On the Eccentricity Matrices of Certain Bi-Block Graphs

Abstract

The eccentricity matrix of a simple connected graph G is obtained from the distance matrix of G by retaining the largest nonzero 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 the diameter of $G\in {\mathscr {B}}$ is more than three, then we show that the eigenvalues of the eccentricity matrix of G are symmetric with respect to the origin if and only if the diameter of G is odd. Further, we prove that the eccentricity matrices of graphs in ${\mathscr {B}}$ are irreducible.

1 Introduction

Let G be a simple connected graph on n vertices with the vertex set $V_G=\{v_1, v_2,\ldots , v_n\}$. In the literature, many properties and structural characteristics of G have been studied by associating several matrices corresponding to G, see [3]. Let us recall some of them which are pertinent to our discussion here. The adjacency matrix of G, denoted by A(G), is an $n \times n$ matrix whose (i, j)-th entry is one if $v_i$ and $v_j$ are adjacent and zero elsewhere. Let $d(v_i,v_j)$ denote the length of a shortest path between the vertices $v_i$ and $v_j$. The distance matrix $D(G):=(d_{ij})$ of G is an $n \times n$ matrix such that $d_{ij}=d(v_i,v_j)$ for all i and j. Clearly, the adjacency matrix can be obtained from the distance matrix by retaining the smallest nonzero 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 [20] associated a new matrix corresponding to G, namely $D_{\text {MAX}}$, which is derived from the distance matrix D(G) by keeping the largest nonzero distance in each row and each column of D(G) and defining the remaining entries equal to zero. Later, Wang et al. [23] obtained an equivalent definition of $D_{\text {MAX}}$ using the notion of eccentricity of a vertex and called it the eccentricity matrix. Recall that the eccentricity of a vertex $v_i$ is defined by $e(v_i)$= max $\{d(v_i,v_j): v_j\in V_G\}$. The eccentricity matrix E(G) of G is an $n \times n$ matrix whose (i, j)-th entry is given by

$$\begin{aligned} E (G)_{ij}= {\left\{ \begin{array}{ll} d(v_i,v_j) &{} \text {if } d(v_i,v_j)=\text {min}\{e(v_i),e(v_j)\},\\ 0 &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

This matrix has been well studied in the literature (see [1, 5, 8, 10,11,12,13,14,15,16,17,18,19,20, 22,23,24,25,26,27,28,29,30,31]) and has applications in Chemistry, see [20, 23, 25].

In the following, we provide a brief survey of the eccentricity matrix. In [5], some bounds for the second, third, and fourth largest eigenvalues of E(G) are given in terms of its diameter. The least and the largest eigenvalues of the eccentricity matrices of trees with odd diameters are studied in [8, 30], and some of these results are extended in [18]. In [23], the authors analyzed various spectral properties of the eccentricity matrices of trees and posed two conjectures related to the eigenvalues of these matrices. These conjectures are addressed in [29]. It is proved in [10] that the graphs with exactly one positive eccentricity eigenvalue are determined by their eccentricity spectra and these graphs are characterized in [28]. The problem of characterizing graphs with a small number of distinct eccentricity eigenvalues is investigated in [22]. Certain properties of the eigenvalues of the eccentricity matrices of threshold graphs are explored in [1, 19]. The eccentricity energies of certain graphs are examined in [11, 16, 17, 24]. Many pairs of $\varepsilon$-cospectral graphs (non-isomorphic graphs having the same eccentricity spectrum) are constructed in [27] using some graph operations. The eccentricity matrices of digraphs are introduced and studied in [31].

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 $K_{l,m}$ is reducible for all $l,m\ge 2$, see [23]. However, for some classes of graphs, the associated eccentricity matrices are irreducible. It is proved in [23] that the eccentricity matrix of a tree with at least two vertices is irreducible, and an alternative proof of this result is given in [15]. This result has been extended to some larger classes of graphs in [12] and [26]. It is shown in [17] 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. [23], 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 G is a maximal connected subgraph of G which has no cut-vertex. A connected graph G 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 [15], 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 Sect. 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 [6], 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 [21]. The inverse formula given for the distance matrix of a tree was extended to a subclass of bi-block graphs in [9]. 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 G is a bipartite graph if and only if the eigenvalues of the adjacency matrix A(G) are symmetric with respect to the origin (i.e., if $\mu$ is an eigenvalue of A(G) with multiplicity l, then -$\mu$ is also an eigenvalue of A(G) with multiplicity l). In [14], 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) of a tree T are symmetric with respect to the origin if and only if ${{\,\textrm{diam}\,}}(T)$ is odd. An analogous result has been proved for a subclass of block graphs (namely, clique trees) in [12]. The problem of characterizing the graphs whose eigenvalues of the eccentricity matrices are symmetric with respect to the origin, posed in [14], 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) are symmetric with respect to the origin if and only if ${{\,\textrm{diam}\,}}(G)$ is odd, whenever $G\in {\mathscr {B}}$ with ${{\,\textrm{diam}\,}}(G)\ge 4$. By means of examples, we show that this equivalence does not hold for a general bi-block graph and for $G\in {\mathscr {B}}$ with ${{\,\textrm{diam}\,}}(G)=3$.

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 A be an $n \times n$ real symmetric matrix and let $i_+(A)$, $i_-(A)$ and $i_0(A)$ denote the number of positive, negative, and zero eigenvalues of A, respectively, including the multiplicities. The inertia of A is the ordered triple $(i_+(A), i_{-}(A), i_0(A))$ and is denoted by ${{\,\textrm{In}\,}}(A)$. It is known that $i_+(A)+i_{-}(A)= {{\,\textrm{rank}\,}}(A)$. An interesting and challenging problem in spectral graph theory is finding the eigenvalues and inertias of graph matrices, see [3, 4, 15,16,17, 23, 25, 26] and the references therein. Among other results, it has been proved that the inertia of the distance matrix of a tree on $n(\ge 2)$ vertices is $(1,n-1,0)$, see [3]. 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 [12, 14, 15, 17]. 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 [12, 14], 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 Sect. 3, we associate a tree $T_G$ for each $G\in {\mathscr {B}}$ 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 $G\in {\mathscr {B}}$. Among other things, we give an equivalent condition for the spectrum of E(G) to be symmetric with respect to the origin where $G\in {\mathscr {B}}$. 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 G by $V_G$ and $E_G$, respectively. A graph $G^{\prime }=(V_{G^{\prime }},E_{G^{\prime }})$ is said to be a subgraph of G if $V_{G^{\prime }}\subseteq V_G$ and $E_{G^{\prime }}\subseteq E_G$. A subgraph $G^{\prime }$ of G is called an induced subgraph of G if the edges of $G^{\prime }$ are precisely the edges of G whose ends are in $V_{G^{\prime }}$. We denote the induced subgraph $G^{\prime }$ by $G[V_{G^{\prime }}]$ and call it as the subgraph induced by $V_{G^{\prime }}$. A vertex $v\in V_G$ is said to be a cut-vertex of G if $G\setminus \{v\}$ is a disconnected graph. The notation $C_G$ stands for the collection of all cut-vertices of G. A maximal connected subgraph of a graph G is known as a component of G. The radius and the diameter of G are, respectively, denoted by ${{\,\textrm{rad}\,}}(G)$ and ${{\,\textrm{diam}\,}}(G)$ and are defined by ${{\,\textrm{rad}\,}}(G):=\min \{e(v): v\in V_G\}$ and ${{\,\textrm{diam}\,}}(G):=\max \{e(v): v\in V_G\}$. A path P in G is a subgraph of G whose vertices are arranged in a sequence such that two vertices are adjacent in P if and only if they are consecutive in the sequence. We denote a path of length k between two vertices a and b in G by $P_G(a,b)=au_1u_2\ldots u_{k-1}b$, where $u_i\in V_G$ for all i. We write the length of the path $P_G(a,b)$ by $l\left( P_G(a,b)\right)$. A diametrical path in G is a shortest path between two vertices u and v such that $d(u,v) = {{\,\textrm{diam}\,}}(G)$. A vertex $v\in V_G$ is said to be a central vertex if $e(v)={{\,\textrm{rad}\,}}(G)$. The center of G, denoted by C(G), is the collection of all central vertices of G.

