skew spectra of graphs without even cycles
TRANSCRIPT
Linear Algebra and its Applications 444 (2014) 67–80
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Linear Algebra and its Applications
www.elsevier.com/locate/laa
Skew spectra of graphs without even cycles
A. Anuradha a, R. Balakrishnan a, Wasin So b,∗a Department of Mathematics, Bharathidasan University, Tiruchirappalli, 620024, Indiab Department of Mathematics, San Jose State University, San Jose, CA 95192-0103, United States
a r t i c l e i n f o a b s t r a c t
Article history:Received 19 March 2013Accepted 20 November 2013Available online 7 December 2013Submitted by R. Brualdi
MSC:05C50
Keywords:Skew spectrumGraph without even cyclesFriendship graphHosoya index
The study of the invariance of the skew spectrum of an orientedgraph under different orientations leads to the special family G ofgraphs without even cycles. Basic graphical and spectral propertiesof G are observed. As a result, the Hosoya index of a graphin G can be computed via its skew spectrum. We also focus onfriendship graphs, a subfamily of G . In particular, the conditionunder which the skew spectrum of a friendship graph is actually itimes the spectrum of some graph is determined, where i = √−1.
© 2013 Elsevier Inc. All rights reserved.
1. Introduction
Let G = (V , E) be a finite simple undirected graph of order n with V = {v1, v2, . . . , vn} as itsvertex set and E as its edge set. The adjacency matrix of G is the n × n matrix A(G) = (aij), whereaij = 1 = a ji if (vi, v j) ∈ E and aij = 0 otherwise. Clearly A(G) is a real symmetric matrix and henceits eigenvalues are all real. An orientation σ of E makes G into an oriented graph Gσ = (V ,Γ ), whereΓ stands for the arc set of Gσ . Recall that (vi, v j) ∈ Γ denotes the arc with head v j and tail vi . Theskew adjacency matrix of the oriented graph Gσ is the n × n matrix S(Gσ ) = (si j), where si j = 1 =−s ji whenever (vi, v j) ∈ Γ and si j = 0 otherwise. Clearly S(Gσ ) is a real skew symmetric matrixand hence its eigenvalues are all pure imaginary. We suppose that the characteristic polynomials ofA(G) and S(Gσ ) are given by χ(G : x) = xn + a1xn−1 + · · · + an and χ(Gσ : x) = xn + b1xn−1 + · · · + bnrespectively. The spectrum of the matrix A(G) is the set of its distinct eigenvalues with multiplicitiesand it is denoted by Sp(G), while SpS(Gσ ) stands for the spectrum of the matrix S(Gσ ) and is calledthe skew spectrum of the oriented graph Gσ . If λ1, λ2, . . . , λs are the distinct eigenvalues of a graph G
* Corresponding author.E-mail addresses: [email protected] (A. Anuradha), [email protected] (R. Balakrishnan), [email protected] (W. So).
0024-3795/$ – see front matter © 2013 Elsevier Inc. All rights reserved.http://dx.doi.org/10.1016/j.laa.2013.11.030
68 A. Anuradha et al. / Linear Algebra and its Applications 444 (2014) 67–80
with corresponding multiplicities m1,m2, . . . ,ms , then its spectrum is denoted by the set Sp(G) ={λ(m1)
1 , λ(m2)2 , . . . , λ
(ms)s }. For basic properties of the spectrum of a graph, the reader may refer to [2],
and for skew spectrum of an oriented graph, to [1,3,12].Given a graph G , different orientations of G may give different skew spectra. It is interesting to
study how skew spectra change with orientations. In Section 2, we characterize an oriented graphGσ whose skew spectrum remains unchanged for the reversal of any arc of Gσ in terms of thecoefficients of the characteristic polynomial of Gσ . The following stronger result is obtained earlier byCavers et al. [3]
Theorem 1.1. SpS (Gσ ) remains invariant under any orientation σ if and only if G is a graph without evencycles.
This result motivates our study of G , which denotes the family of finite simple graphs withouteven cycles. Trees, friendship graphs and unicyclic graphs of odd girth belong to G . In Section 3, weexplore some special graphical properties enjoyed by all the graphs in G . The following problem wasraised in [1]:
Problem 1.2. Is it possible to interpret the skew spectrum of an oriented graph in chemistry and otherdisciplines?
In Section 4, a connection between skew spectra of graphs and chemistry is established by provingthat the Hosoya index of all graphs of G can be computed via its skew spectrum. In addition, wepresent a simple proof of a result of Gutman and Shalabi [5] on the Hosoya index of a tree.
The friendship graph (also called the Dutch windmill graph) F p is constructed by concatenatingp copies of the 3-cycle graph at a common vertex. All friendship graphs form a subfamily of G . InSection 5, we determine the skew spectrum of F p which is independent of its orientation. We use thisresult to show that F p is skew integral if and only if 2p + 1 is a perfect square. Since the friendshipgraph F p is not bipartite, there is no orientation σ of F p satisfying the property given in the followingresult by Shader and So [12].
Theorem 1.3. G is a bipartite graph if and only if there is an orientation σ of G such that SpS(Gσ ) = iSp(G)
where i = √−1.
A natural question is
Problem 1.4. For p � 1, does there exist a graph H such that SpS (F σp ) = iSp(H)?
In Section 6 we settle the above problem by showing that this is the case if and only if p = 7, orp is of the form k(k−1)
2 or k(k−2)2 for k � 3.
Throughout this paper, we follow [2] for standard graph theoretic notation. In particular, we usen and m respectively to denote the order (= number of vertices) and size (= number of edges) of ageneral graph G .
2. Invariance of skew spectrum
By a cycle in Gσ , we refer to not necessarily a directed cycle. An even cycle C of Gσ is said to beevenly oriented or oddly oriented according as the number of arcs of C in each direction is even orodd.
