,x,k≥, and at most one of them is . The degree of a vertexvinG, denoted bydG(v), is the number of edges incident to the
vertex. A vertex is called a pendant vertex if its degree is . A lollipop is a graph obtained by identifying a vertex on a cycle and an end vertex on a path. The following is a list of known results about nullity that we will need in the proofs.
Lemma .([]) Let G=G∪G∪ · · · ∪Gω,where G,G, . . . ,Gωare connected
compo-nents of G.Then
η(G) =
ω
i= η(Gi).
Lemma .([]) Let G be a graph containing a pendant vertex,and let H be the induced subgraph of G obtained by deleting the pendant vertex together with the vertex adjacent to it.Thenη(G) =η(H).
Lemma .([])
η(Pn) =
, n is odd,
, n is even; η(Cn) =
, n is a multiple of,
otherwise.
Lemma .([]) For any vertex x of graph G,η(G) – ≤η(G–x)≤η(G) + .
Lemma .([]) Let x be a cut-vertex of a graph G,and G be a component of G–x.If η(G) =η(G+x) – ,thenη(G) =η(G) +η(G–G).
Lemma .([]) Let G be a graph with at least one cycle.Then the cycles of G are pairwise vertex-disjoint if and only if for any vertex u of G that lies on a cycle,c(G–u) =c(G) – .
Lemma .([]) Let u be a pendant vertex of a graph G,and v be the vertex adjacent to u.Denote G=G–{u,v}.Then,m(TG) =m(TG–WG)if and only if v does not lie on a
cycle and m(TG) =m(TG–WG).
Lemma .([]) For every acyclic graph F with at least one vertex,
η(F) =n(F) – m(F).
equal number of vertices and edges). Guoet al.[] characterized all unicyclic graphs that satisfy the minimal nullity condition.
Lemma .([]) Let G be a unicyclic graph,and let Clbe the unique cycle of G of length l.
Thenη(G) =n(G) – m(G) – if and only if l is odd and m(G) =l–
+m(G–Cl).
The next lemma follows immediately from Lemma ..
Lemma . Among all cycles,only odd cycles satisfy the minimal nullity condition.
Lemma . Let u be a pendant vertex of a graph G and v be the vertex adjacent to u. Denote G–{u,v}by G.Then,G satisfies the minimal nullity condition if and only if v does
not lie on a cycle of G and Gsatisfies the minimal nullity condition.
Proof By Lemma ., we know thatη(G) =η(G). Clearly,m(G) =m(G) – .
‘⇐’ Sincevdoes not lie on a cycle ofG, we havec(G) =c(G). Moreover,Gsatisfies the
minimal nullity condition, then we have
η(G) =η(G) =n(G) – m(G) –c(G)
=n(G) – – m(G) – –c(G)
=n(G) – m(G) –c(G),
which means thatGsatisfies the minimal nullity condition.
‘⇒’ Ifvlies on a cycle, then we havec(G)≤c(G) – . SinceGsatisfies the minimal nullity
condition, we have
η(G) =η(G) =n(G) – m(G) –c(G)
≤n(G) + –
m(G) + –
c(G) +
=n(G) – m(G) –c(G) – ,
which is contradict withη(G)≥n(G) – m(G) –c(G). Therefore,vdoes not lie on any
cycle ofG, thenc(G) =c(G). We have
η(G) =η(G) =n(G) – m(G) –c(G)
=n(G) + –
m(G) + –c(G)
=n(G) – m(G) –c(G)≤η(G),
which means thatGsatisfies the minimal nullity condition.
Lemma . Let G be a graph with at least one cycle. Suppose that all cycles of G are pairwise vertex-disjoint and each cycle of G is odd.Then m(TG) =m(TG–WG)if and only
if m(G) =C∈Gm(C) +m(G–C∈GC).
Proof Let C, . . . ,Cc(G) be all cycles of G. Suppose that G–
c(G)
i= Ci has q connected
components, sayH, . . . ,Hq. ThenH, . . . ,Hq are also all components ofTG–WG since
TG–WG=G–
Fork= , . . . ,q, we can obtain the subgraphH˜kinTGinduced byV(Hk) and its all
neigh-bor vertices. InG, we also can obtain the subgraphHˆkinduced byV(Hk) and its all
neigh-bor vertices. There is at most one edge betweenV(Hk) andV(Ci) sinceHk is connected
and the cycles ofGare pairwise vertex-disjoint. Therefore, for eachk, the graphsH˜kand
ˆ
Hkare isomorphic.