A graph G is said to be bipartite if $V_G$ can be partitioned into two non-empty subsets $V_1$ and $V_2$ such that each edge of G has one end in $V_1$ and the other end in $V_2$. The pair $(V_1,V_2)$ is called a bipartition of the bipartite graph G, and the sets $V_1$ and $V_2$ are referred to as the partite sets of G. A bipartite graph with bipartition $(V_1,V_2)$ is said to be a complete bipartite graph if every vertex of $V_1$ is adjacent to all the vertices of $V_2$, and is denoted by $K_{|V_1|,|V_2|}$ where $|V_1|$ stands for the cardinality of $V_1$. For more details on graph-theoretic notions and terminologies, we refer to the book [2].

Let $G\in {\mathscr {B}}$ and let B be a block of G. Then B is a complete bipartite graph $K_{l,m}$. Throughout this article, we assume that $\left( V_1(B), V_2(B)\right)$ is the bipartition of B. The block B of G is said to be a bridge block if $|V_B\cap C_G|=2$, and a leaf block if $|V_B\cap C_G |=1$.

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

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

Theorem 2.1

( [32]) Let A and D be $r\times r$ and $s\times s$ real matrices, respectively, and let $M=\left( {\begin{matrix} A &{} B\\ C &{} D\\ \end{matrix}}\right)$ be a symmetric partitioned matrix of order n. If A is nonsingular, then

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

• (ii) ${{\,\textrm{In}\,}}(M)= {{\,\textrm{In}\,}}(A)+{{\,\textrm{In}\,}}(D-CA^{-1}B)$.

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

Theorem 2.2

( [32, P.25]) If M is an $n \times n$ real matrix, then the characteristic polynomial of M is given by $\chi (x)=x^n-\delta _1{x}^{n-1}+\delta _2{x}^{n-2}+\ldots +(-1)^{n-1}\delta _{n-1}x+ (-1)^{n}\delta _{n}$ where $\delta _r$ denotes the sum of all principal minors of order r for all $r=1,2,\ldots ,n$.

Theorem 2.3

([32]) Let M be a symmetric matrix of order n and A be a principal submatrix of M order m where $1\le m \le n$. If the eigenvalues of M and A are $\lambda _1\ge \lambda _2\ge \cdots \ge \lambda _n$ and $\beta _1\ge \beta _2\ge \cdots \ge \beta _m$, respectively, then $\lambda _i\ge \beta _i \ge \lambda _{n-m+i},$ for all $i=1,2,\ldots ,m$. Moreover, $i_+(M)\ge i_+(A)~ \text {and}~i_-(M)\ge i_-(A).$

Lemma 2.4

[see [3, 14]] Suppose that $p(x) = {x}^n+\alpha _1{x}^{n-1}+\alpha _2{x}^{n-2}+\ldots +\alpha _{n-1}x+\alpha _n$ is a polynomial whose roots are all non-zero real numbers. If there exists $i \in \{1, \dots , n-1\}$ such that $\alpha _i \ne 0$ and $\alpha _{i+1}\ne 0$, then the roots of p(x) are not symmetric with respect to the origin (i.e., $p(a) =0$ but $p(-a)\ne 0$ for some real number a).

3 Relations between $G\in {\mathscr {B}}$ and its associated tree $T_G$

In this section, we associate a tree $T_G$ for each graph $G\in {\mathscr {B}}$ and obtain some interconnections between G and $T_G$ by employing the properties of trees. In particular, we show that ${{\,\textrm{diam}\,}}(G)={{\,\textrm{diam}\,}}(T_G)$ and $C(T_G)\subseteq C(G)$. 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 $G\in {\mathscr {B}}$ through the associated tree $T_G$ is motivated by [12, 14]. In [12], 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 $G\in {\mathscr {B}}$ in order to obtain a tree satisfying specific properties, and thus, the construction is different from [12].

Let us begin this section by associating a subgraph $T_G$ for each $G\in {\mathscr {B}}$. It will be shown later that $T_G$ is a tree. We collect some selected vertices from each block B of G to define $T_G$. If $B= K_{1,1}$, then we collect both the vertices of B. Suppose that $B\ne K_{1,1}$ and B is a bridge block with $V_B\cap C_G=\{z_1,z_2\}$. If both the cut-vertices $z_1$ and $z_2$ lie on the same partite set, say $V_1(B)$, then we choose three vertices from $V_B$, which are $z_1$, $z_2$ and a non-cut-vertex in $V_2(B)$ with the minimum vertex label; otherwise, we collect exactly two vertices $z_1$ and $z_2$ from $V_B$. In the case of leaf block $B\ne K_{1,1}$, we collect exactly three vertices, which are the cut-vertex in $V_B$, and non-cut-vertices with the minimum vertex label in each partite sets $V_1(B)$ and $V_2(B)$. The graph $T_G$ is defined as the subgraph induced by the vertex set which is the union of all selected vertices in each block of G. The precise construction of $T_G$ is given below.

Definition 3.1

Let $G\in {\mathscr {B}}$ and let B be a block of G with bipartition $(V_1(B), V_2(B))$. For $i =1, 2$, we denote the non-cut-vertex with the minimum vertex label in $V_i(B)$, if it exists, by $u_i$. If $B\ne K_{1,1}$, we define

$$\begin{aligned} NC_B:= {\left\{ \begin{array}{ll} \{u_2\} &{} \text {if }|V_1(B)\cap C_G|=2,\\ \{u_1\} &{} \text {if }|V_2(B)\cap C_G|=2, \\ \{u_1,u_2\}&{} \text {if }B \text {is a leaf block},\\ \emptyset &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

If $B= K_{1,1}$, then we set $NC_B:= \{u_1,u_2\}\cap V_B$. Let $\displaystyle S:=\cup _{B} NC_B$ where B runs over all blocks of G. Define $T_G$ as the subgraph induced by the vertex subset $S\cup C_G$. That is,

$$\begin{aligned} T_G:=G\left[ S\cup C_G\right] . \end{aligned}$$

(3.1)

Remark 3.2

We mention that the choice of a non-cut-vertex with the minimum vertex label ensures the uniqueness of $T_G$.

Example 3.3

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

In the following, we mention a few properties of $G\in {\mathscr {B}}$ that are needed in the sequel. For more details, we refer to [2, 7].

$(P_1)$:

Suppose that B is a block of G with bipartition $(V_1(B),V_2(B))$. Then $|V_1(B)|=1$ if and only if $|V_2(B)|=1$.

$(P_2)$:

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

$(P_3)$:

Cycles of G are exactly cycles of its blocks.

$(P_4)$:

Let $v\in V_G$. Then v is a cut-vertex of G if and only if it lies in at least two blocks of G.

Using the property $(P_4)$ and the fact that each edge of G lies on exactly one block of G (P. 59 in [2]), we obtain the following remark.

Remark 3.4

Let $G\in {\mathscr {B}}$ and B be a block of G. If xy and yz are edges in G such that $x,y\in V_B$ and $z\not \in V_B$, then $y\in C_G$. Moreover, if $P_G(u_0,u_{k+1})=u_0u_1\ldots u_ku_{k+1}$ $(k\ge 2)$ is a shortest path between $u_0$ and $u_{k+1}$ in G such that the edge $u_{i-1}u_{i}$ belongs to the block B for some $1 \le i \le k$ and $u_{i+1}\not \in V_B$, then $u_i\in C_G$. That is, a shortest path leaves a block B of G and enters into another block $B^{\prime }$ of G through a cut-vertex of G which lies in $V_B\cap V_{B^{\prime }}$.

Many results of this section deal the relationship between $G\in {\mathscr {B}}$ and its associated graph $T_G$. We first show that the graph $T_G$ is a tree. In order to distinguish, we use the notations $d_G(a,b)$ and $d_{T_G}(a,b)$, respectively, to denote the distance between the vertices a and b with respect to G and with respect to $T_G$.

Lemma 3.5

Let $G\in {\mathscr {B}}$ and $T_G$ be the subgraph of G defined in (3.1). Then $T_G$ is a tree and $d_G(a,b)=d_{T_G}(a,b)$ for all $a,b \in V_{T_G}$.

