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Inertia and spectral symmetry of the eccentricity matrices of a class of bi-block graphs

T. Divyadevi  and  I. Jeyaraman

Department of Mathematics
National Institute of Technology Tiruchirappalli-620 015, India
tdivyadevi@gmail.comjeyaraman@nitt.edu
Abstract

The eccentricity matrix of a simple connected graph GG is obtained from the distance matrix of GG by retaining the largest non-zero distance in each row and column, and the remaining entries are defined to be zero. A bi-block graph is a simple connected graph whose blocks are all complete bipartite graphs with possibly different orders. In this paper, we study the eccentricity matrices of a subclass \mathscr{B} (which includes trees) of bi-block graphs. We first find the inertia of the eccentricity matrices of graphs in \mathscr{B}, and thereby we characterize graphs in \mathscr{B} with odd diameters. Precisely, if GG\in\mathscr{B} with diameter of GG greater than three, then we show that the eigenvalues of the eccentricity matrix of GG are symmetric with respect to the origin if and only if the diameter of GG is odd. Further, we prove that the eccentricity matrices of graphs in \mathscr{B} are irreducible.

AMS Subject Classification (2010): 05C12, 05C50.

Keywords. Eccentricity matrix, bi-block graph, inertia, spectral symmetry.

1 Introduction

Let GG be a simple connected graph on nn vertices with the vertex set VG={v1,v2,,vn}V_{G}=\{v_{1},v_{2},\ldots,v_{n}\}. In literature, many properties and structural characteristics of GG have been studied by associating several matrices corresponding to GG, see [2]. Let us recall some of them which are pertinent to our discussion here. The adjacency matrix of GG, denoted by A(G)A(G), is an n×nn\times n matrix whose (i,j)(i,j)-th entry is one if viv_{i} and vjv_{j} are adjacent and zero elsewhere. Let d(vi,vj)d(v_{i},v_{j}) denote the length of a shortest path between the vertices viv_{i} and vjv_{j}. The distance matrix D(G):=(dij)D(G)\mathrel{\mathop{\mathchar 58\relax}}=(d_{ij}) of GG, is an n×nn\times n matrix such that dij=d(vi,vj)d_{ij}=d(v_{i},v_{j}) for all ii and jj. Clearly, the adjacency matrix can be obtained from the distance matrix by retaining the smallest non-zero distance (which is equal to one) in each row and column and by setting the rest of the entries equal to zero. Inspired by this, Randić [12] associated a new matrix corresponding to GG, namely DMAXD_{\text{MAX}}, which is derived from the distance matrix D(G)D(G) by keeping the largest non-zero distance in each row and each column of D(G)D(G) and defining the remaining entries equal to zero. Later, Wang et al. [15] obtained an equivalent definition of DMAXD_{\text{MAX}} using the notion of eccentricity of a vertex and called it the eccentricity matrix. Recall that the eccentricity of a vertex viv_{i} is defined by e(vi)e(v_{i})= max {d(vi,vj):vjVG}\{d(v_{i},v_{j})\mathrel{\mathop{\mathchar 58\relax}}v_{j}\in V_{G}\}. The eccentricity matrix E(G)E(G) of GG is an n×nn\times n matrix whose (i,j)(i,j)-th entry is given by

E(G)ij={d(vi,vj)if d(vi,vj)=min{e(vi),e(vj)},0otherwise.\displaystyle E(G)_{ij}=\begin{cases}d(v_{i},v_{j})&\text{if }d(v_{i},v_{j})=\text{min}\{e(v_{i}),e(v_{j})\},\\ 0&\text{otherwise}.\end{cases}

This matrix has been well studied in the literature (see [7, 10, 8, 9, 11, 12, 15, 14, 16]) and has applications in Chemistry, see [12, 15, 14].

It is significant to note that the adjacency and distance matrices of connected graphs are always irreducible, but the eccentricity matrix fails to satisfy this property for a general connected graph. For example, the eccentricity matrix of a complete bipartite graph Kl,mK_{l,m} is reducible for all l,m2l,m\geq 2, see [15]. However, for some classes of graphs, the associated eccentricity matrices are irreducible. It is proved in [15] that the eccentricity matrix of a tree with at least two vertices is irreducible, and an alternative proof of this result is given in [8]. This result has been extended to some larger classes of graphs in [7] and [16]. It is shown in [11] that the eccentricity matrices of the coalescence of complete graphs are irreducible. Characterizing the graphs whose eccentricity matrices are irreducible, posed by Wang et al. [15], remains an open problem. In this paper, we provide a class of bi-block graphs \mathscr{B}, defined below, whose eccentricity matrices are irreducible.

To introduce the graph class \mathscr{B}, let us recall the following definitions. A block of a graph GG is a maximal connected subgraph of GG which has no cut-vertex. A connected graph GG is called a bi-block graph (block graph) if all its blocks are complete bipartite graphs (respectively, complete graphs) of possibly varying orders. Note that the complete bipartite graphs are bi-block graphs with exactly one block. Since the spectral properties of the eccentricity matrices of complete bipartite graphs are already explored in [8], it is sufficient to consider the bi-block graphs with at least two blocks. The main focus of this article is to analyze the spectral symmetry of the eccentricity matrices of bi-block graphs. In Section 4, it is shown that the spectral symmetry result is not true for a bi-block graph having more than two cut-vertices in a block. In view of these, we define the class \mathscr{B} which is the collection of all bi-block graphs with at least two blocks and at most two cut-vertices in each block. Clearly, \mathscr{B} contains all trees with at least three vertices.

In what follows, we present a brief survey of the literature on bi-block graphs. In [4], the authors studied the spectral radius of the adjacency matrix of a class of bi-block graphs with a given independence number. The permanent, determinant, and the rank of the adjacency matrices of bi-block graphs were computed in [13]. The inverse formula given for the distance matrix of a tree was extended to a subclass of bi-block graphs in [6]. Inspired by these, in this article, we study the eccentricity matrices of the subclass \mathscr{B} of bi-block graphs.

A primary motivation for studying this paper comes from the following spectral symmetry results. In related to this, we first recall a well-known characterization for bipartite graphs. A graph GG is a bipartite graph if and only if the eigenvalues of the adjacency matrix A(G)A(G) are symmetric with respect to the origin (i.e., if μ\mu is an eigenvalue of A(G)A(G) with multiplicity ll then -μ\mu is also an eigenvalue of A(G)A(G) with multiplicity ll). In [10], the authors studied a similar equivalence for the eccentricity matrices of trees and gave a characterization for trees of odd diameters. Specifically, the eigenvalues of E(T)E(T) of a tree TT are symmetric with respect to the origin if and only if diam(T)\operatorname{diam}(T) is odd. An analogous result has been proved for a subclass of block graphs (namely, clique trees) in [7]. The problem of characterizing the graphs whose eigenvalues of the eccentricity matrices are symmetric with respect to the origin, posed in [10], remains open. Motivated by these, in this article, we consider the spectral symmetry equivalence for the eccentricity matrices of graphs in \mathscr{B}. Precisely, we prove that the eigenvalues of E(G)E(G) are symmetric with respect to the origin if and only if diam(G)\operatorname{diam}(G) is odd, whenever GG\in\mathscr{B} with diam(G)4\operatorname{diam}(G)\geq 4. By means of examples, we show that this equivalence does not hold for a general bi-block graph and for GG\in\mathscr{B} with diam(G)=3\operatorname{diam}(G)=3.

To present another motivation for this paper, we now turn our attention to the inertias of graph matrices (i.e., matrices that arise from graphs). Let us recall the notion of inertia. Let AA be an n×nn\times n real symmetric matrix and let i+(A)i_{+}(A), i(A)i_{-}(A) and i0(A)i_{0}(A) denote the number of positive, negative and zero eigenvalues of AA, respectively, including the multiplicities. The inertia of AA is the ordered triple (i+(A),i(A),i0(A))(i_{+}(A),i_{-}(A),i_{0}(A)), and is denoted by In(A)\operatorname{In}(A). It is known that i+(A)+i(A)=rank(A)i_{+}(A)+i_{-}(A)=\operatorname{rank}(A). An interesting and challenging problem in spectral graph theory is finding the eigenvalues and inertias of graph matrices, see [2, 3, 8, 9, 11, 15, 14, 16] and the references therein. Among other results, it has been proved that the inertia of the distance matrix of a tree on n(2)n(\geq 2) vertices is (1,n1,0)(1,n-1,0), see [2]. The inertias of the eccentricity matrices of lollipop graphs, coalescence of complete graphs, coalescence of two cycles, trees, and clique trees have been computed in [7, 10, 8, 11]. One of the objectives of this article is to find the inertias of the eccentricity matrices of the graphs in \mathscr{B}. Some of the results obtained and the ideas used in this paper are similar to those in [7, 10] but the proofs are different in many cases.

The outline of this article is as follows. In the next section, we recall some results and notation used in this article. In Section 3, we associate a tree TGT_{G} for each GG\in\mathscr{B} and study their relationships where \mathscr{B} is a subclass of bi-block graphs. Using these results, we derive the centers of graphs in \mathscr{B}. Section 4 deals with the inertia and the spectral symmetry of eccentricity matrix of GG\in\mathscr{B}. Among other things, we give an equivalent condition for the spectrum of E(G)E(G) to be symmetric with respect to the origin where GG\in\mathscr{B}. Finally, we prove the irreducibility of the eccentricity matrices of graphs in \mathscr{B}.

2 Preliminaries

In this section, we recall basic definitions and notation that will be used in the sequel.

We denote the vertex set and the edge set of a graph GG by VGV_{G} and EGE_{G}, respectively. A graph G=(VG,EG)G^{\prime}=(V_{G^{\prime}},E_{G^{\prime}}) is said to be a subgraph of GG if VGVGV_{G^{\prime}}\subseteq V_{G} and EGEGE_{G^{\prime}}\subseteq E_{G}. A subgraph GG^{\prime} of GG is called an induced subgraph of GG if the edges of GG^{\prime} are precisely the edges of GG whose ends are in VGV_{G^{\prime}}. We denote the induced subgraph GG^{\prime} by G[VG]G[V_{G^{\prime}}] and call it as the subgraph induced by VGV_{G^{\prime}}. A vertex vVGv\in V_{G} is said to be a cut-vertex of GG if G{v}G\setminus\{v\} is a disconnected graph. The notation CGC_{G} stands for the collection of all cut-vertices of GG. A maximal connected subgraph of a graph GG is known as a component of GG. The radius and the diameter of GG are, respectively, denoted by rad(G)\operatorname{rad}(G) and diam(G)\operatorname{diam}(G), and are defined by rad(G):=min{e(v):vVG}\operatorname{rad}(G)\mathrel{\mathop{\mathchar 58\relax}}=\min\{e(v)\mathrel{\mathop{\mathchar 58\relax}}v\in V_{G}\} and diam(G):=max{e(v):vVG}\operatorname{diam}(G)\mathrel{\mathop{\mathchar 58\relax}}=\max\{e(v)\mathrel{\mathop{\mathchar 58\relax}}v\in V_{G}\}. A path PP in GG is a subgraph of GG whose vertices are arranged in a sequence such that two vertices are adjacent in PP if and only if they are consecutive in the sequence. We denote a path of length kk between two vertices aa and bb in GG by PG(a,b)=au1u2uk1bP_{G}(a,b)=au_{1}u_{2}\ldots u_{k-1}b, where uiVGu_{i}\in V_{G} for all ii. We write the length of the path PG(a,b)P_{G}(a,b) by l(PG(a,b))l\left(P_{G}(a,b)\right). A diametrical path in GG is a shortest path between two vertices uu and vv such that d(u,v)=diam(G)d(u,v)=\operatorname{diam}(G). A vertex vVGv\in V_{G} is said to be a central vertex if e(v)=rad(G)e(v)=\operatorname{rad}(G). The center of GG, denoted by C(G)C(G), is the collection of all central vertices of GG.

A graph GG is said to be bipartite if VGV_{G} can be partitioned into two non-empty subsets V1V_{1} and V2V_{2} such that each edge of GG has one end in V1V_{1} and the other end in V2V_{2}. The pair (V1,V2)(V_{1},V_{2}) is called a bipartition of the bipartite graph GG and the sets V1V_{1} and V2V_{2} are referred to as the partite sets of GG. A bipartite graph with bipartition (V1,V2)(V_{1},V_{2}) is said to be a complete bipartite graph if every vertex of V1V_{1} is adjacent to all the vertices of V2V_{2}, and is denoted by K|V1|,|V2|K_{\lvert V_{1}\rvert,\lvert V_{2}\rvert} where |V1|\lvert V_{1}\rvert stands for the cardinality of V1V_{1}. For more details on graph-theoretic notions and terminologies, we refer to the book [1].

Let GG\in\mathscr{B} and let BB be a block of GG. Then BB is a complete bipartite graph Kl,mK_{l,m}. Throughout this article, we assume that (V1(B),V2(B))\left(V_{1}(B),V_{2}(B)\right) is the bipartition of BB. The block BB of GG is said to be a bridge block if |VBCG|=2\lvert V_{B}\cap C_{G}\rvert=2, and a leaf block if |VBCG|=1\lvert V_{B}\cap C_{G}\rvert=1.

Let AA be an m×nm\times n matrix. We write the transpose of AA, ii-th row of AA, ii-th column of AA and the rank of AA by AA^{\prime}, AiA_{i*}, AiA_{*i} and rank(A)\operatorname{rank}(A), respectively. We denote the principal submatrix of AA whose rows and columns are indexed, respectively, by the sets U{1,2,,m}U\subseteq\{1,2,\ldots,m\} and V{1,2,,n}V\subseteq\{1,2,\ldots,n\} by A([UV])A\left([U\mid V]\right). The notations JJ and OO are used to denote the matrices with all elements equal to 11 and 0, respectively, and the orders of the matrices are clear from the context. The determinant of a square matrix AA is written by det(A)\det(A).

In the following, we collect some known results which are needed in this paper.

Theorem 2.1 ([17]).

Let AA and DD be r×rr\times r and s×ss\times s real matrices, respectively, and let M=(ABCD)M=\left(\begin{smallmatrix}A&B\\[3.0pt] C&D\\ \end{smallmatrix}\right) be a symmetric partitioned matrix of order nn. If AA is nonsingular, then

  • (i)

    det(M)=det(A)det(DCA1B)\det(M)=\det(A)\det(D-CA^{-1}B).

  • (ii)

    In(M)=In(A)+In(DCA1B)\operatorname{In}(M)=\operatorname{In}(A)+\operatorname{In}(D-CA^{-1}B).

In fact, the above result holds for any non-singular principal submatrix AA of MM.

Theorem 2.2 ([17, P.25]).

If MM is an n×nn\times n real matrix, then the characteristic polynomial of MM is given by χ(x)=xnδ1xn1+δ2xn2++(1)n1δn1x+(1)nδn\chi(x)=x^{n}-\delta_{1}{x}^{n-1}+\delta_{2}{x}^{n-2}+\ldots+(-1)^{n-1}\delta_{n-1}x+(-1)^{n}\delta_{n} where δr\delta_{r} denotes the sum of all principal minors of order rr for all r=1,2,,nr=1,2,\ldots,n.

Theorem 2.3 ([17]).

Let MM be a symmetric matrix of order nn and AA be a principal submatrix of MM order mm where 1mn1\leq m\leq n. If the eigenvalues of MM and AA are λ1λ2λn\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{n} and β1β2βm\beta_{1}\geq\beta_{2}\geq\cdots\geq\beta_{m}, respectively, then λiβiλnm+i,for all i=1,2,,m\lambda_{i}\geq\beta_{i}\geq\lambda_{n-m+i},~{}\text{for all $i=1,2,\ldots,m$}. Moreover, i+(M)i+(A)andi(M)i(A).i_{+}(M)\geq i_{+}(A)~{}\text{and}~{}i_{-}(M)\geq i_{-}(A).

Lemma 2.1 (see [2, 10]).

Suppose that p(x)=xn+α1xn1+α2xn2++αn1x+αnp(x)={x}^{n}+\alpha_{1}{x}^{n-1}+\alpha_{2}{x}^{n-2}+\ldots+\alpha_{n-1}x+\alpha_{n} is a polynomial whose roots are all non-zero real numbers. If there exists i{1,,n1}i\in\{1,\dots,n-1\} such that αi0\alpha_{i}\neq 0 and αi+10\alpha_{i+1}\neq 0, then the roots of p(x)p(x) are not symmetric with respect to the origin (i.e., p(a)=0p(a)=0 but p(a)0p(-a)\neq 0 for some real number aa).

