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; zhangxling04@yahoo.cn
Received 25 November 2012; Revised 29 May 2013; Accepted 30 May 2013
Academic Editor: Maurizio Porfiri
Copyright © 2013 X. Zhang and C. Song. This 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 𝐷 denote the distance matrix of a connected graph 𝐺. The inertia of 𝐷 is the triple of integers (𝑛+ (𝐷), 𝑛0 (𝐷), 𝑛− (𝐷)), where
𝑛+ (𝐷), 𝑛0 (𝐷), and 𝑛− (𝐷) denote the number of positive, 0, and negative eigenvalues of 𝐷, 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 𝐺 = (𝑉, 𝐸) consists of 𝑉, a nonempty set of
vertices, and 𝐸, a set of unordered pairs of distinct elements
of 𝑉 called edges. All graphs considered here are simple and
connected. Let 𝐺 be a simple connected graph with vertex set
𝑉(𝐺) and edge set 𝐸(𝐺). The distance between two vertices 𝑢,
V ∈ 𝑉(𝐺) is denoted by 𝑑𝑢V and is defined as the length of the
shortest path between 𝑢 and V in 𝐺. The distance matrix of 𝐺 is
denoted by 𝐷(𝐺) and is defined by 𝐷(𝐺) = (𝑑𝑢V )𝑢,V∈𝑉(𝐺) . Since
𝐷(𝐺) is a symmetric matrix, its inertia is the triple of integers
(𝑛+ (𝐷(𝐺)), 𝑛0 (𝐷(𝐺)), 𝑛− (𝐷(𝐺))), where 𝑛+ (𝐷(𝐺)), 𝑛0 (𝐷(𝐺)),
and 𝑛− (𝐷(𝐺)) denote the number of positive, 0, and negative
eigenvalues of 𝐷(𝐺), respectively.
The 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 computation 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]. The
distance matrix of a graph contains more structural information 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 complicated 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 𝐷-eigenvalue, such
as trees [8], connected unicyclic graphs [9], the polyacenes,
honeycomb and square lattices [10], complete bipartite graphs
[11], 𝐾𝑛 , and iterated line graphs of some regular graphs [12],
and cacti [13]. This inspires us to find more graphs whose
distance matrices have exactly one positive eigenvalue.
The wheel graph of 𝑛 vertices 𝑊𝑛 is a graph that contains a
cycle of length 𝑛−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.
2
Journal of Applied Mathematics
Lemma 1 (see [14]). Let 𝐴 be a Hermitian matrix with
eigenvalues 𝜆 1 ⩾ ⋅ ⋅ ⋅ ⩾ 𝜆 𝑛 and 𝐵 one of its principal
submatrices. Let 𝐵 have eigenvalues 𝜇1 ⩾ ⋅ ⋅ ⋅ ⩾ 𝜇𝑚 . Then the
inequalities 𝜆 𝑛−𝑚+𝑖 ⩽ 𝜇𝑖 ⩽ 𝜆 𝑖 (𝑖 = 1, . . . , 𝑚) hold.
Comparing 𝐶𝑛 to 𝐷𝑛 , we get the following:
For a square matrix, let cof (𝐴) denote the sum of
̃ by subtracting the first
cofactors of 𝐴. Form the matrix 𝐴
row from all other rows then the first column from all other
̃11 denote the principle submatrix obtained
columns and let 𝐴
̃
from 𝐴 by deleting the first row and first column.
Expanding the determinant 𝐷𝑛 according to the last column
and then the last line, we get the following incursion:
̃11 .
Lemma 2 (see [15]). 𝑐𝑜𝑓(𝐴) = det 𝐴
A cut vertex is a vertex the removal of which would
disconnect the remaining graph; a block of a graph is defined
to be a maximal subgraph having no cut vertices.
that is,
Lemma 3 (see [15]). If 𝐺 is a strongly connected directed graph
with blocks 𝐺1 , 𝐺2 , . . . , 𝐺𝑟 , then
𝐶𝑛 = 𝐷𝑛 +
(1)
det 𝐷 (𝐺) = ∑ det 𝐷 (𝐺𝑖 ) ∏𝑐𝑜𝑓 𝐷 (𝐺𝑖 ) .
