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Linear and Multilinear Algebra, 2007, 1–17, iFirst
Maximizing spectral radius of unoriented Laplacian matrix over bicyclic graphs of a given order YI-ZHENG FANy, BIT-SHUN TAM*z and JUN ZHOUy ySchool of Mathematics and Computation Sciences, Anhui University, Heifei, Anhui 230039, P.R. China zDepartment of Mathematics, Tamkang University, Tamsui, Taiwan 251, ROC Communicated by J. Y. Shao (Received 12 November 2006; in final form 27 February 2007) For every integer n 4, it is proved that there is a unique graph of order n which maximizes the spectral radius of the unoriented Laplacian matrix over all bicyclic graphs of order n, namely, the graph obtained from the cycle C4 by first adding a chord and then attaching n 4 pendant edges to one end of the chord. Keywords: Unoriented Laplacian matrix; Spectral radius; Bicyclic graph 2000 Mathematics Subject Classifications: 05C50; 15A18
1. Introduction In this article, we consider the problem of determining bicyclic graphs of a given order that maximize the spectral radius of the associated unoriented Laplacian matrix. Unless specified otherwise, by a graph we always mean a simple undirected graph (i.e., one without multi-edges or loops). The (oriented) Laplacian matrix of a graph has been a subject of considerable investigation [10]. In recent years, to extend the classical work, attention has been turned to the study of the Laplacian matrix of a mixed graph or of a signed graph [5,9,11,12]. (The relevant definitions will be given later.) In particular, in [6] the first-named author has determined all of the unicyclic mixed graphs (i.e., those whose underlying graphs are connected and have exactly one cycle) that maximize the largest eigenvalue (or equivalently, the spectral radius) of the Laplacian matrix over the class of all unicyclic mixed graphs of a given order. Since a connected graph is unicyclic if and only if its nullity is one, it is natural
*Corresponding author. Email:
[email protected] Linear and Multilinear Algebra ISSN 0308-1087 print/ISSN 1563-5139 online ß 2007 Taylor & Francis http://www.tandf.co.uk/journals DOI: 10.1080/03081080701306589
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to consider the same problem for mixed graphs of a given order and with fixed nullity. It turns out that to treat the latter problem we need only confine ourselves to the unoriented Laplacian matrix of graphs, a topic which has gained some attention in recent years [1,7]. In this article, we give an answer to our question for the special case when the graphs under consideration are bicyclic, that is, when the fixed nullity is two. The following is our main result: THEOREM For every positive integer n 4, there is a unique bicyclic graph which maximizes the spectral radius of the unoriented Laplacian matrix over all bicyclic graphs of order n, namely, the graph obtained from the cycle C4 by first adding a chord and then attaching n 4 pendant edges to one end of the chord. In section 2, we give most of the definitions and some preparatory results. In particular, we give in Lemma B a specialization of a result on the Laplacian matrix of a mixed graph to the unoriented Laplacian matrix of a graph. The definitions of a mixed graph, a signed graph, their Laplacian matrices, and related concepts (such as a quasi-bipartite graph, etc.) will be postponed till section 4, the final section, as they are not needed in understanding the proof of our main result (provided that the validity of Lemma B is assumed). There has been much study of the Laplacian matrix of a mixed graph and the Laplacian matrix of a signed graph. Actually the two studies are ‘‘equivalent’’ — but we are not aware of a mention of this in the literature — and we will give an explanation in section 4. It will be readily seen that one outcome of this work is that we settle the problem of determining signed graphs (or equivalently, mixed graphs) of a given order and nullity two that have maximum spectral radius for their Laplacian matrix. Another by-product is that, we also settle the problem of determining the graph of a given order and nullity two such that the spectral radius of the adjacency matrix of its line graph is maximized.
2. Preliminaries For a graph G with m edges, n vertices, and k components, by the nullity of G we mean the quantity m n þ k. It is known that the nullity of G is equal to the dimension of the cycle space of G (see, for instance, [2, Chapter 12]). A connected graph with n vertices is also referred to as unicyclic if it has n edges, bicyclic if it has n þ 1 edges, and tricyclic if it has n þ 2 edges (see, for instance, [4, p. 4]). Let G be a graph of order n with vertices v1 , . . . , vn and edges e1 , . . . , em . The (vertex-edge, unoriented) incidence matrix of G is the n m matrix MðGÞ ¼ ðmij Þ given by: mij equals 1 if vertex vi is on edge ej and equals 0 otherwise; if we assign a direction to each of the edges of G, set qij ¼ 1 if vi is the initial vertex of ej, set qij ¼ 1 if vi is the terminal vertex of ej, and set qij ¼ 0 if vi is not of ej, then the n m matrix QðGÞ ¼ ðqij Þ is the (vertex-edge) oriented incidence matrix of G; the adjacency matrix of G is the n n matrix AðGÞ ¼ ðaij Þ given by: aij equals 1 if vertices vi , vj are adjacent and equals 0 otherwise; and the diagonal matrix of vertex degrees of G is the n n diagonal matrix D(G) whose i-th diagonal entry is the degree of the vertex vi. The matrix LðGÞ ¼ DðGÞ AðGÞ, or equivalently, QðGÞQðGÞT , is the well-known (oriented) Laplacian matrix associated with G. Here we are interested in
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The spectral radius of unoriented Laplacian matrix