3 Relations between GG\in\mathscr{B} and its associated tree TGT_{G}

In this section, we associate a tree TGT_{G} for each graph GG\in\mathscr{B} and obtain some interconnections between GG and TGT_{G} by employing the properties of trees. In particular, we show that diam(G)=diam(TG)\operatorname{diam}(G)=\operatorname{diam}(T_{G}) and C(TG)C(G)C(T_{G})\subseteq C(G). Making use of these relations, we explicitly find the centers of graphs in \mathscr{B}. These results will be used in the next section to study the inertia and the spectral symmetry of the eccentricity matrix of a graph in \mathscr{B}.

The idea of analyzing a graph GG\in\mathscr{B} through the associated tree TGT_{G} is motivated by [7, 10]. In [7], the authors studied the eccentricity matrices of clique trees by constructing trees of particular types. While defining trees, they considered non-cut-vertices only from the leaf blocks of clique trees. In our case, we have to include non-cut-vertices from both leaf blocks and bridge blocks of GG\in\mathscr{B} in order to obtain a tree satisfying specific properties and thus the construction is different from [7].

Let us begin this section by associating a subgraph TGT_{G} for each GG\in\mathscr{B}. It will be shown later that TGT_{G} is a tree. We collect some selected vertices from each block BB of GG to define TGT_{G}. If B=K1,1B=K_{1,1}, then we collect both the vertices of BB. Suppose that BK1,1B\neq K_{1,1} and BB is a bridge block with VBCG={z1,z2}V_{B}\cap C_{G}=\{z_{1},z_{2}\}. If both the cut-vertices z1z_{1} and z2z_{2} lie on the same partite set, say V1(B)V_{1}(B), then we choose three vertices from VBV_{B}, which are z1z_{1}, z2z_{2} and a non-cut-vertex in V2(B)V_{2}(B) with the minimum vertex label; otherwise, we collect exactly two vertices z1z_{1} and z2z_{2} from VBV_{B}. In the case of leaf block BK1,1B\neq K_{1,1}, we collect exactly three vertices, which are the cut-vertex in VBV_{B}, and non-cut-vertices with the minimum vertex label in each partite sets V1(B)V_{1}(B) and V2(B)V_{2}(B). The graph TGT_{G} is defined as the subgraph induced by the vertex set which is the union of all selected vertices in each block of GG. The precise construction of TGT_{G} is given below.

Definition 3.1.

Let GG\in\mathscr{B} and let BB be a block of GG with bipartition (V1(B),V2(B))(V_{1}(B),V_{2}(B)). For i=1,2i=1,2, we denote the non-cut-vertex with the minimum vertex label in Vi(B)V_{i}(B), if it exists, by uiu_{i}. If BK1,1B\neq K_{1,1}, we define

NCB:={{u2}if |V1(B)CG|=2,{u1}if |V2(B)CG|=2,{u1,u2}if B is a leaf block,otherwise.NC_{B}\mathrel{\mathop{\mathchar 58\relax}}=\begin{cases}\{u_{2}\}&\text{if $\lvert V_{1}(B)\cap C_{G}\rvert=2$},\\ \{u_{1}\}&\text{if $\lvert V_{2}(B)\cap C_{G}\rvert=2$},\\ \{u_{1},u_{2}\}&\text{if $B$ is a leaf block},\\ \emptyset&\text{otherwise}.\end{cases}

If B=K1,1B=K_{1,1}, then we set NCB:={u1,u2}VBNC_{B}\mathrel{\mathop{\mathchar 58\relax}}=\{u_{1},u_{2}\}\cap V_{B}. Let S:=BNCB\displaystyle S\mathrel{\mathop{\mathchar 58\relax}}=\cup_{B}NC_{B} where BB runs over all blocks of GG. Define TGT_{G} as the subgraph induced by the vertex subset SCGS\cup C_{G}. That is,

TG:=G[SCG].T_{G}\mathrel{\mathop{\mathchar 58\relax}}=G\left[S\cup C_{G}\right]. (1)
Remark 3.1.

We mention that the choice of a non-cut-vertex with the minimum vertex label ensures the uniqueness of TGT_{G}.

Example 3.1.

We illustrate the construction of TGT_{G} for a given graph GG\in\mathscr{B} through the following figures. The graph GG has three leaf blocks (K1,1,K1,1K_{1,1},K_{1,1} and K2,3K_{2,3}) and three bridge blocks (K2,2,K1,1K_{2,2},K_{1,1} and K3,3K_{3,3}).

v2v_{2}v4v_{4}v6v_{6}v5v_{5}v3v_{3}v1v_{1}v7v_{7}v10v_{10}v11v_{11}v12v_{12}v8v_{8}v9v_{9}v15v_{15}v14v_{14}v13v_{13}v16v_{16}
(a) A graph GG\in\mathscr{B}
v2v_{2}v5v_{5}v3v_{3}v1v_{1}v7v_{7}v10v_{10}v9v_{9}v15v_{15}v13v_{13}
(b) The tree TGT_{G} associated with GG

In the following, we mention a few properties of GG\in\mathscr{B} that are needed in the sequel. For more details, we refer to [1, 5].

  • (P1)(P_{1})

    Suppose that BB is a block of GG with bipartition (V1(B),V2(B))(V_{1}(B),V_{2}(B)). Then |V1(B)|=1\lvert V_{1}(B)\rvert=1 if and only if |V2(B)|=1\lvert V_{2}(B)\rvert=1.

  • (P2)(P_{2})

    Any shortest path in GG can contain at most three consecutive vertices from a single block BB of GG.

  • (P3)(P_{3})

    Cycles of GG are exactly cycles of its blocks.

  • (P4)(P_{4})

    Let vVGv\in V_{G}. Then vv is a cut-vertex of GG if and only if it lies in at least two blocks of GG.

Using the property (P4)(P_{4}) and the fact that each edge of GG lies on exactly one block of GG (P. 5959 in [1]), we obtain the following remark.

Remark 3.2.

Let GG\in\mathscr{B} and BB be a block of GG. If xyxy and yzyz are edges in GG such that x,yVBx,y\in V_{B} and zVBz\not\in V_{B}, then yCGy\in C_{G}. Moreover, if PG(u0,uk+1)=u0u1ukuk+1P_{G}(u_{0},u_{k+1})=u_{0}u_{1}\ldots u_{k}u_{k+1} (k2)(k\geq 2) is a shortest path between u0u_{0} and uk+1u_{k+1} in GG such that the edge ui1uiu_{i-1}u_{i} belongs to the block BB for some 1ik1\leq i\leq k and ui+1VBu_{i+1}\not\in V_{B}, then uiCGu_{i}\in C_{G}. That is, a shortest path leaves a block BB of GG and enters into another block BB^{\prime} of GG through a cut-vertex of GG which lies in VBVBV_{B}\cap V_{B^{\prime}}.

Many results of this section deals the relationship between GG\in\mathscr{B} and its associated graph TGT_{G}. We first show that the graph TGT_{G} is a tree. In order to distinguish, we use the notations dG(a,b)d_{G}(a,b) and dTG(a,b)d_{T_{G}}(a,b), respectively, to denote the distance between the vertices aa and bb with respect to GG, and with respect to TGT_{G}.

Lemma 3.1.

Let GG\in\mathscr{B} and TGT_{G} be the subgraph of GG defined in (1). Then TGT_{G} is a tree and dG(a,b)=dTG(a,b)d_{G}(a,b)=d_{T_{G}}(a,b) for all a,bVTGa,b\in V_{T_{G}}.

Proof.

Note that each block of GG is a complete bipartite graph. Since TGT_{G} has at most three vertices from each block of GG, by (P3)(P_{3}), TGT_{G} contains no cycle. We claim that TGT_{G} is connected. Let a,bVTGa,b\in V_{T_{G}}. Since GG is connected, there is a path between the vertices aa and bb in GG. Let PG(a,b)=au1u2ukbP_{G}(a,b)=au_{1}u_{2}\ldots u_{k}b be a shortest path in GG. If uiVTGu_{i}\in V_{T_{G}} for all i=1,2,,ki=1,2,\ldots,k then the claim follows. Suppose that ujVTGu_{j}\not\in V_{T_{G}} for some j{1,2,,k}j\in\{1,2,\ldots,k\}. Let j0j_{0} be the smallest index such that uj0VTGu_{j_{0}}\not\in V_{T_{G}}. Then uj0u_{j_{0}} is not a cut-vertex of GG because all the cut-vertices of GG belong to TGT_{G}. Assume that the edge uj01uj0u_{j_{0}-1}u_{j_{0}} belongs to the block BB of GG. If j0=1j_{0}=1 then take u0=au_{0}=a. Without loss of generality, we assume that uj0V1(B)u_{j_{0}}\in V_{1}(B). Then uj01V2(B)u_{j_{0}-1}\in V_{2}(B). Since uj0CGu_{j_{0}}\not\in C_{G}, by Remark 3.2, uj0+1V2(B)u_{j_{0}+1}\in V_{2}(B) where we take uj0+1=bu_{j_{0}+1}=b if j0=kj_{0}=k. By the construction of TGT_{G}, it is clear that V1(B)VTGV_{1}(B)\cap V_{T_{G}}\neq\emptyset. Choose uj0V1(B)VTGu_{j_{0}}^{\prime}\in V_{1}(B)\cap V_{T_{G}}. We now show that uj0PG(a,b)u_{j_{0}}^{\prime}\not\in P_{G}(a,b). Clearly, uj0uj01u_{j_{0}}^{\prime}\neq u_{j_{0}-1} and uj0uj0+1u_{j_{0}}^{\prime}\neq u_{j_{0}+1}. If there exists i{1,2,,j02}i\in\{1,2,\ldots,j_{0}-2\} such that uj0=uiu_{j_{0}}^{\prime}=u_{i} then au1uiuj0+1bau_{1}\ldots u_{i}u_{j_{0}+1}\ldots b is a path between aa and bb whose length is strictly less than l(PG(a,b))l\left(P_{G}(a,b)\right) which is not possible. Therefore uj0uiu_{j_{0}}^{\prime}\neq u_{i} for all i{1,2,,j02}i\in\{1,2,\ldots,j_{0}-2\}. Similarly, we see that uj0uiu_{j_{0}}^{\prime}\neq u_{i} for all i{j0+2,,k}i\in\{j_{0}+2,\ldots,k\}. Hence uj0PG(a,b)u_{j_{0}}^{\prime}\not\in P_{G}(a,b). We obtain a new path PG(a,b)P^{\prime}_{G}(a,b) from PG(a,b)P_{G}(a,b) by replacing the edges uj01uj0u_{j_{0}-1}u_{j_{0}} and uj0uj0+1u_{j_{0}}u_{j_{0}+1}, respectively, by uj01uj0u_{j_{0}-1}u_{j_{0}}^{\prime} and uj0uj0+1u_{j_{0}}^{\prime}u_{j_{0}+1}. Clearly, l(PG(a,b))=l(PG(a,b))l(P^{\prime}_{G}(a,b))=l(P_{G}(a,b)). We replace PG(a,b)P_{G}(a,b) by PG(a,b)P^{\prime}_{G}(a,b). Repeating the above argument leads to obtain a path PTG(a,b)P_{T_{G}}(a,b) in TGT_{G} such that l(PTG(a,b))=l(PG(a,b))l(P_{T_{G}}(a,b))=l(P_{G}(a,b)). This implies that TGT_{G} is connected and dTG(a,b)l(PTG(a,b))=l(PG(a,b))=dG(a,b)d_{T_{G}}(a,b)\leq l(P_{T_{G}}(a,b))=l(P_{G}(a,b))=d_{G}(a,b). Since TGT_{G} is a subgraph of GG, we have dTG(a,b)dG(a,b)d_{T_{G}}(a,b)\geq d_{G}(a,b). Hence dTG(a,b)=dG(a,b)d_{T_{G}}(a,b)=d_{G}(a,b). ∎

The following lemma is useful in establishing the result diam(G)=diam(TG)\operatorname{diam}(G)=\operatorname{diam}(T_{G}). For notational simplicity, we also use the notation d(a,b)d(a,b) to denote the distance between two vertices aa and bb in GG instead of dG(a,b)d_{G}(a,b).

Lemma 3.2.

Let GG\in\mathscr{B}. Given aa and bb in VGV_{G}, there exist aa^{\prime} and bb^{\prime} in VTGV_{T_{G}} such that d(a,b)=d(a,b)d(a,b)=d(a^{\prime},b^{\prime}).

Proof.

Let a,bGa,b\in G. If a,bVTGa,b\in V_{T_{G}} then the result follows. Consider the case bVTGb\not\in V_{T_{G}}. We claim that there exists bVTGb^{\prime}\in V_{T_{G}} such that d(a,b)=d(a,b)d(a,b)=d(a,b^{\prime}). Let PG(a,b)=au1u2ukbP_{G}(a,b)=au_{1}u_{2}\ldots u_{k}b be a path such that d(a,b)=l(PG(a,b))d(a,b)=l\left(P_{G}(a,b)\right) where ukV1(B)u_{k}\in V_{1}(B) and bV2(B)b\in V_{2}(B) for some block BB of GG. We first prove the claim for k2k\geq 2.