‘⇒’ Since m(TG) =m(TG –WG) and m(TG –WG) =
q
k=m(Hk), we have m(TG) =
q
k=m(Hk). LetMT be a maximum matching ofTG. Then|MT|=
q
k=m(Hk), which
implies thatm(Hk) =m(H˜k) for k= , . . . ,q. Otherwise, there isk∈ {, . . . ,q} such that
m(Hk) <m(H˜k). Then we can easily get the matchingMTofTGobtained fromMTby
re-placing the maximum matching ofHkby the maximum matching ofH˜k. Then we have
MT=m(H˜k) +
k=k
m(Hk)
>m(Hk) +
k=k
m(Hk)
=|MT|,
which leads to a contradiction. Thus, m(Hk) =m(Hˆk),k= , . . . ,q, sinceH˜k andHˆk are
isomorphic.
LetMbe a maximum matching ofG. Then we have
|M|=
c(G)
i=
M∩E(Ci)+ q
k=
M∩E(Hˆk)
≤
c(G)
i=
m(Ci) + q
k=
m(Hˆk).
On the other hand, |M| = m(G) ≥ c(G)i= m(Ci) +
q
k=m(Hk). Since
q
k=m(Hk) =
q
k=m(Hˆk), we then have
|M|=
c(G)
i=
m(Ci) + q
k=
m(Hk)
=
c(G)
i=
m(Ci) +m
G– c(G) i= Ci .
‘⇐’ From m(G) = C∈Gm(C) + m(G – C∈GC) we have m(G) = c(G)i= m(Ci) +
q
k=m(Hk). LetMbe a maximum matching ofG. Then|M|=
c(G) i= m(Ci) +
q
k=m(Hk).
Assume to the contrary thatm(TG)=m(TG–WG). Thenm(TG) >m(TG–WG). Since
m(TG–WG) =qk=m(Hk), we have m(TG) >qk=m(Hk). It means that there is k∈
{, . . . ,q}such thatm(H˜k) >m(Hk). SinceH˜kandHˆkare isomorphic, we havem(Hˆk) =
m(H˜k) >m(Hk).
Since there is at most one edge betweenV(Hk) andV(Ci) (i= , . . . ,c(G)), without loss
of generality, we can assume that fori= , . . . ,x, there is exactly one edge betweenV(Hk)
andV(Ci). Letvi∈V(Ci) (i= , . . . ,x) be the unique vertex adjacent toHk. Then we have
by the maximum matching ofHˆkand by adjusting the maximum matching of eachCito
the one without covering the vertexvi,i= , . . . ,x. Then,
M=m(Hˆk) +
k=k
m(Hk) + x
i=
m(Ci) + c(G)
i=x+
m(Ci)
>m(Hk) +
k=k
m(Hk) + c(G)
i=
m(Ci) (∵eachCiis an odd cycle)
=
q
k=
m(Hk) + c(G)
i=
m(Ci)
=|M|,
which leads to a contradiction.
3 Main results
Proposition . Let G be a connected graph without pendant vertices,and c(G)≥.Then
G does not satisfy the minimal nullity condition.
Proof Sincec(G)≥, there are at least two different cycles inG. We distinguish the fol-lowing two cases.
Case. There exists a vertexuon a cycle such thatc(G–u)≤c(G) – . IfGsatisfies the minimal nullity condition, then we have
η(G) =n(G) – m(G) –c(G)
≤n(G–u) + – m(G–u) –c(G–u) + ∵m(G–u)≤m(G)
≤n(G–u) – m(G–u) –c(G–u) –
≤η(G–u) – (by Theorem .).
By Lemma . we know thatη(G)≥η(G–u) – , a contradiction. Case. For any vertexuon a cycle,c(G–u)≥c(G) – . We first prove the following claim.
Claim Let G be a connected graph without pendant vertices.Suppose that for any vertex
u on a cycle of G,c(G–u)≥c(G) – (c(G)≥).Then there are at most c(G) – vertices of
G not covered by its maximum matching.
