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A characterization of graphs G with nullity |V(G)|−2m(G)+2c(G)
Linear Algebra and its Applications 465 (2015) 363–375
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Linear Algebra and its Applications www.elsevier.com/locate/laa
A characterization of graphs G with nullity |V (G)| − 2m(G) + 2c(G) Ya-zhi Song a , Xiao-qiu Song a,∗,1 , Bit-Shun Tam b,2 a
Department of Mathematics, China University of Mining and Technology, Xuzhou 221116, China b Department of Mathematics, Tamkang University, New Taipei City 25137, Taiwan, ROC
a r t i c l e
i n f o
Article history: Received 29 January 2014 Accepted 23 September 2014 Available online xxxx Submitted by R. Brualdi MSC: 05C50 Keywords: Nullity Rank Matching number
a b s t r a c t By |V (G)|, |E(G)|, η(G), and m(G) we denote respectively the order, the number of edges, the nullity, and the matching number of a (simple) graph G. Recently Wang and Wong have proved that for every graph G, η(G) ≤ |V (G)| − 2m(G) + 2c(G), where c(G) = |E(G)| − |V (G)| + θ(G), θ(G) being the number of connected components of G. In this paper graphs G that satisfy the equality η(G) = |V (G)| − 2m(G) + 2c(G) are characterized. © 2014 Elsevier Inc. All rights reserved.
1. Introduction The nullity η(G) of a (simple) graph G is the (algebraic) multiplicity of the eigenvalue 0 in the spectrum of the adjacency matrix A(G) of G. There has been much interest in the nullity of graphs in chemistry. According to the Hückel molecular orbital theory, if * Corresponding author. 1 2
E-mail addresses: [email protected] (X.-q. Song), [email protected] (B.-S. Tam). Supported by National Natural Science Foundation of China (No. 11171343). Supported by National Science Council of the Republic of China (NSC 102-2115-M-032-002).
http://dx.doi.org/10.1016/j.laa.2014.09.034 0024-3795/© 2014 Elsevier Inc. All rights reserved.
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the nullity η(G) of a molecular graph G is positive, then the corresponding chemical compound is highly reactive and unstable, or nonexistent (see, for instance, [3]). Interest in the nullity of graphs has also gained attention among mathematicians. In particular, there has been much work on bounds for η(G). It is known that 0 ≤ η(G) ≤ n − 2 if G is a graph on n vertices and G is not isomorphic to nK1 . Cheng and Liu [1] have characterized the extremal graphs attaining the upper bound n − 2 and the second upper bound n − 3. Subsequently, this has led to the study of the nullity set (i.e., the set of nullities attained by graphs) of various classes of graphs (see [13,11,2,7,12,10]). The present work also involves an upper bound for η(G), but in a different way. Recently, Wang and Wong [14] obtained the following bounds for the matching number of a graph G: r(G)−c(G) ≤ m(G) ≤ r(G)+2c(G) , where r(G) and m(G) denote respectively 2 2 the rank and the matching number of G and c(G) = |E(G)| − |V (G)| + θ(G), θ(G) being the number of (connected) components of G. They also showed that their bounds are best possible by providing examples. The upper bound for the matching number can be rewritten in an equivalent form as an upper bound for the nullity of G: Lemma 1.1. For any graph G, η(G) ≤ |V (G)| − 2m(G) + 2c(G). The purpose of this work is to characterize graphs G that satisfy η(G) = |V (G)| − 2m(G) + 2c(G). For convenience, hereafter in this paper we say a graph G satisfies the maximal nullity condition if η(G) = |V (G)| − 2m(G) + 2c(G). It should not be mixed up with graphs G of order n that satisfy η(G) = n − 2. In view of the following known result, every acyclic graph satisfies the maximal nullity condition: Lemma 1.2. (See [6].) For every acyclic graph T with at least one vertex, η(T ) = |V (T )| − 2m(T ). By the recent work of Guo, Yan and Yeh [9], unicyclic graphs that satisfy the maximal nullity condition are also characterized. The following result is implicit in [9]: Lemma 1.3. A unicyclic graph G has nullity |V (G)| − 2m(G) + 2 if and only if the length of the cycle of G is divisible by 4 and M ∩ E1 = ∅ for any maximum matching M of G, where E1 denotes the set of edges of G with one end vertex in the cycle and one end vertex outside the cycle. In order to state our main result, we need to define an acyclic graph TG for a graph G whose cycles are pairwise vertex-disjoint. Let G be a graph with pairwise vertex-disjoint cycles. The acyclic graph TG is obtained from G by contracting cycles. More specifically, the vertex set V (TG ) is taken to be U ∪ WG , where U consists of all vertices of G that do not lie on a cycle and WG = {vC : C is a cycle of G}. Two vertices in U are adjacent in TG if and only if they are adjacent in G, a vertex u ∈ U is adjacent to a vertex vC ∈ W if and only if u is adjacent
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(in G) to a vertex in the cycle C, and vertices vC1 , vC2 are adjacent in TG if and only if there exists an edge in G joining a vertex of C1 to a vertex of C2 . It is clear that TG is always acyclic and is a tree when G is connected. Since the cycles of G are pairwise vertex-disjoint, deleting edges from G, one from each cycle, will result in a forest F and as c(F ) = 0 it follows that G has altogether c(G) cycles. Note that the graph TG − WG (obtained from TG by deleting vertices in WG and the incident edges) is the same as the graph obtained from G by deleting all cycles and the incident edges. Below is our main result: Theorem 1.4. For any graph G, η(G) = |V (G)| − 2m(G) + 2c(G) if and only if the following three conditions are all satisfied: (a) Distinct cycles of G (if any) are vertex-disjoint; (b) The length of each cycle of G (if any) is a multiple of 4; (c) m(TG ) = m(TG − WG ) or, equivalently, there exists a maximum matching of TG that does not cover any vertex in WG . To illustrate Theorem 1.4, let us adapt the example given at the end of the proof of [14, Theorem 1.1]. Let S be a star K1,p+1 with vertices x, y1 , . . . , yp+1 (p ≥ 1), where x is the star center. Let G be the graph obtained from S by replacing each pendant vertex yi , i = 1, 2, . . . , p, by a cycle Cli of length li , where each li is a multiple of 4. Then the cycles of G are pairwise vertex-disjoint and are of lengths multiples of 4. Furthermore, TG is a K1,p+1 and TG − WG is a P2 ; so m(TG ) = 1 = m(TG − WG ). By Theorem 1.4, p p η(G) = (2 + i=1 li ) − 2(1 + i=1 l2i ) + 2p = 2p. Notice that when G is unicyclic, Theorem 1.4 does not reduce to Lemma 1.3. For the purpose of proof, we will need the following: Lemma 1.5. Let G be an even-unicyclic graph. The following conditions are equivalent: (a) Every maximum matching of G does not have an edge with one end vertex in the cycle of G and the other end vertex outside the cycle. (b) m(TG ) = m(TG − WG ). In view of Lemmas 1.3 and 1.5, one may suspect that in Theorem 1.4 condition (iii) can be replaced by the condition “every maximum matching of G does not have an edge with one end vertex on a cycle and the other end vertex outside the (same) cycle”. It turns out that the answer is “no”. For instance, let G be the graph obtained from the path P4 by attaching two C4 , one at each end of the path. (So G has 10 vertices.) The two cycles of G are each of even length and they are vertex-disjoint. It is readily seen that m(G) = 5 and G has exactly one maximum matching and this matching has no edge with one end vertex on a cycle and the other end vertex outside the cycle. However,
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m(TG ) = 2 = 1 = m(TG − WG ). Indeed, even the weaker condition “m(TG ) = m(TG − v) for every v ∈ WG ” is not fulfilled. Nevertheless, we have the following positive result: Lemma 1.6. Let G be a graph with at least one cycle. Suppose that each cycle of G is of even length and the cycles of G are pairwise vertex-disjoint. If m(TG ) = m(TG − WG ) then every maximum matching of G does not contain an edge with one end vertex on a cycle and the other end vertex outside the cycle. For convenience, we will say G satisfies the matching condition if every maximum matching of G does not have an edge with one end vertex on a cycle and the other end vertex outside the (same) cycle. We would like to mention that the graph TG has appeared in [14] when the lower bound for the matching number m(G) – not the upper bound, which is equivalent to the inequality given in Lemma 1.1 – is being established, but its use there is inessential. However, the graph TG plays an important role in this work. In Section 2 we give the necessary definitions and notations, together with a list of known results which we will need. In Section 3, we will give the proofs for Theorem 1.4, Lemma 1.5, Lemma 1.6 and other auxiliary results. 2. Notations and preliminaries By a graph we mean a finite simple undirected graph (i.e., one with a finite vertex set and without loops or multiple edges). The rank of a graph G, denoted by r(G), is defined to be the rank of its adjacency matrix. For any U ⊆ V (G), we denote by G − U the graph obtained from G by removing the vertices in U together with all incident edges; when U = {x}, we write G − U simply as G − x. Sometimes we use the notation G − G1 instead of G − V (G1 ) when G1 is a (vertex-)induced subgraph of G. If G1 is an induced subgraph of G and x is a vertex not in G1 , we write the subgraph of G induced by V (G1 ) ∪ {x} simply as G1 + x. A vertex x is called a pendant vertex if its degree is 1. An induced subgraph H of a graph G is called a pendant star of G if H is a star with at least two vertices such that the star center of S is the only vertex of S that is adjacent to some vertex not in S. A matching in a graph is a set of pairwise non-adjacent edges. A vertex is covered by a matching if it is an end vertex of one of the edges in the matching. A maximum matching is a matching that contains the largest possible number of edges. The matching number of a graph G, denoted by m(G), is the size of a maximum matching. We denote a cycle (respectively, a path) on n vertices by Cn (respectively, Pn ). The following is a list of known results that we will need in the proofs.