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K(G), the unoriented Laplacian matrix of G, which is defined to be the n n matrix MðGÞMðGÞT , or equivalently, D(G) þ A(G). It is easy to show that the unoriented Laplacian matrix of a graph is equal to the direct sum of the unoriented Laplacian matrix of its components. So to study the spectral properties of the unoriented Laplacian matrix of a graph, we may restrict our attention to connected graphs. Let G be a connected graph. It is clear from the definition that the unoriented Laplacian matrix K ¼ KðGÞ of G is a symmetric positive semidefinite matrix, which is also an irreducible (entrywise) nonnegative matrix. By the Perron–Frobenius theory, the spectral radius (K) of K is a simple eigenvalue and there is a unique (up to multiples) corresponding positive eigenvector, usually referred to as its Perron vector. On the other hand, since K is positive semidefinite, (K) and 1(K), the largest eigenvalue of K, are the same. By the theory of a symmetric matrix, 1(K) is also equal to the maximum value of the quadratic form yTKy over unit vectors y (i.e., yTy ¼ 1), and moreover the quadratic form attains its maximum value at a unit vector y if and only if y is an eigenvector corresponding to 1(K). So, if x is the unit Perron vector of K, then we have ðKÞ ¼ xT Kx and xT Kx > yT Ky for any unit vector y, different from x. It is known (see, for instance, [4, p. 35]) and not difficult to show that, for any graph G, MðGÞT MðGÞ ¼ 2Im þ AðLG Þ, where A(LG) is the adjacency matrix of the line graph LG of G and m is the number of edges of G. But MðGÞMðGÞT (which is K(G)) and MðGÞT MðGÞ have the same set of nonzero eigenvalues, so we have ðKðGÞÞ ¼ ðAðLG ÞÞ þ 2. Thus, the graphs G of a given order and with fixed nullity that maximize ðAðLG ÞÞ are precisely those that maximize ðKðGÞÞ. By the definition KðGÞ ¼ MðGÞMðGÞT for K(G), we have that for any vector x 2 Rn , where n is the order of G, xT KðGÞx ¼
X
ðxu þ xv Þ2 ,
ð1Þ
uv2EðGÞ
where xu denotes the component of x for the entry that corresponds to the vertex u, and uv denotes the edge joining the adjacent vertices u and v. Besides the above, in our treatment we also rely much on the following graphtheoretic interpretation of the eigenvector equation for the unoriented Laplacian matrix K of G, which is a consequence of the equivalent definition KðGÞ ¼ DðGÞ þ AðGÞ for K(G): If x is an eigenvector of K corresponding to the eigenvalue , then for each vertex u of G we have ð du Þxu ¼
X
xv ,
v2NðuÞ
where du is the degree of the vertex u in G and N(u) is the neighbor set of u in G, i.e., the set of all vertices in G adjacent to u. In particular, if u is a pendant vertex and v is the unique vertex adjacent to u, then, provided that 6¼ 1, we have xu ¼ ð 1Þ1 xv . In dealing with the eigenvector equation, we often make use of the latter fact to eliminate the components of x corresponding to pendant vertices.
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We have the following known bounds for the largest eigenvalue of the Laplacian matrix of a mixed graph: THEOREM A ([9], [12]) Then
Let G be a mixed graph on n vertices which has at least one edge.
ðGÞ þ 1 1 ðLðGÞÞ maxfdðuÞ þ dðvÞ : uv 2 EðGÞg, where (G) is the largest degree of a vertex in G. Moreover, if G is connected, then the first inequality holds as equality if and only if ðGÞ ¼ n 1 and G is quasi-bipartite; and the second inequality holds as equality if and only if G is regular or semi-regular and Gc is quasi-bipartite, where Gc is the mixed graph obtained from G by replacing every oriented edge by an unoriented edge and every unoriented edge by an oriented edge. It can be shown that for a given graph G, the Laplacian matrix of the mixed graph with G as its underlying graph and all of whose edges are unoriented is the same as the unoriented Laplacian matrix K(G) of G. Thus, the above result leads to the following, which is what we need for this work. LEMMA B
Let G be a graph on n vertices which has at least one edge. Then ðGÞ þ 1 1 ðKðGÞÞ maxfdðuÞ þ dðvÞ: uv 2 EðGÞg,
where (G) is the largest degree of a vertex in G. Moreover, if G is connected, then the first inequality holds as equality if and only if G is the star K1, n1 ; and the second inequality holds as equality if and only if G is regular or semi-regular. For convenience, we call a connected graph maximizing, if it maximizes the spectral radius of the unoriented Laplacian matrix over connected graphs of the same order and the same nullity. It is not difficult to see that the star K1, n1 is the unique maximizing graph for connected graphs of order n and nullity 0 (i.e., trees of order n). To see this, let T be a tree of order n. Then for any two adjacent vertices u, v of T we have dðuÞ þ dðvÞ n as u, v cannot have a common adjacent vertex. So by the second inequality (and the equivalent condition for equality) of the preceding lemma, we have 1 ðKðT ÞÞ < n for any tree T of order n that is different from K1, n1 . On the other hand, it is readily checked that for G ¼ K1, n1 , the leftmost side and the rightmost side of the inequalities in the lemma are equal; so we have 1 ðKðGÞÞ ¼ n. This establishes our assertion. As we are going to show, the preceding lemma can also help to narrow down the possible candidates for the maximizing graphs when the fixed nullity is greater than or equal to one. Remark 1 Let G be a connected graph with nullity p (0 p ððn 1Þðn 2ÞÞ=2). For any pair of adjacent vertices u, v of G, the set NðuÞ \ NðvÞ has cardinality at most p. To see this, let n be the order of G and let k be the cardinality of NðuÞ \ NðvÞ. Note that the graph obtained from G by deleting the edges joining vertex u to vertices in NðuÞ \ NðvÞ (k of them in total), is still connected and of order n, so it has at least
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n 1 edges. On the other hand, G has p þ n 1 edges as its nullity is p. So we have k þ ðn 1Þ p þ n 1, which implies k p. LEMMA 1 Let G be a connected graph of order n and with nullity p, 1 p ððn 1Þðn 2ÞÞ=2. If G is maximizing, then the quantity maxfdðuÞ þ dðvÞ: uv 2 EðGÞg takes one of the values n þ 1, n þ 2, . . . , n þ p. Proof
It suffices to show that n þ 1 maxfdðuÞ þ dðvÞ: uv 2 EðGÞg n þ p:
First, by Remark 1, for any pair of adjacent vertices u, v of G, dðuÞ þ dðvÞ ¼ jNðuÞj þ jNðvÞj ¼ jNðuÞ [ NðvÞj þ jNðuÞ \ NðvÞj n þ p: So we have maxfdðuÞ þ dðvÞ: uv 2 EðGÞg n þ p. Let H be a (connected) graph of order of n obtained from K1, n1 by adding p edges. Then ðHÞ ¼ n 1 and H has nullity p. Since H is different from K1, n1 , by Lemma B we have n ¼ ðHÞ þ 1 < 1 ðHÞ 1 ðGÞ maxfdðuÞ þ dðvÞ: uv 2 EðGÞg, g
as desired.