Case 1: Suppose that ukCGu_{k}\not\in C_{G}. Then by Remark 3.2, uk1V2(B)u_{k-1}\in V_{2}(B), and by (P2)(P_{2}), uk2VBu_{k-2}\not\in V_{B} where we take uk2=au_{k-2}=a if k=2k=2. Therefore, by Remark 3.2, uk1CGu_{k-1}\in C_{G}.
Subcase 1.1: Let BB be a leaf block. By the construction of TGT_{G}, there exists a non-cut-vertex bV2(B)VTGb^{\prime}\in V_{2}(B)\cap V_{T_{G}}. Using (P3)(P_{3}) and k2k\geq 2, we see that aVBa\not\in V_{B} and bPG(a,b)b^{\prime}\not\in P_{G}(a,b). Therefore, bukuk1u1ab^{\prime}u_{k}u_{k-1}\ldots u_{1}a is a path. It is clear that uk1P¯G(b,a)u_{k-1}\in\bar{P}_{G}(b^{\prime},a) for all shortest paths P¯G(b,a)\bar{P}_{G}(b^{\prime},a) between the vertices bb^{\prime} and aa. Hence d(b,a)=d(b,uk1)+d(uk1,a)=d(b,a)d(b^{\prime},a)=d(b^{\prime},u_{k-1})+d(u_{k-1},a)=d(b,a).
Subcase 1.2: Assume that BB is a bridge block and zz is a cut-vertex of GG in BB other than uk1u_{k-1}.
Subcase 1.2.1: If zV2(B)z\in V_{2}(B) then, by (P3)(P_{3}), zPG(a,b)z\not\in P_{G}(a,b). Therefore, PG(z,a)=zukuk1u1aP_{G}(z,a)=zu_{k}u_{k-1}\ldots u_{1}a is a path. Again by (P3)(P_{3}), uk1P¯G(z,a)u_{k-1}\in\bar{P}_{G}(z,a) for all shortest paths P¯G(z,a)\bar{P}_{G}(z,a) between zz and aa. This fact leads to d(z,a)=d(z,uk1)+d(uk1,a)=d(b,a)d(z,a)=d(z,u_{k-1})+d(u_{k-1},a)=d(b,a). Since zCGz\in C_{G}, we have zVTGz\in V_{T_{G}}. Therefore, in this case, choose b=zb^{\prime}=z.
Subcase 1.2.2: Suppose that zV1(B)z\in V_{1}(B). Then zuk2z\neq u_{k-2} because uk2VBu_{k-2}\not\in V_{B}. So, by (P3)(P_{3}), zPG(a,b)z\not\in P_{G}(a,b), and uk1P¯G(z,a)u_{k-1}\in\bar{P}_{G}(z,a) for all shortest paths P¯G(z,a)\bar{P}_{G}(z,a) between zz and aa. This implies that d(z,a)=1+d(uk1,a)=d(b,a)1d(z,a)=1+d(u_{k-1},a)=d(b,a)-1. Since zCGV1(B)z\in C_{G}\cap V_{1}(B), by (P4)(P_{4}), zVBz\in V_{B^{\prime}} for some block BBB^{\prime}\neq B. By the construction of TGT_{G}, there exists bVBVTGb^{\prime}\in V_{B^{\prime}}\cap V_{T_{G}} such that bb^{\prime} is adjacent to zz. Again by (P3)(P_{3}), bP(z,a)=zuk1u1ab^{\prime}\not\in P(z,a)=zu_{k-1}\ldots u_{1}a and zP¯G(b,a)z\in\bar{P}_{G}(b^{\prime},a) for all shortest paths P¯G(b,a)\bar{P}_{G}(b^{\prime},a) between bb^{\prime} and aa. Therefore, d(b,a)=d(b,z)+d(z,a)=1+d(z,a)=d(b,a)d(b^{\prime},a)=d(b^{\prime},z)+d(z,a)=1+d(z,a)=d(b,a).
Case 2: Let ukCGu_{k}\in C_{G}. Since bV2(B)b\in V_{2}(B) with bVTGb\not\in V_{T_{G}}, we have |V2(B)|2\lvert V_{2}(B)\rvert\geq 2 because V2(B)VTGV_{2}(B)\cap V_{T_{G}}\neq\emptyset. Therefore, BK1,1B\neq K_{1,1}.
Subcase 2.1: If BB is a leaf block, then choose bV2(B)VTGb^{\prime}\in V_{2}(B)\cap V_{T_{G}}. We show that uk1VBu_{k-1}\not\in V_{B}. On the contrary, assume that uk1VBu_{k-1}\in V_{B}. Then, by (P2)(P_{2}), uk2VBu_{k-2}\not\in V_{B}. Also, by Remark 3.2, uk1CGu_{k-1}\in C_{G} which is not possible because BB is a leaf block with ukVBCGu_{k}\in V_{B}\cap C_{G}. Hence uk1VBu_{k-1}\not\in V_{B} and this yields buk1b^{\prime}\neq u_{k-1}. Therefore, by (P3P_{3}), we observe that bPG(a,b)b^{\prime}\not\in P_{G}(a,b), and by Remark 3.2, ukP¯G(b,a)u_{k}\in\bar{P}_{G}(b^{\prime},a) for all shortest paths P¯G(b,a)\bar{P}_{G}(b^{\prime},a) between bb^{\prime} and aa. Clearly, bukuk1u1ab^{\prime}u_{k}u_{k-1}\ldots u_{1}a is a path, and we have d(b,a)=d(b,uk)+d(uk,a)=d(b,a)d(b^{\prime},a)=d(b^{\prime},u_{k})+d(u_{k},a)=d(b,a).
Subcase 2.2: Suppose that BB is a bridge block. Let zVBCGz\in V_{B}\cap C_{G} with zukz\neq u_{k}.
Subcase 2.2.1: If zV1(B)z\in V_{1}(B) then take bV2(B)VTGb^{\prime}\in V_{2}(B)\cap V_{T_{G}}. Rest of the proof in this case is same as that of subcase 2.12.1.
Subcase 2.2.2: Let zV2(B)z\in V_{2}(B). If zuk1z\neq u_{k-1}, then uk1VBu_{k-1}\not\in V_{B}. By (P3)(P_{3}), zPG(a,b)z\not\in P_{G}(a,b), and ukP¯G(z,a)u_{k}\in\bar{P}_{G}(z,a) for all shortest paths P¯G(z,a)\bar{P}_{G}(z,a) between zz and aa. Therefore, zukuk1u1azu_{k}u_{k-1}\ldots u_{1}a is a shortest path and hence d(z,a)=d(b,a)d(z,a)=d(b,a). Suppose that z=uk1z=u_{k-1}. Since ukCGu_{k}\in C_{G}, there exists bVBVTGb^{\prime}\in V_{B^{\prime}}\cap V_{T_{G}} such that bb^{\prime} is adjacent to uku_{k} where BB^{\prime} is a block of GG different from BB. Then, by (P3)(P_{3}), bPG(uk,a)b^{\prime}\not\in P_{G}(u_{k},a), and ukP¯G(b,a)u_{k}\in\bar{P}_{G}(b^{\prime},a) for all shortest paths P¯G(b,a)\bar{P}_{G}(b^{\prime},a). By considering the path PG(b,a)=bukuk1u1aP_{G}(b^{\prime},a)=b^{\prime}u_{k}u_{k-1}\ldots u_{1}a, we obtain d(b,a)=d(b,a)d(b^{\prime},a)=d(b,a).
Hence, from the above cases, we have d(a,b)=d(a,b)d(a,b)=d(a,b^{\prime}) for some bVTGb^{\prime}\in V_{T_{G}}. The proof for the case k=1k=1 can be verified similarly. If aVTGa\in V_{T_{G}} then the result follows. If aVTGa\not\in V_{T_{G}} then repeat the above argument to the newly obtained path PG(b,a)P_{G}(b^{\prime},a) to find aVTGa^{\prime}\in V_{T_{G}} such that d(a,b)=d(a,b)d(a,b)=d(a^{\prime},b^{\prime}). This completes the proof. ∎

The next two lemmas are the consequences of the above result.

Lemma 3.3.

For each GG\in\mathscr{B}, diam(G)=diam(TG)\operatorname{diam}(G)=\operatorname{diam}(T_{G}).

Proof.

Let PG(a,b)P_{G}(a,b) be a diametrical path in GG. By Lemma 3.2, dG(a,b)=dG(a,b)d_{G}(a,b)=d_{G}(a^{\prime},b^{\prime}) for some a,bVTGa^{\prime},b^{\prime}\in V_{T_{G}}. Therefore, by Lemma 3.1, dG(a,b)=dTG(a,b)d_{G}(a^{\prime},b^{\prime})=d_{T_{G}}(a^{\prime},b^{\prime}). This implies that diam(G)=dG(a,b)=dTG(a,b)diam(TG)\operatorname{diam}(G)=d_{G}(a,b)=d_{T_{G}}(a^{\prime},b^{\prime})\leq\operatorname{diam}(T_{G}). To prove the other inequality, let x,yVTGx,y\in V_{T_{G}} such that dTG(x,y)=diam(TG)d_{T_{G}}(x,y)=\operatorname{diam}(T_{G}). Using Lemma 3.1, we get dTG(x,y)=dG(x,y)diam(G)d_{T_{G}}(x,y)=d_{G}(x,y)\leq\operatorname{diam}(G). ∎

Let HH be a connected graph and let uVHu\in V_{H}. We denote the eccentricity of a vertex uu with respect to HH by eH(u)e_{H}(u), and the subscript is omitted if it is clear from the context.

Lemma 3.4.

If GG\in\mathscr{B}, then eG(a)=eTG(a)e_{G}(a)=e_{T_{G}}(a) for all aVTGa\in V_{T_{G}}.

Proof.

Let aVTGa\in V_{T_{G}}. By Lemma 3.1, {dTG(a,x):xVTG}{dG(a,x):xVG}\{d_{T_{G}}(a,x)\mathrel{\mathop{\mathchar 58\relax}}x\in V_{T_{G}}\}\subseteq\{d_{G}(a,x)\mathrel{\mathop{\mathchar 58\relax}}x\in V_{G}\}. This implies eTG(a)eG(a)e_{T_{G}}(a)\leq e_{G}(a). Let bVGb\in V_{G} be such that eG(a)=dG(a,b)e_{G}(a)=d_{G}(a,b). If bVTGb\in V_{T_{G}} then again by Lemma 3.1, dG(a,b)=dTG(a,b)eTG(a)d_{G}(a,b)=d_{T_{G}}(a,b)\leq e_{T_{G}}(a). Suppose that bVTGb\not\in V_{T_{G}}. Then from the proof of Lemma 3.2, we have dG(a,b)=dG(a,b)d_{G}(a,b)=d_{G}(a,b^{\prime}) for some bVTGb^{\prime}\in V_{T_{G}}. So, eTG(a)eG(a)e_{T_{G}}(a)\geq e_{G}(a) follows by Lemma 3.1. ∎

Since most of the results of this paper hold for graphs in \mathscr{B} with diameters at least four (for instance, see Example 4.2), we deal only with those graphs hereafter. In the following lemma, the eccentricity of a non-cut-vertex (when it exists) of GG\in\mathscr{B} is presented in terms of the eccentricities of the cut-vertices of GG. In Section 4, we see the usefulness of this lemma in proving the spectral symmetry of E(G)E(G).

Lemma 3.5.

Let GG\in\mathscr{B} be such that diam(G)4\operatorname{diam}(G)\geq 4. Let BB be a block of GG and let uVBu\in V_{B} be a non-cut-vertex of GG.

  • (i)

    Suppose that BB is a bridge block of GG and v1v_{1} and v2v_{2} are cut-vertices of GG belong to BB. If e(v1)e(v2)e(v_{1})\leq e(v_{2}), then

    e(u)={e(v2)if v1,v2,uV1(B),e(v2)1if v1,v2V1(B)anduV2(B),e(v2)+1if v1,uV1(B)andv2V2(B),e(v1)+1if v1V1(B)andv2,uV2(B).\displaystyle e(u)=\begin{cases}e(v_{2})&\text{if }v_{1},v_{2},u\in V_{1}(B),\\ e(v_{2})-1&\text{if }v_{1},v_{2}\in V_{1}(B)~{}\text{and}~{}u\in V_{2}(B),\\ e(v_{2})+1&\text{if }v_{1},u\in V_{1}(B)~{}\text{and}~{}v_{2}\in V_{2}(B),\\ e(v_{1})+1&\text{if }v_{1}\in V_{1}(B)~{}\text{and}~{}v_{2},u\in V_{2}(B).\end{cases}
  • (ii)

    If BB is a leaf block and vv is a cutvertex of GG belongs to BB, then

    e(u)={e(v)+2if u,vV1(B)oru,vV2(B),e(v)+1otherwise.\displaystyle e(u)=\begin{cases}e(v)+2&\text{if }u,v\in V_{1}(B)~{}\text{or}~{}u,v\in V_{2}(B),\\ e(v)+1&\text{otherwise}.\end{cases}
Proof.

(i)(i) Since BB is a bridge block with v1,v2VBCGv_{1},v_{2}\in V_{B}\cap C_{G}, there exists a vertex xVGVBx\in V_{G}\setminus V_{B} such that e(v2)=d(v2,x)e(v_{2})=d(v_{2},x). Let P(v2,x)P(v_{2},x) be a shortest path between v2v_{2} and xx in GG. We claim that v1P(v2,x)v_{1}\in P(v_{2},x). On the contrary, assume that v1P(v2,x)v_{1}\not\in P(v_{2},x). Then, by (P3)(P_{3}) and Remark 3.2, v2P¯(v1,x)v_{2}\in\bar{P}(v_{1},x) for all paths P¯(v1,x)\bar{P}(v_{1},x). Therefore, e(v1)d(v1,x)=d(v1,v2)+d(v2,x)1+e(v2)>e(v2)e(v_{1})\geq d(v_{1},x)=d(v_{1},v_{2})+d(v_{2},x)\geq 1+e(v_{2})>e(v_{2}), which contradicts the assumption e(v1)e(v2)e(v_{1})\leq e(v_{2}). Hence the claim v1P(v2,x)v_{1}\in P(v_{2},x) follows. This implies that d(x,v2)=d(x,v1)+d(v1,v2)d(x,v_{2})=d(x,v_{1})+d(v_{1},v_{2}). Let uVBCGu\in V_{B}\setminus C_{G}. Note that v1P¯(x,u)v_{1}\in\bar{P}(x,u) for all shortest paths P¯(x,u)\bar{P}(x,u) by (P3)(P_{3}) and Remark 3.2. Therefore, d(x,u)=d(x,v1)+d(v1,u)=[d(x,v2)d(v1,v2)]+d(v1,u)d(x,u)=d(x,v_{1})+d(v_{1},u)=[d(x,v_{2})-d(v_{1},v_{2})]+d(v_{1},u), and hence

e(u)d(x,u)=d(x,v2)+[d(v1,u)d(v1,v2)]=e(v2)+[d(v1,u)d(v1,v2)].e(u)\geq d(x,u)=d(x,v_{2})+[d(v_{1},u)-d(v_{1},v_{2})]=e(v_{2})+[d(v_{1},u)-d(v_{1},v_{2})]. (2)

Since v1,v2VBCGv_{1},v_{2}\in V_{B}\cap C_{G}, there always exists yVGVBy\in V_{G}\setminus V_{B} such that e(u)=d(u,y)e(u)=d(u,y). Let P(u,y)P(u,y) be a shortest path between uu and yy. Suppose that v1P(u,y)v_{1}\in P(u,y) and v2P(u,y)v_{2}\not\in P(u,y). Then, by (P3P_{3}) and Remark 3.2, any shortest path P¯(v2,y)\bar{P}(v_{2},y) contains v1v_{1}. This implies that d(v2,y)=d(v2,v1)+d(v1,y).d(v_{2},y)=d(v_{2},v_{1})+d(v_{1},y). Therefore, we have

e(u)=d(u,y)\displaystyle e(u)=d(u,y) =d(u,v1)+d(v1,y)\displaystyle=d(u,v_{1})+d(v_{1},y) (3)
=d(u,v1)+[d(v2,y)d(v1,v2)]\displaystyle=d(u,v_{1})+[d(v_{2},y)-d(v_{1},v_{2})]
e(v2)+[d(u,v1)d(v1,v2)].\displaystyle\leq e(v_{2})+[d(u,v_{1})-d(v_{1},v_{2})]. (4)

If v1P(u,y)v_{1}\not\in P(u,y), then by Remark 3.2, v2P(u,y)v_{2}\in P(u,y). Also, by (P3)(P_{3}), v2P¯(v1,y)v_{2}\in\bar{P}(v_{1},y) for all shortest paths P¯(v1,y)\bar{P}(v_{1},y). Then d(y,v1)=d(y,v2)+d(v2,v1)d(y,v_{1})=d(y,v_{2})+d(v_{2},v_{1}). Hence

e(u)=d(u,y)=d(u,v2)+d(v2,y)\displaystyle e(u)=d(u,y)=d(u,v_{2})+d(v_{2},y) =d(u,v2)+[d(y,v1)d(v1,v2)]\displaystyle=d(u,v_{2})+[d(y,v_{1})-d(v_{1},v_{2})] (5)
e(v1)+[d(u,v2)d(v1,v2)]\displaystyle\leq e(v_{1})+[d(u,v_{2})-d(v_{1},v_{2})] (6)
e(v2)+[d(u,v2)d(v1,v2)].\displaystyle\leq e(v_{2})+[d(u,v_{2})-d(v_{1},v_{2})]. (7)

Case 1: Let v1,v2V1(B)v_{1},v_{2}\in V_{1}(B). Since P(u,y)P(u,y) is a shortest path between uu and yy, by Remark 3.2, we have |VP(u,y){v1,v2}|=1\lvert V_{P(u,y)}\cap\{v_{1},v_{2}\}\rvert=1 where VP(u,y)V_{P(u,y)} is the set of all vertices in the path P(u,y)P(u,y). Now, the result follows from (2), (4) and (7), case-by-case.
Case 2: Assume that v1V1(B)v_{1}\in V_{1}(B) and v2V2(B)v_{2}\in V_{2}(B).
Subcase 2.1: Let uV1(B)u\in V_{1}(B). From (2), e(u)e(v2)+1e(u)\geq e(v_{2})+1. To prove the other inequality, we first show that v1P(y,u)v_{1}\in P(y,u). On the contrary, assume that v1P(y,u)v_{1}\not\in P(y,u). Then by Remark 3.2, v2P(y,u)v_{2}\in P(y,u) and so, by (P3)(P_{3}), v2P¯(y,v1)v_{2}\in\bar{P}(y,v_{1}) for all shortest paths P¯(y,v1)\bar{P}(y,v_{1}). Therefore, e(u)=d(y,u)=d(y,v2)+d(v2,u)=d(y,v2)+d(v2,v1)=d(y,v1)e(v1)e(u)=d(y,u)=d(y,v_{2})+d(v_{2},u)=d(y,v_{2})+d(v_{2},v_{1})=d(y,v_{1})\leq e(v_{1}). That is, e(v1)e(u)e(v2)+1e(v_{1})\geq e(u)\geq e(v_{2})+1 where the last inequality follows from (2). This contradicts the assumption e(v1)e(v2)e(v_{1})\leq e(v_{2}). Hence v1P(y,u)v_{1}\in P(y,u). If v1P(y,u)v_{1}\in P(y,u) and v2P(y,u)v_{2}\not\in P(y,u) then by (4), e(u)e(v2)+1e(u)\leq e(v_{2})+1. Suppose that v1,v2P(y,u)v_{1},v_{2}\in P(y,u). Then e(u)=d(y,u)=d(y,v2)+d(v2,u)e(v2)+1e(u)=d(y,u)=d(y,v_{2})+d(v_{2},u)\leq e(v_{2})+1. This completes the proof in this case.
Subcase 2.2: Suppose that uV2(B)u\in V_{2}(B). Let e(v1)=d(z,v1)e(v_{1})=d(z,v_{1}) for some zVGVBz\in V_{G}\setminus V_{B} and let P(z,v1)P(z,v_{1}) be a shortest path between zz and v1v_{1} in GG. If v2P(z,v1)v_{2}\in P(z,v_{1}) then by (P3)(P_{3}), v2P¯(z,u)v_{2}\in\bar{P}(z,u) for all shortest paths P¯(z,u)\bar{P}(z,u). We have d(z,u)=d(z,v2)+d(v2,u)d(z,u)=d(z,v_{2})+d(v_{2},u) and d(z,v1)=d(z,v2)+1d(z,v_{1})=d(z,v_{2})+1. This implies that e(u)d(z,u)=d(z,v1)+1=e(v1)+1e(u)\geq d(z,u)=d(z,v_{1})+1=e(v_{1})+1. Suppose that v2P(z,v1)v_{2}\not\in P(z,v_{1}). Then again by (P3)(P_{3}), v2P¯(z,u)v_{2}\not\in\bar{P}(z,u), and by Remark 3.2, v1P¯(z,u)v_{1}\in\bar{P}(z,u) for all paths P¯(z,u)\bar{P}(z,u). So, d(z,u)=d(z,v1)+d(v1,u)=e(v1)+1d(z,u)=d(z,v_{1})+d(v_{1},u)=e(v_{1})+1. Therefore, e(u)e(v1)+1e(u)\geq e(v_{1})+1. If |VP(u,y){v1,v2}|=1\lvert V_{P(u,y)}\cap\{v_{1},v_{2}\}\rvert=1 then the other inequality, e(u)e(v1)+1e(u)\leq e(v_{1})+1 follows from (3) and (6). If v1,v2P(y,u)v_{1},v_{2}\in P(y,u) then e(u)=d(y,u)=d(y,v1)+d(v1,u)e(v1)+1e(u)=d(y,u)=d(y,v_{1})+d(v_{1},u)\leq e(v_{1})+1.