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Lemma 2.1. (See [6].) η(Pn ) equals 1 if n is odd and equals 0 if n is even. η(Cn ) equals 2 if n is a multiple of 4 and equals 0, otherwise. Lemma 2.2. (See [4].) If x is a vertex of a graph G, then η(G) − 1 ≤ η(G − x) ≤ η(G) + 1. Lemma 2.3. (See [5].) Let G be a graph containing a pendant vertex, and let H be the graph obtained from G by deleting the pendant vertex and the vertex adjacent to it. Then η(G) = η(H). Lemma 2.4. (See [8].) Let x be a cut-vertex of a graph G. If there exists a component of G − x, say G1 , such that η(G1 ) = η(G1 + x) + 1, then η(G) = η(G − x) − 1. Lemma 2.5. (See [8].) Let x be a cut-vertex of a graph G and let G1 be a component of G − x. If η(G1 ) = η(G1 + x) − 1, then η(G) = η(G1 ) + η(G − G1 ). Lemma 2.6. (See [9].) If S is a pendant star of a graph G, then m(G) = 1 + m(G − S). Furthermore, G has a maximum matching which consists of edges in a maximum matching for G − S together with an edge of S. 3. Proofs of results Lemma 3.1. For a graph G with pairwise vertex-disjoint cycles, the following conditions are equivalent: (a) There exists a maximum matching of TG that does not cover any vertex in WG . (b) m(TG ) = m(TG − WG ). (c) η(TG ) = η(TG − WG ) + c(G). Proof. We always have the inequality m(TG − WG ) ≤ m(TG ) (as TG − WG is an induced subgraph of TG ), with equality if and only if TG has a maximum matching that is also a matching for TG − WG . But a matching for TG − WG is the same as a matching for TG that does not cover any vertices in WG , so conditions (a) and (b) are equivalent. Since TG and TG − WG are acyclic, by Lemma 1.2 we have η(TG ) = V (TG ) − 2m(TG ) = V (TG − WG ) + c(G) − 2m(TG ) and η(TG − WG ) = V (TG − WG ) − 2m(TG − WG ); hence, η(TG ) − η(TG − WG ) = c(G) − 2 m(TG ) − m(TG − WG ) ≤ c(G).
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Thus, we always have η(TG ) ≤ η(TG − WG ) + c(G), where the equality holds if and only if m(TG ) = m(TG − WG ). This establishes the equivalence of conditions (b) and (c). 2 Lemma 3.2. 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) −1. Proof. We may assume that the graph G is connected. “Only if” part: Let u be a vertex of G that lies on a cycle C, and suppose that G − u has p (≥ 1) components. Then C is the only cycle of G that contains vertex u, vertices in C − u all lie in the same component P of G − u, P has precisely two vertices adjacent (in G) to u, and each of the remaining p − 1 components has precisely one vertex adjacent to u. Hence, dG (u) = p + 1 and we have |E(G − u)| = |E(G)| − p − 1. As |V (G − u)| = |V (G)| − 1 and θ(G − u) = p, a little calculation yields c(G − u) = c(G) − 1. “If” part: Suppose that G has two distinct cycles C1 , C2 that have at least one common vertex. Choose a common vertex u such that the degree of u in the subgraph of G induced by V (C1 ) ∪ V (C2 ) is at least 3, and suppose that G − u has p components. Then either there is a component of G − u that has at least three vertices adjacent to u or there are at least two components, each of which has two vertices adjacent to u. In any case, we have |E(G − u)| ≤ |E(G) − p − 2| and a little calculation yields c(G − u) < c(G) − 1. 2 The argument given in the proof for Lemma 3.2 also yields the following: Remark 3.3. Let G be a graph and let u ∈ V (G). (i) If u lies on a cycle of G, then c(G − u) ≤ c(G) − 1, where the equality holds if and only if dG (u) = θ(G − u) + 1; (ii) If u does not lie on a cycle of G, then c(G − u) = c(G). In connection with Remark 3.3(i), we would like to point out that a sufficient, but not necessary, condition for the equality dG (u) = θ(G − u) + 1 to hold is that u lies on a unique cycle of G. In the proof of [14, Theorem 1.1] it is asserted (without proof) that distinct cycles of a graph G are vertex-disjoint if and only if for every vertex x of G, c(G −x) ≥ c(G) −1. This assertion, which motivated Lemma 3.2, follows from the lemma itself and Remark 3.3(ii). Examining the proof for the inequality given in Lemma 1.1, we obtain the following result. Lemma 3.4. Let G be a graph that has at least one cycle. If G satisfies the maximal nullity condition, then for any vertex u that lies on a cycle of G, we have (a) η(G − u) = η(G) − 1; (b) G − u satisfies the maximal nullity condition;
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(c) m(G − u) = m(G) − 1; (d) c(G − u) = c(G) − 1; (e) dG (u) = θ(G − u) + 1. Proof. We may assume that G is connected. Suppose that G − u has p components. By the argument given in the proof for Lemma 3.2 we have |E(G − u)| ≤ |E(G)| − p − 1, and hence c(G − u) = E(G − u) − V (G − u) + p ≤ E(G) − p − 1 − V (G) − 1 + p = c(G) − 1. On the other hand, clearly m(G − u) ≥ m(G) − 1 and by Lemma 2.2 we have η(G − u) ≥ η(G) − 1. As G satisfies the maximal nullity condition, we obtain η(G − u) ≥ η(G) − 1 = V (G) − 2m(G) + 2c(G) − 1 ≥ V (G − u) + 1 − 2 m(G − u) + 1 + 2 c(G − u) + 1 − 1 = V (G − u) − 2m(G − u) + 2c(G − u) ≥ η(G − u), where the last inequality follows from Lemma 1.1. The above derivation forces each inequality involved to become an equality. Hence, conditions (a)–(d) all hold and, in addition, we have |E(G − u)| = |E(G)| − p − 1, which is equivalent to (e). 2 For possible future use, we take note of the following two observations, whose proofs we omit. Remark 3.5. Let G be a graph that satisfies the condition η(G) = |V (G)|−2m(G)+2c(G). Let u be a vertex of G that does not lie on a cycle. If m(G − u) = m(G), then conditions (a), (b) of Lemma 3.4 are fulfilled. Remark 3.6. A graph G satisfies the condition η(G) = |V (G)| − 2m(G) + 2c(G) if either one of the following two conditions holds: (1) There exists a vertex u that lies on a cycle of G such that conditions (a)–(d) of Lemma 3.4 are fulfilled. (2) There exists a vertex u that does not lie on a cycle of G such that m(G − u) = m(G) and conditions (a), (b) of Lemma 3.4 are fulfilled. Corollary 3.7. Let G be a graph with nullity |V (G)| −2m(G) +2c(G). Then distinct cycles of G (if any) are vertex-disjoint. Thus G has precisely c(G) cycles.