By Lemma 1 and its proof, if G is a connected graph of order n and with nullity 1 and if G is maximizing, then there exists a pair of adjacent vertices u, v such that jNðuÞ \ NðvÞj ¼ 1 and NðuÞ [ NðvÞ ¼ VðGÞ or, in other words, G is necessarily a graph of the form U0 ðr, s; nÞ, as given by figure 1. Here, r, s are nonnegative integers with r s such that n ¼ r þ s þ 3. (These graphs were introduced in [6], where they were denoted differently.) By [6], which is done in the context of mixed graphs, the maximizing graph for the class of unicyclic graphs of order n is (up to graph isomorphism) unique and is equal to U0 ðn 3, 0; nÞ. Indeed, as we will indicate, the argument given in the beginning part of the proof for our main result can be adapted to yield the latter result. We will also need the following general result concerning graphs attaching. For two vertex-disjoint graphs H and H0 , if we attach the graph H0 to the graph H so that some vertex w0 of H0 merges with a vertex w of H and also edges of H0 that are incident on w0 and edges of H that are incident on w are now incident on the new vertex, then
Figure 1.
Candidates for maximizing unicyclic graphs.
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we say the resulting graph is obtained from H by attaching H0 (with the vertex w0 ) at the vertex w. e be the graph obtained from H by attaching LEMMA 2 Let H be a connected graph. Let G the star Kk1, 1 with the vertex of degree k 1 at a vertex u of H. Then for any graph G obtained from H by attaching a tree T of order k with a vertex of degree less than k 1 at the vertex u, we have e > 1 ðKðGÞÞ: 1 ðKðGÞÞ e are clearly of the same order. They also have the same Proof The graphs G and G number of edges, because the edges of T and those of Kk1, 1 are in one-to-one correspondence and the edges of H are their common edges. With a suitable choice e the adjacency matrix of the line graph L of the orderings of edges in G and G, e G is (entrywise) greater than or equal to, but not equal to, that of LG. But both matrices are irreducible, symmetric and nonnegative, as the graphs are connected. So by the Perron–Frobenius theory for an irreducible nonnegative matrix, the desired strict inequality follows. g
3. Proofs Since our graphs are not allowed to have loops, the smallest bicyclic graph is of order 4 and (up to isomorphism) there is only one such graph, namely, the graph obtained from the cycle C4 by adding a chord. (Recall that a chord of a cycle is an edge that joins two nonconsecutive vertices of the cycle.) So for n ¼ 4, we have no problem. For n 5, we introduce nine graphs of order n that, as we will see shortly, are the only possible candidates for maximizing bicyclic graph(s) of order n, namely, U1 ðr, s; nÞ, U12 ðr, s; nÞ, U22 ðr, s; nÞ, U3 ðr, s; nÞ, U14 ðr, s; nÞ, U24 ðr, s; nÞ, U5 ðr, s; nÞ, U16 ðr, s; nÞ, U26 ðr, s; nÞ. (See figure 2 below.) Here n specifies the order of the graph, r, s are nonnegative integers with r s; the parameters r, s, n are related by n ¼ r þ s þ 4 if the graph is U1 ðr, s; nÞ or Uj4 ðr, s; nÞ ( j ¼ 1, 2Þ, and by n ¼ r þ s þ 5 if the graph is Uj2 ðr, s; nÞ, Uj6 ðr, s; nÞ ( j ¼ 1, 2), U3 ðr, s; nÞ or U5 ðr, s; nÞ, and in case the graph is U16 ðr, s; nÞ (respectively, U26 ðr, s; nÞ) the parameter r (respectively, s) takes positive integral values only. The graphs U1 ðr, s; nÞ, Uj4 ðr, s; nÞ, U5 ðr, s; nÞ, and Uj6 ðr, s; nÞ ( j ¼ 1, 2) can all be obtained from U1 ð0, 0; 4Þ, the unique bicyclic graph of order 4, as follows. The graph U1 ðr, s; nÞ is obtained from U1 ð0, 0; 4Þ by attaching r pendant edges to one end of the chord and s pendant edges to the other end (where r, s are nonnegative integers such that r þ s ¼ n 4). The graph U14 ðr, s; nÞ (respectively, U24 ðr, s; nÞ) is obtained from U1 ð0, 0; 4Þ by attaching r (respectively, s) pendant edges to one of the two ends of the chord and s (respectively, r) pendant edges to one of the two vertices that are not incident with the chord. (Note that for r > s, the graphs U14 ðr, s; nÞ and U24 ðr, s; nÞ are different.) The graph U5 ðr, s; nÞ is obtained from U1 ðr, s; n 1Þ by attaching a pendant edge to one of the two vertices that are of degree two. The graph U16 ðr, s; nÞ (respectively, U26 ðr, s; nÞ) is obtained from U1 ðr, s; n 1Þ by attaching a pendant edge to a pendant vertex which is adjacent to the end of the chord that is of degree r þ 3 (respectively, of degree s þ 3).
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The spectral radius of unoriented Laplacian matrix
Figure 2.
Candidates for maximizing bicyclic graphs.
