The Distance Matrices of Some Graphs Related to Wheel Graphs

Abstract

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 𝑑_𝑢𝑣 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)=(𝑑_𝑢𝑣)_{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], 𝐾_𝑛, 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 𝑊_𝑛 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.

Lemma 1 (see [14]). Let 𝐴 be a Hermitian matrix with eigenvalues 𝜆₁⩾⋯⩾𝜆_𝑛 and 𝐵 one of its principal submatrices. Let 𝐵 have eigenvalues 𝜇₁⩾⋯⩾𝜇_𝑚. Then the inequalities 𝜆_{𝑛−𝑚+𝑖}⩽𝜇_𝑖⩽𝜆_𝑖(𝑖=1,…,𝑚) hold.

For a square matrix, let cof(𝐴) denote the sum of cofactors of 𝐴. Form the matrix 𝐴 by subtracting the first row from all other rows then the first column from all other columns and let 𝐴₁₁ denote the principle submatrix obtained from 𝐴 by deleting the first row and first column.

Lemma 2 (see [15]). 𝑐𝑜𝑓(𝐴)=det𝐴₁₁.
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.

Lemma 3 (see [15]). If𝐺 is a strongly connected directed graph with blocks 𝐺₁,𝐺₂,…,𝐺_𝑟, then 𝑐𝑜𝑓𝐷(𝐺)=^𝑟_𝑖=1𝑐𝑜𝑓𝐷𝐺_𝑖,det𝐷(𝐺)=^𝑟_𝑖=1det𝐷𝐺_{𝑖𝑗≠𝑖}𝑐𝑜𝑓𝐷𝐺_𝑖. (1)

Lemma 4. Let 𝐶=1111⋯11−1−2−10⋯000−1−2−1⋯0000−1−2⋯00⋮⋮⋮⋮⋱⋮⋮0000⋯−2−10000⋯−1−2_𝑛×𝑛. (2) Then −𝑛det𝐶=2,𝑖𝑓𝑛𝑖𝑠𝑒𝑣𝑒𝑛,𝑛+12,𝑖𝑓𝑛𝑖𝑠𝑜𝑑𝑑. (3)

Proof. Let 𝐶_𝑛=1111⋯11−1−2−10⋯000−1−2−1⋯0000−1−2⋯00⋮⋮⋮⋮⋱⋮⋮0000⋯−2−10000⋯−1−2_𝑛×𝑛,𝐷_𝑛=12−100⋯001−2−10⋯001−1−2−1⋯0010−1−2⋯00⋮⋮⋮⋮⋱⋮⋮1000⋯−2−11000⋯−1−2_𝑛×𝑛. (4) Comparing 𝐶_𝑛 to 𝐷_𝑛, we get the following: 𝐶_𝑛=𝐷_𝑛+𝑛2×(−1)^𝑛−1. (5) Expanding the determinant 𝐷_𝑛 according to the last column and then the last line, we get the following incursion: 𝐷_𝑛=−2𝐷_𝑛−1−𝐷_𝑛−2+1; (6) that is, 𝐷_𝑛+𝐷_𝑛−1=−𝐷_𝑛−1+𝐷_𝑛−2+1. (7) Since 𝐷₁=1/2, 𝐷₂=0, and 𝐷₃=1/2, from the above incursion, we get the following: 𝐷_𝑛=10,if𝑛iseven,2,if𝑛isodd. (8) So, we have the following: 𝐶_𝑛=−𝑛2,if𝑛iseven,𝑛+12,if𝑛isodd. (9) This completes the proof.

3. Main Results

In the following, we always assume that 𝑉(𝑊_𝑛)={𝑣₀,𝑣₁,…,𝑣_𝑛−1}, where 𝑣₀ is the hub of 𝑊_𝑛.