𝑖=1
Then
1
−2
−1
0
..
.
0
0
1
−1
−2
−1
..
.
0
0
1
0
−1
−2
..
.
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
1
0
0
0
.
d ..
0 ⋅ ⋅ ⋅ −2
0 ⋅ ⋅ ⋅ −1
1
0]
]
0]
]
0]
] .
.. ]
. ]
]
−1]
−2]𝑛×𝑛
𝑛
𝑖𝑓 𝑛 𝑖𝑠 𝑒V𝑒𝑛,
− ,
{
{
{ 2
det 𝐶 = {
{
{ 𝑛 + 1 , 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑.
{ 2
(6)
𝐷𝑛 + 𝐷𝑛−1 = − (𝐷𝑛−1 + 𝐷𝑛−2 ) + 1.
(7)
if 𝑛 is even,
(8)
if 𝑛 is odd.
So, we have the following:
𝑗 ≠ 𝑖
Lemma 4. Let
1
[−1
[
[0
[
[
𝐶=[0
[ ..
[ .
[
[0
[0
𝐷𝑛 = −2𝐷𝑛−1 − 𝐷𝑛−2 + 1;
0,
{
{
𝐷𝑛 = {
{1
,
{2
𝑟
𝑟
(5)
Since 𝐷1 = 1/2, 𝐷2 = 0, and 𝐷3 = 1/2, from the above
incursion, we get the following:
𝑐𝑜𝑓 𝐷 (𝐺) = ∏𝑐𝑜𝑓 𝐷 (𝐺𝑖 ) ,
𝑖=1
𝑛
× (−1)𝑛−1 .
2
𝑛
− ,
{
{
{ 2
𝐶𝑛 = {
{
{𝑛 + 1,
{ 2
(2)
if 𝑛 is even,
(9)
if 𝑛 is odd.
This completes the proof.
3. Main Results
(3)
Proof. Let
In the following, we always assume that 𝑉(𝑊𝑛 ) = {V0 , V1 , . . . ,
V𝑛−1 }, where V0 is the hub of 𝑊𝑛 .
Theorem 5. Let 𝑒 = V𝑖 V𝑖+1 mod
(𝑛−1)
(1 ⩽ 𝑖 ⩽ 𝑛 − 1). Then
2
󵄨󵄨 1
󵄨󵄨
󵄨󵄨−1
󵄨󵄨
󵄨󵄨 0
󵄨󵄨
󵄨󵄨
𝐶𝑛 = 󵄨󵄨󵄨 0
󵄨󵄨 ..
󵄨󵄨 .
󵄨󵄨
󵄨󵄨 0
󵄨󵄨
󵄨󵄨 0
󵄨
1
−2
−1
0
..
.
0
0
1
−1
−2
−1
..
.
0
0
1
0
−1
−2
..
.
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
1
0
0
0
.
d ..
0 ⋅ ⋅ ⋅ −2
0 ⋅ ⋅ ⋅ −1
󵄨󵄨 1
󵄨󵄨 −1 0 0 ⋅ ⋅ ⋅
󵄨󵄨 2
󵄨󵄨
󵄨󵄨
󵄨󵄨 1 −2 −1 0 ⋅ ⋅ ⋅
󵄨󵄨
󵄨󵄨 1 −1 −2 −1 ⋅ ⋅ ⋅
󵄨
𝐷𝑛 = 󵄨󵄨󵄨 1 0 −1 −2 ⋅ ⋅ ⋅
󵄨󵄨
󵄨󵄨 . . . .
󵄨󵄨 . . . .
󵄨󵄨 . . . . d
󵄨󵄨
󵄨󵄨󵄨 1 0 0 0 ⋅ ⋅ ⋅
󵄨󵄨 1 0 0 0 ⋅ ⋅ ⋅
󵄨
0
0
0
0
..
.