The graph U3 ðr, s; nÞ is obtained from a 5-cycle by first adding a chord and then attaching r pendant edges at one end of the chord and s pendant edges at the other end. The graph U12 ðr, s; nÞ (respectively, U22 ðr, s; nÞ) is obtained from a graph consisting of two 3-cycles with a common vertex by attaching r (respectively, s) pendant edges to the common vertex and s (respectively, r) pendant edges to any other vertex of the two 3-cycles. Our main theorem amounts to say that U1 ðn 4, 0; 4Þ is the unique maximizing graph for bicyclic graphs of order n ( 4). LEMMA 3 For a bicyclic graph G of order n 5, maxfdðuÞ þ dðvÞ : uv 2 EðGÞg equals n þ 2 if and only if G ¼ U1 ðr, s; nÞ and equals n þ 1 if and only if G is one of the graphs Uj2 ðr, s; nÞ, U3 ðr, s; nÞ, Uj4 ðr, s; nÞ, U5 ðr, s; nÞ or Uj6 ðr, s; nÞ ð j ¼ 1, 2Þ. Proof It is straightforward to verify the ‘‘if ’’ parts. To prove the ‘‘only if ’’ parts, let G be a bicyclic graph of order n 5 such that the maximum value of the sum of degrees of pairs of adjacent vertices of G is n þ 2 or n þ 1. Suppose that the maximum value is attained at the pair of vertices u, v. By the proof of Lemma 2, if d(u) þ d(v) equals n þ 2, then jNðuÞ [ NðvÞj ¼ n and jNðuÞ \ NðvÞj ¼ 2; if d(u) þ d(v) equals n þ 1,
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then we have either jNðuÞ [ NðvÞj ¼ n and jNðuÞ \ NðvÞj ¼ 1 or jNðuÞ [ NðvÞj ¼ n 1, and jNðuÞ \ NðvÞj ¼ 2. It is readily seen that if jNðuÞ [ NðvÞj ¼ n and jNðuÞ \ NðvÞj ¼ 2, then G must be of the form U1 ðr, s; nÞ for some nonnegative integers r, s with r s such that n ¼ r þ s þ 4. Consider the case when jNðuÞ [ NðvÞj ¼ n and jNðuÞ \ NðvÞj ¼ 1. Say, NðuÞ \ NðvÞ ¼ fwg, R :¼ NðuÞnfv, wg has r0 elements, S :¼ NðvÞnfu, wg has s0 elements, and r0 s0 . Then G has n ¼ r0 þ s0 þ 3 vertices and there are altogether n edges that are incident with either u or v. As G is bicyclic, there is a remaining edge, which is not incident with u or v. The remaining edge may join two vertices of R or two vertices of S, or it may join a vertex of R to a vertex of S, or join the vertex w to a vertex of R or S. If it joins two vertices of R, then G is equal to U12 ðr, s; nÞ or U22 ðs, r; nÞ, where r ¼ r0 2 and s ¼ s0 , depending on whether r s or r s. Likewise, if the edge joins two vertices of S, G is also of the form Uj2 ðr, s; nÞ ( j ¼ 1, 2). If the edge joins a vertex of R to a vertex of S, then G equals U3 ðr, s; nÞ where r ¼ r0 1 and s ¼ s0 1. If the edge joins w to a vertex of R, then G equals U14 ðr, s; nÞ or U24 ðs, r; nÞ, where r ¼ r0 1 and s ¼ s0 , depending on whether r s or r s. (In the diagram for U14 ðr, s; nÞ as given in figure 1, the vertices u, w, v play respectively the role of u, v, w here.) Likewise, if it joins w to a vertex of S, then G is of the form Uj4 ðr, s; nÞ for some j ¼ 1, 2. It remains to consider the case when jNðuÞ [ NðvÞj ¼ n 1 and jNðuÞ \ NðvÞj ¼ 2. In this case, it is not difficult to see that the subgraph of G induced by NðuÞ [ NðvÞ is of the form U1 ðr, s; n 1Þ where r, s are nonnegative integers with r s such that n 1 ¼ r þ s þ 4. Furthermore, G is obtained from this induced subgraph by adding a pendant edge which is not incident with u or v. If the pendant edge is attached to a vertex adjacent to u, then G is U16 ðr, s; nÞ; if it is attached to a vertex adjacent to v, then G is U26 ðr, s; nÞ; for the remaining case, G is U5 ðr, s; nÞ. The proof is complete. g Proof of Theorem By Lemmas 1 and 3, the maximizing bicyclic graphs of order n (5) are among Ui ðr, s; nÞ, i ¼ 1, 3, 5 and Uji ðr, s; nÞ, i ¼ 2, 4, 6 and j ¼ 1, 2. We need to show that the largest eigenvalue of the unoriented Laplacian matrix of U1 ðn 4, 0; nÞ is greater than those of the other graphs in the preceding list. ASSERTION 1 1 ðKðU1 ðn 4, 0; nÞÞÞ > 1 ðKðU1 ðr, s; nÞÞÞ for all positive integers r, s with r s and n ¼ r þ s þ 4. It suffices to show that for any positive integers r, s with r s and n ¼ r þ s þ 4, we have 1 ðKðU1 ðr þ 1, s 1; nÞÞÞ > 1 ðKðU1 ðr, s; nÞÞÞ: We label the vertices of the graph U1 ðr, s; nÞ as in figure 2. The graph U1 ðr þ 1, s 1; nÞ can be obtained from U1 ðr, s; nÞ by replacing the edge vvs by the edge vvs. For the unit Perron vector x of KðU1 ðr, s; nÞÞ, we have 1 ðKðU1 ðr þ 1, s 1; nÞÞÞ xT KðU1 ðr þ 1, s 1; nÞÞx, and xT KðU1 ðr, s; nÞÞx ¼ 1 ðKðU1 ðr, s; nÞÞÞ:
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In view of (1), we readily obtain xT KðU1 ðr þ 1, s 1; nÞÞx xT KðU1 ðr, s; nÞÞx ¼ ðxu þ xvs Þ2 ðxv þ xvs Þ2 ¼ ðxu þ xv þ 2xvs Þðxu xv Þ, where in evaluating the value of xT KðU1 ðr þ 1, s 1; nÞÞx (respectively, of xT KðU1 ðr, s; nÞÞx), we index the components of x by the vertices of U1 ðr þ 1, s 1; nÞ (respectively, by the vertices of U1 ðr, s; nÞ). The desired inequality 1 ðK1 ðUðr þ 1, s 1; nÞÞÞ > 1 ðK1 ðUðr, s; nÞÞÞ will follow if we have xu > xv . For convenience, denote 1 ðKðU1 ðr, s; nÞÞÞ by . From the eigenvector equation for the Perron vector x of KðU1 ðr, s; nÞÞ, we have xui ¼ ð 1Þ1 xu
for i ¼ 1, . . . , r,
xvj ¼ ð 1Þ1 xv
for j ¼ 1, . . . , s,
and
together with ð ðr þ 3ÞÞxu ¼
rxu þ xw þ xz þ xv , 1
ð2Þ
ð ðs þ 3ÞÞxv ¼
sxv þ x w þ xz þ xu : 1
ð3Þ
and
Subtracting (3) from (2) and rearranging terms, we obtain ðr þ 2Þ
r s xu ¼ ðs þ 2Þ xv : 1 1
ð4Þ
By Lemma B, we have ðU1 ðr, s; nÞÞ þ 1 ¼ dðuÞ þ 1 ¼ r þ 4; so the coefficient of xu in (4) is positive. When r > s, it is readily seen that the coefficient of xv in (4) is greater than that of xu; hence xu > xv , as desired. When r ¼ s, we have and 1 ðKðU1 ðr þ 1, s 1; nÞÞÞ xu ¼ xv , xT KðU1 ðr þ 1, s 1; nÞÞx ¼ xT KðU1 ðr, sÞÞx, 1 ðKðU1 ðr, s; nÞÞÞ. If the latter weak inequality is an equality, then x must be the unit Perron vector of KðU1 ðr þ 1, s 1; nÞÞ, and by the eigenvector equation for x as the Perron vector of KðU1 ðr þ 1, s 1; nÞÞ we have ð ðr þ 4ÞÞxu ¼
ðr þ 1Þxu þ xw þ xz þ xv : 1
ð5Þ
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Subtracting (5) from (2), we obtain xu ¼ xu =ð 1Þ, which is a contradiction. So when r ¼ s, we still have the desired strict inequality 1 ðK1 ðUðr þ 1, s 1; nÞÞÞ > 1 ðK1 ðUðr, s; nÞÞÞ. We take a digression. To complete the discussion about maximizing the spectral radius of the unoriented Laplacian matrix over unicyclic graphs that we began near the end of section 2 (before Lemma 2), note that a slight modification of the proof for Assertion 1 yields the inequality 1 ðKðU0 ðn 3, 0; nÞÞÞ > 1 ðKðU0 ðr, s; nÞÞÞ for all positive integers r, s with r s and n ¼ r þ s þ 3. Now back to the proof of our main result. ASSERTION 2 1 ðKðU1 ðn 4, 0; nÞÞÞ > 1 ðKðUj6 ðr, s; nÞÞÞ for j ¼ 1, 2 and any positive integers r, s (where s is allowed to take the value 0 when j ¼ 1) with r s such that n ¼ r þ s þ 5. Again we refer to the labeled graphs given in figure 2. Note that the graph U16 ðr, s; nÞ is obtained from the graph U1 ðr 1, s; n 2Þ (which should be U1 ðs, r 1; n 2Þ when r ¼ s) by attaching the star K2, 1 with one of the pendant vertices at the vertex u. However, if we attach the star K2, 1 to Uðr 1, s; n 2Þ in a different manner, namely, with the vertex of degree 2 at the vertex u, then we obtain the graph U1 ðr þ 1, s; nÞ. So by Assertion 1 and Lemma 2, we have 1 ðKðU1 ðn 4, 0; nÞÞÞ 1 ðKðU1 ðr þ 1, s; nÞ > 1 ðKðU16 ðr, s; nÞÞÞÞ: In a similar manner, we can also establish the inequality 1 ðKðU1 ðn 4, 0; nÞÞÞ > 1 ðKðU26 ðr, s; nÞÞÞ. ASSERTION 3 1 ðKðU1 ðn 4, 0; nÞÞÞ > 1 ðKðUj2 ðr, s; nÞÞÞ for nonnegative integers r, s with r s such that n ¼ r þ s þ 5.
j ¼ 1, 2
and
for
all
We treat the case j ¼ 1 first. For simplicity, we denote 1 ðKðU12 ðr, s; nÞÞÞ by . Note that if we replace the edge pq of U12 ðr, s; nÞ by the edge qv (and relabel the vertices p, q respectively as urþ1 and z), we obtain the graph U1 ðr þ 1, s; nÞ. In view of Assertion 1, it suffices to show that 1 ðKðU1 ðr þ 1, s; nÞÞÞ > 1 ðKðU12 ðr, s; nÞÞÞ. The idea of the proof is similar to that for Assertion 1. Denoting by x the unit Perron vector of KðU12 ðr, s; nÞÞ, we have xT KðU1 ðr þ 1, s; nÞÞx xT KðU12 ðr, s; nÞÞx ¼ ðxv þ xq Þ2 ðxp þ xq Þ2 ¼ ðxv þ xp þ 2xq Þðxv xp Þ: For the graph U12 ðr, s; nÞ, we have dðuÞ ¼ r þ 4, dðvÞ ¼ s þ 2 and n ¼ r þ s þ 5. In this case, ðGÞ ¼ dðuÞ and by Lemma B, dðuÞ þ 1 5. From the eigenvector equation for the Perron vector x of U12 ðr, s; nÞ we have ð 2Þxw ¼ xu þ xv , s xv þ xu þ xw , ð s 2Þxv ¼ 1 ð 2Þxq ¼ xp þ xu , ð 2Þxp ¼ xq þ xu :
