Research Article The Distance Matrices of Some Graphs...
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Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2013, Article ID 707954, 5 pages http://dx.doi.org/10.1155/2013/707954 Research Article The Distance Matrices of Some Graphs Related to Wheel Graphs Xiaoling Zhang and Chengyuan Song School of Mathematics and Information Science, Yantai University, Yantai, Shandong 264005, China Correspondence should be addressed to Xiaoling Zhang; [email protected] Received 25 November 2012; Revised 29 May 2013; Accepted 30 May 2013 Academic Editor: Maurizio Porfiri Copyright © 2013 X. Zhang and C. Song. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Let denote the distance matrix of a connected graph . e inertia of is the triple of integers ( + (), 0 (), − ()), where + (), 0 (), and − () denote the number of positive, 0, and negative eigenvalues of , respectively. In this paper, we mainly study the inertia of distance matrices of some graphs related to wheel graphs and give a construction for graphs whose distance matrices have exactly one positive eigenvalue. 1. Introduction A simple graph = (,) consists of , a nonempty set of vertices, and , a set of unordered pairs of distinct elements of called edges. All graphs considered here are simple and connected. Let be a simple connected graph with vertex set () and edge set (). e distance between two vertices , V ∈ () is denoted by V and is defined as the length of the shortest path between and V in . e distance matrix of is denoted by () and is defined by () = ( V ) ,V∈() . Since () is a symmetric matrix, its inertia is the triple of integers ( + (()), 0 (()), − (())), where + (()), 0 (()), and − (()) denote the number of positive, 0, and negative eigenvalues of (), respectively. e distance matrix of a graph has numerous applications to chemistry [1]. It contains information on various walks and self-avoiding walks of chemical graphs. Moreover, the distance matrix is not only immensely useful in the compu- tation of topological indices such as the Wiener index [1] but also useful in the computation of thermodynamic properties such as pressure and temperature virial coefficients [2]. e distance matrix of a graph contains more structural informa- tion compared to a simple adjacency matrix. Consequently, it seems to be a more powerful structure discriminator than the adjacency matrix. In some cases, it can differentiate isospectral graphs although there are nonisomorphic trees with the same distance polynomials [3]. In addition to such applications in chemical sciences, distance matrices find applications in music theory, ornithology [4], molecular biology [5], psychology [4], archeology [6], sociology [7], and so forth. For more information, we can see [1] which is an excellent recent review on the topic and various uses of distance matrices. Since the distance matrix of a general graph is a compli- cated matrix, it is very difficult to compute its eigenvalues. People focus on studying the inertia of the distance matrices of some graphs. Unfortunately, up to now, only few graphs are known to have exactly one positive -eigenvalue, such as trees [8], connected unicyclic graphs [9], the polyacenes, honeycomb and square lattices [10], complete bipartite graphs [11], , and iterated line graphs of some regular graphs [12], and cacti [13]. is inspires us to find more graphs whose distance matrices have exactly one positive eigenvalue. e wheel graph of vertices is a graph that contains a cycle of length −1 plus a vertex V (sometimes called the hub) not in the cycle such that V is connected to every other vertex. In this paper, we first study the inertia of the distance matrices in wheel graphs if one or more edges are removed from the graph, and then, with the help of the structural characteristics of wheel graphs, we give a construction for graphs whose distance matrices have exactly one positive eigenvalue. 2. Preliminaries We first give some lemmas that will be used in the main results.
Transcript of Research Article The Distance Matrices of Some Graphs...
Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 707954 5 pageshttpdxdoiorg1011552013707954
Research ArticleThe Distance Matrices of Some Graphs Related to Wheel Graphs
Xiaoling Zhang and Chengyuan Song
School of Mathematics and Information Science Yantai University Yantai Shandong 264005 China
Correspondence should be addressed to Xiaoling Zhang zhangxling04yahoocn
Received 25 November 2012 Revised 29 May 2013 Accepted 30 May 2013
Academic Editor Maurizio Porfiri
Copyright copy 2013 X Zhang and C Song This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
Let 119863 denote the distance matrix of a connected graph 119866 The inertia of 119863 is the triple of integers (119899+(119863) 119899
0(119863) 119899
minus(119863)) where
119899+(119863) 119899
0(119863) and 119899
minus(119863) denote the number of positive 0 and negative eigenvalues of 119863 respectively In this paper we mainly
study the inertia of distance matrices of some graphs related to wheel graphs and give a construction for graphs whose distancematrices have exactly one positive eigenvalue
1 Introduction
A simple graph 119866 = (119881 119864) consists of 119881 a nonempty set ofvertices and 119864 a set of unordered pairs of distinct elementsof 119881 called edges All graphs considered here are simple andconnected Let119866 be a simple connected graph with vertex set119881(119866) and edge set 119864(119866)The distance between two vertices 119906V isin 119881(119866) is denoted by 119889
119906V and is defined as the length of theshortest path between 119906 and V in119866The distancematrix of119866 isdenoted by119863(119866) and is defined by119863(119866) = (119889
119906V)119906Visin119881(119866) Since119863(119866) is a symmetric matrix its inertia is the triple of integers(119899+(119863(119866)) 119899
0(119863(119866)) 119899
minus(119863(119866))) where 119899
+(119863(119866)) 119899
0(119863(119866))
and 119899minus(119863(119866)) denote the number of positive 0 and negative
eigenvalues of119863(119866) respectivelyThe distancematrix of a graph has numerous applications
to chemistry [1] It contains information on various walksand self-avoiding walks of chemical graphs Moreover thedistance matrix is not only immensely useful in the compu-tation of topological indices such as the Wiener index [1] butalso useful in the computation of thermodynamic propertiessuch as pressure and temperature virial coefficients [2] Thedistance matrix of a graph contains more structural informa-tion compared to a simple adjacency matrix Consequentlyit seems to be a more powerful structure discriminator thanthe adjacency matrix In some cases it can differentiateisospectral graphs although there are nonisomorphic treeswith the same distance polynomials [3] In addition tosuch applications in chemical sciences distance matrices
find applications in music theory ornithology [4] molecularbiology [5] psychology [4] archeology [6] sociology [7]and so forth For more information we can see [1] which isan excellent recent review on the topic and various uses ofdistance matrices
Since the distance matrix of a general graph is a compli-cated matrix it is very difficult to compute its eigenvaluesPeople focus on studying the inertia of the distance matricesof some graphs Unfortunately up to now only few graphsare known to have exactly one positive 119863-eigenvalue suchas trees [8] connected unicyclic graphs [9] the polyaceneshoneycomb and square lattices [10] complete bipartite graphs[11] 119870
119899 and iterated line graphs of some regular graphs [12]
and cacti [13] This inspires us to find more graphs whosedistance matrices have exactly one positive eigenvalue
Thewheel graph of 119899 vertices119882119899is a graph that contains a
cycle of length 119899minus1 plus a vertex V (sometimes called the hub)not in the cycle such that V is connected to every other vertexIn this paper we first study the inertia of the distancematricesin wheel graphs if one or more edges are removed from thegraph and then with the help of the structural characteristicsof wheel graphs we give a construction for graphs whosedistance matrices have exactly one positive eigenvalue
2 Preliminaries
We first give some lemmas that will be used in the mainresults
2 Journal of Applied Mathematics
Lemma 1 (see [14]) Let 119860 be a Hermitian matrix witheigenvalues 120582
1⩾ sdot sdot sdot ⩾ 120582
119899and 119861 one of its principal
submatrices Let 119861 have eigenvalues 1205831⩾ sdot sdot sdot ⩾ 120583
119898 Then the
inequalities 120582119899minus119898+119894
⩽ 120583119894⩽ 120582119894(119894 = 1 119898) hold
For a square matrix let cof (119860) denote the sum ofcofactors of 119860 Form the matrix 119860 by subtracting the firstrow from all other rows then the first column from all othercolumns and let119860
11denote the principle submatrix obtained
from 119860 by deleting the first row and first column
Lemma 2 (see [15]) 119888119900119891(119860) = det11986011
A cut vertex is a vertex the removal of which woulddisconnect the remaining graph a block of a graph is definedto be a maximal subgraph having no cut vertices