In fact, if a vertexulies on a cycle ofG, thenc(G–u)≤c(G) – . We distinguish the following two cases.
Casea. For any vertexuthat lies on a cycle ofG,c(G–u) =c(G) – .
We assume that the claim holds for a connected graphGwithout pendant vertices and c(G) <land all cycles ofGare pairwise vertex-disjoint. Then we consider the casec(G) =l (l> ).
Since all cycles ofGare pairwise vertex-disjoint, there must exist a subgraphGthat is
a lollipop graph such thatG–Gis a graph without pendant vertices. The cycles ofG–G
are also pairwise vertex-disjoint since it is a subgraph ofG. Clearly,c(G–G) =c(G) – =
l– <l. By the induction assumption there are at mostc(G–G) – vertices ofG–Gnot
covered by the maximum matching ofG–G. It is easy to see that in the lollipop graphG,
there is at most one vertex not covered by the maximum matching ofG. Then, there are
at mostc(G–G) – + =c(G) – vertices ofGnot covered by its maximum matching.
Caseb. There is a vertexuthat lies on a cycle ofGsuch thatc(G–u) =c(G) – . Similarly to Case a, by induction on c(G), whenc(G) = ,GbecomesB(l, ,k) orθ(l,x,k). Since B(l, ,k) andθ(l,x,k) have a path as their spanning subgraph, there is at most =c(G) – vertex not covered by their maximum matchings.
We assume that the claim holds for a connected graphGwithout pendant vertices and c(G) <land there is a vertexuthat lies on a cycle ofGsuch thatc(G–u) =c(G) – . It is suffices to prove the claim in the casec(G) =l(l> ).
Suppose thatG–uhaspconnected components, sayH,H, . . . ,Hp. Letn(Hi) ande(Hi)
be the number of vertices and size ofHi,i= , , . . . ,p, respectively. Obviously,
n(G) =
p
i=
n(Hi) + , e(G) = p
i=
e(Hi) +dG(u),
c(Hi) =e(Hi) –n(Hi) + , c(G) =e(G) –n(G) + ,
c(G–u) =
p
i=
c(Hi).
For eachHi, we havec(Hi) <c(G),i= , . . . ,p. Moreover, we havedG(u) =p+ . In fact,
dG(u) =e(G) – p
i=
e(Hi)
=c(G) +n(G) – –
p
i=
e(Hi)
=c(G–u) + +
p
i=
n(Hi) +
– –
p
i=
e(Hi)
=
p
i=
c(Hi) +
+
p
i=
n(Hi) – p
i=
e(Hi)
=
p
i=
e(Hi) –n(Hi) + + + p
i=
n(Hi) – p
i=
e(Hi)
=
p
i=
e(Hi) – p
i=
n(Hi) +p+ + p
i=
n(Hi) – p
i=
e(Hi)
Figure 2 GandG.
Thus, by the incident relation betweenuandHi,Gmust be isomorphic to one of the two
graphs shown in Figure .
Now, ifGis the graphG shown in Figure , then we distinguish the following two cases. Caseb.. Each componentHihas no pendant vertices,i= , , . . . ,p.
Now assume that fori= , . . . ,s,c(Hi) = and fori=s+ , . . . ,p,c(Hi)≥. Thenc(G–u) =
p
i=c(Hi) =s+c(Hs+) +· · ·+c(Hp). Wheni∈ {, . . . ,s}, sincec(Hi) = ,Hiis a cycle and
there is at most one vertex not covered by a maximum matching of Hi; wheni∈ {s+
, . . . ,p}, sincec(Hi)≥, by the induction assumption there are at mostc(Hi) – vertices
not covered by a maximum matching ofHi. LetVbe a set of the vertices that are not be
covered by a maximum matching ofG. Then
|V| ≤ + · · ·+ s
+c(Hs+) – +· · ·+
c(Hp) – +{u}
=s+c(Hs+) +· · ·+c(Hp) – (p–s) +
=c(G) – (p–s) – ∵c(G) =c(G–u) + =s+c(Hs+) +· · ·+c(Hp) +
≤c(G) – (∵p≥s).
Thus, there are at mostc(G) – vertices not overed by a maximum matching ofG. Caseb.. There exists a component, sayHi(i∈ {, , . . . ,p}) that has a pendant vertex.