ð6Þ ð7Þ ð8Þ ð9Þ
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From (8) and (9), we obtain xu ¼ ð 3Þxp (and xp ¼ xq). Add ð 2Þ times (7) to (6). In the resulting equation, express xu in terms of xp as given above. After rearranging terms, we have xv ¼ xp , where ¼ s2
s ð 2Þ 1 1
and ¼ ð 1Þð 3Þ:
A little algebraic manipulation shows that >
if and only if
s ð 2Þ > 0: 1
So when s > 0, we always have > and hence xv > xp. Then the desired inequality 1 ðKðU1 ðr þ 1, s; nÞÞÞ > 1 ðKðU12 ðr, s; nÞÞÞ follows. When s ¼ 0, we have ¼ ð 2Þ2 1 ¼ and hence xv ¼ xp . By what we have done above, the latter implies that xT KðU1 ðr þ 1, s; nÞÞx ¼ xT KðU12 ðr, s; nÞÞx. In this case, we have U12 ðr, s; nÞ ¼ U12 ðn 5, 0; nÞ and U1 ðr þ 1, s; nÞ ¼ U1 ðn 4, 0; nÞ. Moreover, 1 ðKðU1 ðn 4, 0; nÞÞÞ xT KðU1 ðn 4, 0; nÞÞx ¼ xT KðU12 ðn 5, 0; nÞÞx ¼ 1 ðKðU12 ðn 5, 0; nÞÞÞ: Should the weak inequality be an equality, x would be the unit Perron vector of KðU1 ðn 4, 0; nÞÞ. From the eigenvector equation for x as the Perron vector of KðU1 ðn 4, 0; nÞÞ we obtain ð 3Þxv ¼ xw þ xu þ xq : From (6) and (7), we also have ð 1Þðxw xv Þ ¼ 0 and hence xw ¼ xv. Since we have already shown that xv ¼ xp and xp ¼ xq , the above relation can be rewritten as ð 5Þxv ¼ xu . But by (6), we also obtain ð 3Þxv ¼ xu , which is a contradiction. This proves that 1 ðKðU1 ðn 4, 0; nÞÞÞ > 1 ðKðU12 ðn 5, 0; nÞÞÞ. For the case j ¼ 2, if we replace the edge pq by the edge qv, the resulting graph is U1 ðr, s þ 1; nÞ or U1 ðs þ 1, r; nÞ depending on whether r s þ 1 or not. Using an argument similar to the one for the case j ¼ 1, we obtain the four equations (6)–(9) except that in (7) the parameter s is to be replaced by r. Then we get 0 xv ¼ xp , where is the same as before and 0 is obtained from by replacing s by r. Also, we have > 0 if and only if ðr=ð 1ÞÞð 2Þ > 0 if and only if r > 0. When r > 0 we have xv > xp and we are done. When r ¼ 0, necessarily s ¼ 0. Then the graph U22 ðr, s; nÞ becomes U22 ð0, 0; 5Þ, which is the same as U12 ð0, 0; 5Þ. By what we have just proved, 1 ðKðU12 ð0, 0; 5ÞÞÞ < 1 ðKðU1 ð1, 0; 5ÞÞÞ. So the assertion follows. ASSERTION 4 1 ðKðU1 ðn 4, 0; nÞÞÞ > 1 ðKðU3 ðr, s; nÞÞÞ for all nonnegative integers r, s with r s such that n ¼ r þ s þ 5. To change from U3 ðr, s; nÞ to U1 ðr þ 1, s; nÞ, we replace the edge pq by the edge uq. In this case, it suffices to show xu > xp , where x is the unit Perron vector of
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KðU3 ðr, s; nÞÞ. Denote 1 ðKðU3 ðr, s; nÞÞÞ by . By considering the eigenvector equation for the Perron vector x of KðU3 ðr, s; nÞÞ, we have ð 2Þxq ¼ xp þ xv ,
ð10Þ
ð 2Þxp ¼ xu þ xq ,
ð11Þ
ð 2Þxw ¼ xu þ xv , s xv ¼ xu þ xq þ xw : 3s 1
ð12Þ ð13Þ
Adding ð 2Þ times (13) to (12), we have h
ð 2Þ 3 s
i s 1 xv ¼ ð 1Þxu þ ð 2Þxq : 1
ð14Þ
Solving for xq and xv in terms of xp and xu from (10) and (11) and substituting them into (14), we obtain xp ¼ xu , where h ¼ ð 2Þ 3 s
i s 1 ð 2Þ2 1 ð 2Þ2 1
and ¼ 1 þ ð 2Þ
h
3s
i s ð 2Þ 1 : 1
A little calculation yields h ¼ ð 2Þ ð2 5 þ 5Þ 3 s
i s þ 2 5 : 1
In view of Lemma B, > maxfdðuÞ, dðvÞg þ 1 ¼ maxfr þ 4, s þ 4g 4. So we have 2 > 0 and 2 5 > 0. Since the largest root of the quadratic polynomial t2 5t þ 5 is approximately 3.62, we also have 2 5 þ 5 > 0. In addition, we have 3s
s s > ð4 þ sÞ 3 þ s þ 1 1 1s ¼ 1 > 0:
Therefore, we have > and so xu > xp, as desired. ASSERTION 5 1 ðKðU1 ðn 4, 0; nÞÞÞ is greater than 1 ðKðU24 ðn 4, 0; nÞÞÞ and 1 ðKðUj4 ðr, s; nÞÞÞ for j ¼ 1, 2 and every pair of positive integers r, s with r s such that n ¼ r þ s þ 4. Note that we need not consider the graph U14 ðn 4, 0; nÞ as it is the same as U1 ðn 4, 0; nÞ.