Lemma3 (see [15]) If 119866 is a strongly connected directed graphwith blocks 119866
1 1198662 119866
119903 then
119888119900119891119863 (119866) =
119903
prod
119894=1
119888119900119891119863 (119866119894)
det119863 (119866) =119903
sum
119894=1
det 119863(119866119894)prod
119895 = 119894
119888119900119891119863 (119866119894)
(1)
Lemma 4 Let
119862 =
[[[[[[[[[[
[
1 1 1 1 sdot sdot sdot 1 1
minus1 minus2 minus1 0 sdot sdot sdot 0 0
0 minus1 minus2 minus1 sdot sdot sdot 0 0
0 0 minus1 minus2 sdot sdot sdot 0 0
d
0 0 0 0 sdot sdot sdot minus2 minus1
0 0 0 0 sdot sdot sdot minus1 minus2
]]]]]]]]]]
]119899times119899
(2)
Then
det 119862 =
minus119899
2 119894119891 119899 119894119904 119890V119890119899
119899 + 1
2 119894119891 119899 119894119904 119900119889119889
(3)
Proof Let
119862119899=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1 1 1 1 sdot sdot sdot 1 1
minus1 minus2 minus1 0 sdot sdot sdot 0 0
0 minus1 minus2 minus1 sdot sdot sdot 0 0
0 0 minus1 minus2 sdot sdot sdot 0 0
d
0 0 0 0 sdot sdot sdot minus2 minus1
0 0 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
119863119899=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1
2minus1 0 0 sdot sdot sdot 0 0
1 minus2 minus1 0 sdot sdot sdot 0 0
1 minus1 minus2 minus1 sdot sdot sdot 0 0
1 0 minus1 minus2 sdot sdot sdot 0 0
d
1 0 0 0 sdot sdot sdot minus2 minus1
1 0 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
(4)
Comparing 119862119899to119863119899 we get the following
119862119899= 119863119899+119899
2times (minus1)
119899minus1
(5)
Expanding the determinant 119863119899according to the last column
and then the last line we get the following incursion
119863119899= minus2119863
119899minus1minus 119863119899minus2
+ 1 (6)
that is
119863119899+ 119863119899minus1
= minus (119863119899minus1
+ 119863119899minus2) + 1 (7)
Since 1198631= 12 119863
2= 0 and 119863
3= 12 from the above
incursion we get the following
119863119899=
0 if 119899 is even
1
2 if 119899 is odd
(8)
So we have the following
119862119899=
minus119899
2 if 119899 is even
119899 + 1
2 if 119899 is odd
(9)
This completes the proof
3 Main Results
In the following we always assume that 119881(119882119899) = V
0 V1
V119899minus1 where V
0is the hub of119882
119899
Theorem 5 Let 119890 = V119894V119894+1 mod (119899minus1) (1 ⩽ 119894 ⩽ 119899 minus 1) Then
det 119863(119882119899minus 119890) =
minus1198992
4 119894119891 119899 119894119904 119890V119890119899
1198992minus 1
4 119894119891 119899 119894119904 119900119889119889
(10)
where 119899 ⩾ 3
Proof Without loss of generality we may assume that 119890 =V1V119899minus1
Let
119860119899= det119863(119882
119899minus 119890) =
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 1 1 1 sdot sdot sdot 1 1
1 0 1 2 sdot sdot sdot 2 2
1 1 0 1 sdot sdot sdot 2 2
1 2 1 0 sdot sdot sdot 2 2
d
1 2 2 2 sdot sdot sdot 0 1
1 2 2 2 sdot sdot sdot 1 0
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
(11)
Journal of Applied Mathematics 3
Then
119860119899=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 1 1 minus1
21 sdot sdot sdot 1 1
1 minus2 0 0 sdot sdot sdot 0 0
1 minus1
20 minus2 +
1
2minus1 sdot sdot sdot 0 0
1 0 minus1 minus2 sdot sdot sdot 0 0
d
1 0 0 0 sdot sdot sdot minus2 minus1
1 0 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
(12)
Expanding the determinant 119860119899according to the second line
we get the following incursion
119860119899= (minus1)
119899minus1 119899
2minus 2119860119899minus1
minus 119862119899minus2
minus 119863119899minus2
minus 119860119899minus2 (13)
where 119862119899and119863
119899are defined as in Lemma 4
By Lemma 4 we get the following
119860119899=
minus1 minus 2119860119899minus1
minus 119860119899minus2 if 119899 is even
minus2119860119899minus1
minus 119860119899minus2 if 119899 is odd
(14)
Since 1198603= 2 119860
4= minus4 119860
5= 6 and 119860
6= minus9 according to
the above incursion we get the following
119860119899=
minus1198992
4 if 119899 is even
1198992minus 1
4 if 119899 is odd
(15)
where 119899 ⩾ 3 This completes the proof
Corollary 6 Let 119890 = V119894V119894+1 mod (119899minus1) (1 ⩽ 119894 ⩽ 119899 minus 1) Then
119899+(119863 (119882
119899minus 119890)) = 1 119899
0(119863 (119882
119899minus 119890)) = 0
119899minus(119863 (119882
119899minus 119890)) = 119899 minus 1
(16)
Proof We will prove the result by induction on 119899If 119899 = 3119882
3minus 119890 cong 119875
3is obviously true
Suppose that the result is true for 119899minus1 that is 119899+(119863(119882
119899minus1minus
119890)) = 1 1198990(119863(119882
119899minus1minus 119890)) = 0 119899
minus(119863(119882
119899minus1minus 119890)) = 119899 minus 2
Since 119863(119882119899minus1
minus 119890) is a principle submatrix of 119863(119882119899minus
119890) by Lemma 1 the eigenvalues of 119863(119882119899minus1
minus 119890) interlacethe eigenvalues of 119863(119882
119899minus 119890) By Theorem 5 det119863(119882
119899minus1minus
119890) det119863(119882119899minus 119890) lt 0 So 119863(119882
119899minus 119890) has one negative
eigenvaluemore than119863(119882119899minus1minus119890) According to the induction
hypothesis we get 119899+(119863(119882
119899minus119890)) = 1 119899
0(119863(119882
119899minus119890)) = 0 and
119899minus(119863(119882
119899minus 119890)) = 119899 minus 1 This completes the proof
Theorem 7 One has
det 119863(119882119899) =
1 minus 119899 119894119891 119899 119894119904 119890V119890119899
0 119894119891 119899 119894119904 119900119889119889(17)
where 119899 ⩾ 3
Proof Consider the following
det119863(119882119899) =
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 1 1 1 sdot sdot sdot 1 1
1 0 1 2 sdot sdot sdot 2 1
1 1 0 1 sdot sdot sdot 2 2
1 2 1 0 sdot sdot sdot 2 2
d
1 2 2 2 sdot sdot sdot 0 1
1 1 2 2 sdot sdot sdot 1 0
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 1 1 1 sdot sdot sdot 1 1
1 minus2 minus1 0 sdot sdot sdot 0 minus1
1 minus1 minus2 minus1 sdot sdot sdot 0 0
1 0 minus1 minus2 sdot sdot sdot 0 0
d
1 0 0 0 sdot sdot sdot minus2 minus1
1 minus1 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 119899 minus 1 1 1 sdot sdot sdot 1 1
119899 minus 1 minus4 (119899 minus 1) minus4 minus4 sdot sdot sdot minus4 minus4
1 minus4 minus2 minus1 sdot sdot sdot 0 0
1 minus4 minus1 minus2 sdot sdot sdot 0 0
d
1 minus4 0 0 sdot sdot sdot minus2 minus1
1 minus4 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 119899 minus 1 1 1 sdot sdot sdot 1 1
119899 minus 1 4 (119899 minus 1) 0 0 sdot sdot sdot 0 0
1 0 minus2 minus1 sdot sdot sdot 0 0
1 0 minus1 minus2 sdot sdot sdot 0 0
d
1 0 0 0 sdot sdot sdot minus2 minus1
1 0 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
(18)
Expanding the above determinant according to the secondline we get the following
det119863(119882119899) = (minus1)
119899minus1
(119899 minus 1)3
+ 4 (119899 minus 1)119860119899minus1 (19)
where 119860119899is defined as inTheorem 5
ByTheorem 5 when 119899 ⩾ 3 we get the following
det119863(119882119899) =
1 minus 119899 if 119899 is even0 if 119899 is odd
(20)
This completes the proof
Similar to Corollary 6 we can get the following corollary
Corollary 8 (i) If 119899 is even 119899+(119863(119882
119899)) = 1 119899
0(119863(119882
119899)) = 0
119899minus(119863(119882
119899)) = 119899 minus 1
(ii) If 119899 is odd 119899+(119863(119882
119899)) = 1 119899
0(119863(119882
119899)) = 1
119899minus(119863(119882
119899)) = 119899 minus 2
Denote by119882119899minus V0the graph obtained from119882
119899by deleting
the vertex V0and all the edges adjacent to V
0 that is119882
119899minus V0cong
119862119899minus1
Let 119864119896(1 ⩽ 119896 ⩽ 119899) be any subset of 119864(119882
119899minus V0) with
|119864119896| = 119896 In the following we always denote by 119882
119899minus 119864119896the
graph obtained from119882119899by deleting all the edges in 119864
119896
4 Journal of Applied Mathematics
Theorem 9 One has 119899+(119863(119882
119899minus 119864119896)) = 1 119899
0(119863(119882
119899minus 119864119896)) =
0 119899minus(119863(119882
119899minus 119864119896)) = 119899 minus 1
Proof Denote the components of119882119899minus 119864119896minus V0by 1198621 119862
119904
Let 119866119894denote the graph that contains 119862
119894plus the vertex V
0
such that V0is connected to every other vertex 1 ⩽ 119894 ⩽ 119904
Then each 119866119894(1 ⩽ 119894 ⩽ 119904) is isomorphism to119882
|119881(119866119894)|minus 119890119894or
1198702 where 119890
119894is an edge of119882
|119881(119866119894)|minus V0 By Lemma 2 and some
direct calculations we get the following
cof (119866119894) = det119863
(119882|119881(119866119894)|
minus 119890119894)11
= (minus1)|119881(119866119894)|minus1
1003816100381610038161003816119881 (119866119894)1003816100381610038161003816
(21)
It is easy to check that cof (119866119894) = (minus1)
|119881(119866119894)|minus1|119881(119866119894)| is also
true when 119866119894is isomorphism to 119870
2
In the following we will prove the theorem by introduc-tion on 119904
For 119904 = 1 119866 cong 119882119899minus 119890 where 119890 is an edge of119882
119899minus V0 by