Proof

Note that each block of G is a complete bipartite graph. Since $T_G$ has at most three vertices from each block of G, by $(P_3)$, $T_G$ contains no cycle. We claim that $T_G$ is connected. Let $a,b \in V_{T_G}$. Since G is connected, there is a path between the vertices a and b in G. Let $P_G(a,b)=au_1u_2\ldots u_{k}b$ be a shortest path in G. If $u_i \in V_{T_G}$ for all $i=1,2,\ldots ,k$ then the claim follows. Suppose that $u_j \not \in V_{T_G}$ for some $j \in \{1,2,\ldots ,k\}$. Let $j_0$ be the smallest index such that $u_{j_0} \not \in V_{T_G}$. Then $u_{j_0}$ is not a cut-vertex of G because all the cut-vertices of G belong to $T_G$. Assume that the edge $u_{j_0-1}u_{j_0}$ belongs to the block B of G. If $j_0=1$ then take $u_0=a$. Without loss of generality, we assume that $u_{j_0} \in V_1(B)$. Then $u_{j_0-1} \in V_2(B)$. Since $u_{j_0}\not \in C_G$, by Remark 3.4, $u_{j_0+1}\in V_2(B)$ where we take $u_{j_0+1}=b$ if $j_0=k$. By the construction of $T_G$, it is clear that $V_1(B)\cap V_{T_G}\ne \emptyset$. Choose $u_{j_0}^{\prime } \in V_1(B)\cap V_{T_G}$. We now show that $u_{j_0}^{\prime } \not \in P_G(a,b)$. Clearly, $u_{j_0}^{\prime } \ne u_{j_0-1}$ and $u_{j_0}^{\prime } \ne u_{j_0+1}$. If there exists $i\in \{1,2,\ldots ,j_0-2\}$ such that $u_{j_0}^{\prime } = u_i$, then $au_1\ldots u_{i}u_{j_0+1}\ldots b$ is a path between a and b whose length is strictly less than $l\left( P_G(a,b)\right)$ which is not possible. Therefore, $u_{j_0}^{\prime } \ne u_i$ for all $i\in \{1,2,\ldots ,j_0-2\}$. Similarly, we see that $u_{j_0}^{\prime } \ne u_i$ for all $i\in \{j_0+2, \ldots ,k\}$. Hence $u_{j_0}^{\prime } \not \in P_G(a,b)$. We obtain a new path $P^{\prime }_G(a,b)$ from $P_G(a,b)$ by replacing the edges $u_{j_0-1}u_{j_0}$ and $u_{j_0}u_{j_0+1}$, respectively, by $u_{j_0-1}u_{j_0}^{\prime }$ and $u_{j_0}^{\prime }u_{j_0+1}$. Clearly, $l( P^{\prime }_G(a,b))=l( P_G(a,b))$. We replace $P_G(a,b)$ by $P^{\prime }_G(a,b)$. Repeating the above argument leads to obtain a path $P_{T_G}(a,b)$ in $T_G$ such that $l(P_{T_G}(a,b))=l( P_G(a,b))$. This implies that $T_G$ is connected and $d_{T_G}(a,b)\le l(P_{T_G}(a,b))=l( P_G(a,b))=d_G(a,b)$. Since $T_G$ is a subgraph of G, we have $d_{T_G}(a,b)\ge d_G(a,b)$. Hence, $d_{T_G}(a,b)= d_G(a,b)$. $\square$

The following lemma is useful in establishing the result ${{\,\textrm{diam}\,}}(G)={{\,\textrm{diam}\,}}(T_G)$. For notational simplicity, we also use the notation d(a, b) to denote the distance between two vertices a and b in G instead of $d_{G}(a,b)$.

Lemma 3.6

Let $G\in {\mathscr {B}}$. Given a and b in $V_G$, there exist $a^{\prime }$ and $b^{\prime }$ in $V_{T_G}$ such that $d(a,b)=d(a^{\prime },b^{\prime })$.

Proof

Let $a,b \in G$. If $a,b\in V_{T_G}$, then the result follows. Consider the case $b\not \in V_{T_G}$. We claim that there exists $b^{\prime }\in V_{T_G}$ such that $d(a,b)=d(a,b^{\prime })$. Let $P_G(a,b)=au_1u_2\ldots u_{k}b$ be a path such that $d(a,b)=l\left( P_G(a,b)\right)$ where $u_{k}\in V_1(B)$ and $b\in V_2(B)$ for some block B of G. We first prove the claim for $k\ge 2$.

Case 1: Suppose that $u_{k} \not \in C_G$. Then by Remark 3.4, $u_{k-1}\in V_2(B)$, and by $(P_2)$, $u_{k-2} \not \in V_B$ where we take $u_{k-2}=a$ if $k=2$. Therefore, by Remark 3.4, $u_{k-1} \in C_G$.

Subcase 1.1: Let B be a leaf block. By the construction of $T_G$, there exists a non-cut-vertex $b^\prime \in V_2(B)\cap V_{T_G}$. Using $(P_3)$ and $k\ge 2$, we see that $a \not \in V_B$ and $b^\prime \not \in P_G(a,b)$. Therefore, $b^{\prime } u_{k} u_{k-1}\ldots u_1a$ is a path. It is clear that $u_{k-1}\in {\bar{P}}_G(b^{\prime },a)$ for all shortest paths ${\bar{P}}_G(b^{\prime },a)$ between the vertices $b^{\prime }$ and a. Hence, $d(b^{\prime },a)=d(b^{\prime },u_{k-1})+d(u_{k-1},a)=d(b,a)$.

Subcase 1.2: Assume that B is a bridge block and z is a cut-vertex of G in B other than $u_{k-1}$.

Subcase 1.2.1: If $z \in V_2(B)$, then, by $(P_3)$, $z \not \in P_G(a,b)$. Therefore, $P_G(z,a)=zu_{k} u_{k-1}\ldots u_1a$ is a path. Again by $(P_3)$, $u_{k-1}\in {\bar{P}}_G(z,a)$ for all shortest paths ${\bar{P}}_G(z,a)$ between z and a. This fact leads to $d(z,a)=d(z,u_{k-1})+d(u_{k-1},a)=d(b,a)$. Since $z\in C_G$, we have $z\in V_{T_G}$. Therefore, in this case, choose $b^{\prime }=z$.

Subcase 1.2.2: Suppose that $z \in V_1(B)$. Then $z\ne u_{k-2}$ because $u_{k-2}\not \in V_B$. So, by $(P_3)$, $z \not \in P_G(a,b)$, and $u_{k-1}\in {\bar{P}}_G(z,a)$ for all shortest paths ${\bar{P}}_G(z,a)$ between z and a. This implies that $d(z,a)=1+d(u_{k-1},a)=d(b,a)-1$. Since $z\in C_G\cap V_1(B)$, by $(P_4)$, $z\in V_{B^{\prime }}$ for some block $B^{\prime }\ne B$. By the construction of $T_G$, there exists $b^{\prime } \in V_{B^{\prime }}\cap V_{T_G}$ such that $b^{\prime }$ is adjacent to z. Again by $(P_3)$, $b^{\prime } \not \in P(z,a)=zu_{k-1}\ldots u_1a$ and $z\in {\bar{P}}_G(b^{\prime },a)$ for all shortest paths ${\bar{P}}_G(b^{\prime },a)$ between $b^{\prime }$ and a. Therefore, $d(b^{\prime },a)=d(b^{\prime },z)+d(z,a)=1+d(z,a)=d(b,a)$.

Case 2: Let $u_{k} \in C_G$. Since $b\in V_2(B)$ with $b \not \in V_{T_G}$, we have $|V_2(B)|\ge 2$ because $V_2(B)\cap V_{T_G}\ne \emptyset$. Therefore, $B\ne K_{1,1}$.