−2
−1
𝑛
{
{
𝑖𝑓 𝑛 𝑖𝑠 𝑒V𝑒𝑛,
{
{− 4 ,
det 𝐷 (𝑊𝑛 − 𝑒) = {
{ 𝑛2 − 1
{
{
, 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑,
{ 4
1 󵄨󵄨󵄨
󵄨
0 󵄨󵄨󵄨󵄨
0 󵄨󵄨󵄨󵄨
0 󵄨󵄨󵄨󵄨 ,
.. 󵄨󵄨󵄨󵄨
. 󵄨󵄨
󵄨
−1󵄨󵄨󵄨󵄨
−2󵄨󵄨󵄨𝑛×𝑛
󵄨󵄨
0 󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨
0 󵄨󵄨󵄨
󵄨
0 󵄨󵄨󵄨
󵄨
0 󵄨󵄨󵄨 .
󵄨
.. 󵄨󵄨󵄨
. 󵄨󵄨󵄨
󵄨
−1󵄨󵄨󵄨
󵄨
−2󵄨󵄨󵄨𝑛×𝑛
(10)
where 𝑛 ⩾ 3.
(4)
Proof. Without loss of generality, we may assume that 𝑒 =
V1 V𝑛−1 . Let
󵄨󵄨0
󵄨󵄨
󵄨󵄨1
󵄨󵄨
󵄨󵄨1
󵄨󵄨
󵄨󵄨
𝐴 𝑛 = det 𝐷 (𝑊𝑛 − 𝑒) = 󵄨󵄨󵄨1
󵄨󵄨 ..
󵄨󵄨 .
󵄨󵄨
󵄨󵄨1
󵄨󵄨
󵄨󵄨1
󵄨
1
0
1
2
..
.
1
1
0
1
..
.
1
2
1
0
..
.
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
1
2
2
2
.
d ..
2 2 2 ⋅⋅⋅ 0
2 2 2 ⋅⋅⋅ 1
1󵄨󵄨󵄨
󵄨
2󵄨󵄨󵄨󵄨
2󵄨󵄨󵄨󵄨
2󵄨󵄨󵄨󵄨 .
.. 󵄨󵄨󵄨󵄨
. 󵄨󵄨
󵄨
1󵄨󵄨󵄨󵄨
0󵄨󵄨󵄨𝑛×𝑛
(11)
Journal of Applied Mathematics
3
Then
Proof. Consider the following
󵄨󵄨
1
󵄨󵄨 0
1 1−
1 ⋅⋅⋅
󵄨󵄨
2
󵄨󵄨
󵄨󵄨
󵄨󵄨 1 −2
0
0 ⋅⋅⋅
󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨󵄨
1
1
󵄨1 −
0 −2 +
−1 ⋅ ⋅ ⋅
𝐴 𝑛 = 󵄨󵄨󵄨󵄨
2
2
󵄨󵄨󵄨
󵄨󵄨 1
0
−1 −2 ⋅ ⋅ ⋅
󵄨󵄨
..
..
..
󵄨󵄨 ..
󵄨󵄨 .
.
.
. d
󵄨󵄨
󵄨󵄨 1
0
0
0 ⋅⋅⋅
󵄨󵄨
󵄨󵄨 1
0
0
0 ⋅⋅⋅
󵄨
1
0
0
0
..
.
−2
−1
󵄨󵄨
1 󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨
0 󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨󵄨
0 󵄨󵄨󵄨󵄨
󵄨󵄨 .
󵄨󵄨
0 󵄨󵄨󵄨󵄨
.. 󵄨󵄨󵄨
. 󵄨󵄨󵄨
󵄨
−1󵄨󵄨󵄨
󵄨
−2󵄨󵄨󵄨𝑛×𝑛
(12)
Expanding the determinant 𝐴 𝑛 according to the second line,
we get the following incursion:
𝐴 𝑛 = (−1)
𝑛−1 𝑛
2
− 2𝐴 𝑛−1 − 𝐶𝑛−2 − 𝐷𝑛−2 − 𝐴 𝑛−2 ,
(13)
where 𝐶𝑛 and 𝐷𝑛 are defined as in Lemma 4.
By Lemma 4, we get the following:
𝐴𝑛 = {
−1 − 2𝐴 𝑛−1 − 𝐴 𝑛−2 , if 𝑛 is even,
−2𝐴 𝑛−1 − 𝐴 𝑛−2 ,
if 𝑛 is odd.