Recall that a linear subgraph H of a graph G is a subgraph (not necessarily spanning) of G whosecomponents are single edges or cycles. Following Hou and Lei [9], we say that a linear subgraph Hof G is even linear if H contains no odd cycle. Let E H j(G) denote the set of all even linear subgraphsof G with j vertices.
A. Anuradha et al. / Linear Algebra and its Applications 444 (2014) 67–80 69
In [9], Hou and Lei interpreted the coefficients of the characteristic polynomial of S(Gσ ) in termsof G thereby answering one of the questions raised by Adiga et al. in [1]. A parallel result can also befound in [4, Corollary 2.4].
Theorem 2.1. (See [9].) Let χ(Gσ : x) = xn + b1xn−1 + · · · + bn be the characteristic polynomial of the skewadjacency matrix S(Gσ ) of an oriented graph Gσ . Then
b j =∑
H∈E H j(Gσ )
(−1)β(H)2ce(H),
where the summation is over all even linear subgraphs H of order j of Gσ , β(H) denotes the number of evenlyoriented cycles in H and ce(H) denotes the number of even cycles in H.
Throughout this paper, χ(Gσ : x), β(H) and ce(H) will stand respectively for the polynomial andnumbers mentioned in the statement of Theorem 2.1.
We begin by determining the conditions under which the skew spectrum of Gσ remains invariantunder the reversal of an arc, i.e., reverse the direction of a chosen arc.
Lemma 2.2. Suppose Gσ is an oriented graph with G as its underlying graph. Let e ∈ Γ (Gσ ). If SpS (Gσ )
remains invariant under the reversal of e, then for each even j,∑
H (−1)β(H)2ce(H) = 0, where the summationis over all even linear subgraphs H of order j of Gσ that contain e in an even cycle of Gσ .
Proof. Let e ∈ Γ (Gσ ) and assume that SpS (Gσ ) remains invariant under the reversal of e. Then foreach even j, by Theorem 2.1,
b j =∑
H
(−1)β(H)2ce(H). (1)
The expression on the right of (1) can be split into two sums by making e to vary over all even linearsubgraphs H1 of order j containing e in an even cycle and over all other even linear subgraphs H2 oforder j as:
b j =∑H1
(−1)β(H1)2ce(H1) +∑H2
(−1)β(H2)2ce(H2). (2)
An even cycle C of H1 containing e may either be of evenly or oddly oriented. Hence the aboveexpression can be further split as:
b j =∑H A
(−1)β(H A)2ce(H A) +∑H B
(−1)β(H B )2ce(H B ) +∑H2
(−1)β(H2)2ce(H2) (3)
where H A (respectively H B ) is an even linear subgraph of Gσ which contains e in an evenly (respec-tively oddly) oriented even cycle and H2 does not contain e in an even cycle (H A, H B and H2 areall of order j). Reversal of the orientation of the arc e changes σ into a new orientation σ e , and Hinto the corresponding even linear subgraph He of Gσ e
. Further for even linear subgraphs HeA and He
B
of Gσ e, β(He
A) = β(H A) − 1 and β(HeB) = β(H B) + 1 while for the even linear subgraph He
2 of Gσ e
corresponding to the even linear subgraph H2 of Gσ , β(He2) = β(H2). However, by hypothesis, b j
remains unchanged under σ e . Therefore
b j =∑H A
(−1)β(H A)−12ce(H A) +∑H B
(−1)β(H B )+12ce(H B ) +∑H2
(−1)β(H2)2ce(H2). (4)
Eqs. (3) and (4) show that if b j remains invariant under the reversal of the arc e of Gσ , then∑H A
(−1)β(H A)2ce(H A) +∑H B
(−1)β(H B )2ce(H B ) =∑H1
(−1)β(H1)2ce(H1) = 0, (5)
where the summation is over all even linear subgraphs H1 that contain the arc e in an even cycle. �
70 A. Anuradha et al. / Linear Algebra and its Applications 444 (2014) 67–80
We apply Lemma 2.2 to obtain a necessary and sufficient condition for SpS(Gσ ) to remain invariantunder the reversal of an arc e of Gσ .
Theorem 2.3. Suppose Gσ is an oriented graph of order n with G as its underlying graph. Let e ∈ Γ (Gσ ).Then SpS (Gσ ) remains invariant under the reversal of e if and only if b j = ∑
H (−1)β(H)2ce(H) , where thesummation is over all even linear subgraphs H of order j of Gσ not containing e in an even cycle of Gσ ,1 � j � n.
Proof. Let e ∈ Γ (Gσ ) and assume that SpS (Gσ ) remains invariant under the reversal of e. Expres-sion (2) in the proof of Lemma 2.2 reduces to:
b j =∑H2
(−1)β(H2)2ce(H2) (6)
where the summation runs through all even linear subgraphs H2 of order j of Gσ not containing ein an even cycle of Gσ .
Conversely assume that b j = ∑H (−1)β(H)2ce(H) where the summation runs through all even linear
subgraphs H of order j of Gσ not containing the arc e in an even cycle. Then any such H will eithercontain e as a matching edge, or does not contain e at all. Let Gσ e
be the oriented graph obtainedfrom Gσ by reversal of the arc e. Let He in Gσ e
be the corresponding even linear subgraph of Hin Gσ . Since e is not in any even cycle in He , β(He) = β(H). Hence b j remains unchanged andtherefore SpS (Gσ e
) = SpS (Gσ ). �Corollary 2.4. SpS (Gσ ) remains unchanged under any orientation if and only if for each even j, 1 � j � n,b j is equal to the number of matchings of G of size j/2.
Proof. Apply Theorem 2.3 to each arc e that lies in an even cycle. �3. Graphical properties of GGG
We now establish a few graphical properties of G .