Subcase 2.1: If B is a leaf block, then choose $b^{\prime }\in V_2(B)\cap V_{T_G}$. We show that $u_{k-1} \not \in V_B$. On the contrary, assume that $u_{k-1} \in V_B$. Then, by $(P_2)$, $u_{k-2}\not \in V_B$. Also, by Remark 3.4, $u_{k-1}\in C_G$ which is not possible because B is a leaf block with $u_k\in V_B\cap C_G$. Hence $u_{k-1} \not \in V_B$ and this yields $b^{\prime }\ne u_{k-1}$. Therefore, by ($P_3$), we observe that $b^{\prime }\not \in P_G(a,b)$, and by Remark 3.4, $u_k\in {\bar{P}}_G(b^{\prime },a)$ for all shortest paths ${\bar{P}}_G(b^{\prime },a)$ between $b^{\prime }$ and a. Clearly, $b^{\prime }u_ku_{k-1}\ldots u_1a$ is a path, and we have $d(b^{\prime },a)=d(b^{\prime }, u_k)+d(u_k,a)=d(b,a)$.

Subcase 2.2: Suppose that B is a bridge block. Let $z\in V_B\cap C_G$ with $z\ne u_k$.

Subcase 2.2.1: If $z \in V_1(B)$, then take $b^{\prime }\in V_2(B)\cap V_{T_G}$. Rest of the proof in this case is same as that of subcase 2.1.

Subcase 2.2.2: Let $z \in V_2(B)$. If $z\ne u_{k-1}$, then $u_{k-1} \not \in V_B$. By $(P_3)$, $z \not \in P_G(a,b)$, and $u_{k}\in {\bar{P}}_G(z,a)$ for all shortest paths ${\bar{P}}_G(z,a)$ between z and a. Therefore, $zu_ku_{k-1}\ldots u_1a$ is a shortest path and hence $d(z,a)=d(b,a)$. Suppose that $z= u_{k-1}$. Since $u_k \in C_G$, there exists $b^{\prime }\in V_{B^{\prime }}\cap V_{T_G}$ such that $b^{\prime }$ is adjacent to $u_k$ where $B^{\prime }$ is a block of G different from B. Then, by $(P_3)$, $b^{\prime } \not \in P_G(u_k,a)$, and $u_{k}\in {\bar{P}}_G(b^{\prime },a)$ for all shortest paths ${\bar{P}}_G(b^{\prime },a)$. By considering the path $P_G(b^{\prime },a)=b^{\prime }u_ku_{k-1}\ldots u_1a$, we obtain $d(b^{\prime },a)=d(b,a)$.

Hence, from the above cases, we have $d(a,b)=d(a,b^{\prime })$ for some $b^{\prime }\in V_{T_G}$. The proof for the case $k=1$ can be verified similarly. If $a\in V_{T_G}$, then the result follows. If $a \not \in V_{T_G}$, then repeat the above argument to the newly obtained path $P_G(b^{\prime },a)$ to find $a^{\prime }\in V_{T_G}$ such that $d(a,b)=d(a^{\prime },b^{\prime })$. This completes the proof. $\square$

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

Lemma 3.7

For each $G\in {\mathscr {B}}$, ${{\,\textrm{diam}\,}}(G)={{\,\textrm{diam}\,}}(T_G)$.

Proof

Let $P_G(a,b)$ be a diametrical path in G. By Lemma 3.6, $d_G(a,b)=d_G(a^{\prime },b^{\prime })$ for some $a^{\prime },b^{\prime } \in V_{T_G}$. Therefore, by Lemma 3.5, $d_G(a^{\prime },b^{\prime })=d_{T_G}(a^{\prime },b^{\prime })$. This implies that ${{\,\textrm{diam}\,}}(G)=d_G(a,b)=d_{T_G}(a^{\prime },b^{\prime })\le {{\,\textrm{diam}\,}}(T_G)$. To prove the other inequality, let $x,y \in V_{T_G}$ such that $d_{T_G}(x,y)= {{\,\textrm{diam}\,}}(T_G)$. Using Lemma 3.5, we get $d_{T_G}(x,y)= d_{G}(x,y)\le {{\,\textrm{diam}\,}}(G)$. $\square$

Let H be a connected graph and let $u\in V_H$. We denote the eccentricity of a vertex u with respect to H by $e_H(u)$, and the subscript is omitted if it is clear from the context.

Lemma 3.8

If $G\in {\mathscr {B}}$, then $e_G(a)=e_{T_G}(a)$ for all $a\in V_{T_G}$.

Proof

Let $a\in V_{T_G}$. By Lemma 3.5, $\{d_{T_G}(a,x): x\in V_{T_G}\} \subseteq \{d_G(a,x): x\in V_{G}\}$. This implies $e_{T_G}(a)\le e_G(a)$. Let $b \in V_G$ be such that $e_G(a)=d_G(a,b)$. If $b\in V_{T_G}$, then again by Lemma 3.5, $d_G(a,b)=d_{T_G}(a,b)\le e_{T_G}(a)$. Suppose that $b\not \in V_{T_G}$. Then from the proof of Lemma 3.6, we have $d_G(a,b)=d_G(a,b^{\prime })$ for some $b^{\prime }\in V_{T_G}$. So, $e_{T_G}(a)\ge e_G(a)$ follows by Lemma 3.5. $\square$

Since most of the results of this paper hold for graphs in ${\mathscr {B}}$ with diameters at least four (for instance, see Example 4.3), we deal only with those graphs hereafter. In the following lemma, the eccentricity of a non-cut-vertex (when it exists) of $G\in {\mathscr {B}}$ is presented in terms of the eccentricities of the cut-vertices of G. In Sect. 4, we see the usefulness of this lemma in proving the spectral symmetry of E(G).

Lemma 3.9

Let $G\in {\mathscr {B}}$ be such that ${{\,\textrm{diam}\,}}(G) \ge 4$. Let B be a block of G and let $u\in V_B$ be a non-cut-vertex of G.

• (i) Suppose that B is a bridge block of G and $v_1$ and $v_2$ are cut-vertices of G belong to B. If $e(v_1)\le e(v_2)$, then

  $$\begin{aligned} e(u)= {\left\{ \begin{array}{ll} e(v_2) &{} \text {if } v_1,v_2,u\in V_1(B),\\ e(v_2)-1 &{} \text {if } v_1,v_2\in V_1(B)~\text {and}~ u\in V_2(B),\\ e(v_2)+1 &{} \text {if } v_1,u\in V_1(B)~\text {and}~ v_2\in V_2(B),\\ e(v_1)+1 &{} \text {if } v_1\in V_1(B)~\text {and}~ v_2,u\in V_2(B). \end{array}\right. } \end{aligned}$$

• (ii) If B is a leaf block and v is a cut-vertex of G belongs to B, then

  $$\begin{aligned} e(u)= {\left\{ \begin{array}{ll} e(v)+2 &{} \text {if } u,v\in V_1(B)~\text {or}~ u,v\in V_2(B),\\ e(v)+1 &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Proof

(i) Since B is a bridge block with $v_1,v_2\in V_B\cap C_G$, there exists a vertex $x\in V_G\setminus V_B$ such that $e(v_2)=d(v_2,x)$. Let $P(v_2,x)$ be a shortest path between $v_2$ and x in G. We claim that $v_1\in P(v_2,x)$. On the contrary, assume that $v_1 \not \in P(v_2,x)$. Then, by $(P_3)$ and Remark 3.4, $v_2\in {\bar{P}}(v_1,x)$ for all paths ${\bar{P}}(v_1,x)$. Therefore, $e(v_1)\ge d(v_1,x)=d(v_1,v_2)+d(v_2,x)\ge 1+e(v_2)>e(v_2)$, which contradicts the assumption $e(v_1)\le e(v_2)$. Hence, the claim $v_1\in P(v_2,x)$ follows. This implies that $d(x,v_2)=d(x,v_1)+d(v_1,v_2)$. Let $u\in V_B{\setminus } C_G$. Note that $v_1 \in {\bar{P}}(x,u)$ for all shortest paths ${\bar{P}}(x,u)$ by $(P_3)$ and Remark 3.4. Therefore, $d(x,u)=d(x,v_1)+d(v_1,u)=[d(x,v_2)-d(v_1,v_2)]+d(v_1,u)$, and hence,

$$\begin{aligned} e(u)\ge d(x,u)=d(x,v_2)+[d(v_1,u)-d(v_1,v_2)]=e(v_2)+[d(v_1,u)-d(v_1,v_2)].\nonumber \\ \end{aligned}$$