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Proof. Since G satisfies the maximal nullity condition, by Lemma 3.4 we have c(G −u) = c(G) − 1 for every vertex u that lies on a cycle of G. By Lemma 3.2 our assertion follows. 2 Lemma 3.8. Let u be a pendant vertex of a graph G and let v be the vertex adjacent to u. Denote G − {u, v} by G0 . Suppose that v does not lie on a cycle of G. Then: (i) G satisfies the maximal nullity condition if and only if G0 satisfies the maximal nullity condition. (ii) TG has a maximum matching that does not cover any vertex in WG if and only if TG0 has a maximum matching that does not cover any vertex in WG0 . Proof. (i) By Lemma 2.3 (respectively, Lemma 2.6) we obtain η(G0 ) = η(G) (respectively, m(G0 ) = m(G) −1). As v does not lie on a cycle, by Remark 3.3(ii), c(G0 ) = c(G). So we have η(G) − V (G) + 2m(G) − 2c(G) = η(G0 ) − V (G0 ) + 2 + 2 1 + m(G0 ) − 2c(G0 ) = η(G0 ) − V (G0 ) + 2m(G0 ) − 2c(G0 ). It follows that G0 satisfies the maximal nullity condition if and only if G satisfies the maximal nullity condition. (ii) Since G and G0 have the same set of cycles, TG is well-defined if and only if TG0 is well-defined, and also whenever they are both well-defined, WG = WG0 . Since u, v each does not lie on a cycle of G, u, v are both vertices of TG . Noting that TG0 = TG − {u, v}, by Lemma 2.6 we have m(TG ) = 1 + m(TG0 ). If TG has a maximum matching M that does not cover any vertex in WG , then necessarily M has an edge e that covers vertex v. In that case, M \ {e} is a maximum matching for TG0 that does not cover any vertex in WG0 . Conversely, suppose that TG0 has a maximum matching M0 that does not cover any vertex in WG0 . Then M0 ∪ {uv} is a maximum matching for TG that does not cover any vertex in WG . 2 Corollary 3.9. Let u be a pendant vertex of a graph G and let v be the vertex adjacent to u. Denote G − {u, v} by G0 . Then: (i) G satisfies the maximal nullity condition if and only if v does not lie on a cycle and G0 satisfies the maximal nullity condition. (ii) m(TG ) = m(TG − WG ) if and only if v does not lie on a cycle and m(TG0 ) = m(TG0 − WG0 ). Proof. (i) In view of Lemma 3.8(i), it remains to show that if G satisfies the maximal nullity condition, then v does not lie on a cycle. Assume to the contrary that v lies on a
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cycle. As G0 has one cycle fewer than G, c(G0 ) = c(G) −1. A little calculation shows that the maximal nullity condition for G leads to the condition η(G0 ) = |V (G0 )| − 2m(G0 ) + 2c(G0 ) + 2, which contradicts Lemma 1.1. (ii) By Lemma 3.8(ii) it suffices to show that if m(TG ) = m(TG − WG ) then v does not lie on a cycle. Assume to the contrary that v lies on a cycle C. Then TG − WG is an induced subgraph of TG − vC , and since TG − vC is also an induced subgraph of TG the condition m(TG ) = m(TG − WG ) forces m(TG ) = m(TG − vC ). On the other hand, we also have m(TG − vC ) = m(TG − {u, vC }) = m(TG ) − 1, where the second equality holds by Lemma 2.6. So we arrive at a contradiction. 2 Lemma 3.10. Given an even cycle C and a connected graph K that are vertex-disjoint, let G be obtained from C ∪ K by adding an edge with one end vertex in C and the other end vertex in K. Then m(G) = m(C) + m(K). Proof. It is clear that m(G) ≥ m(C) + m(K). Let e be the edge with one end vertex in C and the other end vertex in K. Let M be a maximum matching for G. If e ∈ / M , then m(G) = |M | = |M ∩ E(C)| + |M ∩ E(K)| ≤ m(C) + m(K) and so the desired equality holds. So suppose e ∈ / M . Since e is incident with C and C is an even cycle, necessarily |M ∩E(C)| ≤ 12 |E(C)|−1. Take any maximum matching M1 for C. Then M1 ∪(M ∩E(K)) is a matching for G with cardinality at least |M |. So we have |M | ≤ m(C) + m(K) and in this case the desired inequality also holds. 2 Proof of Lemma 1.5. Let C denote the unique cycle of G. We have WG = {vC }. Let H1 , . . . , Hp denote the components of TG − vC . Then H1 , . . . , Hp are also all the components of G − C. (a) ⇒ (b): Assume to the contrary that m(TG ) = m(TG − vC ). Take any maximum matching M of TG . Then M must consist of edges in a maximum matching for TG − vC , together with an edge incident with vC , say vC y. Suppose y ∈ V (Hj ). Let x be the unique vertex in C adjacent to y. Clearly, M ∩ E(Hi ) is a maximum matching for Hi for i = 1, . . . , p. Moreover, m(Hj + x) = m(Hj ) + 1. By condition (a) we have p 1 1 m(G) = m(G − C) + m(C) = m(TG − vC ) + E(C) = m(Hi ) + E(C). 2 2 i=1
Let M be obtained from M by replacing the edge vC y by xy. Then M is a matching of G with the same cardinality as M . Choose a matching M of C with cardinality 1 2 |E(C)| − 1 that does not cover vertex x. Then M ∪ M is a maximum matching for G that has an edge with one end vertex in C and the other end vertex outside C. So we arrive at a contradiction. p (b) ⇒ (a): By condition (b) we have m(TG ) = m(TG − vC ) = i=1 m(Hi ). Suppose that for some j, m(Hj + vC ) > m(Hi ). For each i = 1, . . . , p, i = j, take a maximum ˜ j for Hj +vC . Then Mi ∪ M ˜j matching Mi for Hi . Also, take a maximum matching M i=j
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p is a matching for TG with cardinality greater than i=1 m(Hi ). This contradicts our beginning observation on the value of m(TG ). So we have m(Hi + vC ) = m(Hi ) for each i. Now suppose that G has a maximum matching M which has an edge with one end vertex in the cycle C and the other end vertex outside C. As C is an even cycle, clearly M ∩ E(C) is not a maximum matching for C, i.e., |M ∩ E(C)| < 12 |E(C)|. On the other hand, since C is the only cycle of G, for each i = 1, . . . , p, there is precisely one edge between V (C) and V (Hi ); let xi be the unique vertex in C that is adjacent to a vertex in p p Hi . Then |M | = |M ∩ E(C)| + i=1 |M ∩ E(Hi + xi )|. As |M | ≥ 12 |E(C)| + i=1 m(Hi ), it follows that there exists j such that m(Hj + vC ) ≥ |M ∩ (Hj + xj )| > m(Hj ). This contradicts what we have obtained above. 2 Proof of Lemma 1.6. Let C 1 , . . . , C p be all the cycles of G. Let H1 , . . . , Hq be the comp ponents of TG − WG . Then H1 , . . . , Hq are also all the components of G − i=1 V (C i ). ˜ k (respectively, H ˆ k ) be the subgraph of TG (respectively, of G) For k = 1, . . . , q, let H induced by V (Hk ) together with all vertices of TG in WG that are adjacent to a vertex of Hk (respectively, together with vertices of G that lie on a cycle and are adjacent to a vertex of Hk ). Note that for each pair k, i, 1 ≤ k ≤ q, 1 ≤ i ≤ p, there is at most one edge between V (Hk ) and V (C i ); else, since Hk is connected, we would obtain a cycle ˜ k and H ˆ k are of G different from C 1 , . . . , C p . It follows that for each k, the graphs H isomorphic. p q For any maximum matching M of G, we have |M | = i=1 |M ∩ E(C i )| + k=1 |M ∩ ˆ k )|. If M has an edge with one end vertex on a cycle, say on C i , and the other E(H end vertex outside C i then since C i is of even length, |M ∩ E(Ci )| < 12 |E(C i )|. As q p ˆj) ≥ m(G) ≥ 12 i=1 |E(C i )| + k=1 m(Hk ), it follows that there exists j such that m(H ˆ j )| > m(Hj ). On the other hand, we also have m(TG ) = m(TG − WG ) = |M ∩ E(H q ˜ k=1 m(Hk ), which implies that m(Hk ) = m(Hk ) for k = 1, . . . , q and, in particular, ˜ ˆ ˜ j and H ˆ j are isomorphic. m(Hj ) = m(Hj ) = m(Hj ), where the last equality holds as H So we arrive at a contradiction. 2 Now, we are ready to prove the main result of this paper. Proof of Theorem 1.4. First of all, note that G satisfies the maximal nullity condition if and only if each component of G satisfies the maximal nullity condition, and G satisfies conditions (a)–(c) if and only if each component of G satisfies the corresponding conditions (or, by abuse of language, we say each component of G satisfies conditions (a)–(c)). So it suffices to prove the theorem for the case when G is connected. Next, note that if G is a tree then conditions (a)–(c) are satisfied trivially and the equality relation η(G) = |V (G)| − 2m(G) + 2c(G) holds by Lemma 1.2. It remains to deal with the case when c(G) ≥ 1. When c(G) = 1, G is unicyclic. If G satisfies conditions (a)–(c), then by Lemma 1.5 G satisfies the matching condition. In view of Lemma 1.3, the matching condition for