(14)
Since 𝐴 3 = 2, 𝐴 4 = −4, 𝐴 5 = 6, and 𝐴 6 = −9, according to
the above incursion, we get the following:
2
𝑛
{
{
if 𝑛 is even,
{
{− 4 ,
𝐴𝑛 = {
2
{
{
{ 𝑛 − 1 , if 𝑛 is odd,
{ 4
(15)
where 𝑛 ⩾ 3. This completes the proof.
Corollary 6. Let 𝑒 = V𝑖 V𝑖+1 mod
𝑛+ (𝐷 (𝑊𝑛 − 𝑒)) = 1,
(𝑛−1)
󵄨󵄨0
󵄨󵄨
󵄨󵄨1
󵄨󵄨
󵄨󵄨1
󵄨󵄨
󵄨󵄨
= 󵄨󵄨󵄨1
󵄨󵄨 ..
󵄨󵄨 .
󵄨󵄨
󵄨󵄨1
󵄨󵄨
󵄨󵄨1
󵄨
1
0
1
2
..
.
1
2
2
2
.
d ..
2 2 2 ⋅⋅⋅ 0
1 2 2 ⋅⋅⋅ 1
1
−2
−1
0
..
.
1
1
0
1
..
.
1
2
1
0
..
.
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
1
−1
−2
−1
..
.
1
0
−1
−2
..
.
0 0
−1 0
1󵄨󵄨󵄨
󵄨
1󵄨󵄨󵄨󵄨
2󵄨󵄨󵄨󵄨
2󵄨󵄨󵄨󵄨
.. 󵄨󵄨󵄨󵄨
. 󵄨󵄨
󵄨
1󵄨󵄨󵄨󵄨
0󵄨󵄨󵄨𝑛×𝑛
1 1 󵄨󵄨󵄨
󵄨
0 −1󵄨󵄨󵄨󵄨
0 0 󵄨󵄨󵄨󵄨
0 0 󵄨󵄨󵄨󵄨
. . 󵄨󵄨󵄨
d .. .. 󵄨󵄨󵄨
󵄨
0 ⋅ ⋅ ⋅ −2 −1󵄨󵄨󵄨󵄨
0 ⋅ ⋅ ⋅ −1 −2󵄨󵄨󵄨𝑛×𝑛
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
󵄨󵄨 0
𝑛−1
󵄨󵄨
󵄨󵄨𝑛 − 1 −4 (𝑛 − 1)
󵄨󵄨
󵄨󵄨 1
−4
󵄨󵄨
󵄨󵄨 1
−4
󵄨
= 󵄨󵄨
󵄨󵄨 ..
..
󵄨󵄨 .
.
󵄨󵄨
󵄨󵄨 1
−4
󵄨󵄨
󵄨󵄨 1
−4
󵄨
󵄨󵄨 0
𝑛−1
󵄨󵄨
󵄨󵄨𝑛 − 1 4 (𝑛 − 1)
󵄨󵄨
󵄨󵄨 1
0
󵄨󵄨
󵄨󵄨 1
0
󵄨
= 󵄨󵄨
󵄨󵄨 ..
..
󵄨󵄨 .
.
󵄨󵄨
󵄨󵄨 1
0
󵄨󵄨
󵄨󵄨 1
0
󵄨
1
−4
−2
−1
..
.
0
0
1
0
−2
−1
..
.
0
0
1
−4
−1
−2
..
.
𝑛0 (𝐷 (𝑊𝑛 − 𝑒)) = 0,
1
−4
0
0
.
d ..
0 ⋅ ⋅ ⋅ −2
0 ⋅ ⋅ ⋅ −1
1
0
−1
−2
..
.
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
1
0
0
0
.
d ..
0 ⋅ ⋅ ⋅ −2
0 ⋅ ⋅ ⋅ −1
1 󵄨󵄨󵄨
󵄨
0 󵄨󵄨󵄨󵄨
0 󵄨󵄨󵄨󵄨
0 󵄨󵄨󵄨󵄨 .