Theorem 3.1. Any graph in G is planar.
Proof. Suppose that G ∈ G is nonplanar. Then by Kuratowski’s theorem [2], G contains a subdivisionof K5 or K3,3. In either case, G also contains two cycles sharing an edge, and hence G has an evencycle, a contradiction. �Theorem 3.2. For every connected graph G in G of order n and size m,
m � 3
2(n − 1).
Furthermore equality holds if and only if G is a union of triangles such that any two triangles have at most onevertex in common.
Proof. By induction on n. For n = 1,2, the proof is trivial.Consider a connected graph G in G with n � 3.Case 1: G is acyclic. Then G is a tree and m = n − 1 < 3
2 (n − 1).Case 2: There exists an induced cycle C in G of length k � 3. Consider the graph G ′ of order n′ and
size m′ obtained from G by collapsing the cycle C into a single vertex. (This is done by removing theedges of the cycle C and identifying the k vertices of C into a single vertex.) Then n = n′ + k − 1 andm = m′ + k. Hence, by the induction assumption,
m = m′ + k � 3
2
(n′ − 1
) + k = 3
2
(n − 1
3k
)� 3
2(n − 1)
A. Anuradha et al. / Linear Algebra and its Applications 444 (2014) 67–80 71
and hence the result follows. In the above equations, equality holds when k = 3 and hence, as G ∈ G ,G must consist of a set of triangles joined together at vertices. �Lemma 3.3. Suppose G ∈ G . Then the minimum degree of G, δ(G) � 2.
Proof. Suppose that δ(G) � 3. Then m � 32 n. By Theorem 3.2 we have
3
2(n − 1) � m � 3
2n,
a contradiction. �Theorem 3.4. The chromatic number χ(G) of every graph G ∈ G is at most 3.
Proof. By induction on n. If n = 1, there is nothing to prove. Consider a graph G ∈ G of order n � 2.By Lemma 3.3, δ(G) � 2. Let v be a vertex in G such that degG(v) � 2. The graph G − v , of ordern − 1, also belongs to the family G . Hence, by induction assumption, χ(G − v) � 3. Now the vertex vhas at most two neighbors in G − v , and so there is one available color for v so that the graph G isstill 3-colorable. �Corollary 3.5. For G ∈ G , χ(G) = 1 if and only if G is the trivial graph, χ(G) = 2 if and only if G is a forestgraph, and χ(G) = 3 if and only if G has an odd cycle.
4. Spectral properties of GGG
The Hosoya index Z(G) of a graph G was introduced by Haruo Hosoya [7] in 1971 and it is definedas Z(G) = ∑
k�0 m(G,k), where m(G,k) is the number of k-matchings (that is, matchings of size k)in G . By definition, m(G,0) = 1. This parameter is a molecular-graph based structure descriptor andis used frequently in the study of the boiling points of alkane isomers in chemistry. The Hosoyaindex of a graph has wide applications in chemistry. However, of late, there have appeared a numberof papers exploring it purely from the point of view of graph theory (see, for instance, [7,8,13]).The computation of the Hosoya index is #P-complete even for planar graphs [10]. However, usingour results and the fact that the characteristic polynomial of a matrix with integer entries can becomputed in polynomial time (see Lemma 4.1 below), we show that for graphs in the family G , theHosoya index can be computed in polynomial time.
Lemma 4.1. (See D.E. Knuth [11].) The coefficients of the characteristic polynomial of a square matrix withinteger entries can be computed in O (n3) time.
Theorem 4.2. Let G ∈ G . Then the Hosoya index Z(G) of G is given by Z(G) = ∑ni=1 bi + 1, where bi,1 �
i � n, are the coefficients of the characteristic polynomial of the oriented graph Gσ for some orientation σof G. Consequently, Z(G) is computable in polynomial time.
Proof. We have χ(Gσ : x) = xn + b1xn−1 + · · · + bn . By Theorem 2.1, bi = 0 if i is odd. If i is even,as G ∈ G , β(H) = ce(H) = 0 for every even linear subgraph H ⊆ G . Thus bi = ∑
H 1 = number ofmatchings of size i/2. Hence the Hosoya index Z(G) is given by Z(G) = ∑n
i=0 m(G, i/2) = ∑ni=1 bi +1.
By Lemma 4.1, this is computable in polynomial time. �We now derive a product formula for the sum of the coefficients of the characteristic polynomial
of Gσ , under an arbitrary orientation σ of G , in terms of the eigenvalues of Gσ . This incidentlygeneralizes a classical result of Gutman and Shalabi [5] which gives a product formula for the Hosoyaindex of a tree in terms of its eigenvalues.
72 A. Anuradha et al. / Linear Algebra and its Applications 444 (2014) 67–80
Lemma 4.3. Suppose Gσ is an oriented graph of order n with G as its underlying graph. Let xn + b1xn−1 +· · · + bn be the characteristic polynomial of S(Gσ ) with SpS (Gσ ) = {iμ1, iμ2, . . . , iμn}. Then
n∏j=1
√1 + μ2
j = 1 +n∑
j=1
b j .
Proof. By our assumption,
(x − iμ1)(x − iμ2) . . . (x − iμn) = xn + b1xn−1 + b2xn−2 + b3xn−3 + · · · + bn. (7)
Therefore,
(x + iμ1)(x + iμ2) . . . (x + iμn) = xn − b1xn−1 + b2xn−2 − b3xn−3 + · · · + (−1)nbn. (8)
As bi = 0 for odd i, the two expressions on the right side of (7) and (8) are equal. Now set x = 1 in(7) and (8) and multiply them. This gives the required result. �Theorem 4.4. Let G ∈ G . Then the Hosoya index Z(G) of G is given by
Z(G) =n∏
j=1
√1 + μ2
j ,
where {iμ1, iμ2, . . . , iμn} is the skew spectrum of Gσ for an arbitrary orientation σ of G.