(ii)(ii) We claim that there exists xVGVBx\in V_{G}\setminus V_{B} such that e(v)=d(x,v)e(v)=d(x,v). If e(v)3e(v)\geq 3 then the claim follows. Consider the case e(v)=2e(v)=2. Suppose that the claim does not hold. Then d(x,v)<e(v)d(x,v)<e(v) for all xVGVBx\in V_{G}\setminus V_{B}. That is, vv is adjacent to all xVGVBx\in V_{G}\setminus V_{B}. This leads to diam(G)3\operatorname{diam}(G)\leq 3, a contradiction to the hypothesis. Hence the claim. Let u,vV1(B)u,v\in V_{1}(B). By Remark 3.2, any path P(u,x)P(u,x) passes through vv. So, e(u)d(x,u)=d(x,v)+d(v,u)=e(v)+2e(u)\geq d(x,u)=d(x,v)+d(v,u)=e(v)+2. This implies e(u)4e(u)\geq 4 and hence e(u)=d(u,y)e(u)=d(u,y) for some yVGVBy\in V_{G}\setminus V_{B}. Note that e(u)=d(y,u)=d(y,v)+2e(v)+2e(u)=d(y,u)=d(y,v)+2\leq e(v)+2. Hence e(u)=e(v)+2e(u)=e(v)+2. Similarly, the proof follows for the remaining cases. ∎

Let diam(G)4\operatorname{diam}(G)\geq 4 and let BB be a block of GG such that uVBCGu\in V_{B}\setminus C_{G}. Suppose that BB is a leaf block and vVBCGv\in V_{B}\cap C_{G}. From Lemma 3.5, we have e(u)>e(v)e(u)>e(v). If BB is a bridge block such that the cut-vertices v1v_{1} and v2v_{2} of BB are not in the same partite set then also we have e(u)>e(vi)e(u)>e(v_{i}) for some ii. Suppose that the cut-vertices v1v_{1} and v2v_{2} of the bridge block BB are in the same partite set, say V1(B)V_{1}(B), then e(u)>e(u0)e(u)>e(u_{0}) when uV1(B)u\in V_{1}(B) and u0V2(B)u_{0}\in V_{2}(B). Hence uC(G)u\not\in C(G) in the above cases. Therefore, a non-cut-vertex uu of the block BB can belong to C(G)C(G) only when BB is a bridge block whose cut-vertices are in one partite set of BB and uu lies in the other partite set of BB. Precisely, the result is given in the following remark.

Remark 3.3.

Let GG\in\mathscr{B} with diam(G)4\operatorname{diam}(G)\geq 4 and let xV1(B)x\in V_{1}(B) for some block BB of GG. If xC(G)CGx\in C(G)\setminus C_{G}, then BB is a bridge block with |V2(B)CG|=2\lvert V_{2}(B)\cap C_{G}\rvert=2.

The next lemma is a main tool in finding the central vertices of the graphs in \mathscr{B}.

Lemma 3.6.

Let GG\in\mathscr{B} and let diam(G)4\operatorname{diam}(G)\geq 4. Then C(TG)C(G)C(T_{G})\subseteq C(G).

Proof.

Let zC(TG)z\in C(T_{G}). Then by Lemma 3.4, eG(z)=eTG(z)e_{G}(z)=e_{T_{G}}(z). Let uVGu\in V_{G}. Suppose that uVTGu\in V_{T_{G}}. Then eG(u)=eTG(u)eTG(z)=eG(z)e_{G}(u)=e_{T_{G}}(u)\geq e_{T_{G}}(z)=e_{G}(z). If uVTGu\not\in V_{T_{G}}, then uCGu\not\in C_{G} because CGVTGC_{G}\subseteq V_{T_{G}}. Assume that uVBu\in V_{B} for some block BB. If BB is a leaf block such that vVBCGv\in V_{B}\cap C_{G}, then by Lemma 3.5, we have eG(u)eG(v)=eTG(v)eTG(z).e_{G}(u)\geq e_{G}(v)=e_{T_{G}}(v)\geq e_{T_{G}}(z). Suppose that BB is a bridge block such that v1,v2VBCGv_{1},v_{2}\in V_{B}\cap C_{G}. Consider the case v1,v2V1(B)v_{1},v_{2}\in V_{1}(B) and uV2(B)u\in V_{2}(B). Then, by Lemma 3.5, eG(x)=eG(u)e_{G}(x)=e_{G}(u) for all xV2(B)x\in V_{2}(B). In particular, by the construction of TGT_{G}, there exists u0V2(B)VTGu_{0}\in V_{2}(B)\cap V_{T_{G}} such that eG(u0)=eG(u)e_{G}(u_{0})=e_{G}(u). This implies eG(u)=eG(u0)=eTG(u0)eTG(z).e_{G}(u)=e_{G}(u_{0})=e_{T_{G}}(u_{0})\geq e_{T_{G}}(z). For the remaining cases of BB, it directly follows from Lemma 3.5 that eG(u)eTG(z)=eG(z)e_{G}(u)\geq e_{T_{G}}(z)=e_{G}(z). Hence zC(G)z\in C(G), which completes the proof. ∎

Recall the following result to study some properties of the central vertices of TGT_{G}.

Lemma 3.7 ([1], see [10, 7]).

Let TT be a tree on n3n\geq 3 vertices and diam(T)=m\operatorname{diam}(T)=m. Then the center C(T)C(T) of TT has either a single vertex z1z_{1} or two adjacent vertices z1z_{1} and z2z_{2} such that e(zi)=m2e(z_{i})=\lceil{\frac{m}{2}}\rceil for all i=1,2i=1,2, where m2=m2\lceil{\frac{m}{2}}\rceil=\frac{m}{2} if mm is an even integer, and m2=m+12\lceil{\frac{m}{2}}\rceil=\frac{m+1}{2} if mm is an odd integer. Moreover, |C(T)|=1\lvert C(T)\rvert=1 if diam(T)\operatorname{diam}(T) is even, and |C(T)|=2\lvert C(T)\rvert=2 if diam(T)\operatorname{diam}(T) is odd.

Let GG\in\mathscr{B} be such that diam(G)\operatorname{diam}(G) is odd and greater than three. Then by Lemmas 3.3 and 3.7, |C(TG)|=2\lvert C(T_{G})\rvert=2. Assume that C(TG)={z1,z2}C(T_{G})=\{z_{1},z_{2}\}. Again by Lemma 3.7, the vertices z1z_{1} and z2z_{2} are adjacent, and by Lemma 3.6, z1,z2C(G)z_{1},z_{2}\in C(G). Suppose that both z1z_{1} and z2z_{2} are not in CGC_{G}. Then by Remark 3.3, the block BB of GG containing z1z_{1} and z2z_{2} should be a bridge block such that the cut-vertices of GG in BB are in one partite set (say V1(B)V_{1}(B)), and z1z_{1} and z2z_{2} are in the other partite set (V2(B)V_{2}(B)), which is not possible because z1z_{1} and z2z_{2} are adjacent. Therefore, we arrive at the following remark.

Remark 3.4.

Let GG\in\mathscr{B} with diam(G)4\operatorname{diam}(G)\geq 4. Then |C(TG)CG|1\lvert C(T_{G})\cap C_{G}\rvert\geq 1 whenever |C(TG)|=2\lvert C(T_{G})\rvert=2.

The following result presents the collection of all central vertices of graphs in \mathscr{B}. This will be used frequently in the proofs of the next section.

Theorem 3.1.

Let GG\in\mathscr{B} be such that diam(G)4\operatorname{diam}(G)\geq 4 and let TGT_{G} be its associated tree.

  • (i)

    Suppose that diam(G)\operatorname{diam}(G) is even and C(TG)={z}C(T_{G})=\{z\} where zV1(B)z\in V_{1}(B) for some block BB. Then

    C(G)={{z}if zCG,V1(B)if zCG.\displaystyle C(G)=\begin{cases}\{z\}&\text{if }z\in C_{G},\\ V_{1}(B)&\text{if }z\not\in C_{G}.\end{cases}
  • (ii)

    Let diam(G)\operatorname{diam}(G) be odd and C(TG)={z1,z2}C(T_{G})=\{z_{1},z_{2}\} where z1V1(B)z_{1}\in V_{1}(B) and z2V2(B)z_{2}\in V_{2}(B) for some block BB. Then

    C(G)={{z1,z2}if z1,z2CG,{z1}V2(B)if z1CGandz2CG,{z2}V1(B)if z1CGandz2CG.\displaystyle C(G)=\begin{cases}\{z_{1},z_{2}\}&\text{if }z_{1},z_{2}\in C_{G},\\ \{z_{1}\}\cup V_{2}(B)&\text{if }z_{1}\in C_{G}~{}\text{and}~{}z_{2}\not\in C_{G},\\ \{z_{2}\}\cup V_{1}(B)&\text{if }z_{1}\not\in C_{G}~{}\text{and}~{}z_{2}\in C_{G}.\end{cases}
Proof.

(i)(i) Case 1: Let zCGz\in C_{G}. By Lemma 3.6, it is sufficient to show that C(G){z}C(G)\subseteq\{z\}. Let xC(G)x\in C(G). We first claim that xCGx\in C_{G}. On the contrary, assume that xCGx\not\in C_{G}. Then by Remark 3.3 and Lemma 3.5, xBx\in B^{\prime} for some bridge block BB^{\prime} such that the cut-vertices of GG in BB^{\prime} belong to V1(B)V_{1}(B^{\prime}) and xV2(B)x\in V_{2}(B^{\prime}) with eG(x)=eG(y)e_{G}(x)=e_{G}(y) for all yV2(B)y\in V_{2}(B). In particular, there exists y0V2(B)VTGy_{0}\in V_{2}(B^{\prime})\cap V_{T_{G}} such that eG(x)=eG(y0)e_{G}(x)=e_{G}(y_{0}) where the existence of y0y_{0} is guaranteed by the construction of TGT_{G}. Note that y0CGy_{0}\not\in C_{G} as |CGV1(B)|=2\lvert C_{G}\cap V_{1}(B^{\prime})\rvert=2 and y0V2(B)y_{0}\in V_{2}(B^{\prime}). By Lemma 3.4, eG(y0)=eTG(y0)e_{G}(y_{0})=e_{T_{G}}(y_{0}), and we have eTG(y0)=eG(x)eG(z)=eTG(z)e_{T_{G}}(y_{0})=e_{G}(x)\leq e_{G}(z)=e_{T_{G}}(z). Since C(TG)={z}C(T_{G})=\{z\}, eTG(y0)eTG(z)e_{T_{G}}(y_{0})\geq e_{T_{G}}(z) which implies y0C(TG)y_{0}\in C(T_{G}). That is, y0=zy_{0}=z, which is not possible because zCGz\in C_{G} and y0CGy_{0}\not\in C_{G}. Hence the claim xCGx\in C_{G} follows. Therefore, xVTGx\in V_{T_{G}} and eG(x)=eTG(x)e_{G}(x)=e_{T_{G}}(x). We have eTG(x)eTG(z)=eG(z)eG(x)e_{T_{G}}(x)\geq e_{T_{G}}(z)=e_{G}(z)\geq e_{G}(x), which yields eTG(x)=eTG(z)e_{T_{G}}(x)=e_{T_{G}}(z). That is, xC(TG)={z}x\in C(T_{G})=\{z\} and hence x=zx=z. Therefore, C(G)={z}C(G)=\{z\}.
Case 2: Assume that zCGz\not\in C_{G}. Since zV1(B)z\in V_{1}(B), it follows from Remark 3.3 that BB must be a bridge block whose cut-vertices are in V2(B)V_{2}(B). Also, by Lemma 3.5, we get eG(z)=eG(u)e_{G}(z)=e_{G}(u) for all uV1(B)u\in V_{1}(B). Using the fact that zC(TG)C(G)z\in C(T_{G})\subseteq C(G), we have V1(B)C(G)V_{1}(B)\subseteq C(G). To prove C(G)V1(B)C(G)\subseteq V_{1}(B), let xC(G)x\in C(G). We now show that xCGx\not\in C_{G}. Suppose that xCGx\in C_{G}. Then xVTGx\in V_{T_{G}} and by Lemma 3.4, we have eTG(x)=eG(x)eG(z)=eTG(z)eTG(x)e_{T_{G}}(x)=e_{G}(x)\leq e_{G}(z)=e_{T_{G}}(z)\leq e_{T_{G}}(x). This implies eTG(z)=eTG(x)e_{T_{G}}(z)=e_{T_{G}}(x) and hence xC(TG)={z}x\in C(T_{G})=\{z\}. That is, x=zx=z, which is absurd as xCGx\in C_{G} and zCGz\not\in C_{G}. Therefore, xCGx\not\in C_{G}. Since xC(G)x\in C(G), by Remark 3.3, xV2(B)x\in V_{2}(B^{\prime}) for some bridge block BB^{\prime} with |V1(B)CG|=2\lvert V_{1}(B^{\prime})\cap C_{G}\rvert=2. As in the previous case, we have eG(x)=eG(y0)e_{G}(x)=e_{G}(y_{0}) for some y0V2(B)VTGy_{0}\in V_{2}(B^{\prime})\cap V_{T_{G}} and y0=zy_{0}=z. This yields y0V1(B)y_{0}\in V_{1}(B). Since y0VBVBy_{0}\in V_{B}\cap V_{B^{\prime}} and y0CGy_{0}\not\in C_{G}, it follows by (P3)(P_{3}) that B=BB=B^{\prime} and V2(B)=V1(B)V_{2}(B^{\prime})=V_{1}(B). This implies xV1(B)x\in V_{1}(B), which completes the proof in this case.

(ii)(ii) It is similar to the proof of item (i)(i). ∎

Remark 3.5.

It is clear from Theorem 3.1 that all the central vertices of a graph GG\in\mathscr{B} with diam(G)4\operatorname{diam}(G)\geq 4 lie completely in one specific block BB of GG. In particular, all the central vertices of GG belong to exactly one partite set of BB if and only if diam(G)\operatorname{diam}(G) is even.

4 Inertia, spectral symmetry and irreducibility

The inertias of the distance matrices of trees, and the eccentricity matrices of lollipop graphs, trees, clique trees and coalescence of certain graphs are computed in the literature, see [2, 7, 10, 8, 11]. Along these lines, here we find the inertias of the eccentricity matrices of graphs in \mathscr{B}.

It is shown that the eigenvalues of the adjacency matrix A(G)A(G) of a graph GG are symmetric about the origin if and only if GG is bipartite [2]. In [10], it is proved that the eigenvalues of the eccentricity matrix E(T)E(T) of a tree TT are symmetric about the origin if and only if diam(T)\operatorname{diam}(T) is odd. A similar equivalence is established for a subclass of block graphs (clique trees) in [7]. Motivated by these, in this section, we prove an analogous result for the class \mathscr{B}. In the last part of this section, we prove the irreducibility of the eccentricity matrices of graphs in \mathscr{B}.

4.1 Inertia and spectral symmetry of eccentricity matrices of graphs in \mathscr{B} with odd diameters

In this subsection, we consider graphs in \mathscr{B} with odd diameters and find the inertias of the eccentricity matrices of these graphs. Also, we show the spectral symmetry (with respect to the origin) of the above considered matrices. We begin with an example.

Example 4.1.