Theorem 5. Let 𝑒=𝑣_𝑖𝑣_{𝑖+1mod(𝑛−1)}(1⩽𝑖⩽𝑛−1). Then det𝐷𝑊_𝑛=−𝑛−𝑒4²𝑛,𝑖𝑓𝑛𝑖𝑠𝑒𝑣𝑒𝑛,₂−14,𝑖𝑓𝑛𝑖𝑠𝑜𝑑𝑑, (10) where 𝑛⩾3.

Proof. Without loss of generality, we may assume that 𝑒=𝑣₁𝑣_𝑛−1. Let 𝐴_𝑛=det𝐷𝑊_𝑛−𝑒=0111⋯111012⋯221101⋯221210⋯22⋮⋮⋮⋮⋱⋮⋮1222⋯011222⋯10_𝑛×𝑛. (11)
Then 𝐴_𝑛=1011−211⋯111−200⋯001−210−2+2−1⋯0010−1−2⋯00⋮⋮⋮⋮⋱⋮⋮1000⋯−2−11000⋯−1−2_𝑛×𝑛. (12) Expanding the determinant 𝐴_𝑛 according to the second line, we get the following incursion: 𝐴_𝑛=(−1)^𝑛−1𝑛2−2𝐴_𝑛−1−𝐶_𝑛−2−𝐷_𝑛−2−𝐴_𝑛−2, (13) where 𝐶_𝑛 and 𝐷_𝑛 are defined as in Lemma 4.
By Lemma 4, we get the following: 𝐴_𝑛=−1−2𝐴_𝑛−1−𝐴_𝑛−2,if𝑛iseven,−2𝐴_𝑛−1−𝐴_𝑛−2,if𝑛isodd. (14) Since 𝐴₃=2, 𝐴₄=−4, 𝐴₅=6, and 𝐴₆=−9, according to the above incursion, we get the following: 𝐴_𝑛=−𝑛²4𝑛,if𝑛iseven,₂−14,if𝑛isodd, (15) where 𝑛⩾3. This completes the proof.

Corollary 6. Let 𝑒=𝑣_𝑖𝑣_{𝑖+1mod(𝑛−1)}(1⩽𝑖⩽𝑛−1). Then 𝑛₊𝐷𝑊_𝑛−𝑒=1,𝑛₀𝐷𝑊_𝑛𝑛−𝑒=0,₋𝐷𝑊_𝑛−𝑒=𝑛−1. (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(𝑊_𝑛−1−𝑒))=1, 𝑛₀(𝐷(𝑊_𝑛−1−𝑒))=0, 𝑛₋(𝐷(𝑊_𝑛−1−𝑒))=𝑛−2.
Since 𝐷(𝑊_𝑛−1−𝑒) is a principle submatrix of 𝐷(𝑊_𝑛−𝑒), by Lemma 1, the eigenvalues of 𝐷(𝑊_𝑛−1−𝑒) interlace the eigenvalues of 𝐷(𝑊_𝑛−𝑒). By Theorem 5, det𝐷(𝑊_𝑛−1−𝑒)det𝐷(𝑊_𝑛−𝑒)<0. So, 𝐷(𝑊_𝑛−𝑒) has one negative eigenvalue more than 𝐷(𝑊_𝑛−1−𝑒). According to the induction hypothesis, we get 𝑛₊(𝐷(𝑊_𝑛−𝑒))=1, 𝑛₀(𝐷(𝑊_𝑛−𝑒))=0, and 𝑛₋(𝐷(𝑊_𝑛−𝑒))=𝑛−1. This completes the proof.

Theorem 7. One hasdet𝐷𝑊_𝑛=1−𝑛,𝑖𝑓𝑛𝑖𝑠𝑒𝑣𝑒𝑛,0,𝑖𝑓𝑛𝑖𝑠𝑜𝑑𝑑, (17) where 𝑛⩾3.