Proof. This follows from the fact that for a graph G ∈ G , Z(G) = 1 + ∑ni=1 bi (see Theorem 4.2). Now
apply Lemma 4.3. �Using Theorem 4.4, we deduce the following corollary due to Gutman and Shalabi [5]:
Corollary 4.5. If T is a tree of order n with Sp(T ) = {λ1, λ2, . . . , λn}, then Z(T ) = ∏nj=1
√1 + λ2
j .
Proof. We know that SpS(T σ ) = iSp(T ) for any orientation σ of T [12]. Hence the conclusion followsfrom Theorem 4.4 �
Theorem 4.2 gives the Hosoya index of any graph in the family G . We now determine the Hosoyaindex of graphs G whose cycles, if they exist, are edge-disjoint (the cycles may be of varying lengths)but the number of such cycles is bounded by a fixed k in terms of the coefficients bi of the charac-teristic polynomial of S(Gσ ) for some orientation σ of G .
The following two well-known easy reduction formulas are applied to graphs while calculatingtheir Hosoya indices to reduce the computation from bigger graphs to smaller graphs.
Lemma 4.6. (See [13].) Let G1 and G2 be the components of a graph G. Then the Hosoya index Z(G) of G isgiven by Z(G) = Z(G1)Z(G2).
Lemma 4.7 (Hosoya Composition Theorem). (See [13].) Let G be any graph and e = (u, v) be any edge of G.Let Ge denote the graph obtained from G by deleting all edges of G incident to u and v in G. Then Z(G) =Z(G − e) + Z(Ge).
Theorem 4.8. Suppose G is a connected graph of order n in which the cycles are edge-disjoint and the numberof even cycles is k = O (log n). Then the Hosoya index of G is computable in polynomial time.
Proof. Apply Lemma 4.7 iteratively to each even cycle of G , and Lemma 4.6, if necessary to the result-ing graphs. This would result, in at most 2k steps, in a sequence of O (log n) graphs H all belongingto G . Theorem 4.2 now insures that the Hosoya index of G is computable in polynomial time. �
A. Anuradha et al. / Linear Algebra and its Applications 444 (2014) 67–80 73
Fig. 1. Friendship graph F σ4 .
We next compute the exact value of the Hosoya index of a unicyclic graph of even girth. (Whenthe unicyclic graph is of odd girth, the computation of its Hosoya index is covered by Theorem 4.2.)A similar result can also be found in [14, Corollary 12].
Theorem 4.9. Let G be a unicyclic graph of order n containing a unique cycle C of even length r. Let σ be anyorientation of G. Then the Hosoya index of G is given by
Z(G) =n∑
i=1
ci +n∑
i=1
di + 2,
where ci,di , 1 � i � n (where dn−1 = 0 = dn) are the coefficients of xi in the characteristic polynomials of theoriented graphs (G − e)σ , Gσ
e , and e is an arc of the cycle C .
Proof. Applying Lemma 4.7 to the arc e = (u, v), we get two directed forests (G − e)σ and Gσe which
belong to the family G . Their Hosoya indices can be computed in polynomial time using Theorem 4.2.Hence, when r is even,
Z(G) = Z(G − e) + Z(Ge)
=n∑
i=1
ci +n∑
i=1
di + 2
where ci and di are as defined in the statement of the theorem. �5. Skew spectra of friendship graphs
Let F p be the friendship graph of order 2p +1 with p triangles, and so F p has 3p edges. Accordingto Theorem 3.2, F p has the maximum size among all connected graphs in G of order 2p + 1.
Let V (F p) = {a;b1,b2, . . . ,bp; c1, c2, . . . , cp} denote the vertex set of F p where the central vertexis labeled a and the edge set of F p is defined by E(F p) = {(a,bi): 1 � i � p} ∪ {(a, ci): 1 � i �p} ∪ {(bi, ci): 1 � i � p}. (See Fig. 1.) Clearly F p ∈ G for each p.
Let φ be the orientation of F p with the arc set Γ (F φp ) = {(a,bi): 1 � i � p} ∪ {(ci,a): 1 � i �
p} ∪ {(bi, ci): 1 � i � p}. Then the skew symmetric adjacency matrix of F φp can be taken as:
S(
F φp) =
⎡⎢⎢⎢⎢⎢⎢⎣
0 1 −1 1 . . . 1 −1−1 0 1 0 . . . 0 01 −1 0 0 . . . 0 0...
......
.... . .
......
−1 0 0 0 . . . 0 11 0 0 0 . . . −1 0
⎤⎥⎥⎥⎥⎥⎥⎦
.
74 A. Anuradha et al. / Linear Algebra and its Applications 444 (2014) 67–80
Consider the symmetric matrix
Q = (S(
F φp))T
S(
F φp) =
⎡⎢⎢⎢⎢⎢⎢⎣
2p −1 −1 −1 . . . −1 −1−1 2 −1 1 . . . 1 −1−1 −1 2 −1 . . . −1 1...
......
.... . .
......
−1 1 −1 1 . . . 2 −1−1 −1 1 −1 . . . −1 2
⎤⎥⎥⎥⎥⎥⎥⎦
.