(3.2)

Since $v_1,v_2\in V_B\cap C_G$, there always exists $y\in V_G{\setminus } V_B$ such that $e(u)=d(u,y)$. Let P(u, y) be a shortest path between u and y. Suppose that $v_1 \in P(u,y)$ and $v_2 \not \in P(u,y)$. Then, by ($P_3$) and Remark 3.4, any shortest path ${\bar{P}}(v_2,y)$ contains $v_1$. This implies that $d(v_2,y)=d(v_2,v_1)+d(v_1,y).$ Therefore, we have

$$\begin{aligned} e(u)= d(u,y)&=d(u,v_1)+d(v_1,y)\\&=d(u,v_1)+[d(v_2,y)-d(v_1,v_2)]\nonumber \end{aligned}$$

(3.3)

$$\begin{aligned}&\le e(v_2)+[d(u,v_1)-d(v_1,v_2)]. \end{aligned}$$

(3.4)

If $v_1 \not \in P(u,y)$, then by Remark 3.4, $v_2 \in P(u,y)$. Also, by $(P_3)$, $v_2\in {\bar{P}}(v_1,y)$ for all shortest paths ${\bar{P}}(v_1,y)$. Then $d(y,v_1)=d(y,v_2)+d(v_2,v_1)$. Hence,

$$\begin{aligned} e(u)= d(u,y)=d(u,v_2)+d(v_2,y)&=d(u,v_2)+[d(y,v_1)-d(v_1,v_2)]\end{aligned}$$

(3.5)

$$\begin{aligned}&\le e(v_1)+[d(u,v_2)-d(v_1,v_2)]\end{aligned}$$

(3.6)

$$\begin{aligned}&\le e(v_2)+[d(u,v_2)-d(v_1,v_2)]. \end{aligned}$$

(3.7)

Case 1: Let $v_1,v_2\in V_1(B)$. Since P(u, y) is a shortest path between u and y, by Remark 3.4, we have $|V_{P(u,y)}\cap \{v_1,v_2\}|=1$ where $V_{P(u,y)}$ is the set of all vertices in the path P(u, y). Now, the result follows from (3.2), (3.4) and (3.7), case-by-case.

Case 2: Assume that $v_1 \in V_1(B)$ and $v_2 \in V_2(B)$.

Subcase 2.1: Let $u\in V_1(B)$. From (3.2), $e(u)\ge e(v_2)+1$. To prove the other inequality, we first show that $v_1\in P(y,u)$. On the contrary, assume that $v_1 \not \in P(y,u)$. Then by Remark 3.4, $v_2 \in P(y,u)$ and so, by $(P_3)$, $v_2 \in {\bar{P}}(y,v_1)$ for all shortest paths ${\bar{P}}(y,v_1)$. Therefore, $e(u)=d(y,u)=d(y,v_2)+d(v_2,u)=d(y,v_2)+d(v_2,v_1)=d(y,v_1)\le e(v_1)$. That is, $e(v_1)\ge e(u)\ge e(v_2)+1$ where the last inequality follows from (3.2). This contradicts the assumption $e(v_1)\le e(v_2)$. Hence, $v_1\in P(y,u)$. If $v_1\in P(y,u)$ and $v_2 \not \in P(y,u)$, then by (3.4), $e(u)\le e(v_2)+1$. Suppose that $v_1,v_2 \in P(y,u)$. Then $e(u)=d(y,u)=d(y,v_2)+d(v_2,u)\le e(v_2)+1$. This completes the proof in this case.

Subcase 2.2: Suppose that $u\in V_2(B)$. Let $e(v_1)=d(z,v_1)$ for some $z \in V_G{\setminus } V_B$, and let $P(z,v_1)$ be a shortest path between z and $v_1$ in G. If $v_2\in P(z,v_1)$, then by $(P_3)$, $v_2\in {\bar{P}}(z,u)$ for all shortest paths ${\bar{P}}(z,u)$. We have $d(z,u)=d(z,v_2)+d(v_2,u)$ and $d(z,v_1)=d(z,v_2)+1$. This implies that $e(u)\ge d(z,u)=d(z,v_1)+1=e(v_1)+1$. Suppose that $v_2\not \in P(z,v_1)$. Then again by $(P_3)$, $v_2\not \in {\bar{P}}(z,u)$, and by Remark 3.4, $v_1\in {\bar{P}}(z,u)$ for all paths ${\bar{P}}(z,u)$. So, $d(z,u)=d(z,v_1)+d(v_1,u)=e(v_1)+1$. Therefore, $e(u)\ge e(v_1)+1$. If $|V_{P(u,y)}\cap \{v_1,v_2\}|=1$, then the other inequality, $e(u)\le e(v_1)+1$ follows from (3.3) and (3.6). If $v_1,v_2 \in P(y,u)$, then $e(u)=d(y,u)=d(y,v_1)+d(v_1,u)\le e(v_1)+1$.

(ii) We claim that there exists $x\in V_G\setminus V_B$ such that $e(v)=d(x,v)$. If $e(v)\ge 3$, then the claim follows. Consider the case $e(v)= 2$. Suppose that the claim does not hold. Then $d(x,v)<e(v)$ for all $x \in V_G{\setminus } V_B$. That is, v is adjacent to all $x \in V_G\setminus V_B$. This leads to ${{\,\textrm{diam}\,}}(G)\le 3$, a contradiction to the hypothesis. Hence, the claim follows. Let $u,v \in V_1(B)$. By Remark 3.4, any path P(u, x) passes through v. So, $e(u)\ge d(x,u)=d(x,v)+d(v,u)=e(v)+2$. This implies $e(u)\ge 4$ and hence $e(u)=d(u,y)$ for some $y\in V_G{\setminus } V_B$. Note that $e(u)= d(y,u)=d(y,v)+2\le e(v)+2$. Hence $e(u)=e(v)+2$. Similarly, the proof follows for the remaining cases. $\square$

Let ${{\,\textrm{diam}\,}}(G) \ge 4$ and let B be a block of G such that $u\in V_B{\setminus } C_G$. Suppose that B is a leaf block and $v\in V_B\cap C_G$. From Lemma 3.9, we have $e(u)>e(v)$. If B is a bridge block such that the cut-vertices $v_1$ and $v_2$ of B are not in the same partite set, then also we have $e(u)>e(v_i)$ for some i. Suppose that the cut-vertices $v_1$ and $v_2$ of the bridge block B are in the same partite set, say $V_1(B)$, then $e(u)>e(u_0)$ when $u\in V_1(B)$ and $u_0\in V_2(B)$. Hence, $u \not \in C(G)$ in the above cases. Therefore, a non-cut-vertex u of the block B can belong to C(G) only when B is a bridge block whose cut-vertices are in one partite set of B and u lies in the other partite set of B. Precisely, the result is given in the following remark.

Remark 3.10

Let $G\in {\mathscr {B}}$ with ${{\,\textrm{diam}\,}}(G) \ge 4$ and let $x \in V_1(B)$ for some block B of G. If $x \in C(G) {\setminus } C_G$, then B is a bridge block with $|V_2(B)\cap C_G|=2$.

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

Lemma 3.11

Let $G\in {\mathscr {B}}$ and let ${{\,\textrm{diam}\,}}(G)\ge 4$. Then $C(T_G)\subseteq C(G)$.

Proof

Let $z\in C(T_G)$. Then by Lemma 3.8, $e_{G}(z)= e_{T_G}(z)$. Let $u\in V_G$. Suppose that $u\in V_{T_G}$. Then $e_G(u)=e_{T_G}(u) \ge e_{T_G}(z)=e_{G}(z)$. If $u \not \in V_{T_G}$, then $u \not \in C_G$ because $C_G \subseteq V_{T_G}$. Assume that $u \in V_B$ for some block B. If B is a leaf block such that $v\in V_B\cap C_G$, then by Lemma 3.9, we have $e_G(u)\ge e_G(v)=e_{T_G}(v) \ge e_{T_G}(z).$ Suppose that B is a bridge block such that $v_1,v_2\in V_B\cap C_G$. Consider the case $v_1,v_2\in V_1(B)$ and $u\in V_2(B)$. Then, by Lemma 3.9, $e_G(x)=e_G(u)$ for all $x\in V_2(B)$. In particular, by the construction of $T_G$, there exists $u_0\in V_2(B)\cap V_{T_G}$ such that $e_G(u_0)=e_G(u)$. This implies $e_G(u)=e_G(u_0)= e_{T_G}(u_0)\ge e_{T_G}(z).$ For the remaining cases of B, it directly follows from Lemma 3.9 that $e_G(u)\ge e_{T_G}(z)= e_{G}(z)$. Hence $z\in C(G)$, which completes the proof. $\square$

Recall the following result to study some properties of the central vertices of $T_G$.