Corollary 6 we get the resultSuppose the result is true for 119904 minus 1For 119904 let 1198661015840 = 119866
1cup1198662cup sdot sdot sdot cup119866
119904minus1 Then by the induction
hypothesis 119899+(119863(1198661015840)) = 1 119899
0(119863(1198661015840)) = 0 and 119899
minus(119863(1198661015840)) =
|119881(1198661015840)| minus 1 which implies that
det119863(1198661015840) = (minus1)|119881(1198661015840)|minus1
119886 (22)
where 119886 is a positive integerSince
cof (119866119894) = (minus1)
|119881(119866119894)|minus11003816100381610038161003816119881 (119866119894)
1003816100381610038161003816 1 ⩽ 119894 ⩽ 119904 (23)
by Lemma 3
cof119863(1198661015840) =119904minus1
prod
119895=1
cof119863(119866119895) = (minus1)
|119881(1198661015840)|minus1
119904minus1
prod
119895=1
10038161003816100381610038161003816119881 (119866119895)10038161003816100381610038161003816
(24)
Thendet119863(119882
119899minus 119864119896)
= det119863(1198661015840) cof119863(119866119894) + det119863(119866
119894) cof119863(1198661015840)
= (minus1)|119881(1198661015840)|minus1
119886 times (minus1)|119881(119866119894)|minus1
1003816100381610038161003816119881 (119866119894)1003816100381610038161003816
+ (minus1)|119881(119866119894)|minus1119887 times (minus1)
|119881(1198661015840)|minus1
119904minus1
prod
119895=1
10038161003816100381610038161003816119881 (119866119895)10038161003816100381610038161003816
= (minus1)119899minus1
(1198861003816100381610038161003816119881 (119866119894)
1003816100381610038161003816 + 119887
119904minus1
prod
119895=1
10038161003816100381610038161003816119881 (119866119895)10038161003816100381610038161003816)
(25)
where 119887 = 11989924 if 119899 is even and 119887 = (1198992 minus 1)4 if 119899 is oddIn this case similar to Corollary 6 we can easily get
119899+(119863(119882
119899minus 119864119896)) = 1 119899
0(119863(119882
119899minus 119864119896)) = 0 and 119899
minus(119863(119882
119899minus
119864119896)) = 119899 minus 1Up to now we have proved the result
Let119866119906119894sdotV119895119867 denote the graph formed by only identifying
the vertex 119906119894of 119866 with the vertex V
119895of119867 where 119906
119894and V119895are
arbitrary vertices of 119866 and119867 respectively
Lemma 10 (see [13]) Let 119866times119867 denote the Cartesian productof connected graphs 119866 and119867 where 119881(119866) = 119906
1 119906
119898 and
119881(119867) = V1 V
119899 Then we have
(i) 119899+(119866 times 119867) = 119899
+(119866119906119894sdot V119895119867)
(ii) 1198990(119866 times 119867) = (119898 minus 1)(119899 minus 1) + 119899
0(119866119906119894sdot V119895119867)
(iii) 119899minus(119866 times 119867) = 119899
minus(119866119906119894sdot V119895119867)
Theorem 11 Let 1199060and V
0be the hubs of 119882
119899and 119882
119898
respectively Suppose119864119901(0 ⩽ 119901 ⩽ 119899minus1) and119864
119902(0 ⩽ 119902 ⩽ 119898minus1)
are any subsets of 119864(119882119899minus 1199060) and 119864(119882
119898minus V0) with |119864
119901| = 119901
|119864119902| = 119902 respectively Then the distance matrix of the graph
(119882119899minus 119864119901) times (119882
119898minus 119864119902) has exactly one positive eigenvalue
Proof Since 1199060and V
0are the hubs of 119882
119899and 119882
119898 respec-
tively (119882119899minus119864119901)1199060sdotV0(119882119898minus119864119902)must be isomorphism to some
119882119899+119898minus1
minus119864119901+119902
where1199080is the hub of119882
119899+119898minus1and 119864
119901+119902is any
subset of 119864(119882119899+119898minus1
minus 1199080) with |119864
119901+119902| = 119901 + 119902 By Theorem 9
and Lemma 10 we get the result
Given an arbitrary integer 119898 for 1 ⩽ 119894 ⩽ 119898 let V1198940be
the hub of 119882119899119894and 119864
119901119894any subset of 119864(119882
119899119894minus V1198940) Suppose
119881(119882119899119894) = V
1198940 V1198941 V
119894(119899119894minus1)
Theorem 12 For an arbitrary integer 119898 the distance matrixof the graph 119866 = (119882
1198991minus 1198641199011)V1119896sdot V2119895(1198821198992minus 1198641199012)V2ℎsdot sdot sdot
V(119898minus1)119903
(119882119899119898minus1
minus 119864119901119898minus1
)V(119898minus1)119905
sdot V119898119904(119882119899119898minus 119864119901119898) has exactly one
positive eigenvalue
Proof We will prove the conclusion by induction on119898If119898 = 1 by Theorem 9 the conclusion is trueSuppose the conclusion is true for119898minus1 For convenience
let119867 = (1198821198991minus 1198641199011)V1119896sdot V2119895(1198821198992minus 1198641199012)V2ℎsdot sdot sdot V(119898minus2)119894
(119882119899119898minus2
minus
119864119901119898minus2
) Then 119866 = 119867V(119898minus2)119892
sdot V(119898minus1)119903
(119882119899119898minus1
minus 119864119901119898minus1
)V(119898minus1)119905
sdot
V119898119904(119882119899119898minus 119864119901119898) By Lemma 10 we have the following
119899+(119866) = 119899
+(119867 (119882
119899119898minus1minus 119864119901119898minus1
) V(119898minus1)119905
sdot V119898119904(119882119899119898minus 119864119901119898))
= 119899+(119867 (119882
119899119898minus1minus 119864119901119898minus1
) V(119898minus1)0
sdot V1198980(119882119899119898minus 119864119901119898))
(26)
Since (119882119899119898minus1
minus 119864119901119898minus1
)V(119898minus1)0
sdot V1198980(119882119899119898minus 119864119901119898) = 119882
119899119898minus1+119899119898minus1minus
119864119901119898minus1
minus119864119901119898 we get 119899
+(119866) = 119899
+(119867V(119898minus2)119892
sdotV(119898minus1)119903
(119882119899119898minus1+119899119898minus1
minus
119864119901119898minus1
minus119864119901119898)) By the induction hypothesis we get 119899
+(119866) = 1
This completes the proof
Remark 13 Let 1198661and 119866
2be any two graphs with the same
form as119866 inTheorem 12MakingCartesian product of graphs1198661and 119866
2 by Lemma 10 and Theorem 12 we get a series
of graphs whose distance matrices have exactly one positiveeigenvalue
Acknowledgments
The authors would like to thank the anonymous refereesfor their valuable comments and suggestions This workwas supported by NSFC (11126256) and NSF of ShandongProvince of China (ZR2012AQ022)
Journal of Applied Mathematics 5
References
[1] D H Rouvray ldquoThe role of the topological distance matrixin chemistryrdquo in Mathematical and Computational Conceptsin Chemistry N Trinajstid Ed pp 295ndash306 Ellis HarwoodChichester UK 1986
[2] W Brostow D M McEachern and S Petrez- Guiterrez ldquoPres-sure second virial coefficients of hydrocarbons fluorocarbonsand their mixtures Interactions of walksrdquo Journal of ChemicalPhysics vol 71 pp 2716ndash2722 1979
[3] B D McKay ldquoOn the spectral characterisation of treesrdquo ArsCombinatoria vol 3 pp 219ndash232 1977
[4] D W Bradley and R A Bradley ldquoString edits and macro-moleculesrdquo in Time Wraps D Sankoff and J B Kruskal EdsChapter 6 Addison-Wesley Reading Mass USA 1983
[5] J P Boyd and K N Wexler ldquoTrees with structurerdquo Journal ofMathematical Psychology vol 10 pp 115ndash147 1973
[6] R L Graham and L Lovasz ldquoDistance matrix polynomials oftreesrdquo inTheory and Applications of Graphs vol 642 of LectureNotes in Mathematics pp 186ndash190 1978
[7] M S Waterman T F Smith and H I Katcher ldquoAlgorithms forrestriction map comparisonsrdquo Nucleic Acids Research vol 12pp 237ndash242 1984
[8] R L Graham and H O Pollak ldquoOn the addressing problem forloop switchingrdquo The Bell System Technical Journal vol 50 pp2495ndash2519 1971
[9] R Bapat S J Kirkland and M Neumann ldquoOn distancematrices and Laplaciansrdquo Linear Algebra and Its Applicationsvol 401 pp 193ndash209 2005
[10] K Balasubramanian ldquoComputer generation of distance poly-nomials of graphsrdquo Journal of Computational Chemistry vol 11no 7 pp 829ndash836 1990
[11] D M Cvetkovic M Doob I Gutman and A Torgasev RecentResults in the Theory of Graph Spectra vol 36 North-HollandPublishing Amsterdam The Netherlands 1988
[12] H S Ramane D S Revankar I Gutman and H B WalikarldquoDistance spectra anddistance energies of iterated line graphs ofregular graphsrdquo Institut Mathematique vol 85 pp 39ndash46 2009
[13] X Zhang and C Godsil ldquoThe inertia of distance matrices ofsome graphsrdquo Discrete Mathematics vol 313 no 16 pp 1655ndash1664 2013
[14] D M Cvetkovic M Doob and H Sachs Spectra of Graphsvol 87 Academic Press New York NY USA 1980 Theory andapplication
[15] R L Graham A J Hoffman and H Hosoya ldquoOn the distancematrix of a directed graphrdquo Journal of Graph Theory vol 1 no1 pp 85ndash88 1977
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Journal of Applied Mathematics
Lemma 1 (see [14]) Let 119860 be a Hermitian matrix witheigenvalues 120582
1⩾ sdot sdot sdot ⩾ 120582
119899and 119861 one of its principal
submatrices Let 119861 have eigenvalues 1205831⩾ sdot sdot sdot ⩾ 120583
119898 Then the
inequalities 120582119899minus119898+119894
⩽ 120583119894⩽ 120582119894(119894 = 1 119898) hold
For a square matrix let cof (119860) denote the sum ofcofactors of 119860 Form the matrix 119860 by subtracting the firstrow from all other rows then the first column from all othercolumns and let119860