Since each row sum of Q is zero, 0 is an eigenvalue of Q with 1, the all-1 column vector as aneigenvector. It is easy to verify that[
0 1 −1 1 −1 . . . 1 −1]T
with alternate 1’s and −1’s is an eigenvector of Q for the eigenvalue 2p + 1. The rank of the matrixQ − (2p + 1)I2p+1 is 2p − 1 and hence the multiplicity of 2p + 1 as an eigenvalue of Q is 2. 1 is aneigenvalue of Q with eigenvector[
0 1 1 . . . 1 −1 −1 . . . −1]T
with 1 and −1 each repeated p times whenever 2p ≡ 0 (mod 4), and with eigenvector[ 0 0 0 1 1 . . . 1 −1 −1 . . . −1 ]T with 1 and −1 each repeated p − 1 times whenever 2p ≡2 (mod 4). The rank of the matrix Q − I2p+1 is 3 implying the fact that the multiplicity of 1 as aneigenvalue of Q is 2p−2. Hence the spectrum of the matrix Q is given by {(2p+1)(2),1(2p−2),0}, andso the spectrum of the matrix (S(F φ
p ))2 = −(S(F φp ))T (S(F φ
p )) is given by {−(2p + 1)(2),−1(2p−2),0}.Thus we have the following theorem.
Theorem 5.1. The skew spectrum of the friendship graph F φp is {i
√2p + 1, i(p−1),0,−i(p−1),−i
√2p + 1 }
and as F p ∈ G , it is independent of its orientation.
Proof. Let S = S(F φp ) and p(x) = det(xI − S) be the characteristic polynomial of S . Since S is a skew-
symmetric matrix, we have
p(x)2 = det(xI − S)det(xI − S) = det(xI − S)det(xI − S)T
= det(xI − S)det(xI + S) = det(xI − S)(xI + S)
= det(x2 I − S2).
Using the fact that Sp(S2) = {−(2p + 1)(2),−1(2p−2),0}, we have
p(x)2 = (x2 + 2p + 1
)2(x2 + 1
)2p−2x2.
Consequently, p(x) = (x2 + 2p + 1)(x2 + 1)p−1x. �Corresponding to integral graphs, we call a graph skew integral with respect to some orientation σ
if its skew spectrum is of the form {iμ : μ ∈ Z, the set of integers}. In [6], Harary and Schwenk raisedthe following question: Which graphs have integral (adjacency) spectra? Analogously, we ask: Which ori-ented graphs have integral skew spectra? In partially answering this question, we have the followingcorollary:
Corollary 5.2. The friendship graph F p is skew integral for any orientation if and only if 2p + 1 is a perfectsquare. In particular, if q is odd, the oriented friendship graph of order q2 is skew integral for any orientation σ .
Remark 5.3. For p = 4, the adjacency spectrum of F4 is given by { 1+√33
2 ,1(3),−1(4), 1−√33
2 } andtherefore not integral, while for any orientation σ , the oriented graph F σ
4 has the skew spectrum
A. Anuradha et al. / Linear Algebra and its Applications 444 (2014) 67–80 75
{3i, i(3),0,−i(3),−3i} and hence integral. The friendship graph provides an example for the oppo-site possibility as well. For p = 3, the adjacency spectrum of F3 is {3,1(2),−1(3),−2} and thereforeintegral, while its skew spectrum {i
√7, i(2),0,−i(2),−i
√7} is non-integral.
The following theorem computes the exact value of the Hosoya index of the friendship graph F p .We provide two different proofs using our earlier results.
Theorem 5.4. The Hosoya index of the friendship graph F p is given by
Z(F p) = 2p(p + 1).
First Proof. Since the friendship graph F p is in G , by Corollary 2.4, b j is the number of matchings ofsize j/2 in F p . Computations show that bi = 0 for all odd i and b2 j = 2p
(p−1j−1
)+ (pj
)for j = 1,2, . . . , p,
and this can also be verified directly. By Theorem 4.2, Z(F p) = ∑pj=1 b2 j +1 = ∑p
j=1
(2p
(p−1j−1
)+ (pj
))+1 = ∑p
j=1 2p(p−1
j−1
) + ∑pj=0
(pj
) = 2p(p + 1). �Second Proof. Using Theorem 4.4 and Theorem 5.1, we have
Z(F p) =√
(1 + 2p + 1)2(1 + 1)2p−2(1 + 0) = 2p(p + 1). �6. Graph H for which SpS (F σ
p ) = iSp(H)
We now proceed to address Problem 1.4 by beginning with some preliminaries.
Lemma 6.1. There is a simple graph H with
Sp(H) = {−√2p + 1,−1(p−1),0,1(p−1),
√2p + 1
}if and only if there exists a (p + 1) × p binary matrix X such that
X T X = I p + xxT
for some p-vector x with xT x = 2p.
Proof. (Sufficiency) Consider the bipartite graph H with adjacency matrix
A(H) =[
0 XX T 0
].
Every nonzero eigenvalue of the matrix X T X is an eigenvalue of X X T and vice versa (this follows fromthe fact that if X T X .v = μv for some nonzero vector v , then X X T (X v) = μ(X v)). By assumption,Sp(X T X) = {2p + 1,1(p−1)}, and so
Sp(H) = {0} ∪ ±√
Sp(
X T X) = {−√
2p + 1,−1(p−1),0,1(p−1),√
2p + 1}.
(Necessity) By symmetry of Sp(H), H is a bipartite graph [2] and so its adjacency matrix is of theform
A(H) =[
0 XX T 0
]
for some binary matrix X of size a×b with a � b and a+b = 2p+1. Note that {−√2p + 1,−1(p−1),0,
1(p−1),√
2p + 1 } = Sp(H) = {0(a−b)} ∪ ±√Sp(X T X). Hence a − b = 1 and Sp(X T X) = {2p + 1,1(p−1)}.
It follows that Sp(X T X − I p) = {2p,0(p−1)}, that is, X T X − I p = xxT with xT x = 2p. �To solve Problem 1.4, we need to solve two matrix equations. (See Theorems 6.4 and 6.6.)
76 A. Anuradha et al. / Linear Algebra and its Applications 444 (2014) 67–80
Let 1 � c � k. Consider the graph G0 with the (c+1)-subsets of {1,2, . . . ,k+1} as vertices, and two(c+1)-subsets are adjacent in G0 if and only if their intersection is a c-subset. Then G0 has (k+1)!