Lemma 3.12

( [2], see [12, 14]) Let T be a tree on $n\ge 3$ vertices and ${{\,\textrm{diam}\,}}(T)=m$. Then the center C(T) of T has either a single vertex $z_1$ or two adjacent vertices $z_1$ and $z_2$ such that $e(z_i)= \lceil {\frac{m}{2}} \rceil$ for all $i=1,2$, where $\lceil {\frac{m}{2}}\rceil =\frac{m}{2}$ if m is an even integer, and $\lceil {\frac{m}{2}}\rceil =\frac{m+1}{2}$ if m is an odd integer. Moreover, $|C(T)|=1$ if ${{\,\textrm{diam}\,}}(T)$ is even, and $|C(T)|=2$ if ${{\,\textrm{diam}\,}}(T)$ is odd.

Let $G\in {\mathscr {B}}$ be such that ${{\,\textrm{diam}\,}}(G)$ is odd and greater than three. Then by Lemmas 3.7 and 3.12, $|C(T_G)|=2$. Assume that $C(T_G)=\{z_1,z_2\}$. Again by Lemma 3.12, the vertices $z_1$ and $z_2$ are adjacent, and by Lemma 3.11, $z_1,z_2 \in C(G)$. Suppose that both $z_1$ and $z_2$ are not in $C_G$. Then by Remark 3.10, the block B of G containing $z_1$ and $z_2$ should be a bridge block such that the cut-vertices of G in B are in one partite set (say $V_1(B)$), and $z_1$ and $z_2$ are in the other partite set ($V_2(B)$), which is not possible because $z_1$ and $z_2$ are adjacent. Therefore, we arrive at the following remark.

Remark 3.13

Let $G\in {\mathscr {B}}$ with ${{\,\textrm{diam}\,}}(G)\ge 4$. Then $|C(T_G)\cap C_G|\ge 1$ whenever $|C(T_G)|=2$.

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 3.14

Let $G\in {\mathscr {B}}$ be such that ${{\,\textrm{diam}\,}}(G) \ge 4$ and let $T_G$ be its associated tree.

• (i) Suppose that ${{\,\textrm{diam}\,}}(G)$ is even and $C(T_G)=\{z\}$ where $z\in V_1(B)$ for some block B. Then

  $$\begin{aligned} C(G)= {\left\{ \begin{array}{ll} \{z\} &{} \text {if } z\in C_G,\\ V_1(B)&{} \text {if } z\not \in C_G. \end{array}\right. } \end{aligned}$$

• (ii) Let ${{\,\textrm{diam}\,}}(G)$ be odd and $C(T_G)=\{z_1,z_2\}$ where $z_1\in V_1(B)$ and $z_2\in V_2(B)$ for some block B. Then

  $$\begin{aligned} C(G)= {\left\{ \begin{array}{ll} \{z_1,z_2\} &{} \text {if } z_1,z_2\in C_G,\\ \{z_1\}\cup V_2(B)&{} \text {if } z_1\in C_G ~\text {and}~ z_2\not \in C_G,\\ \{z_2\}\cup V_1(B)&{} \text {if } z_1\not \in C_G ~\text {and}~ z_2\in C_G. \end{array}\right. } \end{aligned}$$

Proof

(i) Case 1: Let $z\in C_G$. Since $C(T_G)=\{z\}$, by Lemma 3.11, $\{z\}\subseteq C(G)$. To prove the result in this case, it is sufficient to show that $C(G)\subseteq \{z\}$. Let $x \in C(G)$. We first claim that $x \in C_G$. On the contrary, assume that $x \not \in C_G$. Then by Remark 3.10 and Lemma 3.9, $x\in B^{\prime }$ for some bridge block $B^{\prime }$ such that the cut-vertices of G in $B^{\prime }$ belong to $V_1(B^{\prime })$ and $x \in V_2(B^{\prime })$ with $e_G(x)=e_G(y)$ for all $y\in V_2(B^{\prime })$. In particular, there exists $y_0\in V_2(B^{\prime })\cap V_{T_G}$ such that $e_G(x)=e_G(y_0)$ where the existence of $y_0$ is guaranteed by the construction of $T_G$. Note that $y_0\not \in C_G$ as $|C_G\cap V_1(B^{\prime })|=2$ and $y_0\in V_2(B^{\prime })$. By Lemma 3.8, $e_G(y_0)=e_{T_G}(y_0)$, and we have $e_{T_G}(y_0)=e_G(x)\le e_{G}(z)=e_{T_G}(z)$. Since $C(T_G)=\{z\}$, $e_{T_G}(y_0)\ge e_{T_G}(z)$ which implies $y_0\in C(T_G)$. That is, $y_0=z$, which is not possible because $z\in C_G$ and $y_0 \not \in C_G$. Hence, the claim $x \in C_G$ follows. Therefore, $x\in V_{T_G}$ and $e_G(x)=e_{T_G}(x)$. We have $e_{T_G}(x)\ge e_{T_G}(z)= e_{G}(z)\ge e_{G}(x)$, which yields $e_{T_G}(x)=e_{T_G}(z)$. That is, $x\in C(T_G)=\{z\}$ and hence $x=z$. Therefore, $C(G)=\{z\}$.

Case 2: Assume that $z\not \in C_G$. Since $z\in V_1(B)$, it follows from Remark 3.10 that B must be a bridge block whose cut-vertices are in $V_2(B)$. Also, by Lemma 3.9, we get $e_G(z)=e_G(u)$ for all $u\in V_1(B)$. Using the fact that $z\in C(T_G)\subseteq C(G)$, we have $V_1(B)\subseteq C(G)$. To prove $C(G)\subseteq V_1(B)$, let $x \in C(G)$. We now show that $x \not \in C_G$. Suppose that $x \in C_G$. Then $x\in V_{T_G}$ and by Lemma 3.8, we have $e_{T_G}(x)=e_G(x)\le e_G(z)=e_{T_G}(z)\le e_{T_G}(x)$. This implies $e_{T_G}(z)= e_{T_G}(x)$ and hence $x\in C(T_G)=\{z\}$. That is, $x=z$, which is absurd as $x\in C_G$ and $z\not \in C_G$. Therefore, $x \not \in C_G$. Since $x\in C(G)$, by Remark 3.10, $x \in V_2(B^{\prime })$ for some bridge block $B^{\prime }$ with $|V_1(B^{\prime })\cap C_G|=2$. As in the previous case, we have $e_G(x)=e_G(y_0)$ for some $y_0\in V_2(B^{\prime })\cap V_{T_G}$ and $y_0=z$. This yields $y_0\in V_1(B)$. Since $y_0\in V_B\cap V_{B^\prime }$ and $y_0\not \in C_G$, it follows by $(P_3)$ that $B=B^{\prime }$ and $V_2(B^{\prime })=V_1(B)$. This implies $x\in V_1(B)$, which completes the proof in this case.