Consider the bi-block graph GG\in\mathscr{B} and the associated tree TGT_{G} given in Example 3.1. For the purpose of illustrating Theorem 4.1, let us relabel the vertices v7v_{7}, v9v_{9} and v10v_{10} by z1z_{1}, zz^{\prime} and z2z_{2}, respectively. Note that diam(G)=diam(TG)=7\operatorname{diam}(G)=\operatorname{diam}(T_{G})=7, and C(TG)={z1,z2}C(T_{G})=\{z_{1},z_{2}\} where z1z_{1} and z2z_{2} are adjacent with z1CGz_{1}\in C_{G} and z2CGz_{2}\not\in C_{G}. Let BB be the block of GG containing the edge z1z2z_{1}z_{2} such that z1V1(B)z_{1}\in V_{1}(B) and z2V2(B)z_{2}\in V_{2}(B). It is clear that BB is a bridge block with zCGV1(B)z^{\prime}\in C_{G}\cap V_{1}(B) and zz1z^{\prime}\neq z_{1}. Let GG^{\prime} be the subgraph of GG obtained by deleting all the edges of the block BB in GG. Let C1C_{1} and C2C_{2} be the components of GG^{\prime} containing the vertices z1z_{1} and zz^{\prime}, respectively. Clearly, the sets

U1\displaystyle U_{1} ={xVC1:d(x,z1)=3}={v1,v2,v6},\displaystyle=\{x\in V_{C_{1}}\mathrel{\mathop{\mathchar 58\relax}}d(x,z_{1})=3\}=\{v_{1},v_{2},v_{6}\},
U2\displaystyle U_{2} ={yVC1:0d(y,z1)<3}={v3,v4,v5,z1},\displaystyle=\{y\in V_{C_{1}}\mathrel{\mathop{\mathchar 58\relax}}0\leq d(y,z_{1})<3\}=\{v_{3},v_{4},v_{5},z_{1}\},
U3\displaystyle U_{3} ={xVC2:d(x,z)=2}={v15,v16},\displaystyle=\{x\in V_{C_{2}}\mathrel{\mathop{\mathchar 58\relax}}d(x,z^{\prime})=2\}=\{v_{15},v_{16}\},
U4\displaystyle U_{4} ={yVC2:0d(y,z)<2}={z,v13,v14},\displaystyle=\{y\in V_{C_{2}}\mathrel{\mathop{\mathchar 58\relax}}0\leq d(y,z^{\prime})<2\}=\{z^{\prime},v_{13},v_{14}\},
U5\displaystyle U_{5} =V1(B){z1,z}={v8},and\displaystyle=V_{1}(B)\setminus\{z_{1},z^{\prime}\}=\{v_{8}\},\quad\text{and}
U6\displaystyle U_{6} =V2(B)={z2,v11,v12} partitions VG.\displaystyle=V_{2}(B)=\{z_{2},v_{11},v_{12}\}~{}\text{ partitions $V_{G}$.}~{}

Then the eccentricity matrix of GG can be written in the following form:

E(G)={blockarray}ccccccc&U1U2U3U4U5U6{block}c(cccccc)U1OO7JP5J4JU2OOQOOOU37JQOOOOU4POOOOOU55JOOOOOU64JOOOOO,E(G)=\blockarray{ccccccc}&U_{1}U_{2}U_{3}U_{4}U_{5}U_{6}\\ \block{c(cccccc)}U_{1}OO7JP5J4J\\ U_{2}OOQOOO\\ U_{3}7J^{\prime}Q^{\prime}OOOO\\ U_{4}P^{\prime}OOOOO\\ U_{5}5J^{\prime}OOOOO\\ U_{6}4J^{\prime}OOOOO\\ ,

where

P={blockarray}cccc&v13v14z{block}c(ccc)v1665v2665v6665,andQ={blockarray}ccc&v15v16{block}c(cc)v366v466v555z144.P=\blockarray{cccc}&v_{13}v_{14}z^{\prime}\\ \block{c(ccc)}v_{1}665\\ v_{2}665\\ v_{6}665\\ ,\quad\text{and}\quad Q=\blockarray{ccc}&v_{15}v_{16}\\ \block{c(cc)}v_{3}66\\ v_{4}66\\ v_{5}55\\ z_{1}44\\ .

Using SAGEMATH, it is computed that the eigenvalues of E(G)E(G) are 0, 30.037530.0375, 11.302511.3025, 11.3025-11.3025 and 30.037530.0375 with respective multiplicities 12,1,1,112,1,1,1 and 11. Thus, the spectrum of E(G)E(G) is symmetric with respect to the origin, and In(E(G))=(2,2,12)\operatorname{In}(E(G))=(2,2,12).

Theorem 4.1.

Let GG\in\mathscr{B} be such that diam(G)=2m+1\operatorname{diam}(G)=2m+1, m2m\geq 2. Then the following statements hold:

  • (i)

    In(E(G))=(2,2,n4)\operatorname{In}(E(G))=(2,2,n-4), where nn is the number of vertices of GG.

  • (ii)

    The spectrum of E(G)E(G) is symmetric with respect to the origin.

Proof.

Consider the tree TGT_{G} associated with GG. By Lemmas 3.3 and 3.7, diam(TG)=2m+1\operatorname{diam}(T_{G})=2m+1 and |C(TG)|=2\lvert C(T_{G})\rvert=2. Let C(TG)={z1,z2}C(T_{G})=\{z_{1},z_{2}\}. Again by Lemma 3.7, z1z_{1} and z2z_{2} are adjacent, and eTG(zi)=m+1e_{T_{G}}(z_{i})=m+1 for i=1,2i=1,2. Also, by Lemmas 3.4 and 3.6, z1,z2C(G)z_{1},z_{2}\in C(G) and eG(zi)=m+1e_{G}(z_{i})=m+1 for i=1,2i=1,2. Let z1V1(B)z_{1}\in V_{1}(B) and z2V2(B)z_{2}\in V_{2}(B) for some block BB of GG. By Remark 3.4, ziCGz_{i}\in C_{G} for at least one ii. Without loss of generality, assume that z1CGz_{1}\in C_{G}.

(i)(i) Case 1: Suppose that z2CGz_{2}\not\in C_{G}. Since z2(C(G)V2(B))CGz_{2}\in\left(C(G)\cap V_{2}(B)\right)\setminus C_{G}, by Remark 3.3, BB is a bridge block whose cut-vertices are in V1(B)V_{1}(B). Let zz1z^{\prime}\neq z_{1} be the another cut-vertex of GG in BB. By (P1)(P_{1}), we have |V2(B)|2\lvert V_{2}(B)\rvert\geq 2 as {z1,z}V1(B)\{z_{1},z^{\prime}\}\subseteq V_{1}(B). Now obtain a subgraph GG^{\prime} from GG by deleting all the edges of BB. Clearly, VG=VGV_{G}=V_{G^{\prime}}. By Remark 3.2 and (P3)(P_{3}), GG^{\prime} contains at least four components as |Vi(B)|2\lvert V_{i}(B)\rvert\geq 2 for all i=1,2i=1,2, and the vertices z1z_{1} and zz^{\prime} lie in different components. Let C1C_{1} and C2C_{2} be the non-trivial components of GG^{\prime} containing the vertices z1z_{1} and zz^{\prime}, respectively. Let xVC1x\in V_{C_{1}} and yVC2y\in V_{C_{2}}. Then by (P3)(P_{3}),

z1,zP¯(x,y)for all paths P¯(x,y) between x and y in G.z_{1},z^{\prime}\in\bar{P}(x,y)~{}\text{for all paths $\bar{P}(x,y)$ between $x$ and $y$ in $G$}. (8)

We prove the following inequalities to find a partition for VGV_{G}:

d(a,z1)mfor all aVC1andd(b,z)m1for all bVC2.d(a,z_{1})\leq m~{}\text{for all $a\in V_{C_{1}}$}\quad\text{and}\quad d(b,z^{\prime})\leq m-1~{}\text{for all $b\in V_{C_{2}}$}. (9)

Since e(z1)=m+1e(z_{1})=m+1, we have d(a,z1)m+1d(a,z_{1})\leq m+1. Suppose d(a,z1)=m+1d(a,z_{1})=m+1 then d(a,z2)=m+2d(a,z_{2})=m+2 which is not possible because e(z2)=m+1e(z_{2})=m+1. Hence d(a,z1)md(a,z_{1})\leq m which proves the first inequality in (9)(\ref{**}). Suppose that the second inequality in (9)(\ref{**}) does not hold. That is, there exists bVC2b\in V_{C_{2}} such that d(b,z)>m1d(b,z^{\prime})>m-1. Since z1VC1z_{1}\in V_{C_{1}} and bVC2b\in V_{C_{2}}, using (8), we write d(b,z1)=d(b,z)+d(z,z1)>(m1)+d(z,z1)=m+1d(b,z_{1})=d(b,z^{\prime})+d(z^{\prime},z_{1})>(m-1)+d(z^{\prime},z_{1})=m+1. This implies e(z1)>m+1e(z_{1})>m+1, which is a contradiction. So, the second inequality in (9)(\ref{**}) holds.

We now show that the non-trivial components of GG^{\prime} are precisely C1C_{1} and C2C_{2}. Suppose that xVGx\in V_{G} and xVC1VBx\not\in V_{C_{1}}\cup V_{B}. Let P(x,z1)P(x,z_{1}) be a shortest path between xx and z1z_{1} in GG. If zP(x,z1)z^{\prime}\not\in P(x,z_{1}) then by Remark 3.2, VP(x,z1)VB={z1}V_{P(x,z_{1})}\cup V_{B}=\{z_{1}\}. Therefore, P(x,z1)P(x,z_{1}) is a path in GG^{\prime}. Since C1C_{1} is a component of GG^{\prime} containing z1z_{1}, we have xVC1x\in V_{C_{1}} which is a contradiction. Hence zP(x,z1)z^{\prime}\in P(x,z_{1}). Note that the subpath P(x,z)P(x,z^{\prime}) obtained from the shortest path P(x,z1)P(x,z_{1}) does not contain any vertex other than zz^{\prime} from the block BB. This implies that P(x,z)P(x,z^{\prime}) is a path in GG^{\prime}. Since zVC2z^{\prime}\in V_{C_{2}}, we have xVC2x\in V_{C_{2}}. Thus, the components of GG^{\prime} other than C1C_{1} and C2C_{2} are simply complete graphs of order one which arise from VB{z1,z}V_{B}\setminus\{z_{1},z^{\prime}\}. We now define the following sets to obtain a partition for VGV_{G^{\prime}}:

U1\displaystyle U_{1} ={xVC1:d(x,z1)=m},U2={yVC1:0d(y,z1)<m},\displaystyle=\{x\in V_{C_{1}}\mathrel{\mathop{\mathchar 58\relax}}d(x,z_{1})=m\},\quad\quad\quad\quad\quad~{}~{}~{}~{}U_{2}=\{y\in V_{C_{1}}\mathrel{\mathop{\mathchar 58\relax}}0\leq d(y,z_{1})<m\},
U3\displaystyle U_{3} ={xVC2:d(x,z)=m1},U4={yVC2:0d(y,z)<m1},\displaystyle=\{x\in V_{C_{2}}\mathrel{\mathop{\mathchar 58\relax}}d(x,z^{\prime})=m-1\},\quad\quad\quad\quad~{}~{}U_{4}=\{y\in V_{C_{2}}\mathrel{\mathop{\mathchar 58\relax}}0\leq d(y,z^{\prime})<m-1\},
U5\displaystyle U_{5} =V1(B){z1,z}whenever |V1(B)|3,andU6=V2(B).\displaystyle=V_{1}(B)\setminus\{z_{1},z^{\prime}\}~{}\text{whenever $\lvert V_{1}(B)\rvert\geq 3$},~{}\text{and}~{}~{}U_{6}=V_{2}(B).

Clearly, VG=l=16UlV_{G^{\prime}}=\cup_{l=1}^{6}U_{l} and UlUk=U_{l}\cap U_{k}=\emptyset for lkl\neq k. Also, z1U2z_{1}\in U_{2}, zU4z^{\prime}\in U_{4} and z2U6z_{2}\in U_{6}. To see U1U_{1}\neq\emptyset and U3U_{3}\neq\emptyset, consider e(z1)e(z_{1}) and e(z)e(z^{\prime}). Let e(z1)=d(b,z1)e(z_{1})=d(b,z_{1}) for some bVGb\in V_{G}. Since e(z1)=m+13e(z_{1})=m+1\geq 3, we get bVBb\not\in V_{B}. Also, by (9), bVC1b\not\in V_{C_{1}}. Therefore bVC2b\in V_{C_{2}}, and by (8), d(b,z1)=d(b,z)+d(z,z1)=d(b,z)+2d(b,z_{1})=d(b,z^{\prime})+d(z^{\prime},z_{1})=d(b,z^{\prime})+2. This implies d(b,z)=m1d(b,z^{\prime})=m-1 and hence bU3b\in U_{3}. Since z2VBCGz_{2}\in V_{B}\setminus C_{G}, by item (i)(i) of Lemma 3.5, we have e(z2)=max{e(z1),e(z)}1=max{m+1,e(z)}1e(z_{2})=\max\{e(z_{1}),e(z^{\prime})\}-1=\max\{m+1,e(z^{\prime})\}-1. Since e(z2)=m+1e(z_{2})=m+1, it follows that e(z)=m+2e(z^{\prime})=m+2. Let e(z)=d(a,z)e(z^{\prime})=d(a,z^{\prime}). Then by (9), aVC1a\in V_{C_{1}}, and by (8), we write d(a,z)=d(a,z1)+d(z1,z)=d(a,z1)+2d(a,z^{\prime})=d(a,z_{1})+d(z_{1},z^{\prime})=d(a,z_{1})+2. This gives d(a,z1)=md(a,z_{1})=m and hence aU1a\in U_{1}.

Subcase 1.1: Assume that |V1(B)|3\lvert V_{1}(B)\rvert\geq 3. Then U5U_{5}\neq\emptyset. Thus {U1,U2,U6}\{U_{1},U_{2},\ldots U_{6}\} partitions VGV_{G^{\prime}} and hence VGV_{G} as well. To find E(G)E(G) explicitly, we now compute the eccentricity of each vertex in GG.

Let aU1U2a\in U_{1}\cup U_{2}. To find e(a)e(a), consider d(a,b0)d(a,b_{0}) where b0U3b_{0}\in U_{3} is fixed. By (8),

e(a)d(a,b0)=d(a,z1)+d(z1,z)+d(z,b0)=d(a,z1)+(m+1).e(a)\geq d(a,b_{0})=d(a,z_{1})+d(z_{1},z^{\prime})+d(z^{\prime},b_{0})=d(a,z_{1})+(m+1). (10)

We claim that e(a)=d(a,z1)+(m+1)e(a)=d(a,z_{1})+(m+1). Let e(a)=d(a,x)e(a)=d(a,x) for some xVGx\in V_{G}. By (10), we have e(a)m+1e(a)\geq m+1. If xVC1x\in V_{C_{1}} then by (9), e(a)=d(a,x)d(a,z1)+d(z1,x)d(a,z1)+m<d(a,z1)+(m+1)e(a)=d(a,x)\leq d(a,z_{1})+d(z_{1},x)\leq d(a,z_{1})+m<d(a,z_{1})+(m+1), a contradiction to (10). So, xVC1x\not\in V_{C_{1}}. Similarly, we see that xBx\not\in B. Therefore, xVC2x\in V_{C_{2}}. By (9), d(x,z)m1d(x,z^{\prime})\leq m-1. This implies d(a,x)d(a,z1)+d(z1,z)+d(z,x)d(a,z1)+2+(m1)=d(a,z1)+(m+1)d(a,x)\leq d(a,z_{1})+d(z_{1},z^{\prime})+d(z^{\prime},x)\leq d(a,z_{1})+2+(m-1)=d(a,z_{1})+(m+1) and hence e(a)d(a,z1)+(m+1)e(a)\leq d(a,z_{1})+(m+1). Therefore, e(a)=d(a,z1)+(m+1)e(a)=d(a,z_{1})+(m+1) for all aU1U2a\in U_{1}\cup U_{2}. In particular, if aU1a\in U_{1} then e(a)=2m+1e(a)=2m+1 and if aU2a\in U_{2} then e(a)<2m+1e(a)<2m+1. Similarly, it can be shown that e(b)=d(b,z)+(m+2)e(b)=d(b,z^{\prime})+(m+2) for all bU3U4b\in U_{3}\cup U_{4}. Since e(z)=m+2e(z^{\prime})=m+2 and e(z1)=m+1e(z_{1})=m+1, by Lemma 3.5, we have e(a)=m+2e(a)=m+2 for all aU5a\in U_{5} and e(a)=m+1e(a)=m+1 for all aU6a\in U_{6}.