(c+1)!(k−c)!vertices, and every vertex has the common degree (c + 1)(k − c), that is, G0 is (c + 1)(k − c)-regular.
Lemma 6.2. Suppose that G0 is the graph as defined above. Let S be the subgraph induced by the neighbors ofthe vertex (1,2, . . . , c, c + 1) in G0 . Then the chromatic number of S, χ(S) = max(k − c, c + 1).
Proof. Note that the neighbors of the vertex (1,2, . . . , c, c + 1) of G0 are the vertices in the array:
(1,2, . . . , c, c + 2), (1,2, . . . , c, c + 3), · · · , (1,2, . . . , c,k + 1),
(1,2, . . . , c − 1, c + 1, c + 2), (1,2, . . . , c − 1, c + 1, c + 3), · · · , (1,2, . . . , c − 1, c + 1,k + 1),...
... · · · ...
(2,3, . . . , c + 1, c + 2), (2,3, . . . , c + 1, c + 3), · · · , (2,3, . . . , c + 1,k + 1).
There are c + 1 rows and k − c columns in the above array. Edges exist among vertices which areeither in the same row or in the same column only. Indeed, all the vertices in the same row form acomplete graph Kk−c , and all the vertices in the same column form a complete graph Kc+1. In fact,the subgraph induced by these vertices is S Kk−c�Kc+1, the Cartesian product of Kk−c and Kc+1and this implies that χ(S) = max(k − c, c + 1) (see [2]). �Lemma 6.3. Let c = k. Then there exists a binary matrix Y of order (k + 1) × k such that
Y T Y = Ik + c Jk
if and only if k = 1, where Ik is the k × k identity matrix and Jk is the k × k all one matrix.
Proof. (Sufficiency) Take Y = [ 11
]. Then Y T Y = 2 = I1 + J1.
(Necessity) Let yi be the i-th column of Y . Then yTi yi = 1 + k, and so yi = 1, the vector with all
entries equal to 1, because yi is a binary vector of length k + 1. Consequently, Y T Y = (k + 1) Jk . Byassumption, Ik = Jk , and so k = 1. �Theorem 6.4. Let 0 � c < k. Then there exists a binary matrix Y of order (k + 1) × k such that
Y T Y = Ik + c Jk
if and only if c = 0,1,k − 2 or k − 1.
Proof. Case 1: Let c = 0 and k � 1. Consider Y = [ Ik0
]. Then Y T Y = Ik .
Case 2: Let c = 1 and k � 2. Take Y = [ Ik
1T
], 1 being the all one column vector of length k. Then
Y T Y = Ik + Jk .Case 3: Let c � 2 and k � 3.(Sufficiency) For c = k − 1, take Y = [ Jk−Ik
1T
], then Y T Y = Ik + (k − 1) Jk = Ik + c Jk . For c = k − 2,
take Y = [ Jk−Ik
0T
], then Y T Y = Ik + (k − 2) Jk = Ik + c Jk .
(Necessity) We prove the contrapositive statement: For 2 � c � k − 3, Y T Y = Ik + c Jk has NOsolution. Suppose that such a matrix Y exists. Then we can interpret the k columns of Y as the(c + 1)-subsets of {1,2, . . . ,k + 1} with the property that every pair-wise intersection is always ac-subset. Consider a graph with (c + 1)-subsets of {1,2, . . . ,k + 1} as vertices, and two (c + 1)-subsetsare adjacent if their intersection is a c-subset. Hence the existence of Y implies that this graph hasa subgraph H isomorphic to the complete graph Kk . Then the induced subgraph S of the neighborsof any vertex v in H contains a subgraph isomorphic to Kk−1. But by Lemma 6.2, χ(S) = max(k − c,c + 1). Since 2 � c � k − 3, we have k − c < k − 1 and c + 1 < k − 1. It follows that χ(S) < k − 1, andso S does not have Kk−1 as a subgraph, a contradiction. �
A. Anuradha et al. / Linear Algebra and its Applications 444 (2014) 67–80 77
Lemma 6.5. For h � 2, let B1, . . . , Bh be distinct subsets of a set B with |Bi | = 2 and |Bi ∩ B j | = 1 for i �= j.Then
(i) if h = 2 then |B1 ∪ B2| = 3;
(ii) if h = 3 then |B1 ∪ B2 ∪ B3| ={
3 if B1 ∩ B2 ∩ B3 = ∅,
4 if B1 ∩ B2 ∩ B3 �= ∅;(iii) if h > 3 then |B1 ∪ B2 ∪ · · · ∪ Bh| = h + 1.
Proof. (i) In this case, |B1 ∪ B2| = |B1| + |B2| − |B1 ∩ B2| = 2 + 2 − 1 = 3.(ii) In this case, |B1 ∪ B2 ∪ B3| = |B1|+|B2|+|B3|−|B1 ∩ B2|−|B2 ∩ B3|−|B3 ∩ B1|+|B1 ∩ B2 ∩ B3| =
2 + 2 + 2 − 1 − 1 − 1 + |B1 ∩ B2 ∩ B3| = 3 + |B1 ∩ B2 ∩ B3|.(iii) Suppose to the contrary that there are 2-subsets B1, . . . , Bh with |Bi| = 2 and |Bi ∩ B j | = 1 for
i �= j such that |B1 ∪ B2 ∪ · · · ∪ Bh| = t � h. Now consider a graph G with the 2-subsets of {1,2, . . . , t}as vertices and in which two 2-subsets are adjacent if their intersection has exactly one element.Hence the supposition implies that this graph has a subgraph H isomorphic to the complete graph Kh .Then the induced subgraph S of the neighbors of any vertex v in H contains a subgraph isomorphic toKh−1. By Lemma 6.2, χ(S) = max(t − 2,2) < h − 1 since t � h and h > 3. This contradicts the fact thatKh−1 is a subgraph of G . On the other hand, for h > 3, if |B1 ∪ B2 ∪ · · · ∪ Bh| � h + 2 then there existBi and B j with |Bi | = |B j| = 2 and Bi ∩ B j = ∅, a contradiction. Consequently, |B1 ∪ B2 ∪ · · · ∪ Bh| =h + 1. �Theorem 6.6. Let rT = [2, . . . ,2,1, . . . ,1] be a vector having x 2’s, k − x 1’s, 1 � x � k − 1, and rT r = 2p �2k. Then there is a (k + 1) × k binary matrix Y with
Y T Y = Ik + rrT
if and only if rT = [2,2,2,1,1].