11denote the principle submatrix obtained
from 119860 by deleting the first row and first column
Lemma 2 (see [15]) 119888119900119891(119860) = det11986011
A cut vertex is a vertex the removal of which woulddisconnect the remaining graph a block of a graph is definedto be a maximal subgraph having no cut vertices
Lemma3 (see [15]) If 119866 is a strongly connected directed graphwith blocks 119866
1 1198662 119866
119903 then
119888119900119891119863 (119866) =
119903
prod
119894=1
119888119900119891119863 (119866119894)
det119863 (119866) =119903
sum
119894=1
det 119863(119866119894)prod
119895 = 119894
119888119900119891119863 (119866119894)
(1)
Lemma 4 Let
119862 =
[[[[[[[[[[
[
1 1 1 1 sdot sdot sdot 1 1
minus1 minus2 minus1 0 sdot sdot sdot 0 0
0 minus1 minus2 minus1 sdot sdot sdot 0 0
0 0 minus1 minus2 sdot sdot sdot 0 0
d
0 0 0 0 sdot sdot sdot minus2 minus1
0 0 0 0 sdot sdot sdot minus1 minus2
]]]]]]]]]]
]119899times119899
(2)
Then
det 119862 =
minus119899
2 119894119891 119899 119894119904 119890V119890119899
119899 + 1
2 119894119891 119899 119894119904 119900119889119889
(3)
Proof Let
119862119899=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1 1 1 1 sdot sdot sdot 1 1
minus1 minus2 minus1 0 sdot sdot sdot 0 0
0 minus1 minus2 minus1 sdot sdot sdot 0 0
0 0 minus1 minus2 sdot sdot sdot 0 0
d
0 0 0 0 sdot sdot sdot minus2 minus1
0 0 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
119863119899=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1
2minus1 0 0 sdot sdot sdot 0 0
1 minus2 minus1 0 sdot sdot sdot 0 0
1 minus1 minus2 minus1 sdot sdot sdot 0 0
1 0 minus1 minus2 sdot sdot sdot 0 0
d
1 0 0 0 sdot sdot sdot minus2 minus1
1 0 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
(4)
Comparing 119862119899to119863119899 we get the following
119862119899= 119863119899+119899
2times (minus1)
119899minus1
(5)
Expanding the determinant 119863119899according to the last column
and then the last line we get the following incursion
119863119899= minus2119863
119899minus1minus 119863119899minus2
+ 1 (6)
that is
119863119899+ 119863119899minus1
= minus (119863119899minus1
+ 119863119899minus2) + 1 (7)
Since 1198631= 12 119863
2= 0 and 119863
3= 12 from the above
incursion we get the following
119863119899=
0 if 119899 is even
1
2 if 119899 is odd
(8)
So we have the following
119862119899=
minus119899
2 if 119899 is even
119899 + 1
2 if 119899 is odd
(9)
This completes the proof
3 Main Results
In the following we always assume that 119881(119882119899) = V
0 V1
V119899minus1 where V
0is the hub of119882
119899
Theorem 5 Let 119890 = V119894V119894+1 mod (119899minus1) (1 ⩽ 119894 ⩽ 119899 minus 1) Then
det 119863(119882119899minus 119890) =
minus1198992
4 119894119891 119899 119894119904 119890V119890119899
1198992minus 1
4 119894119891 119899 119894119904 119900119889119889
(10)
where 119899 ⩾ 3
Proof Without loss of generality we may assume that 119890 =V1V119899minus1
Let
119860119899= det119863(119882
119899minus 119890) =
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 1 1 1 sdot sdot sdot 1 1
1 0 1 2 sdot sdot sdot 2 2
1 1 0 1 sdot sdot sdot 2 2
1 2 1 0 sdot sdot sdot 2 2
d
1 2 2 2 sdot sdot sdot 0 1
1 2 2 2 sdot sdot sdot 1 0
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
(11)
Journal of Applied Mathematics 3
Then
119860119899=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 1 1 minus1
21 sdot sdot sdot 1 1
1 minus2 0 0 sdot sdot sdot 0 0
1 minus1
20 minus2 +
1
2minus1 sdot sdot sdot 0 0
1 0 minus1 minus2 sdot sdot sdot 0 0
d
1 0 0 0 sdot sdot sdot minus2 minus1
1 0 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
(12)
Expanding the determinant 119860119899according to the second line
we get the following incursion
119860119899= (minus1)
119899minus1 119899
2minus 2119860119899minus1
minus 119862119899minus2
minus 119863119899minus2
minus 119860119899minus2 (13)
where 119862119899and119863
119899are defined as in Lemma 4
By Lemma 4 we get the following
119860119899=
minus1 minus 2119860119899minus1
minus 119860119899minus2 if 119899 is even
minus2119860119899minus1
minus 119860119899minus2 if 119899 is odd
(14)
Since 1198603= 2 119860
4= minus4 119860
5= 6 and 119860
6= minus9 according to
the above incursion we get the following
119860119899=
minus1198992
4 if 119899 is even
1198992minus 1
4 if 119899 is odd
(15)
where 119899 ⩾ 3 This completes the proof
Corollary 6 Let 119890 = V119894V119894+1 mod (119899minus1) (1 ⩽ 119894 ⩽ 119899 minus 1) Then
119899+(119863 (119882
119899minus 119890)) = 1 119899
0(119863 (119882
119899minus 119890)) = 0
119899minus(119863 (119882
119899minus 119890)) = 119899 minus 1
(16)
Proof We will prove the result by induction on 119899If 119899 = 3119882
3minus 119890 cong 119875
3is obviously true
Suppose that the result is true for 119899minus1 that is 119899+(119863(119882
119899minus1minus
119890)) = 1 1198990(119863(119882
119899minus1minus 119890)) = 0 119899
minus(119863(119882
119899minus1minus 119890)) = 119899 minus 2
Since 119863(119882119899minus1
minus 119890) is a principle submatrix of 119863(119882119899minus
119890) by Lemma 1 the eigenvalues of 119863(119882119899minus1
minus 119890) interlacethe eigenvalues of 119863(119882
119899minus 119890) By Theorem 5 det119863(119882
119899minus1minus
119890) det119863(119882119899minus 119890) lt 0 So 119863(119882
119899minus 119890) has one negative
eigenvaluemore than119863(119882119899minus1minus119890) According to the induction
hypothesis we get 119899+(119863(119882
119899minus119890)) = 1 119899
0(119863(119882
119899minus119890)) = 0 and
119899minus(119863(119882
119899minus 119890)) = 119899 minus 1 This completes the proof
Theorem 7 One has
det 119863(119882119899) =
1 minus 119899 119894119891 119899 119894119904 119890V119890119899
0 119894119891 119899 119894119904 119900119889119889(17)
where 119899 ⩾ 3
Proof Consider the following
det119863(119882119899) =
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 1 1 1 sdot sdot sdot 1 1
1 0 1 2 sdot sdot sdot 2 1
1 1 0 1 sdot sdot sdot 2 2
1 2 1 0 sdot sdot sdot 2 2
d
1 2 2 2 sdot sdot sdot 0 1
1 1 2 2 sdot sdot sdot 1 0
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 1 1 1 sdot sdot sdot 1 1
1 minus2 minus1 0 sdot sdot sdot 0 minus1
1 minus1 minus2 minus1 sdot sdot sdot 0 0
1 0 minus1 minus2 sdot sdot sdot 0 0
d
1 0 0 0 sdot sdot sdot minus2 minus1
1 minus1 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 119899 minus 1 1 1 sdot sdot sdot 1 1
119899 minus 1 minus4 (119899 minus 1) minus4 minus4 sdot sdot sdot minus4 minus4
1 minus4 minus2 minus1 sdot sdot sdot 0 0
1 minus4 minus1 minus2 sdot sdot sdot 0 0
d
1 minus4 0 0 sdot sdot sdot minus2 minus1
1 minus4 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 119899 minus 1 1 1 sdot sdot sdot 1 1
119899 minus 1 4 (119899 minus 1) 0 0 sdot sdot sdot 0 0
1 0 minus2 minus1 sdot sdot sdot 0 0
1 0 minus1 minus2 sdot sdot sdot 0 0
d
1 0 0 0 sdot sdot sdot minus2 minus1
1 0 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
(18)
Expanding the above determinant according to the secondline we get the following
det119863(119882119899) = (minus1)
119899minus1
(119899 minus 1)3
+ 4 (119899 minus 1)119860119899minus1 (19)
where 119860119899is defined as inTheorem 5
ByTheorem 5 when 119899 ⩾ 3 we get the following
det119863(119882119899) =
1 minus 119899 if 119899 is even0 if 119899 is odd
(20)
This completes the proof
Similar to Corollary 6 we can get the following corollary
Corollary 8 (i) If 119899 is even 119899+(119863(119882
119899)) = 1 119899
0(119863(119882
119899)) = 0
119899minus(119863(119882
119899)) = 119899 minus 1
(ii) If 119899 is odd 119899+(119863(119882
119899)) = 1 119899
0(119863(119882
119899)) = 1
119899minus(119863(119882
119899)) = 119899 minus 2
Denote by119882119899minus V0the graph obtained from119882
119899by deleting
the vertex V0and all the edges adjacent to V
0 that is119882
119899minus V0cong
119862119899minus1
Let 119864119896(1 ⩽ 119896 ⩽ 119899) be any subset of 119864(119882
119899minus V0) with
|119864119896| = 119896 In the following we always denote by 119882
119899minus 119864119896the
graph obtained from119882119899by deleting all the edges in 119864
119896
4 Journal of Applied Mathematics
Theorem 9 One has 119899+(119863(119882
119899minus 119864119896)) = 1 119899
0(119863(119882
119899minus 119864119896)) =
0 119899minus(119863(119882
119899minus 119864119896)) = 119899 minus 1
Proof Denote the components of119882119899minus 119864119896minus V0by 1198621 119862
119904
Let 119866119894denote the graph that contains 119862
119894plus the vertex V
0
such that V0is connected to every other vertex 1 ⩽ 119894 ⩽ 119904
Then each 119866119894(1 ⩽ 119894 ⩽ 119904) is isomorphism to119882
|119881(119866119894)|minus 119890119894or
1198702 where 119890
119894is an edge of119882
|119881(119866119894)|minus V0 By Lemma 2 and some
direct calculations we get the following
cof (119866119894) = det119863
(119882|119881(119866119894)|
minus 119890119894)11
= (minus1)|119881(119866119894)|minus1
1003816100381610038161003816119881 (119866119894)1003816100381610038161003816
(21)