.. 󵄨󵄨󵄨󵄨
. 󵄨󵄨
󵄨
−1󵄨󵄨󵄨󵄨
−2󵄨󵄨󵄨𝑛×𝑛
(18)
Expanding the above determinant according to the second
line, we get the following:
(16)
Proof. We will prove the result by induction on 𝑛.
If 𝑛 = 3, 𝑊3 − 𝑒 ≅ 𝑃3 is obviously true.
Suppose that the result is true for 𝑛−1; that is, 𝑛+ (𝐷(𝑊𝑛−1 −
𝑒)) = 1, 𝑛0 (𝐷(𝑊𝑛−1 − 𝑒)) = 0, 𝑛− (𝐷(𝑊𝑛−1 − 𝑒)) = 𝑛 − 2.
Since 𝐷(𝑊𝑛−1 − 𝑒) is a principle submatrix of 𝐷(𝑊𝑛 −
𝑒), by Lemma 1, the eigenvalues of 𝐷(𝑊𝑛−1 − 𝑒) interlace
the eigenvalues of 𝐷(𝑊𝑛 − 𝑒). By Theorem 5, det 𝐷(𝑊𝑛−1 −
𝑒) det 𝐷(𝑊𝑛 − 𝑒) < 0. So, 𝐷(𝑊𝑛 − 𝑒) has one negative
eigenvalue more than 𝐷(𝑊𝑛−1 −𝑒). According to the induction
hypothesis, we get 𝑛+ (𝐷(𝑊𝑛 − 𝑒)) = 1, 𝑛0 (𝐷(𝑊𝑛 − 𝑒)) = 0, and
𝑛− (𝐷(𝑊𝑛 − 𝑒)) = 𝑛 − 1. This completes the proof.
Theorem 7. One has
1 − 𝑛, 𝑖𝑓 𝑛 𝑖𝑠 𝑒V𝑒𝑛,
det 𝐷 (𝑊𝑛 ) = {
0,
𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑,
1 󵄨󵄨󵄨
󵄨
−4󵄨󵄨󵄨󵄨
0 󵄨󵄨󵄨󵄨
0 󵄨󵄨󵄨󵄨
.. 󵄨󵄨󵄨󵄨
. 󵄨󵄨
󵄨
−1󵄨󵄨󵄨󵄨
−2󵄨󵄨󵄨𝑛×𝑛
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
(1 ⩽ 𝑖 ⩽ 𝑛 − 1). Then
𝑛− (𝐷 (𝑊𝑛 − 𝑒)) = 𝑛 − 1.
where 𝑛 ⩾ 3.
󵄨󵄨0
󵄨󵄨
󵄨󵄨1
󵄨󵄨
󵄨󵄨1
󵄨󵄨
󵄨󵄨
det 𝐷 (𝑊𝑛 ) = 󵄨󵄨󵄨1
󵄨󵄨 ..
󵄨󵄨 .
󵄨󵄨
󵄨󵄨1
󵄨󵄨
󵄨󵄨1
󵄨
(17)
det 𝐷 (𝑊𝑛 ) = (−1)𝑛−1 (𝑛 − 1)3 + 4 (𝑛 − 1) 𝐴 𝑛−1 ,
(19)
where 𝐴 𝑛 is defined as in Theorem 5.
By Theorem 5, when 𝑛 ⩾ 3, we get the following:
1 − 𝑛, if 𝑛 is even,
det 𝐷 (𝑊𝑛 ) = {
0,
if 𝑛 is odd.
(20)
This completes the proof.
Similar to Corollary 6, we can get the following corollary.
Corollary 8. (i) If 𝑛 is even, 𝑛+ (𝐷(𝑊𝑛 )) = 1, 𝑛0 (𝐷(𝑊𝑛 )) = 0,
𝑛− (𝐷(𝑊𝑛 )) = 𝑛 − 1.
(ii) If 𝑛 is odd, 𝑛+ (𝐷(𝑊𝑛 )) = 1, 𝑛0 (𝐷(𝑊𝑛 )) = 1,
𝑛− (𝐷(𝑊𝑛 )) = 𝑛 − 2.