Proof. (Sufficiency) Take Y =
⎡⎢⎢⎣
0 1 1 0 01 0 1 0 01 1 0 0 01 1 1 1 01 1 1 1 11 1 1 0 1
⎤⎥⎥⎦. Then Y T Y = I5 + rrT =
⎡⎢⎣
5 4 4 2 24 5 4 2 24 4 5 2 22 2 2 2 12 2 2 1 2
⎤⎥⎦ where rT = [2,2,2,1,1]
with k = 5, x = 3, p = 7 because rT r = 14.(Necessity) Observe that rT r = 4x + (k − x) = 2p � 2k and so k and x have the same parity, and
3x � k � x + 1.Case 1: x = 1. Hence k = 3. Of course there is no 4 × 3 binary matrix Y such that
Y T Y = I3 +⎡⎣ 2
11
⎤⎦[
2 1 1] =
⎡⎣ 5 2 2
2 2 12 1 2
⎤⎦
because the diagonal entries of Y T Y are at most 4, not 5.Case 2: x � 2. Let the columns yi of Y be represented by the subsets A1, A2, . . . , Ak of {1,2, . . . ,
k + 1}. Then Y T Y = Ik + rrT implies that |Ax+1| = · · · = |Ak| = 2 and |Ai ∩ A j | = 1 for x + 1 � i <
j � k. Since k and x have the same parity, we have k − x �= 3, and so, by Lemma 6.5(i) and (iii),|Ax+1 ∪ · · · ∪ Ak| � (k − x) + 1. Moreover, for 1 � i � x and x + 1 � j � k, yT
i y j = 2 and yTj y j = 2
implies that A j ⊆ Ai , and so Ax+1 ∪ · · · ∪ Ak ⊆ Ai . In particular, Ax+1 ∪ · · · ∪ Ak ⊆ A1 ∩ A2 (becausex � 2), hence
k − x + 1 � |Ax+1 ∪ · · · ∪ Ak|� |A1 ∩ A2| = yT1 y2 = 4.
Therefore k � x + 3, and so k = x + 2 because k and x have the same parity. Now |Ak−1| = |Ak| = 2and |Ak−1 ∩ Ak| = 1. By Lemma 6.5(i), we have |Ak−1 ∪ Ak| = 3. On the other hand, Ak−1 ∪ Ak ⊆A1, . . . , Ak−2 and so Ak−1 ∪ Ak ⊆ Ai ∩ A j for any 1 � i < j � k − 2. Define new sets Bi = Ai − (Ak−1 ∪Ak) ⊆ {1,2, . . . ,k + 1} for 1 � i � k − 2. Then |Bi| = |Ai | − |Ak−1 ∪ Ak| = 5 − 3 = 2, and |Bi ∩ B j| =|Ai ∩ A j | − |Ak−1 ∪ Ak| = 4 − 3 = 1.
78 A. Anuradha et al. / Linear Algebra and its Applications 444 (2014) 67–80
Claim. k = 5. Otherwise, k − 2 �= 3, then by Lemma 6.5(iii), |B1 ∪ · · · ∪ Bk−2| � (k − 2) + 1 = k − 1.Consequently, |A1 ∪ · · · ∪ Ak−2| = |B1 ∪ · · · ∪ Bk−2| + |Ak−1 ∪ Ak| � k − 1 + 3 = k + 2, a contradictionbecause the Ai ’s are subsets of {1,2, . . . ,k + 1}.
Finally, we have k = 5, x = 3 and rT = [2,2,2,1,1]. �Lemma 6.7. (i) If there exists a (p + 1) × p binary matrix X such that
X T X = I p + xxT
for some p-vector x with xT x = 2p, then there exists a (k + 1) × k binary matrix Y such that
Y T Y = Ik + yyT
for some k-vector y with yT y = 2p � 2k and yi �= 0 for all i.(ii) If there exists a (k + 1) × k binary matrix Y such that
Y T Y = Ik + yyT
for some k-vector y with yT y = 2p � 2k and yi �= 0 for all i then either
yyT = c Jk
where 2 � c < k � p and ck = 2p or
yyT = P T
⎡⎢⎢⎢⎣
4 4 4 2 24 4 4 2 24 4 4 2 22 2 2 1 12 2 2 1 1
⎤⎥⎥⎥⎦ P
for some permutation matrix P .
Proof. (i) Let P be a permutation matrix such that P x = [ y0
]where y is a k-vector with k � p and
yi �= 0 for all i. Let X P T be partitioned as [ X1 X2 ]. Then, by the assumption on X , we have[X T
1 X1 X T1 X2
X T2 X1 X T
2 X2
]= P X T X P T = I p + P xxT P T =
[Ik + yyT 0
0 I p−k
].
Now X T2 X2 = I p−k implies that the columns of X2 are p − k distinct standard vectors of length k + 1.
Hence there exists a permutation matrix Q such that Q X2 = [ 0I p−k
]. Write Q X P T = [ Y 0
R I p−k
]. Note
that 0 = X T1 X2 = X T
1 Q T Q X2 = Y T 0 + RT I p−k = RT , that is, R = 0. Consequently Q X P T = [ Y 00 I p−k
].