It is easy to check that cof (119866119894) = (minus1)
|119881(119866119894)|minus1|119881(119866119894)| is also
true when 119866119894is isomorphism to 119870
2
In the following we will prove the theorem by introduc-tion on 119904
For 119904 = 1 119866 cong 119882119899minus 119890 where 119890 is an edge of119882
119899minus V0 by
Corollary 6 we get the resultSuppose the result is true for 119904 minus 1For 119904 let 1198661015840 = 119866
1cup1198662cup sdot sdot sdot cup119866
119904minus1 Then by the induction
hypothesis 119899+(119863(1198661015840)) = 1 119899
0(119863(1198661015840)) = 0 and 119899
minus(119863(1198661015840)) =
|119881(1198661015840)| minus 1 which implies that
det119863(1198661015840) = (minus1)|119881(1198661015840)|minus1
119886 (22)
where 119886 is a positive integerSince
cof (119866119894) = (minus1)
|119881(119866119894)|minus11003816100381610038161003816119881 (119866119894)
1003816100381610038161003816 1 ⩽ 119894 ⩽ 119904 (23)
by Lemma 3
cof119863(1198661015840) =119904minus1
prod
119895=1
cof119863(119866119895) = (minus1)
|119881(1198661015840)|minus1
119904minus1
prod
119895=1
10038161003816100381610038161003816119881 (119866119895)10038161003816100381610038161003816
(24)
Thendet119863(119882
119899minus 119864119896)
= det119863(1198661015840) cof119863(119866119894) + det119863(119866
119894) cof119863(1198661015840)
= (minus1)|119881(1198661015840)|minus1
119886 times (minus1)|119881(119866119894)|minus1
1003816100381610038161003816119881 (119866119894)1003816100381610038161003816
+ (minus1)|119881(119866119894)|minus1119887 times (minus1)
|119881(1198661015840)|minus1
119904minus1
prod
119895=1
10038161003816100381610038161003816119881 (119866119895)10038161003816100381610038161003816
= (minus1)119899minus1
(1198861003816100381610038161003816119881 (119866119894)
1003816100381610038161003816 + 119887
119904minus1
prod
119895=1
10038161003816100381610038161003816119881 (119866119895)10038161003816100381610038161003816)
(25)
where 119887 = 11989924 if 119899 is even and 119887 = (1198992 minus 1)4 if 119899 is oddIn this case similar to Corollary 6 we can easily get
119899+(119863(119882
119899minus 119864119896)) = 1 119899
0(119863(119882
119899minus 119864119896)) = 0 and 119899
minus(119863(119882
119899minus
119864119896)) = 119899 minus 1Up to now we have proved the result
Let119866119906119894sdotV119895119867 denote the graph formed by only identifying
the vertex 119906119894of 119866 with the vertex V
119895of119867 where 119906
119894and V119895are
arbitrary vertices of 119866 and119867 respectively
Lemma 10 (see [13]) Let 119866times119867 denote the Cartesian productof connected graphs 119866 and119867 where 119881(119866) = 119906
1 119906
119898 and
119881(119867) = V1 V
119899 Then we have
(i) 119899+(119866 times 119867) = 119899
+(119866119906119894sdot V119895119867)
(ii) 1198990(119866 times 119867) = (119898 minus 1)(119899 minus 1) + 119899
0(119866119906119894sdot V119895119867)
(iii) 119899minus(119866 times 119867) = 119899
minus(119866119906119894sdot V119895119867)
Theorem 11 Let 1199060and V
0be the hubs of 119882
119899and 119882
119898
respectively Suppose119864119901(0 ⩽ 119901 ⩽ 119899minus1) and119864
119902(0 ⩽ 119902 ⩽ 119898minus1)
are any subsets of 119864(119882119899minus 1199060) and 119864(119882
119898minus V0) with |119864
119901| = 119901
|119864119902| = 119902 respectively Then the distance matrix of the graph
(119882119899minus 119864119901) times (119882
119898minus 119864119902) has exactly one positive eigenvalue
Proof Since 1199060and V
0are the hubs of 119882
119899and 119882
119898 respec-
tively (119882119899minus119864119901)1199060sdotV0(119882119898minus119864119902)must be isomorphism to some
119882119899+119898minus1
minus119864119901+119902
where1199080is the hub of119882
119899+119898minus1and 119864
119901+119902is any
subset of 119864(119882119899+119898minus1
minus 1199080) with |119864
119901+119902| = 119901 + 119902 By Theorem 9
and Lemma 10 we get the result
Given an arbitrary integer 119898 for 1 ⩽ 119894 ⩽ 119898 let V1198940be
the hub of 119882119899119894and 119864
119901119894any subset of 119864(119882
119899119894minus V1198940) Suppose
119881(119882119899119894) = V
1198940 V1198941 V
119894(119899119894minus1)
Theorem 12 For an arbitrary integer 119898 the distance matrixof the graph 119866 = (119882
1198991minus 1198641199011)V1119896sdot V2119895(1198821198992minus 1198641199012)V2ℎsdot sdot sdot
V(119898minus1)119903
(119882119899119898minus1
minus 119864119901119898minus1
)V(119898minus1)119905
sdot V119898119904(119882119899119898minus 119864119901119898) has exactly one
positive eigenvalue
Proof We will prove the conclusion by induction on119898If119898 = 1 by Theorem 9 the conclusion is trueSuppose the conclusion is true for119898minus1 For convenience
let119867 = (1198821198991minus 1198641199011)V1119896sdot V2119895(1198821198992minus 1198641199012)V2ℎsdot sdot sdot V(119898minus2)119894
(119882119899119898minus2
minus
119864119901119898minus2
) Then 119866 = 119867V(119898minus2)119892
sdot V(119898minus1)119903
(119882119899119898minus1
minus 119864119901119898minus1
)V(119898minus1)119905
sdot
V119898119904(119882119899119898minus 119864119901119898) By Lemma 10 we have the following
119899+(119866) = 119899
+(119867 (119882
119899119898minus1minus 119864119901119898minus1
) V(119898minus1)119905
sdot V119898119904(119882119899119898minus 119864119901119898))
= 119899+(119867 (119882
119899119898minus1minus 119864119901119898minus1
) V(119898minus1)0
sdot V1198980(119882119899119898minus 119864119901119898))
(26)
Since (119882119899119898minus1
minus 119864119901119898minus1
)V(119898minus1)0
sdot V1198980(119882119899119898minus 119864119901119898) = 119882
119899119898minus1+119899119898minus1minus
119864119901119898minus1
minus119864119901119898 we get 119899
+(119866) = 119899
+(119867V(119898minus2)119892
sdotV(119898minus1)119903
(119882119899119898minus1+119899119898minus1
minus
119864119901119898minus1
minus119864119901119898)) By the induction hypothesis we get 119899
+(119866) = 1
This completes the proof
Remark 13 Let 1198661and 119866
2be any two graphs with the same
form as119866 inTheorem 12MakingCartesian product of graphs1198661and 119866
2 by Lemma 10 and Theorem 12 we get a series
of graphs whose distance matrices have exactly one positiveeigenvalue
Acknowledgments
The authors would like to thank the anonymous refereesfor their valuable comments and suggestions This workwas supported by NSFC (11126256) and NSF of ShandongProvince of China (ZR2012AQ022)
Journal of Applied Mathematics 5
References
[1] D H Rouvray ldquoThe role of the topological distance matrixin chemistryrdquo in Mathematical and Computational Conceptsin Chemistry N Trinajstid Ed pp 295ndash306 Ellis HarwoodChichester UK 1986
[2] W Brostow D M McEachern and S Petrez- Guiterrez ldquoPres-sure second virial coefficients of hydrocarbons fluorocarbonsand their mixtures Interactions of walksrdquo Journal of ChemicalPhysics vol 71 pp 2716ndash2722 1979
[3] B D McKay ldquoOn the spectral characterisation of treesrdquo ArsCombinatoria vol 3 pp 219ndash232 1977
[4] D W Bradley and R A Bradley ldquoString edits and macro-moleculesrdquo in Time Wraps D Sankoff and J B Kruskal EdsChapter 6 Addison-Wesley Reading Mass USA 1983
[5] J P Boyd and K N Wexler ldquoTrees with structurerdquo Journal ofMathematical Psychology vol 10 pp 115ndash147 1973
[6] R L Graham and L Lovasz ldquoDistance matrix polynomials oftreesrdquo inTheory and Applications of Graphs vol 642 of LectureNotes in Mathematics pp 186ndash190 1978
[7] M S Waterman T F Smith and H I Katcher ldquoAlgorithms forrestriction map comparisonsrdquo Nucleic Acids Research vol 12pp 237ndash242 1984
[8] R L Graham and H O Pollak ldquoOn the addressing problem forloop switchingrdquo The Bell System Technical Journal vol 50 pp2495ndash2519 1971
[9] R Bapat S J Kirkland and M Neumann ldquoOn distancematrices and Laplaciansrdquo Linear Algebra and Its Applicationsvol 401 pp 193ndash209 2005
[10] K Balasubramanian ldquoComputer generation of distance poly-nomials of graphsrdquo Journal of Computational Chemistry vol 11no 7 pp 829ndash836 1990
[11] D M Cvetkovic M Doob I Gutman and A Torgasev RecentResults in the Theory of Graph Spectra vol 36 North-HollandPublishing Amsterdam The Netherlands 1988
[12] H S Ramane D S Revankar I Gutman and H B WalikarldquoDistance spectra anddistance energies of iterated line graphs ofregular graphsrdquo Institut Mathematique vol 85 pp 39ndash46 2009
[13] X Zhang and C Godsil ldquoThe inertia of distance matrices ofsome graphsrdquo Discrete Mathematics vol 313 no 16 pp 1655ndash1664 2013
[14] D M Cvetkovic M Doob and H Sachs Spectra of Graphsvol 87 Academic Press New York NY USA 1980 Theory andapplication
[15] R L Graham A J Hoffman and H Hosoya ldquoOn the distancematrix of a directed graphrdquo Journal of Graph Theory vol 1 no1 pp 85ndash88 1977
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Journal of Applied Mathematics 3
Then
119860119899=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 1 1 minus1
21 sdot sdot sdot 1 1
1 minus2 0 0 sdot sdot sdot 0 0
1 minus1