Denote by 𝑊𝑛 − V0 the graph obtained from 𝑊𝑛 by deleting
the vertex V0 and all the edges adjacent to V0 ; that is, 𝑊𝑛 − V0 ≅
𝐶𝑛−1 . Let 𝐸𝑘 (1 ⩽ 𝑘 ⩽ 𝑛) be any subset of 𝐸(𝑊𝑛 − V0 ) with
|𝐸𝑘 | = 𝑘. In the following, we always denote by 𝑊𝑛 − 𝐸𝑘 the
graph obtained from 𝑊𝑛 by deleting all the edges in 𝐸𝑘 .
4
Journal of Applied Mathematics
Theorem 9. One has 𝑛+ (𝐷(𝑊𝑛 − 𝐸𝑘 )) = 1, 𝑛0 (𝐷(𝑊𝑛 − 𝐸𝑘 )) =
0, 𝑛− (𝐷(𝑊𝑛 − 𝐸𝑘 )) = 𝑛 − 1.
Proof. Denote the components of 𝑊𝑛 − 𝐸𝑘 − V0 by 𝐶1 , . . . , 𝐶𝑠 .
Let 𝐺𝑖 denote the graph that contains 𝐶𝑖 plus the vertex V0
such that V0 is connected to every other vertex, 1 ⩽ 𝑖 ⩽ 𝑠.
Then each 𝐺𝑖 (1 ⩽ 𝑖 ⩽ 𝑠) is isomorphism to 𝑊|𝑉(𝐺𝑖 )| − 𝑒𝑖 or
𝐾2 , where 𝑒𝑖 is an edge of 𝑊|𝑉(𝐺𝑖 )| − V0 . By Lemma 2 and some
direct calculations, we get the following:
̃
󵄨
󵄨
cof (𝐺𝑖 ) = det 𝐷(𝑊|𝑉(𝐺𝑖 )| − 𝑒𝑖 )11 = (−1)|𝑉(𝐺𝑖 )|−1 󵄨󵄨󵄨𝑉 (𝐺𝑖 )󵄨󵄨󵄨 .
(21)
It is easy to check that cof (𝐺𝑖 ) = (−1)|𝑉(𝐺𝑖 )|−1 |𝑉(𝐺𝑖 )| is also
true when 𝐺𝑖 is isomorphism to 𝐾2 .
In the following, we will prove the theorem by introduction on 𝑠.
For 𝑠 = 1, 𝐺 ≅ 𝑊𝑛 − 𝑒, where 𝑒 is an edge of 𝑊𝑛 − V0 , by
Corollary 6, we get the result.
Suppose the result is true for 𝑠 − 1.
For 𝑠, let 𝐺󸀠 = 𝐺1 ∪ 𝐺2 ∪ ⋅ ⋅ ⋅ ∪ 𝐺𝑠−1 . Then by the induction
hypothesis, 𝑛+ (𝐷(𝐺󸀠 )) = 1, 𝑛0 (𝐷(𝐺󸀠 )) = 0, and 𝑛− (𝐷(𝐺󸀠 )) =
|𝑉(𝐺󸀠 )| − 1, which implies that
󸀠
det 𝐷 (𝐺󸀠 ) = (−1)|𝑉(𝐺 )|−1 𝑎,
(22)
where 𝑎 is a positive integer.
Since
󵄨
󵄨
cof (𝐺𝑖 ) = (−1)|𝑉(𝐺𝑖 )|−1 󵄨󵄨󵄨𝑉 (𝐺𝑖 )󵄨󵄨󵄨 ,
1 ⩽ 𝑖 ⩽ 𝑠,
(23)
by Lemma 3,
𝑠−1
𝑠−1
󸀠
󵄨
󵄨
cof 𝐷 (𝐺󸀠 ) = ∏cof𝐷 (𝐺𝑗 ) = (−1)|𝑉(𝐺 )|−1 ∏ 󵄨󵄨󵄨󵄨𝑉 (𝐺𝑗 )󵄨󵄨󵄨󵄨 .