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G together with condition (b) implies that G satisfies the maximal nullity condition. Conversely, if G satisfies the maximal nullity condition, then condition (a) holds trivially, and by Lemma 1.3 and Lemma 1.5, G satisfies conditions (b), (c) (and the matching condition). In below we treat the case when c(G) ≥ 2. Note that when G satisfies the maximal nullity condition, by Corollary 3.7 condition (a) holds and so the graph TG is well-defined. By induction on c(G), the number of cycles G has, we are going to show that if G satisfies the maximal nullity condition then G satisfies condition (b). Consider any cycle C of G. Choose any cycle C , different from C, and take any vertex w on C . It is clear that all vertices on C lie on the same component of G − w, say, H. Since G satisfies the maximal nullity condition and w lies on a cycle of G, by Lemma 3.4 G − w satisfies the maximal nullity condition, and hence so does H. As c(H) < c(G), by the induction assumption, H satisfies condition (b). Hence the length of C is a multiple of 4. It remains to show that under the assumption that conditions (a), (b) both hold, G satisfies the maximal nullity condition if and only if G satisfies condition (c). We will prove the “if” part and the “only if” part both by induction on c(G). We first reduce our proof to the case when G is the graph obtained from a cycle of length a multiple of 4 and a connected graph, vertex-disjoint from the cycle, by adding an edge joining a vertex on the cycle to a vertex in the connected graph. Consider the graph T˜ with WG as its vertex set such that two vertices in WG are adjacent in T˜ if and only if the path in TG between these two vertices does not contain other vertex of WG . Clearly T˜ is a tree. Take any pendant vertex, say vC , of T˜, and let H1 , . . . , Hp (p ≥ 1) be the components of G − C. Our choice of vC guarantees that except for the cycle C, all cycles of G, lie in the same component of G − C, say in H1 . Then all Hi except H1 are trees. Note that for each i, there is precisely one edge between V (C) and V (Hi ). We want to reduce our proof to the case when p = 1. Suppose that p p ≥ 2. Take any pendant vertex u of G that belongs to i=2 V (Hi ) and let v be the vertex adjacent to u. Denote G − {u, v} by G0 . Let P1 , . . . , Pr be the components of G0 and suppose that P1 is the component where the cycle C, and hence also other cycles of G, lie. By Corollary 3.9, when G satisfies the maximal nullity condition or condition (c) – that is, in either direction of the proof for this theorem, v does not lie on a cycle. Furthermore, by Lemma 3.8 G satisfies the maximal nullity condition if and only if G0 satisfies the maximal nullity condition, and G satisfies condition (c) if and only if G0 satisfies condition (c). Note that a graph satisfies the maximal nullity condition (respectively, condition (c)) if and only if each of its components satisfies the maximal nullity condition (respectively, condition (c)), and also that every tree satisfies the maximal nullity condition as well as condition (c). It follows that G satisfies the maximal nullity condition (respectively, condition (c)) if and only if P1 satisfies the maximal nullity condition (respectively, condition (c)). Certainly, P1 inherits conditions (a) and (b) from G. Thus, to prove this theorem, we can work with P1 instead of G. If P1 = G[V (C) ∪ V (H1 )], then our reduction is completed. Otherwise, by repeating the
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preceding argument for P1 and so forth (if necessary), after a finite number of steps, we complete the reduction process. Hereafter, we assume that G is the graph obtained from a cycle C of length a multiple of 4 and a connected graph G1 , vertex-disjoint from C, by adding an edge xy, with x ∈ V (C) and y ∈ V (G2 ). Denote the graph G1 + x by G2 . We deal with the “if” part and the “only if” part of the proof separately. “If” part: Since C is a cycle with length a multiple of 4, by Lemma 2.1 the hypothesis of Lemma 2.5 is met, i.e., η(C −x) = 1 = η(C) −1; so we have η(G) = η(C −x) +η(G2 ) = η(G2 ) + 1. Let M be a maximum matching for TG that does not cover any vertex in WG . Then M is also a maximum matching for TG2 that does not cover any vertex in WG2 . Since G satisfies conditions (a), (b), clearly so does G2 . Thus G2 satisfies conditions (a)–(c), and as c(G2 ) < c(G), by the induction assumption G2 satisfies the maximal nullity condition. Since G satisfies condition (c), by Lemma 1.6 G satisfies the matching condition; so we have m(G) = m(TG − WG ) + 12 l(G) = m(TG ) − 12 l(G), where l(G) denotes the sum of lengths of cycles in G. Similarly, we also have m(G2 ) = m(TG2 ) + 12 l(G2 ), where l(G2 ) is the sum of lengths of cycles in G2 . Noting that m(TG2 ) = m(TG ) (as TG = TG2 ) and l(G) = l(G2 ) + |E(C)|, by a little calculation we obtain m(G2 ) = m(G) − 12 |E(C)|. As we have shown that η(G2 ) = η(G) − 1 at the beginning and it is clear that |V (G2 )| = |V (G)| − |V (C)| + 1 and c(G2 ) = c(G) − 1, the maximal nullity condition for G2 can be rewritten as η(G) − 1 = V (G) − V (C) + 1 − 2m(G) − E(C) + 2 c(G) − 1 ; thus the maximal nullity condition for G follows. “Only if” part: Let C be the cycle x, x2 , . . . , x2s , x. As x lies on a cycle, by Lemma 3.4 G − x satisfies the maximal nullity condition and m(G − x) = m(G) − 1. Since xy is the only edge between V (C) and V (G1 ), G − x has two components, namely, G1 and the path P2s−1 : x2 x3 · · · x2s . So G1 also satisfies the maximal nullity condition and we have m(G1 ) = m(G − x) − m(P2s−1 ) = m(G) − 1 − (s − 1) = m(G) − s. On the other hand, by Lemma 3.4 again G−x2 also satisfies the maximal nullity condition and m(G − x2 ) = m(G) − 1, and as x3 is a pendant vertex of G − x2 and x4 is the unique vertex (in G − x2 ) adjacent to x3 , by Lemma 3.8 and Lemma 2.6 (G − x2 ) − {x3 , x4 } also satisfies the maximal nullity condition and m(G −x2 −{x3 , x4 }) = m(G) −2. By repeated applications of Lemma 3.8 and Lemma 2.6 we deduce that G2 (= G − {x2 , x3 , . . . , x2s }) also satisfies the maximal nullity condition and m(G2 ) = m(G) − s; so m(G1 ) = m(G2 ). Since Gi , i = 1, 2, satisfies the maximal nullity condition and c(Gi ) < c(G), by the induction assumption m(TGi ) = m(TGi − WGi ); hence by Lemma 1.6 Gi satisfies the matching condition, and so we have m(Gi ) = m(TGi ) + 12 l(Gi ), where l(Gi ) denotes the sum of lengths of cycles in Gi . Since G1 , G2 have the same set of cycles, l(G1 ) = l(G2 ),
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and as m(G1 ) = m(G2 ), it follows that we have m(TG1 ) = m(TG2 ). But TG = TG2 , so m(TG ) = m(TG1 ). Now by the induction assumption TG1 has a maximum matching M that does not cover any vertex in WG1 . As m(TG ) = m(TG1 ), it is clear that M is also a maximum matching for TG that does not cover any vertex in WG . This shows that G satisfies condition (c), completing the proof. 2 References [1] B. Cheng, B. Liu, On the nullity of graphs, Electron. J. Linear Algebra 16 (2007) 60–67. [2] B. Chen, B. Liu, On the nullity of tricyclic graphs, Linear Algebra Appl. 434 (2011) 1799–1810. [3] B. Borovićanin, I. Gutman, Nullity of graphs, in: D. Cvetković, I. Gutman (Eds.), Applications of Graph Spectra, Zb. Rad. 13 (21) (2009), Matematički institut SANU. [4] A.E. Brouwer, W.H. Haemers, Spectra of Graphs, Springer, New York, 2011. [5] D. Cvetković, M. Doob, H. Sachs, Spectra of Graphs – Theory and Applications, Academic Press, New York, 1980. [6] D. Cvetković, I. Gutman, The algebraic multiplicity of the number zero in the spectrum of a bipartite graph, Mat. Vesnik (Beograd) 9 (1972) 141–150. [7] Y.-Z. Fan, K.-S. Qian, On the nullity of bipartite graphs, Linear Algebra Appl. 430 (2009) 2943–2949. [8] S.-C. Gong, G.-H. Xu, On the nullity of a graph with cut-points, Linear Algebra Appl. 436 (2012) 135–142. [9] J.-M. Guo, W.-G. Yan, Y.-N. Yeh, On the nullity and the matching number of unicyclic graphs, Linear Algebra Appl. 431 (2009) 1293–1301. [10] I. Gutman, I. Sciriha, On the nullity of line graphs of trees, Discrete Math. 232 (2001) 35–45. [11] S.-B. Hu, X.-Z. Tan, B.-L. Liu, On the nullity of bicyclic graphs, Linear Algebra Appl. 429 (2008) 1387–1391. [12] G.R. Omidi, On the nullity of bipartite graphs, Graphs Combin. 25 (2009) 111–114. [13] X.-Z. Tan, B.-L. Liu, On the nullity of unicyclic graphs, Linear Algebra Appl. 408 (2005) 212–220. [14] L. Wang, D. Wong, Bounds for the matching number, the edge chromatic number and the independence number of a graph in terms of rank, Discrete Appl. Math. 166 (2014) 276–281.