20 minus2 +
1
2minus1 sdot sdot sdot 0 0
1 0 minus1 minus2 sdot sdot sdot 0 0
d
1 0 0 0 sdot sdot sdot minus2 minus1
1 0 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
(12)
Expanding the determinant 119860119899according to the second line
we get the following incursion
119860119899= (minus1)
119899minus1 119899
2minus 2119860119899minus1
minus 119862119899minus2
minus 119863119899minus2
minus 119860119899minus2 (13)
where 119862119899and119863
119899are defined as in Lemma 4
By Lemma 4 we get the following
119860119899=
minus1 minus 2119860119899minus1
minus 119860119899minus2 if 119899 is even
minus2119860119899minus1
minus 119860119899minus2 if 119899 is odd
(14)
Since 1198603= 2 119860
4= minus4 119860
5= 6 and 119860
6= minus9 according to
the above incursion we get the following
119860119899=
minus1198992
4 if 119899 is even
1198992minus 1
4 if 119899 is odd
(15)
where 119899 ⩾ 3 This completes the proof
Corollary 6 Let 119890 = V119894V119894+1 mod (119899minus1) (1 ⩽ 119894 ⩽ 119899 minus 1) Then
119899+(119863 (119882
119899minus 119890)) = 1 119899
0(119863 (119882
119899minus 119890)) = 0
119899minus(119863 (119882
119899minus 119890)) = 119899 minus 1
(16)
Proof We will prove the result by induction on 119899If 119899 = 3119882
3minus 119890 cong 119875
3is obviously true
Suppose that the result is true for 119899minus1 that is 119899+(119863(119882
119899minus1minus
119890)) = 1 1198990(119863(119882
119899minus1minus 119890)) = 0 119899
minus(119863(119882
119899minus1minus 119890)) = 119899 minus 2
Since 119863(119882119899minus1
minus 119890) is a principle submatrix of 119863(119882119899minus
119890) by Lemma 1 the eigenvalues of 119863(119882119899minus1
minus 119890) interlacethe eigenvalues of 119863(119882
119899minus 119890) By Theorem 5 det119863(119882
119899minus1minus
119890) det119863(119882119899minus 119890) lt 0 So 119863(119882
119899minus 119890) has one negative
eigenvaluemore than119863(119882119899minus1minus119890) According to the induction
hypothesis we get 119899+(119863(119882
119899minus119890)) = 1 119899
0(119863(119882
119899minus119890)) = 0 and
119899minus(119863(119882
119899minus 119890)) = 119899 minus 1 This completes the proof
Theorem 7 One has
det 119863(119882119899) =
1 minus 119899 119894119891 119899 119894119904 119890V119890119899
0 119894119891 119899 119894119904 119900119889119889(17)
where 119899 ⩾ 3
Proof Consider the following
det119863(119882119899) =
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 1 1 1 sdot sdot sdot 1 1
1 0 1 2 sdot sdot sdot 2 1
1 1 0 1 sdot sdot sdot 2 2
1 2 1 0 sdot sdot sdot 2 2
d
1 2 2 2 sdot sdot sdot 0 1
1 1 2 2 sdot sdot sdot 1 0
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 1 1 1 sdot sdot sdot 1 1
1 minus2 minus1 0 sdot sdot sdot 0 minus1
1 minus1 minus2 minus1 sdot sdot sdot 0 0
1 0 minus1 minus2 sdot sdot sdot 0 0
d
1 0 0 0 sdot sdot sdot minus2 minus1
1 minus1 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 119899 minus 1 1 1 sdot sdot sdot 1 1
119899 minus 1 minus4 (119899 minus 1) minus4 minus4 sdot sdot sdot minus4 minus4
1 minus4 minus2 minus1 sdot sdot sdot 0 0
1 minus4 minus1 minus2 sdot sdot sdot 0 0
d
1 minus4 0 0 sdot sdot sdot minus2 minus1
1 minus4 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
=
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
0 119899 minus 1 1 1 sdot sdot sdot 1 1
119899 minus 1 4 (119899 minus 1) 0 0 sdot sdot sdot 0 0
1 0 minus2 minus1 sdot sdot sdot 0 0
1 0 minus1 minus2 sdot sdot sdot 0 0
d
1 0 0 0 sdot sdot sdot minus2 minus1
1 0 0 0 sdot sdot sdot minus1 minus2
100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816119899times119899
(18)
Expanding the above determinant according to the secondline we get the following
det119863(119882119899) = (minus1)
119899minus1
(119899 minus 1)3
+ 4 (119899 minus 1)119860119899minus1 (19)
where 119860119899is defined as inTheorem 5
ByTheorem 5 when 119899 ⩾ 3 we get the following
det119863(119882119899) =
1 minus 119899 if 119899 is even0 if 119899 is odd
(20)
This completes the proof
Similar to Corollary 6 we can get the following corollary
Corollary 8 (i) If 119899 is even 119899+(119863(119882
119899)) = 1 119899
0(119863(119882
119899)) = 0
119899minus(119863(119882
119899)) = 119899 minus 1
(ii) If 119899 is odd 119899+(119863(119882
119899)) = 1 119899
0(119863(119882
119899)) = 1
119899minus(119863(119882
119899)) = 119899 minus 2
Denote by119882119899minus V0the graph obtained from119882
119899by deleting
the vertex V0and all the edges adjacent to V
0 that is119882
119899minus V0cong
119862119899minus1
Let 119864119896(1 ⩽ 119896 ⩽ 119899) be any subset of 119864(119882
119899minus V0) with
|119864119896| = 119896 In the following we always denote by 119882
119899minus 119864119896the
graph obtained from119882119899by deleting all the edges in 119864
119896
4 Journal of Applied Mathematics
Theorem 9 One has 119899+(119863(119882
119899minus 119864119896)) = 1 119899
0(119863(119882
119899minus 119864119896)) =
0 119899minus(119863(119882
119899minus 119864119896)) = 119899 minus 1
Proof Denote the components of119882119899minus 119864119896minus V0by 1198621 119862
119904
Let 119866119894denote the graph that contains 119862
119894plus the vertex V
0
such that V0is connected to every other vertex 1 ⩽ 119894 ⩽ 119904
Then each 119866119894(1 ⩽ 119894 ⩽ 119904) is isomorphism to119882
|119881(119866119894)|minus 119890119894or
1198702 where 119890
119894is an edge of119882
|119881(119866119894)|minus V0 By Lemma 2 and some
direct calculations we get the following
cof (119866119894) = det119863
(119882|119881(119866119894)|
minus 119890119894)11
= (minus1)|119881(119866119894)|minus1
1003816100381610038161003816119881 (119866119894)1003816100381610038161003816
(21)
It is easy to check that cof (119866119894) = (minus1)
|119881(119866119894)|minus1|119881(119866119894)| is also
true when 119866119894is isomorphism to 119870
2
In the following we will prove the theorem by introduc-tion on 119904
For 119904 = 1 119866 cong 119882119899minus 119890 where 119890 is an edge of119882
119899minus V0 by
Corollary 6 we get the resultSuppose the result is true for 119904 minus 1For 119904 let 1198661015840 = 119866
1cup1198662cup sdot sdot sdot cup119866
119904minus1 Then by the induction
hypothesis 119899+(119863(1198661015840)) = 1 119899
0(119863(1198661015840)) = 0 and 119899
minus(119863(1198661015840)) =
|119881(1198661015840)| minus 1 which implies that
det119863(1198661015840) = (minus1)|119881(1198661015840)|minus1
119886 (22)
where 119886 is a positive integerSince
cof (119866119894) = (minus1)
|119881(119866119894)|minus11003816100381610038161003816119881 (119866119894)
1003816100381610038161003816 1 ⩽ 119894 ⩽ 119904 (23)
by Lemma 3
cof119863(1198661015840) =119904minus1
prod
119895=1
cof119863(119866119895) = (minus1)
|119881(1198661015840)|minus1
119904minus1
prod
119895=1
10038161003816100381610038161003816119881 (119866119895)10038161003816100381610038161003816
(24)
Thendet119863(119882
119899minus 119864119896)
= det119863(1198661015840) cof119863(119866119894) + det119863(119866
119894) cof119863(1198661015840)
= (minus1)|119881(1198661015840)|minus1
119886 times (minus1)|119881(119866119894)|minus1
1003816100381610038161003816119881 (119866119894)1003816100381610038161003816
+ (minus1)|119881(119866119894)|minus1119887 times (minus1)
|119881(1198661015840)|minus1
119904minus1
prod
119895=1
10038161003816100381610038161003816119881 (119866119895)10038161003816100381610038161003816
= (minus1)119899minus1
(1198861003816100381610038161003816119881 (119866119894)
1003816100381610038161003816 + 119887
119904minus1
prod
119895=1
10038161003816100381610038161003816119881 (119866119895)10038161003816100381610038161003816)
(25)
where 119887 = 11989924 if 119899 is even and 119887 = (1198992 minus 1)4 if 119899 is oddIn this case similar to Corollary 6 we can easily get
119899+(119863(119882
119899minus 119864119896)) = 1 119899
0(119863(119882
119899minus 119864119896)) = 0 and 119899
minus(119863(119882
119899minus
119864119896)) = 119899 minus 1Up to now we have proved the result
Let119866119906119894sdotV119895119867 denote the graph formed by only identifying
the vertex 119906119894of 119866 with the vertex V
119895of119867 where 119906
119894and V119895are
arbitrary vertices of 119866 and119867 respectively
Lemma 10 (see [13]) Let 119866times119867 denote the Cartesian productof connected graphs 119866 and119867 where 119881(119866) = 119906
1 119906
119898 and
119881(119867) = V1 V
119899 Then we have
(i) 119899+(119866 times 119867) = 119899
+(119866119906119894sdot V119895119867)
(ii) 1198990(119866 times 119867) = (119898 minus 1)(119899 minus 1) + 119899
0(119866119906119894sdot V119895119867)
(iii) 119899minus(119866 times 119867) = 119899
minus(119866119906119894sdot V119895119867)
Theorem 11 Let 1199060and V
0be the hubs of 119882
119899and 119882
119898
respectively Suppose119864119901(0 ⩽ 119901 ⩽ 119899minus1) and119864
119902(0 ⩽ 119902 ⩽ 119898minus1)
are any subsets of 119864(119882119899minus 1199060) and 119864(119882
119898minus V0) with |119864
119901| = 119901