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We deal with the case j ¼ 1 first. Replacing the edge zv of U14 ðr, s; nÞ by the edge zw, we obtain a graph isomorphic with U1 ðr, s; nÞ. Considering the eigenvector equation for the Perron vector x of KðU14 ðr, s; nÞÞ, we have
ð 3Þxv ¼ xz þ xu þ xw , s xw ¼ xv þ xu , 2s 1 ð 2Þxz ¼ xv þ xu :
ð15Þ ð16Þ ð17Þ
Subtracting (17) from (16) yields 2s
s xw ¼ ð 2Þxz : 1
ð18Þ
Subtracting (17) from (15), we have ð 2Þxv ¼ ð 1Þxz þ xw :
ð19Þ
From (17) and (18), eliminating xz we obtain xw ¼ xv , where ¼ ð 1Þð 2 sÞ s þ 2
and ¼ ð 2Þ2 :
If s 2, a little calculation will show that ¼ ðs 2Þ þ 4 ðs þ 4Þðs 2Þ þ 4, where the inequality holds as > dðuÞ þ 1 ¼ r þ 4 s þ 4. So we have > and hence xw > xv . Then by an argument that we have already used several times, we obtain 1 ðKðU14 ðr, s; nÞÞÞ < 1 ðKðU1 ðr, s; nÞÞÞ. If s ¼ 1, then ¼ 4 < 0 and so xw < xv . In this case, we obtain the graph U1 ðn 5, 1; nÞ from U14 ðn 5, 1; nÞ by replacing the edge w1 w by w1 v. As xw < xv , we still get 1 ðKðU14 ðr, s; nÞÞÞ < 1 ðKðU1 ðr, s; nÞÞÞ. In any case, in view of Assertion 1, Assertion 5 follows. For the case j ¼ 2, we obtain a graph isomorphic with U1 ðr, s; nÞ from U24 ðr, s; nÞ by replacing the edge zv by zw when r 2, but when r ¼ 1 we replace the edge w1 w by w1 v. (Also recall that in this case s is allowed to take the value 0.) By considering the eigenvector equation for the Perron vector x of KðU24 ðr, s; nÞÞ we still obtain equations (15)–(17), except that in (16) s is to be replaced by r. With r in place of s, the rest of the argument is the same as for the previous case. ASSERTION 6 1 ðKðU1 ðn 4, 0; nÞÞÞ > 1 ðKðU5 ðr, s; nÞÞÞ for all nonnegative integers r, s with r s such that n ¼ r þ s þ 5. For the graph U5 ðr, s; nÞ we have > 1 þ maxfdðuÞ, dðvÞg ¼ maxfr þ 4, s þ 4g 4. In this case we obtain the graph U1 ðr þ 1, s; nÞ from U5 ðr, s; nÞ by replacing the edge wp by the edge up. It suffices to show that xu > xw, where x is the unit Perron vector
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of KðU5 ðr, s; nÞÞ. Considering the eigenvector equation for the Perron vector, we obtain (17) and the following equations: 3
1 xw ¼ xu þ xv , 1 s xv ¼ xu þ xz þ xw , 3s 1 r 3r xu ¼ xv þ xz þ xw : 1
ð20Þ ð21Þ ð22Þ
From (21) and (22), we have 2s
s r xv ¼ 2 r xu : 1 1
ð23Þ
Adding ð 2 s ðs=ð 1ÞÞÞ times (20) to (23), we obtain uðsÞuð1Þxw ¼ ðuðsÞ þ uðrÞÞxu , where uðtÞ :¼ 2 t
t : 1
Since uðsÞ uðrÞ (as r s), we have uðsÞuð1Þ uðsÞ uðrÞ uðsÞðuð1Þ 2Þ ¼
1 uðsÞð2 6 þ 4Þ: 1
As > s þ 4, by the last part of the proof of Assertion 4, we have uðsÞ > 0. So, to establish xu > xw , it suffices to show 2 6 þ 4 > 0:
ð24Þ
If r 2, then > r þ 4 6 and inequality (24) follows. For r ¼ 1 or 0, the graph U5 ðr, s; nÞ becomes U5 ð1, 1; 7Þ, U5 ð1, 0; 6Þ or U5 ð0, 0; 5Þ. By Assertion 5, we can dispense with the latter two cases as U5 ð1, 0; 6Þ ¼ U14 ð1, 1; 6Þ and U5 ð0, 0; 5Þ ¼ U24 ð1, 0; 5Þ. It remains to consider the graph U5 ð1, 1; 7Þ. By (23) (with r ¼ s ¼ 1) and the fact that 3 ð1=ð 1ÞÞ 6¼ 0 (as > r þ 4 ¼ 5) we have xu ¼ xv ; say xu ¼ a. By (17) and (20) respectively, we obtain xz ¼
2a 2
and
xw ¼
2a : 3 ð1=ð 1ÞÞ
Then by (22), we have 2 6 þ 4 2 2ð 1Þ ¼ þ : 1 2 2 4 þ 2
ð25Þ
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As > r þ 4 ¼ 5, the right side of (25) is positive; so (24) holds. This establishes Assertion 6. The proof is complete. g
4. Mixed graphs versus signed graphs A signed graph is a graph with a sign attached to each of its edges. Formally, a signed graph ¼ ðG, Þ consists of a graph G ¼ ðV, E Þ with vertex set V and edge set E, referred to as its underlying graph, and a mapping : E ! fþ, g, the edge labeling. By the Laplacian matrix of , denoted by L(), we mean the matrix DðÞ AðÞ, where D() is the diagonal matrix of vertex degrees of the underlying graph G and AðÞ ¼ ðaij Þ is the signed adjacency matrix of given by aij ¼ ðvi vj Þaij , where ðaij Þ is the adjacency matrix of the underlying graph G. If G is a graph, and if is the all-positive edge labeling signed graph with G as the underlying graph, then the (classical, oriented) Laplacian matrix of G and the Laplacian matrix of the signed graph are equal. On the other hand, if we take to be the all-negative edge labeling signed graph with G as the underlying graph, then the Laplacian matrix of is equal to the unoriented Laplacian matrix of G. For a complex matrix C ¼ ðcij Þ, denote by jCj the matrix ðjcij jÞ. For the signed graph ¼ ðG, Þ, it is clear that jLðGÞj equals DðGÞ þ jAðÞj and the latter, in turn, is the unoriented Laplacian matrix of G. But for any square irreducible complex matrix C, we have ðC Þ ðjCjÞ, where the equality holds if and only if C ¼ DAD1 for some nonnegative nonsingular diagonal matrix D and some complex number with modulus 1 (see Theorem 8.4.5 of [8]); so we have ðLðÞÞ ðKðGÞÞ, and the inequality becomes equality when, in particular, is the all-negative edge labeling signed graph. This explains why in considering the problem of maximizing the spectral radius of the Laplacian matrix over the class of connected signed graphs (or mixed graphs) of a given order and with fixed nullity, we need only confine ourselves to the unoriented Laplacian matrix of graphs. A mixed graph is a graph with some oriented and some unoriented edges. For a mixed graph G with n vertices and m edges, but no multi-edges and loops, by the oriented incidence matrix of G we mean the n m matrix M ¼ MðGÞ ¼ ðmij Þ whose entries are given by: mij ¼ 1 if ej is an unoriented edge incident with vi or ej is an oriented edge with initial vertex vi, mij ¼ 1 if ej is an oriented edge with terminal vertex vi, and mij ¼ 0 otherwise. (In the literature, people have used the term ‘‘incidence matrix’’ instead of ‘‘oriented incidence matrix.’’ We prefer to reserving the former for the incidence matrix of the underlying graph.) Then the Laplacian matrix (also referred to as ‘‘mixed Laplacian matrix’’ in [1]) of G is defined as LðGÞ ¼ MMT . One readily shows that L(G) is also equal to D(G) þ A(G), where D(G) is the diagonal matrix of vertex degrees of the underlying graph of G and AðGÞ ¼ ðaij Þ is the oriented adjacency matrix of G given by: aij equals 0 if there is no edge joining vi and vj, and equals 1 or 1 depending on whether the edge joining vi and vj is unoriented or oriented. For a given graph, if we transform it into a mixed graph by assigning to each edge one of the two possible orientations, then the Laplacian matrix of the resulting mixed graph is equal to the (classical) Laplacian matrix of the graph. If, however, we transform the graph into the mixed graph with all edges unoriented, then the