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Linear Algebra and its Applications www.elsevier.com/locate/laa
A characterization of graphs G with nullity |V (G)| − 2m(G) + 2c(G) Ya-zhi Song a , Xiao-qiu Song a,∗,1 , Bit-Shun Tam b,2 a
Department of Mathematics, China University of Mining and Technology, Xuzhou 221116, China b Department of Mathematics, Tamkang University, New Taipei City 25137, Taiwan, ROC
a r t i c l e
i n f o
Article history: Received 29 January 2014 Accepted 23 September 2014 Available online xxxx Submitted by R. Brualdi MSC: 05C50 Keywords: Nullity Rank Matching number
a b s t r a c t By |V (G)|, |E(G)|, η(G), and m(G) we denote respectively the order, the number of edges, the nullity, and the matching number of a (simple) graph G. Recently Wang and Wong have proved that for every graph G, η(G) ≤ |V (G)| − 2m(G) + 2c(G), where c(G) = |E(G)| − |V (G)| + θ(G), θ(G) being the number of connected components of G. In this paper graphs G that satisfy the equality η(G) = |V (G)| − 2m(G) + 2c(G) are characterized. © 2014 Elsevier Inc. All rights reserved.
1. Introduction The nullity η(G) of a (simple) graph G is the (algebraic) multiplicity of the eigenvalue 0 in the spectrum of the adjacency matrix A(G) of G. There has been much interest in the nullity of graphs in chemistry. According to the Hückel molecular orbital theory, if * Corresponding author. 1 2
E-mail addresses: [email protected] (X.-q. Song), [email protected] (B.-S. Tam). Supported by National Natural Science Foundation of China (No. 11171343). Supported by National Science Council of the Republic of China (NSC 102-2115-M-032-002).
http://dx.doi.org/10.1016/j.laa.2014.09.034 0024-3795/© 2014 Elsevier Inc. All rights reserved.
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the nullity η(G) of a molecular graph G is positive, then the corresponding chemical compound is highly reactive and unstable, or nonexistent (see, for instance, [3]). Interest in the nullity of graphs has also gained attention among mathematicians. In particular, there has been much work on bounds for η(G). It is known that 0 ≤ η(G) ≤ n − 2 if G is a graph on n vertices and G is not isomorphic to nK1 . Cheng and Liu [1] have characterized the extremal graphs attaining the upper bound n − 2 and the second upper bound n − 3. Subsequently, this has led to the study of the nullity set (i.e., the set of nullities attained by graphs) of various classes of graphs (see [13,11,2,7,12,10]). The present work also involves an upper bound for η(G), but in a different way. Recently, Wang and Wong [14] obtained the following bounds for the matching number of a graph G: r(G)−c(G) ≤ m(G) ≤ r(G)+2c(G) , where r(G) and m(G) denote respectively 2 2 the rank and the matching number of G and c(G) = |E(G)| − |V (G)| + θ(G), θ(G) being the number of (connected) components of G. They also showed that their bounds are best possible by providing examples. The upper bound for the matching number can be rewritten in an equivalent form as an upper bound for the nullity of G: Lemma 1.1. For any graph G, η(G) ≤ |V (G)| − 2m(G) + 2c(G). The purpose of this work is to characterize graphs G that satisfy η(G) = |V (G)| − 2m(G) + 2c(G). For convenience, hereafter in this paper we say a graph G satisfies the maximal nullity condition if η(G) = |V (G)| − 2m(G) + 2c(G). It should not be mixed up with graphs G of order n that satisfy η(G) = n − 2. In view of the following known result, every acyclic graph satisfies the maximal nullity condition: Lemma 1.2. (See [6].) For every acyclic graph T with at least one vertex, η(T ) = |V (T )| − 2m(T ). By the recent work of Guo, Yan and Yeh [9], unicyclic graphs that satisfy the maximal nullity condition are also characterized. The following result is implicit in [9]: Lemma 1.3. A unicyclic graph G has nullity |V (G)| − 2m(G) + 2 if and only if the length of the cycle of G is divisible by 4 and M ∩ E1 = ∅ for any maximum matching M of G, where E1 denotes the set of edges of G with one end vertex in the cycle and one end vertex outside the cycle. In order to state our main result, we need to define an acyclic graph TG for a graph G whose cycles are pairwise vertex-disjoint. Let G be a graph with pairwise vertex-disjoint cycles. The acyclic graph TG is obtained from G by contracting cycles. More specifically, the vertex set V (TG ) is taken to be U ∪ WG , where U consists of all vertices of G that do not lie on a cycle and WG = {vC : C is a cycle of G}. Two vertices in U are adjacent in TG if and only if they are adjacent in G, a vertex u ∈ U is adjacent to a vertex vC ∈ W if and only if u is adjacent
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(in G) to a vertex in the cycle C, and vertices vC1 , vC2 are adjacent in TG if and only if there exists an edge in G joining a vertex of C1 to a vertex of C2 . It is clear that TG is always acyclic and is a tree when G is connected. Since the cycles of G are pairwise vertex-disjoint, deleting edges from G, one from each cycle, will result in a forest F and as c(F ) = 0 it follows that G has altogether c(G) cycles. Note that the graph TG − WG (obtained from TG by deleting vertices in WG and the incident edges) is the same as the graph obtained from G by deleting all cycles and the incident edges. Below is our main result: Theorem 1.4. For any graph G, η(G) = |V (G)| − 2m(G) + 2c(G) if and only if the following three conditions are all satisfied: (a) Distinct cycles of G (if any) are vertex-disjoint; (b) The length of each cycle of G (if any) is a multiple of 4; (c) m(TG ) = m(TG − WG ) or, equivalently, there exists a maximum matching of TG that does not cover any vertex in WG . To illustrate Theorem 1.4, let us adapt the example given at the end of the proof of [14, Theorem 1.1]. Let S be a star K1,p+1 with vertices x, y1 , . . . , yp+1 (p ≥ 1), where x is the star center. Let G be the graph obtained from S by replacing each pendant vertex yi , i = 1, 2, . . . , p, by a cycle Cli of length li , where each li is a multiple of 4. Then the cycles of G are pairwise vertex-disjoint and are of lengths multiples of 4. Furthermore, TG is a K1,p+1 and TG − WG is a P2 ; so m(TG ) = 1 = m(TG − WG ). By Theorem 1.4, p p η(G) = (2 + i=1 li ) − 2(1 + i=1 l2i ) + 2p = 2p. Notice that when G is unicyclic, Theorem 1.4 does not reduce to Lemma 1.3. For the purpose of proof, we will need the following: Lemma 1.5. Let G be an even-unicyclic graph. The following conditions are equivalent: (a) Every maximum matching of G does not have an edge with one end vertex in the cycle of G and the other end vertex outside the cycle. (b) m(TG ) = m(TG − WG ). In view of Lemmas 1.3 and 1.5, one may suspect that in Theorem 1.4 condition (iii) can be replaced by the condition “every maximum matching of G does not have an edge with one end vertex on a cycle and the other end vertex outside the (same) cycle”. It turns out that the answer is “no”. For instance, let G be the graph obtained from the path P4 by attaching two C4 , one at each end of the path. (So G has 10 vertices.) The two cycles of G are each of even length and they are vertex-disjoint. It is readily seen that m(G) = 5 and G has exactly one maximum matching and this matching has no edge with one end vertex on a cycle and the other end vertex outside the cycle. However,
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m(TG ) = 2 = 1 = m(TG − WG ). Indeed, even the weaker condition “m(TG ) = m(TG − v) for every v ∈ WG ” is not fulfilled. Nevertheless, we have the following positive result: Lemma 1.6. Let G be a graph with at least one cycle. Suppose that each cycle of G is of even length and the cycles of G are pairwise vertex-disjoint. If m(TG ) = m(TG − WG ) then every maximum matching of G does not contain an edge with one end vertex on a cycle and the other end vertex outside the cycle. For convenience, we will say G satisfies the matching condition if every maximum matching of G does not have an edge with one end vertex on a cycle and the other end vertex outside the (same) cycle. We would like to mention that the graph TG has appeared in [14] when the lower bound for the matching number m(G) – not the upper bound, which is equivalent to the inequality given in Lemma 1.1 – is being established, but its use there is inessential. However, the graph TG plays an important role in this work. In Section 2 we give the necessary definitions and notations, together with a list of known results which we will need. In Section 3, we will give the proofs for Theorem 1.4, Lemma 1.5, Lemma 1.6 and other auxiliary results. 2. Notations and preliminaries By a graph we mean a finite simple undirected graph (i.e., one with a finite vertex set and without loops or multiple edges). The rank of a graph G, denoted by r(G), is defined to be the rank of its adjacency matrix. For any U ⊆ V (G), we denote by G − U the graph obtained from G by removing the vertices in U together with all incident edges; when U = {x}, we write G − U simply as G − x. Sometimes we use the notation G − G1 instead of G − V (G1 ) when G1 is a (vertex-)induced subgraph of G. If G1 is an induced subgraph of G and x is a vertex not in G1 , we write the subgraph of G induced by V (G1 ) ∪ {x} simply as G1 + x. A vertex x is called a pendant vertex if its degree is 1. An induced subgraph H of a graph G is called a pendant star of G if H is a star with at least two vertices such that the star center of S is the only vertex of S that is adjacent to some vertex not in S. A matching in a graph is a set of pairwise non-adjacent edges. A vertex is covered by a matching if it is an end vertex of one of the edges in the matching. A maximum matching is a matching that contains the largest possible number of edges. The matching number of a graph G, denoted by m(G), is the size of a maximum matching. We denote a cycle (respectively, a path) on n vertices by Cn (respectively, Pn ). The following is a list of known results that we will need in the proofs.