𝑗=1
𝑗=1
(24)
Then
Lemma 10 (see [13]). Let 𝐺 × 𝐻 denote the Cartesian product
of connected graphs 𝐺 and 𝐻, where 𝑉(𝐺) = {𝑢1 , . . . , 𝑢𝑚 } and
𝑉(𝐻) = {V1 , . . . , V𝑛 }. Then we have
(i) 𝑛+ (𝐺 × 𝐻) = 𝑛+ (𝐺𝑢𝑖 ⋅ V𝑗 𝐻);
(ii) 𝑛0 (𝐺 × 𝐻) = (𝑚 − 1)(𝑛 − 1) + 𝑛0 (𝐺𝑢𝑖 ⋅ V𝑗 𝐻);
(iii) 𝑛− (𝐺 × 𝐻) = 𝑛− (𝐺𝑢𝑖 ⋅ V𝑗 𝐻).
Theorem 11. Let 𝑢0 and V0 be the hubs of 𝑊𝑛 and 𝑊𝑚 ,
respectively. Suppose 𝐸𝑝 (0 ⩽ 𝑝 ⩽ 𝑛−1) and 𝐸𝑞 (0 ⩽ 𝑞 ⩽ 𝑚−1)
are any subsets of 𝐸(𝑊𝑛 − 𝑢0 ) and 𝐸(𝑊𝑚 − V0 ) with |𝐸𝑝 | = 𝑝,
|𝐸𝑞 | = 𝑞, respectively. Then, the distance matrix of the graph
(𝑊𝑛 − 𝐸𝑝 ) × (𝑊𝑚 − 𝐸𝑞 ) has exactly one positive eigenvalue.
Proof. Since 𝑢0 and V0 are the hubs of 𝑊𝑛 and 𝑊𝑚 , respectively, (𝑊𝑛 −𝐸𝑝 )𝑢0 ⋅V0 (𝑊𝑚 −𝐸𝑞 ) must be isomorphism to some
𝑊𝑛+𝑚−1 − 𝐸𝑝+𝑞 , where 𝑤0 is the hub of 𝑊𝑛+𝑚−1 and 𝐸𝑝+𝑞 is any
subset of 𝐸(𝑊𝑛+𝑚−1 − 𝑤0 ) with |𝐸𝑝+𝑞 | = 𝑝 + 𝑞. By Theorem 9
and Lemma 10, we get the result.
Given an arbitrary integer 𝑚, for 1 ⩽ 𝑖 ⩽ 𝑚, let V𝑖0 be
the hub of 𝑊𝑛𝑖 and 𝐸𝑝𝑖 any subset of 𝐸(𝑊𝑛𝑖 − V𝑖0 ). Suppose
𝑉(𝑊𝑛𝑖 ) = {V𝑖0 , V𝑖1 , . . . , V𝑖(𝑛𝑖 −1) }.
Theorem 12. For an arbitrary integer 𝑚, the distance matrix
of the graph 𝐺 = (𝑊𝑛1 − 𝐸𝑝1 )V1𝑘 ⋅ V2𝑗 (𝑊𝑛2 − 𝐸𝑝2 )V2ℎ ⋅ ⋅ ⋅
V(𝑚−1)𝑟 (𝑊𝑛𝑚−1 − 𝐸𝑝𝑚−1 )V(𝑚−1)𝑡 ⋅ V𝑚𝑠 (𝑊𝑛𝑚 − 𝐸𝑝𝑚 ) has exactly one
positive eigenvalue.
Proof. We will prove the conclusion by induction on 𝑚.
If 𝑚 = 1, by Theorem 9, the conclusion is true.
Suppose the conclusion is true for 𝑚−1. For convenience,
let 𝐻 = (𝑊𝑛1 − 𝐸𝑝1 )V1𝑘 ⋅ V2𝑗 (𝑊𝑛2 − 𝐸𝑝2 )V2ℎ ⋅ ⋅ ⋅ V(𝑚−2)𝑖 (𝑊𝑛𝑚−2 −
𝐸𝑝𝑚−2 ). Then 𝐺 = 𝐻V(𝑚−2)𝑔 ⋅ V(𝑚−1)𝑟 (𝑊𝑛𝑚−1 − 𝐸𝑝𝑚−1 )V(𝑚−1)𝑡 ⋅
V𝑚𝑠 (𝑊𝑛𝑚 − 𝐸𝑝𝑚 ). By Lemma 10, we have the following:
𝑛+ (𝐺) = 𝑛+ (𝐻 (𝑊𝑛𝑚−1 − 𝐸𝑝𝑚−1 ) V(𝑚−1)𝑡 ⋅ V𝑚𝑠 (𝑊𝑛𝑚 − 𝐸𝑝𝑚 ))
det 𝐷 (𝑊𝑛 − 𝐸𝑘 )
= 𝑛+ (𝐻 (𝑊𝑛𝑚−1 − 𝐸𝑝𝑚−1 ) V(𝑚−1)0 ⋅ V𝑚0 (𝑊𝑛𝑚 − 𝐸𝑝𝑚 )) .