Proof. Consider the followingdet𝐷𝑊_𝑛=0111⋯111012⋯211101⋯221210⋯22⋮⋮⋮⋮⋱⋮⋮1222⋯011122⋯10_𝑛×𝑛=0111⋯111−2−10⋯0−11−1−2−1⋯0010−1−2⋯00⋮⋮⋮⋮⋱⋮⋮1000⋯−2−11−100⋯−1−2_𝑛×𝑛=()0𝑛−111⋯11𝑛−1−4𝑛−1−4−4⋯−4−41−4−2−1⋯001−4−1−2⋯00⋮⋮⋮⋮⋱⋮⋮1−400⋯−2−11−400⋯−1−2_𝑛×𝑛=()0𝑛−111⋯11𝑛−14𝑛−100⋯0010−2−1⋯0010−1−2⋯00⋮⋮⋮⋮⋱⋮⋮1000⋯−2−11000⋯−1−2_𝑛×𝑛. (18) Expanding the above determinant according to the second line, we get the following: det𝐷𝑊_𝑛=(−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: det𝐷𝑊_𝑛=1−𝑛,if𝑛iseven,0,if𝑛isodd. (20)
This completes the proof.

Similar to Corollary 6, we can get the following corollary.

Corollary 8. (i)  If 𝑛 is even, 𝑛₊(𝐷(𝑊_𝑛))=1, 𝑛₀(𝐷(𝑊_𝑛))=0, 𝑛₋(𝐷(𝑊_𝑛))=𝑛−1.
(ii)  If 𝑛 is odd, 𝑛₊(𝐷(𝑊_𝑛))=1, 𝑛₀(𝐷(𝑊_𝑛))=1, 𝑛₋(𝐷(𝑊_𝑛))=𝑛−2.
Denote by 𝑊_𝑛−𝑣₀ the graph obtained from 𝑊_𝑛 by deleting the vertex 𝑣₀ and all the edges adjacent to 𝑣₀; that is, 𝑊_𝑛−𝑣₀≅𝐶_𝑛−1. Let 𝐸_𝑘(1⩽𝑘⩽𝑛) be any subset of 𝐸(𝑊_𝑛−𝑣₀) with |𝐸_𝑘|=𝑘. In the following, we always denote by 𝑊_𝑛−𝐸_𝑘 the graph obtained from 𝑊_𝑛 by deleting all the edges in 𝐸_𝑘.

Theorem 9. One has 𝑛₊(𝐷(𝑊_𝑛−𝐸_𝑘))=1, 𝑛₀(𝐷(𝑊_𝑛−𝐸_𝑘))=0, 𝑛₋(𝐷(𝑊_𝑛−𝐸_𝑘))=𝑛−1.