(ii) It is similar to the proof of item (i). $\square$

Remark 3.15

It is clear from Theorem 3.14 that all the central vertices of a graph $G\in {\mathscr {B}}$ with ${{\,\textrm{diam}\,}}(G)\ge 4$ lie completely in one specific block B of G. In particular, all the central vertices of G belong to exactly one partite set of B if and only if ${{\,\textrm{diam}\,}}(G)$ 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 [3, 12, 14, 15, 17]. 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) of a graph G are symmetric about the origin if and only if G is bipartite [3]. In [14], it is proved that the eigenvalues of the eccentricity matrix E(T) of a tree T are symmetric about the origin if and only if ${{\,\textrm{diam}\,}}(T)$ is odd. A similar equivalence is established for a subclass of block graphs (clique trees) in [12]. 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 $G\in {\mathscr {B}}$ and the associated tree $T_G$ given in Example 3.3. For the purpose of illustrating Theorem 4.2, let us relabel the vertices $v_7$, $v_{9}$ and $v_{10}$ by $z_1$, $z^{\prime }$ and $z_2$, respectively. Note that ${{\,\textrm{diam}\,}}(G)={{\,\textrm{diam}\,}}(T_G)=7$, and $C(T_G)=\{z_1,z_2\}$ where $z_1$ and $z_2$ are adjacent with $z_1 \in C_G$ and $z_2 \not \in C_G$. Let B be the block of G containing the edge $z_1z_2$ such that $z_1\in V_1(B)$ and $z_2\in V_2(B)$. It is clear that B is a bridge block with $z^{\prime }\in C_G\cap V_1(B)$ and $z^{\prime }\ne z_1$. Let $G^{\prime }$ be the subgraph of G obtained by deleting all the edges of the block B in G. Let $C_1$ and $C_2$ be the components of $G^{\prime }$ containing the vertices $z_1$ and $z^{\prime }$, respectively. Clearly, the sets

$$\begin{aligned} U_1&= \{ x\in V_{C_1}: d(x,z_1)=3\} =\{v_1,v_2,v_6\}, \\ U_{2}&= \{ y\in V_{C_1}:0 \le d(y,z_1)< 3\} =\{v_3,v_4,v_5,z_1\}, \\ U_3&= \{ x\in V_{C_2}: d(x,z^{\prime })=2\} =\{v_{15},v_{16}\},\\ U_{4}&= \{ y\in V_{C_2}:0 \le d(y,z^{\prime })<2\} =\{z^{\prime },v_{13},v_{14}\},\\ U_{5}&= V_1(B)\setminus \{z_1,z^{\prime }\} =\{v_8\}, \quad \text {and}\\ U_{6}&= V_2(B) =\{z_2,v_{11},v_{12}\}~\text { partitions }V_{G}.~ \end{aligned}$$

Let .

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

Using SAGEMATH, it is computed that the eigenvalues of E(G) are 0, 30.0375, 11.3025, $-11.3025$ and $-30.0375$ with respective multiplicities 12, 1, 1, 1 and 1. Thus, the spectrum of E(G) is symmetric with respect to the origin, and ${{\,\textrm{In}\,}}(E(G))=(2,2,12)$.

Theorem 4.2

Let $G \in {\mathscr {B}}$ be such that ${{\,\textrm{diam}\,}}(G)=2\,m+1$, $m\ge 2$. Then the following statements hold:

• (i) ${{\,\textrm{In}\,}}(E(G))=(2,2,n-4)$, where n is the number of vertices of G.

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

Proof

Consider the tree $T_G$ associated with G. By Lemmas 3.7 and 3.12, ${{\,\textrm{diam}\,}}(T_G)=2m+1$ and $|C(T_G)|=2$. Let $C(T_G)=\{z_1,z_2\}$. Again by Lemma 3.12, $z_1$ and $z_2$ are adjacent, and $e_{T_G}(z_i)=m+1$ for $i=1,2$. Also, by Lemmas 3.8 and 3.11, $z_1,z_2 \in C(G)$ and $e_{G}(z_i)=m+1$ for $i=1,2$. Let $z_1\in V_1(B)$ and $z_2\in V_2(B)$ for some block B of G. By Remark 3.13, $z_i \in C_G$ for at least one i. Without loss of generality, assume that $z_1 \in C_G$.

(i) Case 1: Suppose that $z_2\not \in C_G$. Since $z_2\in \left( C(G)\cap V_2(B)\right) \setminus C_G$, by Remark 3.10, B is a bridge block whose cut-vertices are in $V_1(B)$. Let $z^{\prime }\ne z_1$ be the another cut-vertex of G in B. By $(P_1)$, we have $|V_2(B)|\ge 2$ as $\{z_1,z^{\prime }\}\subseteq V_1(B)$. Now obtain a subgraph $G^{\prime }$ from G by deleting all the edges of B. Clearly, $V_G=V_{G^{\prime }}$. By Remark 3.4 and $(P_3)$, $G^{\prime }$ contains at least four components as $|V_i(B)|\ge 2$ for all $i=1,2$, and the vertices $z_1$ and $z^{\prime }$ lie in different components. Let $C_1$ and $C_2$ be the non-trivial components of $G^{\prime }$ containing the vertices $z_1$ and $z^{\prime }$, respectively. Let $x\in V_{C_1}$ and $y\in V_{C_2}$. Then by $(P_3)$,

$$\begin{aligned} z_1,z^{\prime } \in {\bar{P}}(x,y)~\text {for all paths }{\bar{P}}(x,y) \text {between }x \text {and }y\text { in }G. \end{aligned}$$

(4.1)

We prove the following inequalities to find a partition for $V_G$:

$$\begin{aligned} d(a,z_1)\le m~ \text {for all }a\in V_{C_1}\quad \text {and}\quad d(b,z^{\prime })\le m-1~ \text {for all }b\in V_{C_2}. \end{aligned}$$

(4.2)

Since $e(z_1)=m+1$, we have $d(a,z_1)\le m+1$. Suppose $d(a,z_1)= m+1$ then $d(a,z_2)= m+2$ which is not possible because $e(z_2)=m+1$. Hence $d(a,z_1)\le m$ which proves the first inequality in (4.2). Suppose that the second inequality in (4.2) does not hold. That is, there exists $b\in V_{C_2}$ such that $d(b,z^{\prime })> m-1$. Since $z_1 \in V_{C_1}$ and $b\in V_{C_2}$, using (4.1), we write $d(b,z_1)=d(b,z^{\prime })+d(z^{\prime },z_1) >(m-1)+d(z^{\prime },z_1)=m+1$. This implies $e(z_1)>m+1$, which is a contradiction. So, the second inequality in (4.2) holds.

We now show that the non-trivial components of $G^{\prime }$ are precisely $C_1$ and $C_2$. Suppose that $x\in V_G$ and $x\not \in V_{C_1}\cup V_B$. Let $P(x,z_1)$ be a shortest path between x and $z_1$ in G. If $z^{\prime }\not \in P(x,z_1)$, then by Remark 3.4, $V_{P(x,z_1)}\cup V_B=\{z_1\}$. Therefore, $P(x,z_1)$ is a path in $G^{\prime }$. Since $C_1$ is a component of $G^{\prime }$ containing $z_1$, we have $x\in V_{C_1}$ which is a contradiction. Hence $z^{\prime } \in P(x,z_1)$. Note that the subpath $P(x,z^{\prime })$ obtained from the shortest path $P(x,z_1)$ does not contain any vertex other than $z^{\prime }$ from the block B. This implies that $P(x,z^{\prime })$ is a path in $G^{\prime }$. Since $z^{\prime }\in V_{C_2}$, we have $x\in V_{C_2}$. Thus, the components of $G^{\prime }$ other than $C_1$ and $C_2$ are simply complete graphs of order one which arise from $V_B\setminus \{z_1,z^{\prime }\}$. We now define the following sets to obtain a partition for $V_{G^{\prime }}$:

$$\begin{aligned} U_1&= \{ x\in V_{C_1}: d(x,z_1)=m\},\quad \quad \quad \quad \quad ~~~~ U_{2} = \{ y\in V_{C_1}:0 \le d(y,z_1)< m\},\\ U_3&= \{ x\in V_{C_2}: d(x,z^{\prime })=m-1\},\quad \quad \quad \quad ~~ U_{4} = \{ y\in V_{C_2}:0 \le d(y,z^{\prime })<m-1\},\\ U_{5}&= V_1(B)\setminus \{z_1,z^{\prime }\}~ \text {whenever }|V_1(B)|\ge 3, ~ \text {and}~~ U_{6} = V_2(B). \end{aligned}$$