(26)
= det 𝐷 (𝐺󸀠 ) cof 𝐷 (𝐺𝑖 ) + det 𝐷 (𝐺𝑖 ) cof 𝐷 (𝐺󸀠 )
󵄨
󵄨
= (−1)|𝑉(𝐺 )|−1 𝑎 × (−1)|𝑉(𝐺𝑖 )|−1 󵄨󵄨󵄨𝑉 (𝐺𝑖 )󵄨󵄨󵄨
󸀠
𝑠−1
󸀠
󵄨
󵄨
+ (−1)|𝑉(𝐺𝑖 )|−1 𝑏 × (−1)|𝑉(𝐺 )|−1 ∏ 󵄨󵄨󵄨󵄨𝑉 (𝐺𝑗 )󵄨󵄨󵄨󵄨
(25)
𝑗=1
𝑠−1
󵄨
󵄨
󵄨
󵄨
= (−1)𝑛−1 (𝑎 󵄨󵄨󵄨𝑉 (𝐺𝑖 )󵄨󵄨󵄨 + 𝑏∏ 󵄨󵄨󵄨󵄨𝑉 (𝐺𝑗 )󵄨󵄨󵄨󵄨) ,
𝑗=1
where 𝑏 = 𝑛2 /4, if 𝑛 is even and 𝑏 = (𝑛2 − 1)/4, if 𝑛 is odd.
In this case, similar to Corollary 6, we can easily get
𝑛+ (𝐷(𝑊𝑛 − 𝐸𝑘 )) = 1, 𝑛0 (𝐷(𝑊𝑛 − 𝐸𝑘 )) = 0, and 𝑛− (𝐷(𝑊𝑛 −
𝐸𝑘 )) = 𝑛 − 1.
Up to now, we have proved the result.
Let 𝐺𝑢𝑖 ⋅V𝑗 𝐻 denote the graph formed by only identifying
the vertex 𝑢𝑖 of 𝐺 with the vertex V𝑗 of 𝐻, where 𝑢𝑖 and V𝑗 are
arbitrary vertices of 𝐺 and 𝐻, respectively.
Since (𝑊𝑛𝑚−1 − 𝐸𝑝𝑚−1 )V(𝑚−1)0 ⋅ V𝑚0 (𝑊𝑛𝑚 − 𝐸𝑝𝑚 ) = 𝑊𝑛𝑚−1 +𝑛𝑚 −1 −
𝐸𝑝𝑚−1 −𝐸𝑝𝑚 , we get 𝑛+ (𝐺) = 𝑛+ (𝐻V(𝑚−2)𝑔 ⋅V(𝑚−1)𝑟 (𝑊𝑛𝑚−1 +𝑛𝑚 −1 −
𝐸𝑝𝑚−1 − 𝐸𝑝𝑚 )). By the induction hypothesis, we get 𝑛+ (𝐺) = 1.
This completes the proof.
Remark 13. Let 𝐺1 and 𝐺2 be any two graphs with the same
form as 𝐺 in Theorem 12. Making Cartesian product of graphs
𝐺1 and 𝐺2 , by Lemma 10 and Theorem 12, we get a series
of graphs whose distance matrices have exactly one positive
eigenvalue.
Acknowledgments
The authors would like to thank the anonymous referees
for their valuable comments and suggestions. This work
was supported by NSFC (11126256) and NSF of Shandong
Province of China (ZR2012AQ022).
Journal of Applied Mathematics
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