Letv be the pendant vertex ofHi, andPvv···vxvx+ be a path inHisuch thatvis its
end vertex anddG(v) =· · ·=dG(vx) = ,dG(vx+) > . Clearly,vis an adjacent vertex ofu.
Wheni= , by removing the pathPvv···vx fromGthe resulting graphG–Pvv···vx is a graph without pendant vertex withc(G–Pvv···vx) <c(G). By the induction assumption there are at mostc(G–Pvv···vx) – vertices not covered by a maximum matching ofG–
Pvv···vx. Recall that there is at most one vertex not covered by a maximum matching of the pathPvv···vx. Therefore, there are at mostc(G–Pvv···vx) – + ≤c(G) – vertices not covered by a maximum matching ofG.
When i∈ {, . . . ,p}, by removing the path Pvv···vx from Gthe resulting graph G–
Pvv···vx is a graph without pendant vertex, whereasc(G–Pvv···vx) =c(G). It is obvious thatG–Pvv···vxhas exactly two components, sayGandG, both without pendant ver-tices and withc(Gi) <c(G),i= , .
Ifc(Gi)≥,i= , , by the induction assumption there are at mostc(Gi) – vertices not
covered by a maximum matching ofPvv···vx, there are at mostc(G) – +c(G) – + =
c(G) – vertices not covered by a maximum matching ofG.
If there is one ofGi, sayG, withc(G) = , thenPvv···vx+G (the subgraph induced byv,v, . . . ,vxandV(G)) is a lollipop graph. Therefore, there is at most one vertex not
covered by a maximum matching ofPvv···vx+G. Sincec(G)≥, we havec(G)≥. By
the induction assumption there are at mostc(G) – vertices not covered by a maximum
matching ofG. Thus, there are at mostc(G) – + =c(G) – vertices not covered by a
maximum matching ofG.
In a similar way, we can prove the claim whenGis the graphG shown in Figure . By the previous claim, ifGsatisfies the minimal nullity condition, then we have
η(G) =n(G) – m(G) –c(G)
≤c(G) – –c(G) = –,
a contradiction. This completes the proof of the lemma.
By Proposition . we have that if a connected graphGwithc(G)≥ satisfies the mini-mal nullity condition, thenGmust have a pendant vertex.
We are now ready to prove Theorem ..
Proof It suffices to prove the theorem for the case whereGis connected.
‘⇒’ LetGbe a connected graph that satisfies the minimal nullity condition. Whenc(G) = ,Gis a tree. Then (a), (b), and (c) hold trivially.
Whenc(G) = ,Gis a unicyclic graph. LetClbe the unique cycle ofGof lengthl. By
Lemma .,Clis an odd cycle, andm(G) = l– +m(G–Cl). Thus, (a) and (b) hold obviously.
Moreover, sincem(Cl) =l– andm(G–Cl) =m(TG–vCl), we havem(G) =m(Cl) +m(TG–
vCl). By Lemma .,m(TG) =m(TG–vCl). Thus, (c) holds. It remains to prove (a), (b), and (c) forGwhenc(G)≥.
Assume that Gis a connected graph withc(G)≥ that satisfies the minimal nullity condition. By Proposition .,Gmust have a pendant vertex. Letube a pendant vertex ofG. It follows from Lemma . that the adjacent vertexvofudoes not lie on a cycle ofG, and by removing these two verticesuandvthe resulting graphGalso satisfies the
minimal nullity condition. Clearly,c(G) =c(G)≥.
LetH,H, . . . ,Hpbe the connected components ofG. Then eachHi,i= , . . . ,p, satisfies
the minimal nullity condition. Without loss of generality, we assume that fori∈ {, . . . ,s}, c(Hi)≤ and fori∈ {s+ , . . . ,p},c(Hi)≥. By Proposition ., fori∈ {s+ , . . . ,p}, eachHi
must have a pendant vertex, and by Lemma . the adjacent vertex of each pendant vertex does not lie on a cycle ofHi. By removing a pendant vertex ofHiand its adjacent vertex the
resulting graph also satisfies the minimal nullity condition. For a connected component withc≥ that satisfies the minimal nullity condition, there must be a pendant vertex, and its adjacent vertex does not lie on a cycle. Then we can repeatedly remove a pendant vertex ofHi,i∈ {s+ , . . . ,p}, and its adjacent vertex until all connected components of the
and keepsc(G∗) =c(G). It follows from Lemma . thatG∗is the disjoint union ofc(G) odd cycles, sayC,C, . . . ,Cc(G), and some isolated vertices. Therefore, the cycles ofGare
pairwise vertex-disjoint, and each cycle is odd. Thus, (a) and (b) hold.