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Lemma 2.1. (See [6].) η(Pn ) equals 1 if n is odd and equals 0 if n is even. η(Cn ) equals 2 if n is a multiple of 4 and equals 0, otherwise. Lemma 2.2. (See [4].) If x is a vertex of a graph G, then η(G) − 1 ≤ η(G − x) ≤ η(G) + 1. Lemma 2.3. (See [5].) Let G be a graph containing a pendant vertex, and let H be the graph obtained from G by deleting the pendant vertex and the vertex adjacent to it. Then η(G) = η(H). Lemma 2.4. (See [8].) Let x be a cut-vertex of a graph G. If there exists a component of G − x, say G1 , such that η(G1 ) = η(G1 + x) + 1, then η(G) = η(G − x) − 1. Lemma 2.5. (See [8].) Let x be a cut-vertex of a graph G and let G1 be a component of G − x. If η(G1 ) = η(G1 + x) − 1, then η(G) = η(G1 ) + η(G − G1 ). Lemma 2.6. (See [9].) If S is a pendant star of a graph G, then m(G) = 1 + m(G − S). Furthermore, G has a maximum matching which consists of edges in a maximum matching for G − S together with an edge of S. 3. Proofs of results Lemma 3.1. For a graph G with pairwise vertex-disjoint cycles, the following conditions are equivalent: (a) There exists a maximum matching of TG that does not cover any vertex in WG . (b) m(TG ) = m(TG − WG ). (c) η(TG ) = η(TG − WG ) + c(G). Proof. We always have the inequality m(TG − WG ) ≤ m(TG ) (as TG − WG is an induced subgraph of TG ), with equality if and only if TG has a maximum matching that is also a matching for TG − WG . But a matching for TG − WG is the same as a matching for TG that does not cover any vertices in WG , so conditions (a) and (b) are equivalent. Since TG and TG − WG are acyclic, by Lemma 1.2 we have η(TG ) = V (TG ) − 2m(TG ) = V (TG − WG ) + c(G) − 2m(TG ) and η(TG − WG ) = V (TG − WG ) − 2m(TG − WG ); hence, η(TG ) − η(TG − WG ) = c(G) − 2 m(TG ) − m(TG − WG ) ≤ c(G).
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Thus, we always have η(TG ) ≤ η(TG − WG ) + c(G), where the equality holds if and only if m(TG ) = m(TG − WG ). This establishes the equivalence of conditions (b) and (c). 2 Lemma 3.2. 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) −1. Proof. We may assume that the graph G is connected. “Only if” part: Let u be a vertex of G that lies on a cycle C, and suppose that G − u has p (≥ 1) components. Then C is the only cycle of G that contains vertex u, vertices in C − u all lie in the same component P of G − u, P has precisely two vertices adjacent (in G) to u, and each of the remaining p − 1 components has precisely one vertex adjacent to u. Hence, dG (u) = p + 1 and we have |E(G − u)| = |E(G)| − p − 1. As |V (G − u)| = |V (G)| − 1 and θ(G − u) = p, a little calculation yields c(G − u) = c(G) − 1. “If” part: Suppose that G has two distinct cycles C1 , C2 that have at least one common vertex. Choose a common vertex u such that the degree of u in the subgraph of G induced by V (C1 ) ∪ V (C2 ) is at least 3, and suppose that G − u has p components. Then either there is a component of G − u that has at least three vertices adjacent to u or there are at least two components, each of which has two vertices adjacent to u. In any case, we have |E(G − u)| ≤ |E(G) − p − 2| and a little calculation yields c(G − u) < c(G) − 1. 2 The argument given in the proof for Lemma 3.2 also yields the following: Remark 3.3. Let G be a graph and let u ∈ V (G). (i) If u lies on a cycle of G, then c(G − u) ≤ c(G) − 1, where the equality holds if and only if dG (u) = θ(G − u) + 1; (ii) If u does not lie on a cycle of G, then c(G − u) = c(G). In connection with Remark 3.3(i), we would like to point out that a sufficient, but not necessary, condition for the equality dG (u) = θ(G − u) + 1 to hold is that u lies on a unique cycle of G. In the proof of [14, Theorem 1.1] it is asserted (without proof) that distinct cycles of a graph G are vertex-disjoint if and only if for every vertex x of G, c(G −x) ≥ c(G) −1. This assertion, which motivated Lemma 3.2, follows from the lemma itself and Remark 3.3(ii). Examining the proof for the inequality given in Lemma 1.1, we obtain the following result. Lemma 3.4. Let G be a graph that has at least one cycle. If G satisfies the maximal nullity condition, then for any vertex u that lies on a cycle of G, we have (a) η(G − u) = η(G) − 1; (b) G − u satisfies the maximal nullity condition;
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(c) m(G − u) = m(G) − 1; (d) c(G − u) = c(G) − 1; (e) dG (u) = θ(G − u) + 1. Proof. We may assume that G is connected. Suppose that G − u has p components. By the argument given in the proof for Lemma 3.2 we have |E(G − u)| ≤ |E(G)| − p − 1, and hence c(G − u) = E(G − u) − V (G − u) + p ≤ E(G) − p − 1 − V (G) − 1 + p = c(G) − 1. On the other hand, clearly m(G − u) ≥ m(G) − 1 and by Lemma 2.2 we have η(G − u) ≥ η(G) − 1. As G satisfies the maximal nullity condition, we obtain η(G − u) ≥ η(G) − 1 = V (G) − 2m(G) + 2c(G) − 1 ≥ V (G − u) + 1 − 2 m(G − u) + 1 + 2 c(G − u) + 1 − 1 = V (G − u) − 2m(G − u) + 2c(G − u) ≥ η(G − u), where the last inequality follows from Lemma 1.1. The above derivation forces each inequality involved to become an equality. Hence, conditions (a)–(d) all hold and, in addition, we have |E(G − u)| = |E(G)| − p − 1, which is equivalent to (e). 2 For possible future use, we take note of the following two observations, whose proofs we omit. Remark 3.5. Let G be a graph that satisfies the condition η(G) = |V (G)|−2m(G)+2c(G). Let u be a vertex of G that does not lie on a cycle. If m(G − u) = m(G), then conditions (a), (b) of Lemma 3.4 are fulfilled. Remark 3.6. A graph G satisfies the condition η(G) = |V (G)| − 2m(G) + 2c(G) if either one of the following two conditions holds: (1) There exists a vertex u that lies on a cycle of G such that conditions (a)–(d) of Lemma 3.4 are fulfilled. (2) There exists a vertex u that does not lie on a cycle of G such that m(G − u) = m(G) and conditions (a), (b) of Lemma 3.4 are fulfilled. Corollary 3.7. Let G be a graph with nullity |V (G)| −2m(G) +2c(G). Then distinct cycles of G (if any) are vertex-disjoint. Thus G has precisely c(G) cycles.