|119864119902| = 119902 respectively Then the distance matrix of the graph
(119882119899minus 119864119901) times (119882
119898minus 119864119902) has exactly one positive eigenvalue
Proof Since 1199060and V
0are the hubs of 119882
119899and 119882
119898 respec-
tively (119882119899minus119864119901)1199060sdotV0(119882119898minus119864119902)must be isomorphism to some
119882119899+119898minus1
minus119864119901+119902
where1199080is the hub of119882
119899+119898minus1and 119864
119901+119902is any
subset of 119864(119882119899+119898minus1
minus 1199080) with |119864
119901+119902| = 119901 + 119902 By Theorem 9
and Lemma 10 we get the result
Given an arbitrary integer 119898 for 1 ⩽ 119894 ⩽ 119898 let V1198940be
the hub of 119882119899119894and 119864
119901119894any subset of 119864(119882
119899119894minus V1198940) Suppose
119881(119882119899119894) = V
1198940 V1198941 V
119894(119899119894minus1)
Theorem 12 For an arbitrary integer 119898 the distance matrixof the graph 119866 = (119882
1198991minus 1198641199011)V1119896sdot V2119895(1198821198992minus 1198641199012)V2ℎsdot sdot sdot
V(119898minus1)119903
(119882119899119898minus1
minus 119864119901119898minus1
)V(119898minus1)119905
sdot V119898119904(119882119899119898minus 119864119901119898) has exactly one
positive eigenvalue
Proof We will prove the conclusion by induction on119898If119898 = 1 by Theorem 9 the conclusion is trueSuppose the conclusion is true for119898minus1 For convenience
let119867 = (1198821198991minus 1198641199011)V1119896sdot V2119895(1198821198992minus 1198641199012)V2ℎsdot sdot sdot V(119898minus2)119894
(119882119899119898minus2
minus
119864119901119898minus2
) Then 119866 = 119867V(119898minus2)119892
sdot V(119898minus1)119903
(119882119899119898minus1
minus 119864119901119898minus1
)V(119898minus1)119905
sdot
V119898119904(119882119899119898minus 119864119901119898) By Lemma 10 we have the following
119899+(119866) = 119899
+(119867 (119882
119899119898minus1minus 119864119901119898minus1
) V(119898minus1)119905
sdot V119898119904(119882119899119898minus 119864119901119898))
= 119899+(119867 (119882
119899119898minus1minus 119864119901119898minus1
) V(119898minus1)0
sdot V1198980(119882119899119898minus 119864119901119898))
(26)
Since (119882119899119898minus1
minus 119864119901119898minus1
)V(119898minus1)0
sdot V1198980(119882119899119898minus 119864119901119898) = 119882
119899119898minus1+119899119898minus1minus
119864119901119898minus1
minus119864119901119898 we get 119899
+(119866) = 119899
+(119867V(119898minus2)119892
sdotV(119898minus1)119903
(119882119899119898minus1+119899119898minus1
minus
119864119901119898minus1
minus119864119901119898)) By the induction hypothesis we get 119899
+(119866) = 1
This completes the proof
Remark 13 Let 1198661and 119866
2be any two graphs with the same
form as119866 inTheorem 12MakingCartesian product of graphs1198661and 119866
2 by Lemma 10 and Theorem 12 we get a series
of graphs whose distance matrices have exactly one positiveeigenvalue
Acknowledgments
The authors would like to thank the anonymous refereesfor their valuable comments and suggestions This workwas supported by NSFC (11126256) and NSF of ShandongProvince of China (ZR2012AQ022)
Journal of Applied Mathematics 5
References
[1] D H Rouvray ldquoThe role of the topological distance matrixin chemistryrdquo in Mathematical and Computational Conceptsin Chemistry N Trinajstid Ed pp 295ndash306 Ellis HarwoodChichester UK 1986
[2] W Brostow D M McEachern and S Petrez- Guiterrez ldquoPres-sure second virial coefficients of hydrocarbons fluorocarbonsand their mixtures Interactions of walksrdquo Journal of ChemicalPhysics vol 71 pp 2716ndash2722 1979
[3] B D McKay ldquoOn the spectral characterisation of treesrdquo ArsCombinatoria vol 3 pp 219ndash232 1977
[4] D W Bradley and R A Bradley ldquoString edits and macro-moleculesrdquo in Time Wraps D Sankoff and J B Kruskal EdsChapter 6 Addison-Wesley Reading Mass USA 1983
[5] J P Boyd and K N Wexler ldquoTrees with structurerdquo Journal ofMathematical Psychology vol 10 pp 115ndash147 1973
[6] R L Graham and L Lovasz ldquoDistance matrix polynomials oftreesrdquo inTheory and Applications of Graphs vol 642 of LectureNotes in Mathematics pp 186ndash190 1978
[7] M S Waterman T F Smith and H I Katcher ldquoAlgorithms forrestriction map comparisonsrdquo Nucleic Acids Research vol 12pp 237ndash242 1984
[8] R L Graham and H O Pollak ldquoOn the addressing problem forloop switchingrdquo The Bell System Technical Journal vol 50 pp2495ndash2519 1971
[9] R Bapat S J Kirkland and M Neumann ldquoOn distancematrices and Laplaciansrdquo Linear Algebra and Its Applicationsvol 401 pp 193ndash209 2005
[10] K Balasubramanian ldquoComputer generation of distance poly-nomials of graphsrdquo Journal of Computational Chemistry vol 11no 7 pp 829ndash836 1990
[11] D M Cvetkovic M Doob I Gutman and A Torgasev RecentResults in the Theory of Graph Spectra vol 36 North-HollandPublishing Amsterdam The Netherlands 1988
[12] H S Ramane D S Revankar I Gutman and H B WalikarldquoDistance spectra anddistance energies of iterated line graphs ofregular graphsrdquo Institut Mathematique vol 85 pp 39ndash46 2009
[13] X Zhang and C Godsil ldquoThe inertia of distance matrices ofsome graphsrdquo Discrete Mathematics vol 313 no 16 pp 1655ndash1664 2013
[14] D M Cvetkovic M Doob and H Sachs Spectra of Graphsvol 87 Academic Press New York NY USA 1980 Theory andapplication
[15] R L Graham A J Hoffman and H Hosoya ldquoOn the distancematrix of a directed graphrdquo Journal of Graph Theory vol 1 no1 pp 85ndash88 1977
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Journal of Applied Mathematics
Theorem 9 One has 119899+(119863(119882
119899minus 119864119896)) = 1 119899
0(119863(119882
119899minus 119864119896)) =
0 119899minus(119863(119882
119899minus 119864119896)) = 119899 minus 1
Proof Denote the components of119882119899minus 119864119896minus V0by 1198621 119862
119904
Let 119866119894denote the graph that contains 119862
119894plus the vertex V
0
such that V0is connected to every other vertex 1 ⩽ 119894 ⩽ 119904
Then each 119866119894(1 ⩽ 119894 ⩽ 119904) is isomorphism to119882
|119881(119866119894)|minus 119890119894or
1198702 where 119890
119894is an edge of119882
|119881(119866119894)|minus V0 By Lemma 2 and some
direct calculations we get the following
cof (119866119894) = det119863
(119882|119881(119866119894)|
minus 119890119894)11
= (minus1)|119881(119866119894)|minus1
1003816100381610038161003816119881 (119866119894)1003816100381610038161003816
(21)
It is easy to check that cof (119866119894) = (minus1)
|119881(119866119894)|minus1|119881(119866119894)| is also
true when 119866119894is isomorphism to 119870
2
In the following we will prove the theorem by introduc-tion on 119904
For 119904 = 1 119866 cong 119882119899minus 119890 where 119890 is an edge of119882
119899minus V0 by
Corollary 6 we get the resultSuppose the result is true for 119904 minus 1For 119904 let 1198661015840 = 119866
1cup1198662cup sdot sdot sdot cup119866
119904minus1 Then by the induction
hypothesis 119899+(119863(1198661015840)) = 1 119899
0(119863(1198661015840)) = 0 and 119899
minus(119863(1198661015840)) =
|119881(1198661015840)| minus 1 which implies that
det119863(1198661015840) = (minus1)|119881(1198661015840)|minus1
119886 (22)
where 119886 is a positive integerSince
cof (119866119894) = (minus1)
|119881(119866119894)|minus11003816100381610038161003816119881 (119866119894)
1003816100381610038161003816 1 ⩽ 119894 ⩽ 119904 (23)
by Lemma 3
cof119863(1198661015840) =119904minus1
prod
119895=1
cof119863(119866119895) = (minus1)
|119881(1198661015840)|minus1
119904minus1
prod
119895=1
10038161003816100381610038161003816119881 (119866119895)10038161003816100381610038161003816
(24)
Thendet119863(119882
119899minus 119864119896)
= det119863(1198661015840) cof119863(119866119894) + det119863(119866
119894) cof119863(1198661015840)
= (minus1)|119881(1198661015840)|minus1
119886 times (minus1)|119881(119866119894)|minus1
1003816100381610038161003816119881 (119866119894)1003816100381610038161003816
+ (minus1)|119881(119866119894)|minus1119887 times (minus1)
|119881(1198661015840)|minus1
119904minus1
prod
119895=1
10038161003816100381610038161003816119881 (119866119895)10038161003816100381610038161003816
= (minus1)119899minus1
(1198861003816100381610038161003816119881 (119866119894)
1003816100381610038161003816 + 119887
119904minus1
prod
119895=1
10038161003816100381610038161003816119881 (119866119895)10038161003816100381610038161003816)
(25)
where 119887 = 11989924 if 119899 is even and 119887 = (1198992 minus 1)4 if 119899 is oddIn this case similar to Corollary 6 we can easily get
119899+(119863(119882
119899minus 119864119896)) = 1 119899
0(119863(119882
119899minus 119864119896)) = 0 and 119899
minus(119863(119882
119899minus
119864119896)) = 119899 minus 1Up to now we have proved the result
Let119866119906119894sdotV119895119867 denote the graph formed by only identifying
the vertex 119906119894of 119866 with the vertex V
119895of119867 where 119906
119894and V119895are
arbitrary vertices of 119866 and119867 respectively