Proof. Denote the components of 𝑊_𝑛−𝐸_𝑘−𝑣₀ by 𝐶₁,…,𝐶_𝑠. Let 𝐺_𝑖 denote the graph that contains 𝐶_𝑖 plus the vertex 𝑣₀ such that 𝑣₀ is connected to every other vertex, 1⩽𝑖⩽𝑠. Then each 𝐺_𝑖(1⩽𝑖⩽𝑠) is isomorphism to 𝑊_{|𝑉(𝐺𝑖)|}−𝑒_𝑖 or 𝐾₂, where 𝑒_𝑖 is an edge of 𝑊_{|𝑉(𝐺𝑖)|}−𝑣₀. By Lemma 2 and some direct calculations, we get the following: cof𝐺_𝑖=det𝐷𝑊_{|𝑉(𝐺𝑖)|}−𝑒_𝑖11=(−1)^{|𝑉(𝐺𝑖)|−1}𝑉𝐺_𝑖. (21) It is easy to check that cof(𝐺_𝑖)=(−1)^{|𝑉(𝐺𝑖)|−1}|𝑉(𝐺_𝑖)| is also true when 𝐺_𝑖 is isomorphism to 𝐾₂.
In the following, we will prove the theorem by introduction on 𝑠.
For 𝑠=1, 𝐺≅𝑊_𝑛−𝑒, where 𝑒 is an edge of 𝑊_𝑛−𝑣₀, by Corollary 6, we get the result.
Suppose the result is true for 𝑠−1.
For 𝑠, let 𝐺^′=𝐺₁∪𝐺₂∪⋯∪𝐺_𝑠−1. Then by the induction hypothesis, 𝑛₊(𝐷(𝐺^′))=1, 𝑛₀(𝐷(𝐺^′))=0, and 𝑛₋(𝐷(𝐺^′))=|𝑉(𝐺^′)|−1, which implies that det𝐷𝐺^′=(−1)^{|𝑉(𝐺′)|−1}𝑎, (22) where 𝑎 is a positive integer.
Since cof𝐺_𝑖=(−1)^{|𝑉(𝐺𝑖)|−1}𝑉𝐺_𝑖,1⩽𝑖⩽𝑠, (23) by Lemma 3, cof𝐷𝐺^′=^𝑠−1_𝑗=1cof𝐷𝐺_𝑗=(−1)^{|𝑉(𝐺′)|−1𝑠−1}_𝑗=1𝑉𝐺_𝑗. (24)
Then det𝐷𝑊_𝑛−𝐸_𝑘=det𝐷𝐺^′cof𝐷𝐺_𝑖+det𝐷𝐺_𝑖cof𝐷𝐺^′=(−1)^{|𝑉(𝐺′)|−1}𝑎×(−1)^{|𝑉(𝐺𝑖)|−1}𝑉𝐺_𝑖+(−1)^{|𝑉(𝐺𝑖)|−1}𝑏×(−1)^{|𝑉(𝐺′)|−1𝑠−1}_𝑗=1𝑉𝐺_𝑗=(−1)^𝑛−1𝑎𝑉𝐺_𝑖+𝑏^𝑠−1_𝑗=1𝑉𝐺_𝑗, (25) 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(𝑊_𝑛−𝐸_𝑘))=1, 𝑛₀(𝐷(𝑊_𝑛−𝐸_𝑘))=0, and 𝑛₋(𝐷(𝑊_𝑛−𝐸_𝑘))=𝑛−1.
Up to now, we have proved the result.

Let 𝐺𝑢_𝑖⋅𝑣_𝑗𝐻 denote the graph formed by only identifying the vertex 𝑢_𝑖 of 𝐺 with the vertex 𝑣_𝑗 of 𝐻, where 𝑢_𝑖 and 𝑣_𝑗 are arbitrary vertices of 𝐺 and 𝐻, respectively.

Lemma 10 (see [13]). Let 𝐺×𝐻 denote the Cartesian product of connected graphs 𝐺 and 𝐻, where 𝑉(𝐺)={𝑢₁,…,𝑢_𝑚} and 𝑉(𝐻)={𝑣₁,…,𝑣_𝑛}. Then we have(i)𝑛₊(𝐺×𝐻)=𝑛₊(𝐺𝑢_𝑖⋅𝑣_𝑗𝐻); (ii)𝑛₀(𝐺×𝐻)=(𝑚−1)(𝑛−1)+𝑛₀(𝐺𝑢_𝑖⋅𝑣_𝑗𝐻); (iii)𝑛₋(𝐺×𝐻)=𝑛₋(𝐺𝑢_𝑖⋅𝑣_𝑗𝐻).

Theorem 11. Let 𝑢₀ and 𝑣₀ be the hubs of 𝑊_𝑛 and 𝑊_𝑚, respectively. Suppose 𝐸_𝑝(0⩽𝑝⩽𝑛−1) and 𝐸_𝑞(0⩽𝑞⩽𝑚−1) are any subsets of 𝐸(𝑊_𝑛−𝑢₀) and 𝐸(𝑊_𝑚−𝑣₀) with |𝐸_𝑝|=𝑝, |𝐸_𝑞|=𝑞, respectively. Then, the distance matrix of the graph (𝑊_𝑛−𝐸_𝑝)×(𝑊_𝑚−𝐸_𝑞) has exactly one positive eigenvalue.