On the other hand, since in each step we just remove a pendant vertex and its adjacent vertex, we havem(G) =c(G)i= m(Ci) +m(G–
c(G)
i= V(Ci)). Thus, by Lemma . we have
m(TG) =m(TG–WG), and thus (c) holds.
‘⇐’ LetGbe a connected graph satisfying (a), (b), and (c). We deal with the sufficient part of the proof by induction onc(G). Whenc(G) = ,Gis a tree. Then, by Lemma ., the result is trivial. Whenc(G) = ,Gis a unicyclic graph. LetClbe the unique cycle of
Gof lengthl. ThenClis an odd cycle, andm(TG) =m(TG–vCl). By Lemma . we have
m(G) =m(Cl) +m(TG–vCl) = l–
+m(G–Cl). Furthermore, by Lemma .,η(G) =n(G) –
m(G) –c(G) in this case.
We assume that the result holds whenc(G) <l. Now we consider the casec(G) =l(l≥). Since cycles ofGare pairwise vertex-disjoint, there exists a cycleCinGsuch that all cycles ofG–Clie in the same component, sayH, ofG–C, and the other components
ofG–C, sayHi(i= , . . . ,p), are trees. Moreover,n(Hi)≥ (i= , . . . ,p). If not, there is
i∈ {, . . . ,p}such thatn(Hi) = , which means that there is a pendant vertex ofGsuch
that its adjacent vertex lies onC. But this contradicts Lemma . because (c) holds forG if and only if the adjacent vertex of any pendant vertex does not lie on a cycle.
Ifp≥, then letube a pendant vertex ofGthat belongs topi=V(Hi), and letvbe
its adjacent vertex. Denote by G the graph obtained fromGby removing uandv. By
Lemma .,vdoes not lie on a cycle ofG, and (c) also holds forG. Repeating this
opera-tion, after a finite number of steps, we can obtain a graphG∗that is a disjoint union of a connected componentGinduced byV(C)∪V(H) and some acyclic graphs. Obviously,
(c) also holds forG∗.
SinceGis a connected graph and all cycles are pairwise vertex-disjoint, there is exactly one edge betweenCandH inG∗. Letxyis the unique edge betweenCandH, where
x∈V(C) andy∈V(H). Note that (c) holds for the graphG∗if and only if (c) also holds for
its each connected component. Therefore, (c) holds forG, that is,m(TG) =m(TG–WG).
Certainly, (a) and (b) hold forGsince it is a subgraph ofG. Moreover, bothH+xandH
also satisfy (a) and (b) since they are subgraphs ofG. Now let us show that (c) also holds
forH+xandH, that is,
() m(TH+x) =m(TH+x–WH+x).
In fact, We havem(TG) =m(TH+x) sinceTG∼=TH+x, andWG=WH+x∪ {vC}, where
vC is the vertex inWG contracted from the cycleC. Sincem(TG) =m(TG–WG), we
have
m(TH+x) =m
TH+x–
WH+x∪ {vC}
=m(TH+x–WH+x–vC)
≤m(TH+x–WH+x).
On the other hand, it is clear thatm(TH+x)≥m(TH+x–WH+x). Therefore,m(TH+x) =
m(TH+x–WH+x).
In fact,TGcan be regarded as the graph induced byV(TH) and the vertexvC. Therefore,
TG=TH+vCandWG=WH∪ {vC}. Since (c) holds forG, we havem(TG) =m(TG–
WG). Thus,
m(TH+vC) =m
TH+vC–
WH∪ {vC}
=m(TH–WH)
≤m(TH).
Clearly,m(TH+vC)≥m(TH). Then we havem(TH+vC) =m(TH). The previous
deriva-tion forces the last inequality involved to become an equality. Thus,m(TH) =m(TH–
WH).