Lemma 10 (see [13]) Let 119866times119867 denote the Cartesian productof connected graphs 119866 and119867 where 119881(119866) = 119906
1 119906
119898 and
119881(119867) = V1 V
119899 Then we have
(i) 119899+(119866 times 119867) = 119899
+(119866119906119894sdot V119895119867)
(ii) 1198990(119866 times 119867) = (119898 minus 1)(119899 minus 1) + 119899
0(119866119906119894sdot V119895119867)
(iii) 119899minus(119866 times 119867) = 119899
minus(119866119906119894sdot V119895119867)
Theorem 11 Let 1199060and V
0be the hubs of 119882
119899and 119882
119898
respectively Suppose119864119901(0 ⩽ 119901 ⩽ 119899minus1) and119864
119902(0 ⩽ 119902 ⩽ 119898minus1)
are any subsets of 119864(119882119899minus 1199060) and 119864(119882
119898minus V0) with |119864
119901| = 119901
|119864119902| = 119902 respectively Then the distance matrix of the graph
(119882119899minus 119864119901) times (119882
119898minus 119864119902) has exactly one positive eigenvalue
Proof Since 1199060and V
0are the hubs of 119882
119899and 119882
119898 respec-
tively (119882119899minus119864119901)1199060sdotV0(119882119898minus119864119902)must be isomorphism to some
119882119899+119898minus1
minus119864119901+119902
where1199080is the hub of119882
119899+119898minus1and 119864
119901+119902is any
subset of 119864(119882119899+119898minus1
minus 1199080) with |119864
119901+119902| = 119901 + 119902 By Theorem 9
and Lemma 10 we get the result
Given an arbitrary integer 119898 for 1 ⩽ 119894 ⩽ 119898 let V1198940be
the hub of 119882119899119894and 119864
119901119894any subset of 119864(119882
119899119894minus V1198940) Suppose
119881(119882119899119894) = V
1198940 V1198941 V
119894(119899119894minus1)
Theorem 12 For an arbitrary integer 119898 the distance matrixof the graph 119866 = (119882
1198991minus 1198641199011)V1119896sdot V2119895(1198821198992minus 1198641199012)V2ℎsdot sdot sdot
V(119898minus1)119903
(119882119899119898minus1
minus 119864119901119898minus1
)V(119898minus1)119905
sdot V119898119904(119882119899119898minus 119864119901119898) has exactly one
positive eigenvalue
Proof We will prove the conclusion by induction on119898If119898 = 1 by Theorem 9 the conclusion is trueSuppose the conclusion is true for119898minus1 For convenience
let119867 = (1198821198991minus 1198641199011)V1119896sdot V2119895(1198821198992minus 1198641199012)V2ℎsdot sdot sdot V(119898minus2)119894
(119882119899119898minus2
minus
119864119901119898minus2
) Then 119866 = 119867V(119898minus2)119892
sdot V(119898minus1)119903
(119882119899119898minus1
minus 119864119901119898minus1
)V(119898minus1)119905
sdot
V119898119904(119882119899119898minus 119864119901119898) By Lemma 10 we have the following
119899+(119866) = 119899
+(119867 (119882
119899119898minus1minus 119864119901119898minus1
) V(119898minus1)119905
sdot V119898119904(119882119899119898minus 119864119901119898))
= 119899+(119867 (119882
119899119898minus1minus 119864119901119898minus1
) V(119898minus1)0
sdot V1198980(119882119899119898minus 119864119901119898))
(26)
Since (119882119899119898minus1
minus 119864119901119898minus1
)V(119898minus1)0
sdot V1198980(119882119899119898minus 119864119901119898) = 119882
119899119898minus1+119899119898minus1minus
119864119901119898minus1
minus119864119901119898 we get 119899
+(119866) = 119899
+(119867V(119898minus2)119892
sdotV(119898minus1)119903
(119882119899119898minus1+119899119898minus1
minus
119864119901119898minus1
minus119864119901119898)) By the induction hypothesis we get 119899
+(119866) = 1
This completes the proof
Remark 13 Let 1198661and 119866
2be any two graphs with the same
form as119866 inTheorem 12MakingCartesian product of graphs1198661and 119866
2 by Lemma 10 and Theorem 12 we get a series
of graphs whose distance matrices have exactly one positiveeigenvalue
Acknowledgments
The authors would like to thank the anonymous refereesfor their valuable comments and suggestions This workwas supported by NSFC (11126256) and NSF of ShandongProvince of China (ZR2012AQ022)
Journal of Applied Mathematics 5
References
[1] D H Rouvray ldquoThe role of the topological distance matrixin chemistryrdquo in Mathematical and Computational Conceptsin Chemistry N Trinajstid Ed pp 295ndash306 Ellis HarwoodChichester UK 1986
[2] W Brostow D M McEachern and S Petrez- Guiterrez ldquoPres-sure second virial coefficients of hydrocarbons fluorocarbonsand their mixtures Interactions of walksrdquo Journal of ChemicalPhysics vol 71 pp 2716ndash2722 1979
[3] B D McKay ldquoOn the spectral characterisation of treesrdquo ArsCombinatoria vol 3 pp 219ndash232 1977
[4] D W Bradley and R A Bradley ldquoString edits and macro-moleculesrdquo in Time Wraps D Sankoff and J B Kruskal EdsChapter 6 Addison-Wesley Reading Mass USA 1983
[5] J P Boyd and K N Wexler ldquoTrees with structurerdquo Journal ofMathematical Psychology vol 10 pp 115ndash147 1973
[6] R L Graham and L Lovasz ldquoDistance matrix polynomials oftreesrdquo inTheory and Applications of Graphs vol 642 of LectureNotes in Mathematics pp 186ndash190 1978
[7] M S Waterman T F Smith and H I Katcher ldquoAlgorithms forrestriction map comparisonsrdquo Nucleic Acids Research vol 12pp 237ndash242 1984
[8] R L Graham and H O Pollak ldquoOn the addressing problem forloop switchingrdquo The Bell System Technical Journal vol 50 pp2495ndash2519 1971
[9] R Bapat S J Kirkland and M Neumann ldquoOn distancematrices and Laplaciansrdquo Linear Algebra and Its Applicationsvol 401 pp 193ndash209 2005
[10] K Balasubramanian ldquoComputer generation of distance poly-nomials of graphsrdquo Journal of Computational Chemistry vol 11no 7 pp 829ndash836 1990
[11] D M Cvetkovic M Doob I Gutman and A Torgasev RecentResults in the Theory of Graph Spectra vol 36 North-HollandPublishing Amsterdam The Netherlands 1988
[12] H S Ramane D S Revankar I Gutman and H B WalikarldquoDistance spectra anddistance energies of iterated line graphs ofregular graphsrdquo Institut Mathematique vol 85 pp 39ndash46 2009
[13] X Zhang and C Godsil ldquoThe inertia of distance matrices ofsome graphsrdquo Discrete Mathematics vol 313 no 16 pp 1655ndash1664 2013
[14] D M Cvetkovic M Doob and H Sachs Spectra of Graphsvol 87 Academic Press New York NY USA 1980 Theory andapplication
[15] R L Graham A J Hoffman and H Hosoya ldquoOn the distancematrix of a directed graphrdquo Journal of Graph Theory vol 1 no1 pp 85ndash88 1977
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Journal of Applied Mathematics 5
References
[1] D H Rouvray ldquoThe role of the topological distance matrixin chemistryrdquo in Mathematical and Computational Conceptsin Chemistry N Trinajstid Ed pp 295ndash306 Ellis HarwoodChichester UK 1986
[2] W Brostow D M McEachern and S Petrez- Guiterrez ldquoPres-sure second virial coefficients of hydrocarbons fluorocarbonsand their mixtures Interactions of walksrdquo Journal of ChemicalPhysics vol 71 pp 2716ndash2722 1979
[3] B D McKay ldquoOn the spectral characterisation of treesrdquo ArsCombinatoria vol 3 pp 219ndash232 1977
[4] D W Bradley and R A Bradley ldquoString edits and macro-moleculesrdquo in Time Wraps D Sankoff and J B Kruskal EdsChapter 6 Addison-Wesley Reading Mass USA 1983
[5] J P Boyd and K N Wexler ldquoTrees with structurerdquo Journal ofMathematical Psychology vol 10 pp 115ndash147 1973
[6] R L Graham and L Lovasz ldquoDistance matrix polynomials oftreesrdquo inTheory and Applications of Graphs vol 642 of LectureNotes in Mathematics pp 186ndash190 1978
[7] M S Waterman T F Smith and H I Katcher ldquoAlgorithms forrestriction map comparisonsrdquo Nucleic Acids Research vol 12pp 237ndash242 1984
[8] R L Graham and H O Pollak ldquoOn the addressing problem forloop switchingrdquo The Bell System Technical Journal vol 50 pp2495ndash2519 1971
[9] R Bapat S J Kirkland and M Neumann ldquoOn distancematrices and Laplaciansrdquo Linear Algebra and Its Applicationsvol 401 pp 193ndash209 2005
[10] K Balasubramanian ldquoComputer generation of distance poly-nomials of graphsrdquo Journal of Computational Chemistry vol 11no 7 pp 829ndash836 1990
[11] D M Cvetkovic M Doob I Gutman and A Torgasev RecentResults in the Theory of Graph Spectra vol 36 North-HollandPublishing Amsterdam The Netherlands 1988
[12] H S Ramane D S Revankar I Gutman and H B WalikarldquoDistance spectra anddistance energies of iterated line graphs ofregular graphsrdquo Institut Mathematique vol 85 pp 39ndash46 2009
[13] X Zhang and C Godsil ldquoThe inertia of distance matrices ofsome graphsrdquo Discrete Mathematics vol 313 no 16 pp 1655ndash1664 2013
[14] D M Cvetkovic M Doob and H Sachs Spectra of Graphsvol 87 Academic Press New York NY USA 1980 Theory andapplication
[15] R L Graham A J Hoffman and H Hosoya ldquoOn the distancematrix of a directed graphrdquo Journal of Graph Theory vol 1 no1 pp 85ndash88 1977
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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![Which graphs are determined by their spectrum?weng/references/distance-regular... · Fig. 1. Two graphs with cospectral adjacency matrices. Günthard and Primas [42] raised the question](https://static.fdocuments.us/doc/165x107/5f701c8a56816a73b951d13d/which-graphs-are-determined-by-their-spectrum-wengreferencesdistance-regular.jpg)