Finally,[Y T Y 0
0 I p−k
]= (
Q X P T )T (Q X P T ) = P
(X T X
)P T =
[Ik + yyT 0
0 I p−k
].
Hence Y T Y = Ik + yyT where yT y = xT x = 2p � 2k and yi �= 0 for all i.(ii) Let Y =[ z1 z2 · · · zk ] where zi are binary vectors of length k+1. Also let yT =[ y1 y2 · · · yk ].
Note that for each i, 1 � i � k, zTi zi − 1 = y2
i is a positive integer because yi �= 0, and so write y2i =
αiβ2i for some square-free integer αi and integer βi . Hence yi = √
αiβi . Also note that√
αiα jβiβ j =yi y j = zT
i z j is an integer, and so αi = α j because they are both square-free. Let α = α1 = · · · = αk .Hence yT = √
α[ β1 β2 · · · βk ]. Suppose that βi �= β j . Without loss of generality, let β j = βi +δ, δ �= 0.Since zi, z j are binary vectors, we have zT
i z j � min{zTi zi, zT
j z j} = min{1 + αβ2i ,1 + αβ2
j } = 1 + αβ2i .
On the other hand, zTi z j = αβiβ j > αβ2
i . It follows that αβi(βi + δ) = αβiβ j = 1 + αβ2i , and so α =
δ = βi = 1, and β j = 2.Case 1: y is a constant vector. Then βi = β for all i. Hence yyT = c Jk where c = αβ2.
A. Anuradha et al. / Linear Algebra and its Applications 444 (2014) 67–80 79
Case 2: y is not a constant vector. From the argument above, y contains only 1 and 2 as itsentries. Then by Theorem 6.6, y = Pr where P is a permutation matrix and rT = [2,2,2,1,1], and so
yT y = P T
⎡⎢⎣
4 4 4 2 24 4 4 2 24 4 4 2 22 2 2 1 12 2 2 1 1
⎤⎥⎦ P . �
Theorem 6.8. (i) There is a simple graph H with
Sp(H) = {−√15,−1(6),0,1(6),
√15
}.
(ii) If p = k(k−1)2 for some integer k � 3 then there is a simple graph H with
Sp(H) = {−√2p + 1,−1(p−1),0,1(p−1),
√2p + 1
}.
(iii) If p = k(k−2)2 for some even integer k � 4 (that is 2p + 1 = (k − 1)2 is a perfect square) then there is a
simple integral graph H with
Sp(H) = {−√2p + 1,−1(p−1),0,1(p−1),
√2p + 1
}.
Proof. (i) Take Y =
⎡⎢⎢⎣
0 1 1 0 01 0 1 0 01 1 0 0 01 1 1 1 01 1 1 1 11 1 1 0 1
⎤⎥⎥⎦, then Y T Y = I5 + [2,2,2,1,1]T [2,2,2,1,1]. Take X = [ Y 0
0 I2
], then
X T X = I7 + xxT where x = [2,2,2,1,1,0,0]T with xT x = 14 = 2 · 7. Hence, by Lemma 6.1, there is asimple graph H with Sp(H) = {−√
15,−1(6),0,1(6),√
15 }.
(ii) Take Y = [ Jk−Ik
1T
], then Y T Y = Ik + (k − 1) Jk . Take X = [ Y 0
0 I p−k
], then X T X = I p + xxT where
x = √k − 1
[ 10
]with xT x = (k − 1)k = 2p. Hence, by Lemma 6.1, there is a simple graph H with
Sp(H) = {−√2p + 1,−1(p−1),0,1(p−1),
√2p + 1
}.
(iii) Take Y = [ Jk−Ik
0T
], then Y T Y = Ik + (k − 2) Jk . Take X = [ Y 0
0 I p−k
], then X T X = I p + xxT where
x = √k − 2
[ 10
]with xT x = (k − 2)k = 2p. Hence, by Lemma 6.1, there is a simple graph H with
Sp(H) = {−√2p + 1,−1(p−1),0,1(p−1),
√2p + 1
}. �
Theorem 6.9. If there is a simple graph H with
Sp(H) = {−√2p + 1,−1(p−1),0,1(p−1),
√2p + 1
}then either p = 7 or p is of the form k(k−1)
2 or k(k−2)2 for k � 3.
Proof. Suppose that there exists a simple graph H with
Sp(H) = {−√2p + 1,−1(p−1),0,1(p−1),
√2p + 1
}.
Then, by Lemmas 6.1 and 6.7, there exists a (k + 1) × k binary matrix Y such that either
Y T Y = I5 + yyT
where y = Pr, for some permutation matrix P , rT = [2,2,2,1,1] and 2p = rT r = 14, and hence p = 7;or
Y T Y = Ik + c Jk
where 2 � c < k � p and ck = 2p. In this case, by Theorem 6.4, c = k − 1 or k − 2. Hence p is of theform k(k−1)
2 or k(k−2)2 for k � 3. �
80 A. Anuradha et al. / Linear Algebra and its Applications 444 (2014) 67–80
Corollary 6.10. For p � 1, let F σp be the friendship graph of p triangles with any orientation σ . There exists a
graph H such that SpS(F σp ) = iSp(H) if and only if p = 7, p = k(k−1)
2 or p = k(k−2)2 for k � 3.
Proof. By Theorem 5.1,
SpS(
F σp
) = i{−√
2p + 1,−1(p−1),0,1(p−1),√
2p + 1}.
Hence the result follows from Theorems 6.8 and 6.9. �Acknowledgements
For the first two authors, this research was supported by the Department of Science and Tech-nology, Government of India grant DST:SR/S4/MS:492, dated April 16, 2009. The authors thank thereferee for some suggestions and new references [4,14].
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