We now compute the entries of E(G)E(G). Let aU1a\in U_{1} and bVGb\in V_{G}. Then e(a)e(b)=min{e(a),e(b)}e(a)\geq e(b)=\min\{e(a),e(b)\}. If bU1U2b\in U_{1}\cup U_{2} then d(a,b)d(a,z1)+d(z1,b)=m+d(z1,b)<e(b)d(a,b)\leq d(a,z_{1})+d(z_{1},b)=m+d(z_{1},b)<e(b), and hence E(G)a,b=0E(G)_{a,b}=0. Suppose that bU3U4b\in U_{3}\cup U_{4}. Using (8), we write d(a,b)=d(a,z1)+d(z1,z)+d(z,b)=m+2+d(z,b)=e(b)d(a,b)=d(a,z_{1})+d(z_{1},z^{\prime})+d(z^{\prime},b)=m+2+d(z^{\prime},b)=e(b), and so E(G)a,b=2m+1E(G)_{a,b}=2m+1 if bU3b\in U_{3} and E(G)a,b=e(b)E(G)_{a,b}=e(b) if bU4b\in U_{4}. If bU5U6b\in U_{5}\cup U_{6} then by (P3)(P_{3}) and Remark 3.2, z1P¯(a,b)z_{1}\in\bar{P}(a,b) for all paths P¯(a,b)\bar{P}(a,b). Therefore, d(a,b)=d(a,z1)+d(z1,b)=m+d(z1,b)=e(b)d(a,b)=d(a,z_{1})+d(z_{1},b)=m+d(z_{1},b)=e(b). Thus, E(G)a,bE(G)_{a,b} is m+2m+2 if bU5b\in U_{5} and m+1m+1 if bU6b\in U_{6}. Since m2m\geq 2, we see that E(G)a,b=0E(G)_{a,b}=0 for all a,bU6a,b\in U_{6}. It can be shown that d(x,y)=e(x)=min{e(x),e(y)}d(x,y)=e(x)=\min\{e(x),e(y)\} if xU2x\in U_{2} and yU3y\in U_{3}, and d(x,y)<min{e(x),e(y)}d(x,y)<\min\{e(x),e(y)\} for all other cases. Therefore, the eccentricity matrix E(G)E(G) can be written as

E(G)={blockarray}ccccccc&U1U2U3U4U5U6{block}c(cccccc)U1OO(2m+1)JP(m+2)J(m+1)JU2OOQOOOU3(2m+1)JQOOOOU4POOOOOU5(m+2)JOOOOOU6(m+1)JOOOOO,E(G)=\blockarray{ccccccc}&U_{1}U_{2}U_{3}U_{4}U_{5}U_{6}\\ \block{c(cccccc)}U_{1}OO(2m+1)JP(m+2)J(m+1)J\\ U_{2}OOQOOO\\ U_{3}(2m+1)J^{\prime}Q^{\prime}OOOO\\ U_{4}P^{\prime}OOOOO\\ U_{5}(m+2)J^{\prime}OOOOO\\ U_{6}(m+1)J^{\prime}OOOOO\\ , (11)

where Pa=PbP_{a*}=P_{b*} for all a,bU1a,b\in U_{1} and Qu=QvQ_{*u}=Q_{*v} for all u,vU3u,v\in U_{3}. In fact, if U3={u31,u32,,u3|U3|}U_{3}=\{u_{31},u_{32},\ldots,u_{3\lvert U_{3}\rvert}\} and U4={u41,u42,,u4|U4|}U_{4}=\{u_{41},u_{42},\ldots,u_{4\lvert U_{4}\rvert}\} then Pa=(e(u41),e(u42),,e(u4|U4|))P_{a*}=\left(e(u_{41}),e(u_{42}),\ldots,e(u_{4\lvert U_{4}\rvert)}\right) for all aU1a\in U_{1} and Qv=(e(u31),e(u32),,e(u3|U3|))Q_{*v}=\left(e(u_{31}),e(u_{32}),\ldots,e(u_{3\lvert U_{3}\rvert)}\right)^{\prime} for all vU3v\in U_{3}. The structures of PP and QQ give rank(E(G)([UiVG]))=1\operatorname{rank}(E(G)\left([U_{i}\mid V_{G}]\right))=1 for all i=1,2,3,4i=1,2,3,4. Also, for some fixed xU4x\in U_{4},

E(G)a=1e(x){(m+2)E(G)xfor all aU5,(m+1)E(G)xfor all aU6.E(G)_{a*}=\frac{1}{e(x)}\begin{cases}(m+2)E(G)_{x*}&\text{for all $a\in U_{5}$},\\ (m+1)E(G)_{x*}&\text{for all $a\in U_{6}$}.\end{cases}

Hence rank(E(G))4\operatorname{rank}(E(G))\leq 4. Fix x1U1x_{1}\in U_{1}, x2U3x_{2}\in U_{3}. Let P(x1,z1)P(x_{1},z_{1}) and P(x2,z)P(x_{2},z^{\prime}) be shortest paths. Choose y1P(x1,z1)y_{1}\in P(x_{1},z_{1}) and y2P(x2,z)y_{2}\in P(x_{2},z^{\prime}) such that xix_{i} is adjacent to yiy_{i} for i=1,2i=1,2. To see rank(E(G))4\operatorname{rank}(E(G))\geq 4, consider the principal submatrix RR of E(G)E(G), which is given by

R=x1x2y1y2x1( 02m+102m) x22m+102m0y102m00y22m000.R=\bordermatrix{&x_{1}&x_{2}&y_{1}&y_{2}\cr x_{1}&0&2m+1&0&2m\cr x_{2}&2m+1&0&2m&0\cr y_{1}&0&2m&0&0\cr y_{2}&2m&0&0&0\cr}.

Note that det(R)0\det(R)\neq 0. This implies rank(E(G))4\operatorname{rank}(E(G))\geq 4 and hence rank(E(G))=4\operatorname{rank}(E(G))=4. Applying Theorem 2.1 to RR, we get In(R)=(2,2,0)\operatorname{In}(R)=(2,2,0), where we have taken A=(02m+12m+10)A=\left(\begin{smallmatrix}0&2m+1\\ 2m+1&0\end{smallmatrix}\right). Therefore, by Theorem 2.3, n(E(G))2n_{-}(E(G))\geq 2 and n+(E(G))2n_{+}(E(G))\geq 2. Since rank(E(G))=4\operatorname{rank}(E(G))=4, we have In(E(G))=(2,2,n4)\operatorname{In}(E(G))=(2,2,n-4).

Subcase 1.2: Suppose that |V1(B)|=2\lvert V_{1}(B)\rvert=2. Then U5=U_{5}=\emptyset. Now, it is easy to see that {U1,U2,U3,U4,U6}\{U_{1},U_{2},U_{3},U_{4},U_{6}\} partitions VGV_{G}, and E(G)E(G) is the principal submatrix of the matrix given in (11), obtained by deleting the rows and columns corresponding to the vertices in U5U_{5}. Hereafter, the proof is the same as in subcase 1.1.

Case 2: Assume that z2CGz_{2}\in C_{G}. That is, BB is a bridge block such that z1V1(B)CGz_{1}\in V_{1}(B)\cap C_{G} and z2V2(B)CGz_{2}\in V_{2}(B)\cap C_{G}. We obtain the non-trivial components C1C_{1} and C2C_{2} of GG^{\prime} such that z1VC1z_{1}\in V_{C_{1}} and z2VC2z_{2}\in V_{C_{2}} where GG^{\prime} is the subgraph constructed from GG by deleting all the edges of BB. As in case 1, it is easily seen that d(x,z1)md(x,z_{1})\leq m for all xVC1x\in V_{C_{1}} and d(y,z2)md(y,z_{2})\leq m for all yVC2y\in V_{C_{2}}. Define

U1\displaystyle U_{1} ={xVC1:d(x,z1)=m},U2={yVC1:0d(y,z1)<m},\displaystyle=\{x\in V_{C_{1}}\mathrel{\mathop{\mathchar 58\relax}}d(x,z_{1})=m\},\quad U_{2}=\{y\in V_{C_{1}}\mathrel{\mathop{\mathchar 58\relax}}0\leq d(y,z_{1})<m\},
U3\displaystyle U_{3} ={xVC2:d(x,z2)=m},U4={yVC2:0d(y,z2)<m},\displaystyle=\{x\in V_{C_{2}}\mathrel{\mathop{\mathchar 58\relax}}d(x,z_{2})=m\},\quad U_{4}=\{y\in V_{C_{2}}\mathrel{\mathop{\mathchar 58\relax}}0\leq d(y,z_{2})<m\},
U5\displaystyle U_{5} =V1(B){z1}whenever |V1(B)|2,and\displaystyle=V_{1}(B)\setminus\{z_{1}\}~{}\text{whenever $\lvert V_{1}(B)\rvert\geq 2$},\quad\text{and}
U6\displaystyle U_{6} =V2(B){z2}whenever |V2(B)|2.\displaystyle=V_{2}(B)\setminus\{z_{2}\}~{}\text{whenever $\lvert V_{2}(B)\rvert\geq 2$}.

Subcase 2.1: Suppose that |V1(B)|2\lvert V_{1}(B)\rvert\geq 2. Then, by (P1P_{1}), |V2(B)|2\lvert V_{2}(B)\rvert\geq 2. We now see that {U1,U2,,U6}\{U_{1},U_{2},\ldots,U_{6}\} partitions VGV_{G}. Let a,bVGa,b\in V_{G}. It is easy to verify that

e(a)\displaystyle e(a) ={2m+1if aU1U3,d(a,z1)+(m+1)if aU2,d(a,z2)+(m+1)if aU4,m+2if aU5U6.\displaystyle=\begin{cases}2m+1&\text{if }a\in U_{1}\cup U_{3},\\ d(a,z_{1})+(m+1)&\text{if }a\in U_{2},\\ d(a,z_{2})+(m+1)&\text{if }a\in U_{4},\\ m+2&\text{if }a\in U_{5}\cup U_{6}.\end{cases}

Then E(G)E(G) in this case takes the following form:

E(G)={blockarray}ccccccc&U1U2U3U4U5U6{block}c(cccccc)U1OO(2m+1)JP(m+2)JOU2OOQOOOU3(2m+1)JQOOO(m+2)JU4POOOOOU5(m+2)JOOOOOU6OO(m+2)JOOO,E(G)=\blockarray{ccccccc}&U_{1}U_{2}U_{3}U_{4}U_{5}U_{6}\\ \block{c(cccccc)}U_{1}OO(2m+1)JP(m+2)JO\\ U_{2}OOQOOO\\ U_{3}(2m+1)J^{\prime}Q^{\prime}OOO(m+2)J\\ U_{4}P^{\prime}OOOOO\\ U_{5}(m+2)J^{\prime}OOOOO\\ U_{6}OO(m+2)J^{\prime}OOO\\ , (12)

where all the rows of PP are identical and all the columns of QQ are the same. Hereafter, the proof is similar to that of case 1.

Let E(G)~\widetilde{E(G)} denote the 4×44\times 4 block matrix, which is the leading principal submatrix of E(G)E(G) in (12).

Subcase 2.2: Let |V1(B)|=1\lvert V_{1}(B)\rvert=1. Then by (P1P_{1}), |V2(B)|=1\lvert V_{2}(B)\rvert=1. Note that {U1,U2,U3,U4}\{U_{1},U_{2},U_{3},U_{4}\} partitions VGV_{G}, and E(G)=E(G)~E(G)=\widetilde{E(G)}. The rest of the proof is similar to case 11.

(ii)(ii) Consider the matrix E(G)E(G) given in (12). Suppose that

E(G)v=μvfor some 0μ and v=(𝐯𝟏,𝐯𝟐,,𝐯𝟔)n,E(G)\textbf{v}=\mu\textbf{v}~{}\text{for some $0\neq\mu\in\mathbb{R}$ and }~{}\textbf{v}=(\mathbf{v_{1}}^{\prime},\mathbf{v_{2}}^{\prime},\ldots,\mathbf{v_{6}}^{\prime})^{\prime}\in\mathbb{R}^{n},

where 𝐯\mathbf{v} is partitioned according to the partition of E(G)E(G). Then it is not difficult to verify that E(G)v~=μv~E(G)\tilde{\textbf{v}}=-\mu\tilde{\textbf{v}} where 𝐯~=(𝐯𝟏,𝐯𝟐,𝐯𝟑,𝐯𝟒,𝐯𝟓,𝐯𝟔)\tilde{\mathbf{v}}=(\mathbf{v_{1}}^{\prime},\mathbf{v_{2}}^{\prime},-\mathbf{v_{3}}^{\prime},-\mathbf{v_{4}}^{\prime},-\mathbf{v_{5}}^{\prime},\mathbf{v_{6}}^{\prime})^{\prime}. Also, the multiplicities of μ\mu and μ-\mu are equal.

Similar to the above case, by employing suitable eigenvectors, the result can be verified for the remaining subcases of item (i)(i). ∎

For m=1m=1, Theorem 4.1 need not be true. The following example illustrates this.

Example 4.2.

Consider the following graph HH and its eccentricity matrix E(H)E(H):

u2u_{2}u1u_{1}u3u_{3}u4u_{4}u5u_{5}HH

E(H)=(0200220002000230020022300).E(H)=\begin{pmatrix}0&2&0&0&2\\ 2&0&0&0&2\\ 0&0&0&2&3\\ 0&0&2&0&0\\ 2&2&3&0&0\\ \end{pmatrix}.

Note that diam(H)=3\operatorname{diam}(H)=3 and the rows and columns of E(H)E(H) are indexed by {u1,u2,u3,u4,u5}\{u_{1},u_{2},u_{3},u_{4},u_{5}\}. The eigenvalues of E(H)E(H) are 2,2,4.1394,0.78492,-2,-4.1394,-0.7849 and 4.92434.9243. So, the spectrum of E(G)E(G) is not symmetric about origin.

4.2 Inertia of eccentricity matrices of graphs in \mathscr{B} with even diameters

Let GG\in\mathscr{B} be such that diam(G)\operatorname{diam}(G) is even. In this subsection, we compute the inertia of E(G)E(G) and show that the spectrum of E(G)E(G) is not symmetric with respect to the origin. As a consequence, we obtain the main result of this section which characterizes the spectral symmetry of E(G)E(G) where GG\in\mathscr{B}, see Theorem 4.3.

The following lemma is needed to prove Theorem 4.2.

Lemma 4.1.

If GG\in\mathscr{B} then every diametrical path in GG contains a central vertex of GG.

Proof.

Let P(u,v)P(u,v) be a diametrical path in GG where u,vVGu,v\in V_{G}. We first prove the result for the case diam(G)\operatorname{diam}(G) is even. Assume that diam(G)=2m\operatorname{diam}(G)=2m for some m2m\geq 2. Then by Lemmas 3.3 and 3.7, we have diam(TG)=2m\operatorname{diam}(T_{G})=2m and |C(TG)|=1\lvert C(T_{G})\rvert=1. Let C(TG)={z}C(T_{G})=\{z\}. Since C(TG)C(G)C(T_{G})\subseteq C(G), we have zC(G)z\in C(G). By Lemmas 3.4 and 3.7, eG(z)=eTG(z)=m=eG(x)e_{G}(z)=e_{T_{G}}(z)=m=e_{G}(x) for all xC(G)x\in C(G). Note that there exists a vertex aa in the path P(u,v){P(u,v)} such that d(u,a)=m=d(a,v)d(u,a)=m=d(a,v). Therefore, eG(a)me_{G}(a)\geq m. We prove the result by showing that aC(G)a\in C(G). To do this, we claim that d(a,y)md(a,y)\leq m for all yVGy\in V_{G}. On the contrary, assume that d(a,y0)>md(a,y_{0})>m for some y0VGy_{0}\in V_{G}. Then aa and y0y_{0} do not belong to the same block, and y0y_{0} does not lie on the path P(u,v){P(u,v)}. Clearly, either aa lies in P(u,y0){P(u,y_{0})} for some shortest path P(u,y0)P(u,y_{0}) or aa lies in P(v,y0){P(v,y_{0})} for some shortest path P(v,y0)P(v,y_{0}), otherwise (P3)(P_{3}) fails. So, d(u,y0)=d(u,a)+d(a,y0)>d(u,a)+m=2md(u,y_{0})=d(u,a)+d(a,y_{0})>d(u,a)+m=2m or d(v,y0)>2md(v,y_{0})>2m, which is a contradiction. Hence the result follows in this case. The proof is similar when diam(G)\operatorname{diam}(G) is odd. ∎

The notion of diametrically distinguished vertex is introduced in [10] for a tree with even diameter which has exactly one central vertex. Motivated by this, in the following definition, we study this notion for the graph class \mathscr{B} where the graphs can have more than one central vertices.