Sincec(H+x) <c(G) =c(G) (respectively,c(H) <c(G) =c(G)), by the induction
as-sumption,H+x(respectively,H) satisfies the minimal nullity condition, that is,η(H+
x) =n(H+x) – m(H+x) –c(H+x) andη(H) =n(H) – m(H) –c(H).
It follows from () and () that m(TH+x) =m(TG) =m(TH +vC) =m(TH), that is,
m(TH+x) = m(TH). Since m(TH+x) =m(TH+x–WH+x) and m(TH) =m(TH –WH),
by Lemma . we have m(H +x) =
C∈H+xm(C) +m(TH+x –WH+x) and m(H) =
C∈Hm(C) +m(TH–WH). Then,
m(H+x) =
C∈H+x
m(C) +m(TH+x–WH+x)
=
C∈H+x
m(C) +m(TH+x)
=
C∈H
m(C) +m(TH)
=
C∈H
m(C) +m(TH–WH)
=m(H),
that is, m(H+x) =m(H). Furthermore, since bothH+xandH satisfy the minimal
nullity condition, we have
η(H+x) =n(H+x) – m(H+x) –c(H+x)
=n(H) + – m(H) –c(H)
=η(H) + .
Then by Lemma . we have
η(G) =η(H) +η(G–H)
=n(H) – m(H) –c(H) +η(C)
=n(G) –n(C) –
m(G) –m(C) –c(G) – +
that is,Gsatisfies the minimal nullity condition. On the other hand, each acyclic graph
satisfies the minimal nullity condition. ThenG∗also satisfies the minimal nullity condition. It follows from Lemma . thatGsatisfies the minimal nullity condition sinceG∗does.
The proof is completed.
Proposition . Let G be a graph with at least one cycle.If G satisfies the minimal nullity condition,then for any vertex u that lies on a cycle of G,we have that G–u also satisfies the minimal nullity condition andη(G–u) =η(G).
Proof It suffices to prove the result for the case whereGis connected. LetGbe a
con-nected graph satisfying the minimal nullity condition. By Theorem . all cycles ofG, say C,C, . . . ,Cc(G), are pairwise vertex-disjoint, and eachCi(i= , . . . ,c(G)) is an odd cycle.
Letube any vertex that lies on a cycle ofG, sayCi (i∈ {, . . . ,c(G)}). SinceGsatisfies
the minimal nullity condition, by Theorem . we havem(TG) =m(TG–WG). Then by
Lemma . we have
m(G) =
c(G)
i=
m(Ci) +m
G–
c(G)
i=
Ci
=
i=i
m(Ci) +m(Ci) +m
(G–u) –
i=i
Ci– (Ci–u)
=
i=i
m(Ci) +m(Ci) +m
(G–u) –
i=i
Ci
–m(Ci–u)
=
i=i
m(Ci) +m
(G–u) –
i=i
Ci
=
CiinG–u
m(Ci) +m
(G–u) –
CiinG–u
Ci
≤m(G–u).
However,m(G)≥m(G–u). Thus, we havem(G) =m(G–u). Furthermore,
m(G–u) =
CiinG–u
m(Ci) +m
(G–u) –
CiinG–u
Ci
.
By Lemma . we havem(TG–u) =m(TG–u–WG–u), that is,G–usatisfies (c) of
Theo-rem .. Moreover,G–ualso satisfies both (a) and (b) of Theorem . since G–uis a subgraph ofG. SoG–usatisfies the minimal nullity condition. Then we have
η(G) =n(G) – m(G) –c(G)
=n(G–u) + – m(G–u) –c(G–u) + ∵m(G–u) =m(G)
=n(G–u) – m(G–u) –c(G–u)
=η(G–u),
4 Conclusions
All graphsGwithη(G) =n(G) – m(G) –c(G) are completely characterized. This means that we can easily obtain the nullity η(G) of each of those graphs from its order n(G), matching number m(G), and elementary cyclic number c(G), and, thus, we can know whether the graph is singular or not.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
SR carried out the proofs of the main results in the manuscript. AC and YZ participated in the design of the study and drafted the manuscripts. All the authors read and approved the final manuscripts.
Author details
1Center for Discrete Mathematics, Fuzhou University, Fuzhou, Fujian, P.R. China.2School of Applied Mathematics, Xiamen
University of Technology, Xiamen, 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. 11331003, 11471077), China.
Received: 12 November 2015 Accepted: 3 February 2016
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