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Laplacian matrix of the resulting mixed graph is equal to the unoriented Laplacian matrix of the original graph. The study of the Laplacian matrix of a mixed graph and that of a signed graph are equivalent. This can be seen as follows. To communicate better, let us be informal. (The formal definition of a mixed graph as given in [11] is subject to criticism. To give a formal definition, one can introduce an incidence function that associates to each edge of the mixed graph an unordered pair or an ordered pair, depending on whether the edge is unoriented or oriented (cf. the definitions of a graph and of a directed graph as given in [2]). However, such formal definition is not really needed in practice.) Given a mixed graph G, one can transform it into a signed graph by keeping the same underlying graph and defining a signing on the edge set by (e) equals þ1 or 1 depending on whether e is an oriented or unoriented edge of the mixed graph. Then the oriented adjacency matrix of G and the signed adjacency matrix of are related by AðGÞ ¼ AðÞ, and G and have the same Laplacian matrix. (Here we are using the definitions of Laplacian matrix for signed graphs and mixed graphs as appeared in [9] and [1] respectively.) Conversely, given any signed graph , one can transform it into a mixed graph G by keeping the same underlying graph and replacing each negative edge by an unoriented edge and each positive edge by an oriented edge (in one of the two possible ways). Then and G have the same Laplacian matrix. The concepts of a quasi-bipartite graph for mixed graphs and a balanced graph for signed graphs also correspond to each other. This can also be seen as follows. By definition, a cycle in a mixed graph is nonsingular if the cycle, as a mixed graph in its own right, has nonsingular incidence matrix (or equivalently, nonsingular Laplacian matrix). It is known ([1], Lemma 1) that a cycle in a mixed graph is nonsingular if and only if it contains an odd number of unoriented edges. If we transform a mixed graph into a signed graph in the above manner, then a nonsingular cycle in the mixed graph will become a cycle with an odd number of negative edges or, in other words, a negative cycle. By definition, a mixed graph is quasi-bipartite if it has no nonsingular cycles. When transformed into a signed graph, a quasi-bipartite graph will become a signed graph all of whose cycles are positive (i.e., having an even number of negative edges), or in other words, a balanced signed graph. So, results on the Laplacian matrix of a mixed graph can be translated into results on the Laplacian matrix of a signed graph, and vice versa. In this sense, there is much overlap between the work of [1,11,12] on mixed graphs and the work of [9] (and other papers) on signed graphs. For conceptual clarity, it seems better to give the formulation of results on the Laplacian matrix in the setting of signed graphs, as the orientation of the oriented edges of a mixed graph does not affect the definition of its Laplacian matrix.
Acknowledgements Yi-Zheng Fan was supported by National Natural Science Foundation of China (Grant No. 10601001), Anhui Provincial Natural Science Foundation (Grant No. 050460102), NSF of Department of Education of Anhui province (Grant No. 2004kj027, 2005kj005zd), the Project of Mathematical Innovation Team of Anhui
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University, and the Project of Talents Group Construction of Anhui University. Bit-Shun Tam was supported by National Science Council of the Republic of China.
References [1] Bapat, R.B., Grossman, J.W. and Kulkarni, J.W., 1999, Generalized matrix tree theorem for mixed graphs. Linear Multilinear Algebra, 46, 299–312. [2] Bondy, J.A. and Murty, U.S.R., 1976, Graph Theory with Applications (New York: American Elsevier Publishing Co.). [3] Brualdi, R. and Ryser, H., 1991, Combinatorial Matrix Theory (New York: Cambridge University Press). [4] Cvetkovic´, D., Rowlinson, P. and Simic´, S., 1997, Eigenspaces of Graphs (Cambridge: Cambridge University Press). [5] Fan, Y.-Z., 2003, On spectral integral variations of mixed graphs. Linear Algebra and its Applications, 374, 307–316. [6] Fan, Y.-Z., 2004, Largest eigenvalue of a unicyclic mixed graph. Applied Mathematics. A Journal of Chinese Universities, Series B, 19, 140–148. [7] Grossman, J.W., Kulkarni, D.M. and Schochetman, I.E., 1994, Algebraic graph theory without orientation. Linear Algebra and its Applications, 212/213, 289–307. [8] Horn, R.A. and Johnson, C.R., 1985, Matrix Analysis (Cambridge: Cambridge University Press). [9] Hou, Y.-P., Li, J.-S. and Pan, Y.-L., 2003, On the Laplacian eigenvalues of signed graphs. Linear Multilinear Algebra, 51, 21–30. [10] Merris, R., 1994, Laplacian matrices of graphs: a survey, Linear Algebra and its Applications, 197/198, 143–176. [11] Zhang, X.-D. and Li, J.-S., 2002, The Laplacian spectrum of a mixed graph. Linear Algebra and its Applications, 353, 11–20. [12] Zhang, X.-D. and Lou, R., 2003, The Laplacian eigenvalues of a mixed graph. Linear Algebra and its Applications, 362, 109–119.