Proof. Since 𝑢₀ and 𝑣₀ are the hubs of 𝑊_𝑛 and 𝑊_𝑚, respectively, (𝑊_𝑛−𝐸_𝑝)𝑢₀⋅𝑣₀(𝑊_𝑚−𝐸_𝑞) must be isomorphism to some 𝑊_{𝑛+𝑚−1}−𝐸_𝑝+𝑞, where 𝑤₀ is the hub of 𝑊_{𝑛+𝑚−1} and 𝐸_𝑝+𝑞 is any subset of 𝐸(𝑊_{𝑛+𝑚−1}−𝑤₀) with |𝐸_𝑝+𝑞|=𝑝+𝑞. By Theorem 9 and Lemma 10, we get the result.

Given an arbitrary integer 𝑚, for 1⩽𝑖⩽𝑚, let 𝑣_𝑖0 be the hub of 𝑊_𝑛𝑖 and 𝐸_𝑝𝑖 any subset of 𝐸(𝑊_𝑛𝑖−𝑣_𝑖0). Suppose 𝑉(𝑊_𝑛𝑖)={𝑣_𝑖0,𝑣_𝑖1,…,𝑣_{𝑖(𝑛𝑖−1)}}.

Theorem 12. For an arbitrary integer 𝑚, the distance matrix of the graph 𝐺=(𝑊_𝑛1−𝐸_𝑝1)𝑣_1𝑘⋅𝑣_2𝑗(𝑊_𝑛2−𝐸_𝑝2)𝑣_2ℎ⋯𝑣_{(𝑚−1)𝑟}(𝑊_𝑛𝑚−1−𝐸_𝑝𝑚−1)𝑣_{(𝑚−1)𝑡}⋅𝑣_𝑚𝑠(𝑊_𝑛𝑚−𝐸_𝑝𝑚) 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=(𝑊_𝑛1−𝐸_𝑝1)𝑣_1𝑘⋅𝑣_2𝑗(𝑊_𝑛2−𝐸_𝑝2)𝑣_2ℎ⋯𝑣_{(𝑚−2)𝑖}(𝑊_𝑛𝑚−2−𝐸_𝑝𝑚−2). Then 𝐺=𝐻𝑣_{(𝑚−2)𝑔}⋅𝑣_{(𝑚−1)𝑟}(𝑊_𝑛𝑚−1−𝐸_𝑝𝑚−1)𝑣_{(𝑚−1)𝑡}⋅𝑣_𝑚𝑠(𝑊_𝑛𝑚−𝐸_𝑝𝑚). By Lemma 10, we have the following: 𝑛₊(𝐺)=𝑛₊𝐻𝑊_𝑛𝑚−1−𝐸_𝑝𝑚−1𝑣_{(𝑚−1)𝑡}⋅𝑣_𝑚𝑠𝑊_𝑛𝑚−𝐸_𝑝𝑚=𝑛₊𝐻𝑊_𝑛𝑚−1−𝐸_𝑝𝑚−1𝑣_(𝑚−1)0⋅𝑣_𝑚0𝑊_𝑛𝑚−𝐸_𝑝𝑚. (26) Since (𝑊_𝑛𝑚−1−𝐸_𝑝𝑚−1)𝑣_(𝑚−1)0⋅𝑣_𝑚0(𝑊_𝑛𝑚−𝐸_𝑝𝑚)=𝑊_{𝑛𝑚−1+𝑛𝑚−1}−𝐸_𝑝𝑚−1−𝐸_𝑝𝑚, we get 𝑛₊(𝐺)=𝑛₊(𝐻𝑣_{(𝑚−2)𝑔}⋅𝑣_{(𝑚−1)𝑟}(𝑊_{𝑛𝑚−1+𝑛𝑚−1}−𝐸_𝑝𝑚−1−𝐸_𝑝𝑚)). 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).