Definition 4.1.

Let GG\in\mathscr{B} and let uVGu\in V_{G}. Then uu is said to be diametrically distinguished if there exists a diametrical path containing the vertex uu and uu is adjacent to zz for some zC(G)z\in C(G).

Theorem 4.2.

Let GG\in\mathscr{B} be such that diam(G)=2m\operatorname{diam}(G)=2m with m2m\geq 2. Let the center of the tree TGT_{G}, assoicated with GG, be {z}\{z\} and kk be the number of elements in the center C(G)C(G). Then the following hold:

  • (i)

    If zCGz\not\in C_{G} then In(E(G))={(3,k+1,nk4)when m=2,(2,2,n4)otherwise,\operatorname{In}(E(G))=\begin{cases}(3,k+1,n-k-4)&\text{when $m=2$},\\ (2,2,n-4)&\text{otherwise},\end{cases}
    where nn is the number of vertices of GG.

  • (ii)

    If zCGz\in C_{G} then In(E(G))=(r,r,n2r)\operatorname{In}(E(G))=(r,r,n-2r) where rr is the number of distinct blocks of GG having a diametrically distinguished vertex.

  • (iii)

    The spectrum of E(G)E(G) is not symmetric with respect to the origin.

Proof.

By Lemma 3.6, we have zC(G)z\in C(G), and by Lemma 3.7, e(z)=me(z)=m.

(i)(i) Suppose that zCGz\not\in C_{G}. Let zV1(B)z\in V_{1}(B) for some block BB of GG. Then by Remark 3.3 and Theorem 3.1, BB is a bridge block such that w1,w2V2(B)CGw_{1},w_{2}\in V_{2}(B)\cap C_{G} and C(G)=V1(B)C(G)=V_{1}(B). Without loss of generality, assume that e(w1)e(w2)e(w_{1})\leq e(w_{2}). By Lemma 3.5, e(z)=e(w2)1e(z)=e(w_{2})-1 which implies e(w2)=m+1e(w_{2})=m+1. Since w1C(G)w_{1}\not\in C(G) and e(z)e(w1)e(w2)e(z)\leq e(w_{1})\leq e(w_{2}), we get e(w1)=m+1e(w_{1})=m+1. Construct a subgraph GG^{\prime} from GG by deleting all the edges of the block BB. Then by (P3)(P_{3}) and (P4)(P_{4}), GG^{\prime} has two non-trivial components C1C_{1} and C2C_{2} containing the vertices w1w_{1} and w2w_{2}, respectively, and the remaining |VB|2\lvert V_{B}\rvert-2 components of GG^{\prime} are complete graphs of order one. We claim that d(x,w1)m1d(x,w_{1})\leq m-1 for all xVC1x\in V_{C_{1}}. Suppose there exists x0VC1x_{0}\in V_{C_{1}} such that d(x0,w1)>m1d(x_{0},w_{1})>m-1. Then by (P3P_{3}) and Remark 3.2 that w1P¯(x0,z)w_{1}\in\bar{P}(x_{0},z) for all paths P¯(x0,z)\bar{P}(x_{0},z). This implies d(x0,z)=d(x0,w1)+d(w1,z)>md(x_{0},z)=d(x_{0},w_{1})+d(w_{1},z)>m which yields e(z)>me(z)>m, a contradiction. Hence, d(x,w1)m1d(x,w_{1})\leq m-1 for all xVC1x\in V_{C_{1}}. Similarly, d(y,w2)m1d(y,w_{2})\leq m-1 for all yVC2y\in V_{C_{2}}. For i=1,2i=1,2, define

Ui\displaystyle U_{i} ={xVCi:d(x,wi)=m1},\displaystyle=\{x\in V_{C_{i}}\mathrel{\mathop{\mathchar 58\relax}}d(x,w_{i})=m-1\},
U2+i\displaystyle U_{2+i} ={yVCi:0d(y,wi)<m1},\displaystyle=\{y\in V_{C_{i}}\mathrel{\mathop{\mathchar 58\relax}}0\leq d(y,w_{i})<m-1\},
U5\displaystyle U_{5} =V1(B)and\displaystyle=V_{1}(B)~{}\text{and}
U6\displaystyle U_{6} =V2(B){w1,w2}whenever |V2(B)|3.\displaystyle=V_{2}(B)\setminus\{w_{1},w_{2}\}~{}\text{whenever $\lvert V_{2}(B)\rvert\geq 3$}.

Let x0VGx_{0}\in V_{G} such that e(w1)=d(x0,w1)e(w_{1})=d(x_{0},w_{1}). This implies x0VC1x_{0}\not\in V_{C_{1}} as e(w1)=m+1e(w_{1})=m+1. Since m2m\geq 2 and w1V2(B)w_{1}\in V_{2}(B), x0VBx_{0}\not\in V_{B}. Therefore, x0VC2x_{0}\in V_{C_{2}}. By (P3P_{3}), w2P¯(x0,w1)w_{2}\in\bar{P}(x_{0},w_{1}) for all paths P¯(x0,w1)\bar{P}(x_{0},w_{1}). Hence d(x0,w1)=d(x0,w2)+d(w2,w1)=d(x0,w2)+2d(x_{0},w_{1})=d(x_{0},w_{2})+d(w_{2},w_{1})=d(x_{0},w_{2})+2 which yields d(x0,w2)=m1d(x_{0},w_{2})=m-1. Therefore, x0U2x_{0}\in U_{2}. Similarly, using the fact that e(w2)=m+1e(w_{2})=m+1, we see that U1U_{1}\neq\emptyset. Assume that |V2(B)|3\lvert V_{2}(B)\rvert\geq 3. Now, it is clear that {Ui:1i6}\{U_{i}\mathrel{\mathop{\mathchar 58\relax}}1\leq i\leq 6\} partitions VGV_{G}.

Since C(G)=V1(B)C(G)=V_{1}(B), e(a)=e(z)=me(a)=e(z)=m for all aU5a\in U_{5}. By Lemma 3.5, e(b)=m+1e(b)=m+1 for all bU6b\in U_{6}. By (P3P_{3}), we see that w1,w2PG(x,y)w_{1},w_{2}\in P_{G}(x,y) for all xVC1x\in V_{C_{1}} and yVC2y\in V_{C_{2}}. Fix a0U1a_{0}\in U_{1} and b0U2b_{0}\in U_{2}. Similar to subcase 1.11.1 of item (i)(i) in Theorem 4.1, we can compute the eccentricity of the remaining vertices of VGV_{G}, using the shortest paths P(a0,w1)P(a_{0},w_{1}) in C1C_{1} and P(b0,w2)P(b_{0},w_{2}) in C2C_{2}, which are given below:

e(a)\displaystyle e(a) ={2mif aU1U2,d(a,w1)+(m+1)if aU3,d(a,w2)+(m+1)if aU4.\displaystyle=\begin{cases}2m&\text{if }a\in U_{1}\cup U_{2},\\ d(a,w_{1})+(m+1)&\text{if }a\in U_{3},\\ d(a,w_{2})+(m+1)&\text{if }a\in U_{4}.\end{cases}

The eccentricity matrix E(G)E(G) is given by

E(G)={blockarray}ccccccc&U1U2U3U4U5U6{block}c(cccccc)U1O(2m)JOPmJ(m+1)JU2(2m)JOQOmJ(m+1)JU3OQOOOOU4POOOOOU5mJmJOOROU6(m+1)J(m+1)JOOOO,E(G)=\blockarray{ccccccc}&U_{1}U_{2}U_{3}U_{4}U_{5}U_{6}\\ \block{c(cccccc)}U_{1}O(2m)JOPmJ(m+1)J\\ U_{2}(2m)J^{\prime}OQOmJ(m+1)J\\ U_{3}OQ^{\prime}OOOO\\ U_{4}P^{\prime}OOOOO\\ U_{5}mJ^{\prime}mJ^{\prime}OORO\\ U_{6}(m+1)J^{\prime}(m+1)J^{\prime}OOOO\\ , (13)

where Pa=PbP_{a*}=P_{b*} for all a,bU1a,b\in U_{1}, Qu=QvQ_{u*}=Q_{v*} for all u,vU2u,v\in U_{2} and R={2(JkIk)if m=2,Oif m3.R=\begin{cases}2(J_{k}-I_{k})&\text{if $m=2$},\\ O&\text{if $m\geq 3$}.\end{cases} The structures of PP and QQ yield that rank(E(G)([UiVG]))=1\operatorname{rank}(E(G)\left([U_{i}\mid V_{G}]\right))=1 for all i=1,2,3,4i=1,2,3,4.

Case 1: Assume that m3m\geq 3. Then the principal submatrix R=OR=O. For each aU5U6a\in U_{5}\cup U_{6}, observe that E(G)a=αE(G)w1+βE(G)w2E(G)_{a*}=\alpha E(G)_{w_{1}*}+\beta E(G)_{w_{2}*} for some real numbers α\alpha and β\beta. Hence rank(E(G))4\operatorname{rank}(E(G))\leq 4. The rest of the proof is similar to that of Theorem 4.1 by considering the principal submatrix

(02m02m12m02m1002m1002m1000).\begin{pmatrix}0&2m&0&2m-1\\ 2m&0&2m-1&0\\ 0&2m-1&0&0\\ 2m-1&0&0&0\end{pmatrix}.

If |V2(B)|=2|V_{2}(B)\rvert=2 then U6=U_{6}=\emptyset. Therefore, E(G)E(G) is a 5×55\times 5 block matrix. The result in this subcase can be verified similarly.

Case 2: Let m=2m=2. Then U3={w1}U_{3}=\{w_{1}\} and U4={w2}U_{4}=\{w_{2}\}. Note that the principal submatrix RR of E(G)E(G) in (13) is 2(JkIk)2(J_{k}-I_{k}) where k=|C(G)|=|U5|k=\lvert C(G)\rvert=\lvert U_{5}\rvert. Let C=(2Jk×|U1|2Jk×|U2|𝟎𝟎)C=\begin{pmatrix}2J_{k\times\lvert U_{1}\rvert}&2J_{k\times\lvert U_{2}\rvert}&\mathbf{0}&\mathbf{0}\end{pmatrix} and D=2(JkIk)D=2(J_{k}-I_{k}). Then D1=12(k1)Jk12IkD^{-1}=\frac{1}{2(k-1)}J_{k}-\frac{1}{2}I_{k}. Let AA denote the 4×44\times 4 block leading principal submatrix of E(G)E(G) in (13) and let M=ACD1CM=A-C^{\prime}D^{-1}C. We have

M=(μJ|U1|(μ+4)J|U1|×|U2|𝟎3𝐞(μ+4)JμJ|U2|3𝐞𝟎𝟎3𝐞003𝐞𝟎00)whereμ=2kk1.M=\begin{pmatrix}\mu J_{\lvert U_{1}\rvert}&(\mu+4)J_{\lvert U_{1}\rvert\times\lvert U_{2}\rvert}&\mathbf{0}&3\mathbf{e}\\ (\mu+4)J^{\prime}&\mu J_{\lvert U_{2}\rvert}&3\mathbf{e}&\mathbf{0}\\ \mathbf{0}^{\prime}&3\mathbf{e}^{\prime}&0&0\\ 3\mathbf{e}^{\prime}&\mathbf{0}^{\prime}&0&0\\ \end{pmatrix}~{}\text{where}~{}\mu=\frac{-2k}{k-1}.

Consider the principal submatrix N=(μμ+403μ+4μ3003003000)N=\left(\begin{smallmatrix}\mu&\mu+4&0&3\\ \mu+4&\mu&3&0\\ 0&3&0&0&\\ 3&0&0&0\end{smallmatrix}\right) of MM. Apply Theorem 2.1 to NN, by taking AA as 2×22\times 2 leading principal submatrix of NN, we get In(N)=(2,2,0)\operatorname{In}(N)=(2,2,0). Therefore, by Theorem 2.3, n+(M)2n_{+}(M)\geq 2 and n(M)2n_{-}(M)\geq 2.
Subcase 2.1: Suppose that |V2(B)|=2\lvert V_{2}(B)\rvert=2. Then U6=U_{6}=\emptyset and so E(G)E(G) is a 5×55\times 5 block matrix. Since rank(M)=4\operatorname{rank}(M)=4, we have In(M)=(2,2,nk4)\operatorname{In}(M)=(2,2,n-k-4). Using Theorem 2.1 to E(G)E(G), we get In(E(G))=In(2Jk2Ik)+In(M)=(1,k1,0)+(2,2,nk4)=(3,k+1,nk4)\operatorname{In}(E(G))=\operatorname{In}(2J_{k}-2I_{k})+\operatorname{In}(M)=(1,k-1,0)+(2,2,n-k-4)=(3,k+1,n-k-4).
Subcase 2.2: If |V2(B)|3\lvert V_{2}(B)\rvert\geq 3 then U6U_{6}\neq\emptyset. By subcase 2.12.1, we have rank(E(G))k+4\operatorname{rank}(E(G))\geq k+4. Since each column in U6U_{6} is a linear combination of columns in U3U_{3} and U4U_{4}, we have rank(E(G))=k+4\operatorname{rank}(E(G))=k+4. Therefore, the result follows using In(X)\operatorname{In}(X) and Theorem 2.3 where XX is a 5×55\times 5 block leading principal submatrix of E(G)E(G).

(ii)(ii) Assume that zCGz\in C_{G}. Then by Theorem 3.1, C(G)={z}C(G)=\{z\}. Let deg(z)=p\deg(z)=p. Then by (P4)(P_{4}), p2p\geq 2. Let w1,w2,,wpw_{1},w_{2},\ldots,w_{p} be the vertices in GG that are adjacent to zz. Now obtain the subgraph GG^{\prime} from GG by deleting all the edges that are incident with zz. By the construction of GG^{\prime}, it is clear that VG=VGV_{G}=V_{G^{\prime}}, and GG^{\prime} has a component K1K_{1} with the vertex set {z}\{z\}. Let the components of GG^{\prime} be C1,C2,,CqC_{1},C_{2},\ldots,C_{q}. Among these, let C1,C2,,CrC_{1},C_{2},\ldots,C_{r} be the components containing the vertices x1,x2,,xrx_{1},x_{2},\ldots,x_{r}, respectively, such that d(xi,z)=md(x_{i},z)=m for all i=1,2,,ri=1,2,\ldots,r. Since e(z)=me(z)=m, existence of CiC_{i} containing such xix_{i} is guaranteed. We claim that r2r\geq 2. Let PG(a,b)P_{G}(a,b) be a diametrical path in GG where a,bVGa,b\in V_{G}. Then, by Lemma 4.1, zPG(a,b)z\in P_{G}(a,b), and hence d(a,z)=m=d(b,z)d(a,z)=m=d(b,z). Therefore, a,bi=1rVCia,b\in\cup_{i=1}^{r}V_{C_{i}}. If aVCia\in V_{C_{i}} and bVCjb\in V_{C_{j}} with CiCjC_{i}\neq C_{j} then the claim follows. Using (P3)(P_{3}), we observe that any path PG(a,b)P^{\prime}_{G}(a,b) between aa and bb in GG contains the central vertex zz as zPG(a,b)z\in P_{G}(a,b). This implies aa and bb do not belong to a single component CiC_{i} and hence r2r\geq 2.

We now find the possible components of GG^{\prime} containing a diametrically distinguished vertex. Let SS be the set of all diametrically distinguished vertices of GG and let wSw\in S. Then ww is adjacent to zz and there exists a diametrical path PG(a0,b0)P_{G}(a_{0},b_{0}) containing ww. By Lemma 4.1, zPG(a0,b0)z\in P_{G}(a_{0},b_{0}). Then either d(a0,w)=m1d(a_{0},w)=m-1 or d(b0,w)=m1d(b_{0},w)=m-1. Without loss of generality, assume that d(a0,w)=m1d(a_{0},w)=m-1. That is, the path PG(a0,b0)P_{G}(a_{0},b_{0}) yields a path PG(a0,w)P_{G}(a_{0},w) with zPG(a0,w)z\not\in P_{G}(a_{0},w), and d(a0,z)=md(a_{0},z)=m. This implies a0VCia_{0}\in V_{C_{i}} for some i=1,2,,ri=1,2,\ldots,r. By the maximality of CiC_{i}, the path PG(a0,w)P_{G}(a_{0},w) is completely contained in CiC_{i}. That is, wVCiw\in V_{C_{i}}. Hence diametrically distinguished vertices are necessarily belong to i=1rCi\cup_{i=1}^{r}C_{i}.