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Proof. Since G satisfies the maximal nullity condition, by Lemma 3.4 we have c(G −u) = c(G) − 1 for every vertex u that lies on a cycle of G. By Lemma 3.2 our assertion follows. 2 Lemma 3.8. Let u be a pendant vertex of a graph G and let v be the vertex adjacent to u. Denote G − {u, v} by G0 . Suppose that v does not lie on a cycle of G. Then: (i) G satisfies the maximal nullity condition if and only if G0 satisfies the maximal nullity condition. (ii) TG has a maximum matching that does not cover any vertex in WG if and only if TG0 has a maximum matching that does not cover any vertex in WG0 . Proof. (i) By Lemma 2.3 (respectively, Lemma 2.6) we obtain η(G0 ) = η(G) (respectively, m(G0 ) = m(G) −1). As v does not lie on a cycle, by Remark 3.3(ii), c(G0 ) = c(G). So we have η(G) − V (G) + 2m(G) − 2c(G) = η(G0 ) − V (G0 ) + 2 + 2 1 + m(G0 ) − 2c(G0 ) = η(G0 ) − V (G0 ) + 2m(G0 ) − 2c(G0 ). It follows that G0 satisfies the maximal nullity condition if and only if G satisfies the maximal nullity condition. (ii) Since G and G0 have the same set of cycles, TG is well-defined if and only if TG0 is well-defined, and also whenever they are both well-defined, WG = WG0 . Since u, v each does not lie on a cycle of G, u, v are both vertices of TG . Noting that TG0 = TG − {u, v}, by Lemma 2.6 we have m(TG ) = 1 + m(TG0 ). If TG has a maximum matching M that does not cover any vertex in WG , then necessarily M has an edge e that covers vertex v. In that case, M \ {e} is a maximum matching for TG0 that does not cover any vertex in WG0 . Conversely, suppose that TG0 has a maximum matching M0 that does not cover any vertex in WG0 . Then M0 ∪ {uv} is a maximum matching for TG that does not cover any vertex in WG . 2 Corollary 3.9. Let u be a pendant vertex of a graph G and let v be the vertex adjacent to u. Denote G − {u, v} by G0 . Then: (i) G satisfies the maximal nullity condition if and only if v does not lie on a cycle and G0 satisfies the maximal nullity condition. (ii) m(TG ) = m(TG − WG ) if and only if v does not lie on a cycle and m(TG0 ) = m(TG0 − WG0 ). Proof. (i) In view of Lemma 3.8(i), it remains to show that if G satisfies the maximal nullity condition, then v does not lie on a cycle. Assume to the contrary that v lies on a
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cycle. As G0 has one cycle fewer than G, c(G0 ) = c(G) −1. A little calculation shows that the maximal nullity condition for G leads to the condition η(G0 ) = |V (G0 )| − 2m(G0 ) + 2c(G0 ) + 2, which contradicts Lemma 1.1. (ii) By Lemma 3.8(ii) it suffices to show that if m(TG ) = m(TG − WG ) then v does not lie on a cycle. Assume to the contrary that v lies on a cycle C. Then TG − WG is an induced subgraph of TG − vC , and since TG − vC is also an induced subgraph of TG the condition m(TG ) = m(TG − WG ) forces m(TG ) = m(TG − vC ). On the other hand, we also have m(TG − vC ) = m(TG − {u, vC }) = m(TG ) − 1, where the second equality holds by Lemma 2.6. So we arrive at a contradiction. 2 Lemma 3.10. Given an even cycle C and a connected graph K that are vertex-disjoint, let G be obtained from C ∪ K by adding an edge with one end vertex in C and the other end vertex in K. Then m(G) = m(C) + m(K). Proof. It is clear that m(G) ≥ m(C) + m(K). Let e be the edge with one end vertex in C and the other end vertex in K. Let M be a maximum matching for G. If e ∈ / M , then m(G) = |M | = |M ∩ E(C)| + |M ∩ E(K)| ≤ m(C) + m(K) and so the desired equality holds. So suppose e ∈ / M . Since e is incident with C and C is an even cycle, necessarily |M ∩E(C)| ≤ 12 |E(C)|−1. Take any maximum matching M1 for C. Then M1 ∪(M ∩E(K)) is a matching for G with cardinality at least |M |. So we have |M | ≤ m(C) + m(K) and in this case the desired inequality also holds. 2 Proof of Lemma 1.5. Let C denote the unique cycle of G. We have WG = {vC }. Let H1 , . . . , Hp denote the components of TG − vC . Then H1 , . . . , Hp are also all the components of G − C. (a) ⇒ (b): Assume to the contrary that m(TG ) = m(TG − vC ). Take any maximum matching M of TG . Then M must consist of edges in a maximum matching for TG − vC , together with an edge incident with vC , say vC y. Suppose y ∈ V (Hj ). Let x be the unique vertex in C adjacent to y. Clearly, M ∩ E(Hi ) is a maximum matching for Hi for i = 1, . . . , p. Moreover, m(Hj + x) = m(Hj ) + 1. By condition (a) we have p 1 1 m(G) = m(G − C) + m(C) = m(TG − vC ) + E(C) = m(Hi ) + E(C). 2 2 i=1
Let M be obtained from M by replacing the edge vC y by xy. Then M is a matching of G with the same cardinality as M . Choose a matching M of C with cardinality 1 2 |E(C)| − 1 that does not cover vertex x. Then M ∪ M is a maximum matching for G that has an edge with one end vertex in C and the other end vertex outside C. So we arrive at a contradiction. p (b) ⇒ (a): By condition (b) we have m(TG ) = m(TG − vC ) = i=1 m(Hi ). Suppose that for some j, m(Hj + vC ) > m(Hi ). For each i = 1, . . . , p, i = j, take a maximum ˜ j for Hj +vC . Then Mi ∪ M ˜j matching Mi for Hi . Also, take a maximum matching M i=j
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p is a matching for TG with cardinality greater than i=1 m(Hi ). This contradicts our beginning observation on the value of m(TG ). So we have m(Hi + vC ) = m(Hi ) for each i. Now suppose that G has a maximum matching M which has an edge with one end vertex in the cycle C and the other end vertex outside C. As C is an even cycle, clearly M ∩ E(C) is not a maximum matching for C, i.e., |M ∩ E(C)| < 12 |E(C)|. On the other hand, since C is the only cycle of G, for each i = 1, . . . , p, there is precisely one edge between V (C) and V (Hi ); let xi be the unique vertex in C that is adjacent to a vertex in p p Hi . Then |M | = |M ∩ E(C)| + i=1 |M ∩ E(Hi + xi )|. As |M | ≥ 12 |E(C)| + i=1 m(Hi ), it follows that there exists j such that m(Hj + vC ) ≥ |M ∩ (Hj + xj )| > m(Hj ). This contradicts what we have obtained above. 2 Proof of Lemma 1.6. Let C 1 , . . . , C p be all the cycles of G. Let H1 , . . . , Hq be the comp ponents of TG − WG . Then H1 , . . . , Hq are also all the components of G − i=1 V (C i ). ˜ k (respectively, H ˆ k ) be the subgraph of TG (respectively, of G) For k = 1, . . . , q, let H induced by V (Hk ) together with all vertices of TG in WG that are adjacent to a vertex of Hk (respectively, together with vertices of G that lie on a cycle and are adjacent to a vertex of Hk ). Note that for each pair k, i, 1 ≤ k ≤ q, 1 ≤ i ≤ p, there is at most one edge between V (Hk ) and V (C i ); else, since Hk is connected, we would obtain a cycle ˜ k and H ˆ k are of G different from C 1 , . . . , C p . It follows that for each k, the graphs H isomorphic. p q For any maximum matching M of G, we have |M | = i=1 |M ∩ E(C i )| + k=1 |M ∩ ˆ k )|. If M has an edge with one end vertex on a cycle, say on C i , and the other E(H end vertex outside C i then since C i is of even length, |M ∩ E(Ci )| < 12 |E(C i )|. As q p ˆj) ≥ m(G) ≥ 12 i=1 |E(C i )| + k=1 m(Hk ), it follows that there exists j such that m(H ˆ j )| > m(Hj ). On the other hand, we also have m(TG ) = m(TG − WG ) = |M ∩ E(H q ˜ k=1 m(Hk ), which implies that m(Hk ) = m(Hk ) for k = 1, . . . , q and, in particular, ˜ ˆ ˜ j and H ˆ j are isomorphic. m(Hj ) = m(Hj ) = m(Hj ), where the last equality holds as H So we arrive at a contradiction. 2 Now, we are ready to prove the main result of this paper. Proof of Theorem 1.4. First of all, note that G satisfies the maximal nullity condition if and only if each component of G satisfies the maximal nullity condition, and G satisfies conditions (a)–(c) if and only if each component of G satisfies the corresponding conditions (or, by abuse of language, we say each component of G satisfies conditions (a)–(c)). So it suffices to prove the theorem for the case when G is connected. Next, note that if G is a tree then conditions (a)–(c) are satisfied trivially and the equality relation η(G) = |V (G)| − 2m(G) + 2c(G) holds by Lemma 1.2. It remains to deal with the case when c(G) ≥ 1. When c(G) = 1, G is unicyclic. If G satisfies conditions (a)–(c), then by Lemma 1.5 G satisfies the matching condition. In view of Lemma 1.3, the matching condition for
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G together with condition (b) implies that G satisfies the maximal nullity condition. Conversely, if G satisfies the maximal nullity condition, then condition (a) holds trivially, and by Lemma 1.3 and Lemma 1.5, G satisfies conditions (b), (c) (and the matching condition). In below we treat the case when c(G) ≥ 2. Note that when G satisfies the maximal nullity condition, by Corollary 3.7 condition (a) holds and so the graph TG is well-defined. By induction on c(G), the number of cycles G has, we are going to show that if G satisfies the maximal nullity condition then G satisfies condition (b). Consider any cycle C of G. Choose any cycle C , different from C, and take any vertex w on C . It is clear that all vertices on C lie on the same component of G − w, say, H. Since G satisfies the maximal nullity condition and w lies on a cycle of G, by Lemma 3.4 G − w satisfies the maximal nullity condition, and hence so does H. As c(H) < c(G), by the induction assumption, H satisfies condition (b). Hence the length of C is a multiple of 4. It remains to show that under the assumption that conditions (a), (b) both hold, G satisfies the maximal nullity condition if and only if G satisfies condition (c). We will prove the “if” part and the “only if” part both by induction on c(G). We first reduce our proof to the case when G is the graph obtained from a cycle of length a multiple of 4 and a connected graph, vertex-disjoint from the cycle, by adding an edge joining a vertex on the cycle to a vertex in the connected graph. Consider the graph T˜ with WG as its vertex set such that two vertices in WG are adjacent in T˜ if and only if the path in TG between these two vertices does not contain other vertex of WG . Clearly T˜ is a tree. Take any pendant vertex, say vC , of T˜, and let H1 , . . . , Hp (p ≥ 1) be the components of G − C. Our choice of vC guarantees that except for the cycle C, all cycles of G, lie in the same component of G − C, say in H1 . Then all Hi except H1 are trees. Note that for each i, there is precisely one edge between V (C) and V (Hi ). We want to reduce our proof to the case when p = 1. Suppose that p p ≥ 2. Take any pendant vertex u of G that belongs to i=2 V (Hi ) and let v be the vertex adjacent to u. Denote G − {u, v} by G0 . Let P1 , . . . , Pr be the components of G0 and suppose that P1 is the component where the cycle C, and hence also other cycles of G, lie. By Corollary 3.9, when G satisfies the maximal nullity condition or condition (c) – that is, in either direction of the proof for this theorem, v does not lie on a cycle. Furthermore, by Lemma 3.8 G satisfies the maximal nullity condition if and only if G0 satisfies the maximal nullity condition, and G satisfies condition (c) if and only if G0 satisfies condition (c). Note that a graph satisfies the maximal nullity condition (respectively, condition (c)) if and only if each of its components satisfies the maximal nullity condition (respectively, condition (c)), and also that every tree satisfies the maximal nullity condition as well as condition (c). It follows that G satisfies the maximal nullity condition (respectively, condition (c)) if and only if P1 satisfies the maximal nullity condition (respectively, condition (c)). Certainly, P1 inherits conditions (a) and (b) from G. Thus, to prove this theorem, we can work with P1 instead of G. If P1 = G[V (C) ∪ V (H1 )], then our reduction is completed. Otherwise, by repeating the
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preceding argument for P1 and so forth (if necessary), after a finite number of steps, we complete the reduction process. Hereafter, we assume that G is the graph obtained from a cycle C of length a multiple of 4 and a connected graph G1 , vertex-disjoint from C, by adding an edge xy, with x ∈ V (C) and y ∈ V (G2 ). Denote the graph G1 + x by G2 . We deal with the “if” part and the “only if” part of the proof separately. “If” part: Since C is a cycle with length a multiple of 4, by Lemma 2.1 the hypothesis of Lemma 2.5 is met, i.e., η(C −x) = 1 = η(C) −1; so we have η(G) = η(C −x) +η(G2 ) = η(G2 ) + 1. Let M be a maximum matching for TG that does not cover any vertex in WG . Then M is also a maximum matching for TG2 that does not cover any vertex in WG2 . Since G satisfies conditions (a), (b), clearly so does G2 . Thus G2 satisfies conditions (a)–(c), and as c(G2 ) < c(G), by the induction assumption G2 satisfies the maximal nullity condition. Since G satisfies condition (c), by Lemma 1.6 G satisfies the matching condition; so we have m(G) = m(TG − WG ) + 12 l(G) = m(TG ) − 12 l(G), where l(G) denotes the sum of lengths of cycles in G. Similarly, we also have m(G2 ) = m(TG2 ) + 12 l(G2 ), where l(G2 ) is the sum of lengths of cycles in G2 . Noting that m(TG2 ) = m(TG ) (as TG = TG2 ) and l(G) = l(G2 ) + |E(C)|, by a little calculation we obtain m(G2 ) = m(G) − 12 |E(C)|. As we have shown that η(G2 ) = η(G) − 1 at the beginning and it is clear that |V (G2 )| = |V (G)| − |V (C)| + 1 and c(G2 ) = c(G) − 1, the maximal nullity condition for G2 can be rewritten as η(G) − 1 = V (G) − V (C) + 1 − 2m(G) − E(C) + 2 c(G) − 1 ; thus the maximal nullity condition for G follows. “Only if” part: Let C be the cycle x, x2 , . . . , x2s , x. As x lies on a cycle, by Lemma 3.4 G − x satisfies the maximal nullity condition and m(G − x) = m(G) − 1. Since xy is the only edge between V (C) and V (G1 ), G − x has two components, namely, G1 and the path P2s−1 : x2 x3 · · · x2s . So G1 also satisfies the maximal nullity condition and we have m(G1 ) = m(G − x) − m(P2s−1 ) = m(G) − 1 − (s − 1) = m(G) − s. On the other hand, by Lemma 3.4 again G−x2 also satisfies the maximal nullity condition and m(G − x2 ) = m(G) − 1, and as x3 is a pendant vertex of G − x2 and x4 is the unique vertex (in G − x2 ) adjacent to x3 , by Lemma 3.8 and Lemma 2.6 (G − x2 ) − {x3 , x4 } also satisfies the maximal nullity condition and m(G −x2 −{x3 , x4 }) = m(G) −2. By repeated applications of Lemma 3.8 and Lemma 2.6 we deduce that G2 (= G − {x2 , x3 , . . . , x2s }) also satisfies the maximal nullity condition and m(G2 ) = m(G) − s; so m(G1 ) = m(G2 ). Since Gi , i = 1, 2, satisfies the maximal nullity condition and c(Gi ) < c(G), by the induction assumption m(TGi ) = m(TGi − WGi ); hence by Lemma 1.6 Gi satisfies the matching condition, and so we have m(Gi ) = m(TGi ) + 12 l(Gi ), where l(Gi ) denotes the sum of lengths of cycles in Gi . Since G1 , G2 have the same set of cycles, l(G1 ) = l(G2 ),
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and as m(G1 ) = m(G2 ), it follows that we have m(TG1 ) = m(TG2 ). But TG = TG2 , so m(TG ) = m(TG1 ). Now by the induction assumption TG1 has a maximum matching M that does not cover any vertex in WG1 . As m(TG ) = m(TG1 ), it is clear that M is also a maximum matching for TG that does not cover any vertex in WG . This shows that G satisfies condition (c), completing the proof. 2 References [1] B. Cheng, B. Liu, On the nullity of graphs, Electron. J. Linear Algebra 16 (2007) 60–67. [2] B. Chen, B. Liu, On the nullity of tricyclic graphs, Linear Algebra Appl. 434 (2011) 1799–1810. [3] B. Borovićanin, I. Gutman, Nullity of graphs, in: D. Cvetković, I. Gutman (Eds.), Applications of Graph Spectra, Zb. Rad. 13 (21) (2009), Matematički institut SANU. [4] A.E. Brouwer, W.H. Haemers, Spectra of Graphs, Springer, New York, 2011. [5] D. Cvetković, M. Doob, H. Sachs, Spectra of Graphs – Theory and Applications, Academic Press, New York, 1980. [6] D. Cvetković, I. Gutman, The algebraic multiplicity of the number zero in the spectrum of a bipartite graph, Mat. Vesnik (Beograd) 9 (1972) 141–150. [7] Y.-Z. Fan, K.-S. Qian, On the nullity of bipartite graphs, Linear Algebra Appl. 430 (2009) 2943–2949. [8] S.-C. Gong, G.-H. Xu, On the nullity of a graph with cut-points, Linear Algebra Appl. 436 (2012) 135–142. [9] J.-M. Guo, W.-G. Yan, Y.-N. Yeh, On the nullity and the matching number of unicyclic graphs, Linear Algebra Appl. 431 (2009) 1293–1301. [10] I. Gutman, I. Sciriha, On the nullity of line graphs of trees, Discrete Math. 232 (2001) 35–45. [11] S.-B. Hu, X.-Z. Tan, B.-L. Liu, On the nullity of bicyclic graphs, Linear Algebra Appl. 429 (2008) 1387–1391. [12] G.R. Omidi, On the nullity of bipartite graphs, Graphs Combin. 25 (2009) 111–114. [13] X.-Z. Tan, B.-L. Liu, On the nullity of unicyclic graphs, Linear Algebra Appl. 408 (2005) 212–220. [14] L. Wang, D. Wong, Bounds for the matching number, the edge chromatic number and the independence number of a graph in terms of rank, Discrete Appl. Math. 166 (2014) 276–281.