We claim that there are exactly rr distinct blocks of GG that contain a diametrically distinguished vertex. For each i{1,2,,r}i\in\{1,2,\ldots,r\}, if we prove VCiSV_{C_{i}}\cap S\neq\emptyset and VCiSBiV_{C_{i}}\cap S\subseteq B_{i} for some block BiB_{i} in GG with B1,B2,,BrB_{1},B_{2},\ldots,B_{r} are pairwise distinct then the claim follows. Let 1ir1\leq i\leq r. By the definition of CiC_{i}, there exists xiVCix_{i}\in V_{C_{i}} such that d(xi,z)=md(x_{i},z)=m. Let PG(xi,z)P_{G}(x_{i},z) be a shortest path between xix_{i} and zz. Let wiw_{i} be the vertex in the path PG(xi,z)P_{G}(x_{i},z) such that wiw_{i} is adjacent to zz in GG. Then we see that d(xi,wi)=m1d(x_{i},w_{i})=m-1 which follows from the subpath PG(xi,wi)P_{G}(x_{i},w_{i}) obtained form PG(xi,z)P_{G}(x_{i},z). Since xiVCix_{i}\in V_{C_{i}}, and zPG(xi,wi)z\not\in P_{G}(x_{i},w_{i}), by the maximality of CiC_{i}, PG(xi,wi)P_{G}(x_{i},w_{i}) lies in CiC_{i}. Choose j{1,2,,r}j\in\{1,2,\ldots,r\} such that CjCiC_{j}\neq C_{i}. Then there exist xjx_{j} and wjw_{j} in VCjV_{C_{j}} such that d(xj,wj)=m1d(x_{j},w_{j})=m-1. Also, the subpath PG(xj,wj)P_{G}(x_{j},w_{j}) obtained form a shortest path PG(xj,z)P_{G}(x_{j},z), completely lies in CjC_{j}. Since the components CiC_{i} and CjC_{j} are disjoint, we have PG(xi,wi)PG(xj,wj)=P_{G}(x_{i},w_{i})\cap P_{G}(x_{j},w_{j})=\emptyset. Note that, by (P3P_{3}), zP¯G(xi,xj)z\in\bar{P}_{G}(x_{i},x_{j}) for all paths P¯G(xi,xj)\bar{P}_{G}(x_{i},x_{j}). This implies d(xi,xj)=d(xi,z)+d(z,xj)=m+m=diam(G)d(x_{i},x_{j})=d(x_{i},z)+d(z,x_{j})=m+m=\operatorname{diam}(G). That is, any shortest path between xix_{i} and xjx_{j} in GG is a diametrical path. In particular, the shortest paths PG(xi,z)P_{G}(x_{i},z) and PG(xj,z)P_{G}(x_{j},z) induce a diametrical path containing wiw_{i}. This implies wiSw_{i}\in S and hence VCiSV_{C_{i}}\cap S\neq\emptyset.

We next prove that VCiSBiV_{C_{i}}\cap S\subseteq B_{i} for some block BiB_{i} in GG. Let wVCiSw\in V_{C_{i}}\cap S. Assume that wiVBiw_{i}\in V_{B_{i}} for some block BiB_{i} in GG. Suppose that |VCiS|2\lvert V_{C_{i}}\cap S\rvert\geq 2. Let wVCiSw^{\prime}\in V_{C_{i}}\cap S with www^{\prime}\neq w. Then wBiw^{\prime}\in B_{i}, otherwise (P3P_{3}) fails. Hence VCiSBiV_{C_{i}}\cap S\subseteq B_{i}. Given a block BB in GG there exists l{1,2,,q}l\in\{1,2,\ldots,q\} such that VB{z}VClV_{B}\setminus\{z\}\subseteq V_{C_{l}}, which is clear from the construction of GG^{\prime}. Since components of GG^{\prime} are disjoint, the existed ll for the block BB is unique. This implies BiBjB_{i}\neq B_{j} for all i,j{1,2,,r}i,j\in\{1,2,\ldots,r\} with iji\neq j. Thus, there are exactly rr distinct blocks of GG which have a diametrically distinguished vertex.

We next show that rank(E(G))=2r\operatorname{rank}\left(E(G)\right)=2r by explicitly finding the matrix E(G)E(G). By suitably relabelling the vertices w1,w2,,wpw_{1},w_{2},\ldots,w_{p}, it is assumed that wiVCiw_{i}\in V_{C_{i}} for all 1ir1\leq i\leq r. For each i{1,2,,r}i\in\{1,2,\ldots,r\}, define

Ui\displaystyle U_{i} ={xVCi:d(x,z)=m},\displaystyle=\{x\in V_{C_{i}}\mathrel{\mathop{\mathchar 58\relax}}d(x,z)=m\},
Ur+i\displaystyle U_{r+i} ={yVCi:0<d(y,z)<m}, and\displaystyle=\{y\in V_{C_{i}}\mathrel{\mathop{\mathchar 58\relax}}0<d(y,z)<m\},~{}~{}~{}\text{ and}
U2r+1\displaystyle U_{2r+1} =VGi=1r(UiUr+i).\displaystyle=V_{G}\setminus\cup_{i=1}^{r}\left(U_{i}\cup U_{r+i}\right).

It is clear that xiUix_{i}\in U_{i}, wiUr+iw_{i}\in U_{r+i}, zu2r+1z\in u_{2r+1} and UlUk=U_{l}\cap U_{k}=\emptyset for lkl\neq k. Thus {U1,U2,,U2r+1}\{U_{1},U_{2},\ldots,U_{2r+1}\} partitions VGV_{G}. Let aVGa\in V_{G}. We claim that e(a)=d(a,z)+me(a)=d(a,z)+m. Note that d(a,x)d(a,z)+d(z,x)d(a,z)+md(a,x)\leq d(a,z)+d(z,x)\leq d(a,z)+m for all xVGx\in V_{G}, which yields e(a)d(a,z)+me(a)\leq d(a,z)+m. Choose an element yy such that yy and aa are in different components of GG^{\prime} and d(y,z)=md(y,z)=m. Since any path between aa and yy passes through zz, we have d(a,y)=d(a,z)+d(z,y)=d(a,z)+md(a,y)=d(a,z)+d(z,y)=d(a,z)+m, and hence the claim. This implies e(a)=2me(a)=2m for all aUia\in U_{i}.

Next we find the entries of E(G)E(G) case-by-case. Let a,bVGa,b\in V_{G} and 1i,jr1\leq i,j\leq r with iji\neq j.

  • \bullet

    If aUia\in U_{i} and bUjb\in U_{j} then E(G)a,b=2mE(G)_{a,b}=2m as

    d(a,b)=d(a,z)+d(z,b)=m+m=2m=min{e(a),e(b)}.d(a,b)=d(a,z)+d(z,b)=m+m=2m=\min\{e(a),e(b)\}.
  • \bullet

    Note that E(G)a,b=0E(G)_{a,b}=0 whenever a,bUia,b\in U_{i} because

    d(a,b)d(a,wi)+d(wi,b)=(m1)+(m1)<min{e(a),e(b)}.d(a,b)\leq d(a,w_{i})+d(w_{i},b)=(m-1)+(m-1)<\min\{e(a),e(b)\}.
  • \bullet

    Let aUia\in U_{i} and bUr+jU2r+1b\in U_{r+j}\cup U_{2r+1}. Then e(a)=2me(a)=2m and

    d(a,b)=d(a,z)+d(z,b)=m+d(z,b)=e(b)=min{e(a),e(b)}.d(a,b)=d(a,z)+d(z,b)=m+d(z,b)=e(b)=\min\{e(a),e(b)\}.

    In this case, E(G)a,b=e(b)E(G)_{a,b}=e(b).

Since E(G)E(G) is symmetric, E(G)x,y=e(x)E(G)_{x,y}=e(x) for all xUr+jU2r+1x\in U_{r+j}\cup U_{2r+1} and yUiy\in U_{i} with iji\neq j. Similarly the remaining entries of E(G)E(G) can be easily computed and are zero. Hence the eccentricity matrix E(G)E(G) of GG can be written in the block form

{blockarray}cccccccccU1&U2UrUr+1Ur+2U2rU2r+1{block}(ccccccccc)O(2m)J(2m)JOA1,r+2A1,2rA1,2r+1(2m)JO(2m)JA2,r+1OA2,2rA2,2r+1(2m)J(2m)JOAr,r+1Ar,r+2OAr,2r+1OA2,r+1Ar,r+1OOOOA1,r+2OAr,r+2OOOOA1,2rA2,2rOOOOOA1,2r+1A2,2r+1Ar,2r+1OOOO.\blockarray{ccccccccc}U_{1}&U_{2}\ldots U_{r}U_{r+1}U_{r+2}\ldots U_{2r}U_{2r+1}\\ \block{(ccccccccc)}O(2m)J\ldots(2m)JOA_{1,r+2}\ldots A_{1,2r}A_{1,2r+1}\\ (2m)JO\ldots(2m)JA_{2,r+1}O\ldots A_{2,2r}A_{2,2r+1}\\ \vdots\vdots\ddots\vdots\vdots\vdots\ddots\vdots\vdots\\ (2m)J(2m)J\ldots OA_{r,r+1}A_{r,r+2}\ldots OA_{r,2r+1}\\ \\ OA_{2,r+1}^{\prime}\ldots A_{r,r+1}^{\prime}OO\ldots OO\\ A_{1,r+2}^{\prime}O\ldots A_{r,r+2}^{\prime}OO\ldots OO\\ \vdots\vdots\ddots\vdots\vdots\vdots\ddots\vdots\vdots\\ A_{1,2r}^{\prime}A_{2,2r}^{\prime}\ldots OOO\ldots OO\\ \\ A_{1,2r+1}^{\prime}A_{2,2r+1}^{\prime}\ldots A_{r,2r+1}^{\prime}OO\ldots OO\\ . (14)

Since the matrix obtained in (14) is of the form ε(T)\varepsilon(T) given in [10, Theorem 3.23.2], the remainder of the proof is similar to that of Theorem 3.23.2 in [10].

(iii)(iii) Case 1: Consider the case rank(E(G))=2l\operatorname{rank}(E(G))=2l where l{r,2}l\in\{r,2\}. Then by Theorem 2.2, the characteristic polynomial of E(G)E(G) can be written as

χx(E(G))=xn2l(x2lδ1x2l1+δ2x2l2+(1)2l1δ2l1x+(1)2lδ2l),\chi_{x}(E(G))={x}^{n-2l}({x}^{2l}-\delta_{1}{x}^{2l-1}+\delta_{2}{x}^{2l-2}-\cdots+(-1)^{2l-1}\delta_{2l-1}x+(-1)^{2l}\delta_{2l}),

where δ2l0\delta_{2l}\neq 0 and δi\delta_{i} is the sum of all principal minors of order ii, for all i=1,2,,2li=1,2,\ldots,2l. To prove the spectrum of E(G)E(G) is not symmetric with respect to the origin, by Lemma 2.1, it enough to find some i{1,2,,2l1}i\in\{1,2,\ldots,2l-1\} such that δi0\delta_{i}\neq 0 and δi+10\delta_{i+1}\neq 0. It is clear from (13) and (14) that every 2×22\times 2 and 3×33\times 3 principal sumatrices Q1Q_{1} and Q2Q_{2} of E(G)E(G) are, respectively, in the following form:

Q1=(0αα0)andQ2=(0αβα0γβγ0)where α,β and γ are some non-negative integers.Q_{1}=\begin{pmatrix}0&\alpha\\ \alpha&0\end{pmatrix}~{}\text{and}~{}Q_{2}=\begin{pmatrix}0&\alpha&\beta\\ \alpha&0&\gamma\\ \beta&\gamma&0\end{pmatrix}~{}\text{where $\alpha,\beta$ and $\gamma$ are some non-negative integers.}

Since det(Q1)=α2\det(Q_{1})=-\alpha^{2} and det(Q2)=2αβγ\det(Q_{2})=2\alpha\beta\gamma, we have δ20\delta_{2}\leq 0 and δ30\delta_{3}\geq 0. Also, the following principal submatrices have non-zero determinant:

x1x2x1( 02m) x22m0,andx1x2zx1( 02mm) x22m0mzmm0,\ \bordermatrix{&x_{1}&x_{2}\cr x_{1}&0&2m\cr x_{2}&2m&0\cr},\qquad\text{and}\qquad\ \bordermatrix{&x_{1}&x_{2}&z\cr x_{1}&0&2m&m\cr x_{2}&2m&0&m\cr z&m&m&0\cr},

where xiUix_{i}\in U_{i} for i=1,2i=1,2, and zC(TG)z\in C(T_{G}). Hence δ2<0\delta_{2}<0 and δ3>0\delta_{3}>0.

Case 2: Suppose that rank(E(G))=k+4\operatorname{rank}(E(G))=k+4. Since k=|V1(B)|k=\lvert V_{1}(B)\rvert with cut-vertices w1,w2w_{1},w_{2} in V2(B)V_{2}(B), by (P1P_{1}), we have k2k\geq 2. If k=2k=2 then the proof follows similarly to the previous case. If k3k\geq 3 then by item (ii)(ii), n(E(G))4>n+(E(G))n_{-}\left(E(G)\right)\geq 4>n_{+}\left(E(G)\right) and hence the proof. ∎

We are now ready to state the main result of this section.

Theorem 4.3.

Let GG\in\mathscr{B} with diam(G)4\operatorname{diam}(G)\geq 4. Then the eigenvalues of E(G){E}(G) are symmetric with respect to the origin if and only if diam(G)\operatorname{diam}(G) is odd.

Proof.

It follows from Theorems 4.1 and 4.2. ∎

The following example shows that Theorem 4.1 need not be true for a general bi-block graph.

Example 4.3.

Consider the graph GG\not\in\mathscr{B} and the matrix E(G)E(G) which are given below:

v1v_{1}v2v_{2}v3v_{3}v4v_{4}v8v_{8}v5v_{5}v6v_{6}v9v_{9}v7v_{7}GG

v1v2v3v4v5v6v7v8v9v1( 000000030) v2000030003v3000000003v4000000004v5030000004v6000000040v7000000040v8300004405v9033440050.\bordermatrix{&v_{1}&v_{2}&v_{3}&v_{4}&v_{5}&v_{6}&v_{7}&v_{8}&v_{9}\cr v_{1}&0&0&0&0&0&0&0&3&0\cr v_{2}&0&0&0&0&3&0&0&0&3\cr v_{3}&0&0&0&0&0&0&0&0&3\cr v_{4}&0&0&0&0&0&0&0&0&4\cr v_{5}&0&3&0&0&0&0&0&0&4\cr v_{6}&0&0&0&0&0&0&0&4&0\cr v_{7}&0&0&0&0&0&0&0&4&0\cr v_{8}&3&0&0&0&0&4&4&0&5\cr v_{9}&0&3&3&4&4&0&0&5&0\cr}.

The eigenvalues of E(G)E(G) are 0 with multiplicity 33 and the remaining simple eigenvalues are 9.4967,4.3784,2.9329,1.4150,5.2920,10.1010-9.4967,~{}-4.3784,~{}-2.9329,~{}1.4150,~{}5.2920,~{}10.1010, which are computed using SAGEMATH. Therefore, In(E(G))=(3,3,3)\operatorname{In}(E(G))=(3,3,3) and E(G)E(G) is not symmetric about the origin.

4.3 Irreducibility of eccentricity matrices of graphs in \mathscr{B}

The problem of characterizing graphs whose eccentricity matrices are irreducible remains open. So far, only a few classes of graphs have been identified whose eccentricity matrices are irreducible, see [7, 9, 11, 15, 16]. In this subsection, we prove that the eccentricity matrices of graphs in \mathscr{B} are irreducible.

For our purpose, let us recall the following lemma, which gives an equivalent condition for the irreducibility of a non-negative matrix.

Lemma 4.2.

(see [8]) Let M=(mij)M=(m_{ij}) be an n×nn\times n nonnegative symmetric matrix and G(M)G(M) be the graph on nn vertices such that there is an edge between the vertices ii and jj in G(M)G(M) if and only if mij0m_{ij}\neq 0. Then MM is irreducible if and only if G(M)G(M) is connected.

Theorem 4.4.

Let GG\in\mathscr{B} with diam(G)4\operatorname{diam}(G)\geq 4. Then the eccentricity matrix E(G)E(G) is irreducible.

Proof.

Let GG\in\mathscr{B}. Using Lemma 4.2 and the matrices given in (11),(12),(13) and (14), it is direct to see that E(G)E(G) is irreducible. ∎

Conclusion

The inertias of the eccentricity matrices of a subclass of bi-block graphs \mathscr{B} (which contains trees) are derived by associating a tree TGT_{G} for each GG\in\mathscr{B}. A characterization for the spectrum of the eccentricity matrix E(G)E(G) is symmetric about the origin being given in terms of the diameter of GG. Also, it is proved that the eccentricity matrix of GG\in\mathscr{B} with diam(G)4\operatorname{diam}(G)\geq 4 is irreducible.

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