Clearly, $V_{G^{\prime }}=\cup _{l=1}^{6}U_l$ and $U_l\cap U_k=\emptyset$ for $l\ne k$. Also, $z_1\in U_2$, $z^{\prime }\in U_{4}$ and $z_2 \in U_{6}$. To see $U_1\ne \emptyset$ and $U_3\ne \emptyset$, consider $e(z_1)$ and $e(z^{\prime })$. Let $e(z_1)=d(b,z_1)$ for some $b\in V_G$. Since $e(z_1)=m+1 \ge 3$, we get $b \not \in V_B$. Also, by (4.2), $b \not \in V_{C_1}$. Therefore, $b \in V_{C_2}$, and by (4.1), $d(b,z_1)=d(b,z^{\prime })+d(z^{\prime },z_1)=d(b,z^{\prime })+2$. This implies $d(b,z^{\prime })=m-1$ and hence $b\in U_3$. Since $z_2 \in V_B{\setminus } C_G$, by item (i) of Lemma 3.9, we have $e(z_2)= \max \{e(z_1),e(z^{\prime })\}-1=\max \{m+1,e(z^{\prime })\}-1$. Since $e(z_2)=m+1$, it follows that $e(z^{\prime })=m+2$. Let $e(z^{\prime })=d(a,z^{\prime })$. Then by (4.2), $a \in V_{C_1}$, and by (4.1), we write $d(a,z^{\prime })=d(a,z_1)+d(z_1,z^{\prime })=d(a,z_1)+2$. This gives $d(a,z_1)=m$ and hence $a \in U_1$.

Subcase 1.1: Assume that $|V_1(B)|\ge 3$. Then $U_5\ne \emptyset$. Thus, $\{U_1,U_2,\ldots U_{6}\}$ partitions $V_{G^{\prime }}$ and hence $V_G$ as well. To find E(G) explicitly, we now compute the eccentricity of each vertex in G.

Let $a\in U_1 \cup U_2$. To find e(a), consider $d(a,b_0)$ where $b_0 \in U_3$ is fixed. By (4.1),

$$\begin{aligned} e(a)\ge d(a,b_0)=d(a,z_1)+d(z_1,z^{\prime })+d(z^{\prime },b_0) =d(a,z_1)+(m+1). \end{aligned}$$

(4.3)

We claim that $e(a)=d(a,z_1)+(m+1)$. Let $e(a)=d(a,x)$ for some $x\in V_G$. By (4.3), we have $e(a)\ge m+1$. If $x \in V_{C_1}$ then by (4.2), $e(a)=d(a,x)\le d(a,z_1)+d(z_1,x)\le d(a,z_1)+m<d(a,z_1)+(m+1)$, a contradiction to (4.3). So, $x \not \in V_{C_1}$. Similarly, we see that $x \not \in B$. Therefore, $x\in V_{C_2}$. By (4.2), $d(x,z^{\prime })\le m-1$. This implies $d(a,x)\le d(a,z_1)+d(z_1,z^{\prime })+d(z^{\prime },x) \le d(a,z_1)+2+(m-1)= d(a,z_1)+(m+1)$ and hence $e(a)\le d(a,z_1)+(m+1)$. Therefore, $e(a)= d(a,z_1)+(m+1)$ for all $a\in U_1 \cup U_2$. In particular, if $a\in U_1$ then $e(a)=2\,m+1$ and if $a\in U_2$ then $e(a)<2\,m+1$. Similarly, it can be shown that $e(b)= d(b,z^{\prime })+(m+2)$ for all $b\in U_3 \cup U_4$. Since $e(z^{\prime })=m+2$ and $e(z_1)=m+1$, by Lemma 3.9, we have $e(a)=m+2$ for all $a\in U_5$ and $e(a)=m+1$ for all $a\in U_6$.

We now compute the entries of E(G). Let $a \in U_1$ and $b\in V_G$. Then $e(a)\ge e(b)=\min \{e(a),e(b)\}$. If $b \in U_1 \cup U_2$, then $d(a,b)\le d(a,z_1)+d(z_1,b)= m+d(z_1,b)<e(b)$, and hence $E(G)_{a,b}=0$. Suppose that $b \in U_3 \cup U_4$. Using (4.1), we write $d(a,b)=d(a,z_1)+d(z_1,z^{\prime })+d(z^{\prime },b) =m+2+d(z^{\prime },b)= e(b)$, and so $E(G)_{a,b}=2\,m+1$ if $b\in U_3$ and $E(G)_{a,b}=e(b)$ if $b \in U_4$. If $b \in U_5 \cup U_6$, then by $(P_3)$ and Remark 3.4, $z_1\in {\bar{P}}(a,b)$ for all paths ${\bar{P}}(a,b)$. Therefore, $d(a,b)=d(a,z_1)+d(z_1,b)=m+d(z_1,b)= e(b)$. Thus, $E(G)_{a,b}$ is $m+2$ if $b\in U_5$ and $m+1$ if $b \in U_6$. Since $m\ge 2$, we see that $E(G)_{a,b}=0$ for all $a,b\in U_6$. It can be shown that $d(x,y)=e(x)=\min \{e(x),e(y)\}$ if $x \in U_2$ and $y\in U_3$, and $d(x,y)<\min \{e(x),e(y)\}$ for all other cases. Therefore, the eccentricity matrix E(G) can be written as

(4.4)

where $P_{a*}=P_{b*}$ for all $a,b \in U_1$ and $Q_{*u}=Q_{*v}$ for all $u,v \in U_3$. In fact, if $U_{3}=\{u_{31},u_{32},\ldots ,u_{3|U_{3}|}\}$ and $U_{4}=\{u_{41},u_{42},\ldots ,u_{4|U_{4}|}\}$, then $P_{a*}=\left( e(u_{41}), e(u_{42}), \ldots , e(u_{4|U_{4}|)}\right)$ for all $a \in U_1$ and $Q_{*v}=\left( e(u_{31}), e(u_{32}), \ldots , e(u_{3|U_{3}|)}\right) ^{\prime }$ for all $v \in U_3$. The structures of P and Q give ${{\,\textrm{rank}\,}}(E(G)\left( [U_i\mid V_G]\right) )=1$ for all $i=1,2,3,4$. Also, for some fixed $x\in U_{4}$,

$$\begin{aligned} E(G)_{a*}= \frac{1}{e(x)} {\left\{ \begin{array}{ll} (m+2) E(G)_{x*} &{}\text {for all } a \in U_{5},\\ (m+1) E(G)_{x*} &{}\text {for all } a \in U_{6}. \end{array}\right. } \end{aligned}$$

Hence, ${{\,\textrm{rank}\,}}(E(G)) \le 4$. Fix $x_1\in U_1$, $x_2\in U_3$. Let $P(x_1,z_1)$ and $P(x_2,z^{\prime })$ be shortest paths. Choose $y_1\in P(x_1,z_1)$ and $y_2\in P(x_2,z^{\prime })$ such that $x_i$ is adjacent to $y_i$ for $i=1,2$. To see ${{\,\textrm{rank}\,}}(E(G))\ge 4$, consider the principal submatrix R of E(G), which is given by

Note that $\det (R)\ne 0$. This implies ${{\,\textrm{rank}\,}}(E(G))\ge 4$ and hence ${{\,\textrm{rank}\,}}(E(G))= 4$. Applying Theorem 2.1 to R, we get ${{\,\textrm{In}\,}}(R)=(2,2,0)$, where we have taken $A=\left( {\begin{matrix} 0 &{} 2\,m+1\\ 2\,m+1 &{} 0 \end{matrix}}\right)$. Therefore, by Theorem 2.3, $n_-(E(G))\ge 2$ and $n_+(E(G))\ge 2$. Since ${{\,\textrm{rank}\,}}(E(G))= 4$, we have ${{\,\textrm{In}\,}}(E(G))=(2,2,n-4)$.

Subcase 1.2: Suppose that $|V_1(B)|=2$. Then $U_5= \emptyset$. Now, it is easy to see that $\{U_1,U_2,U_3,U_4,U_6\}$ partitions $V_G$, and E(G) is the principal submatrix of the matrix given in (4.4), obtained by deleting the rows and columns corresponding to the vertices in $U_5$. Hereafter, the proof is the same as in subcase 1.1.

Case 2: Assume that $z_2\in C_G$. That is, B is a bridge block such that $z_1\in V_1(B)\cap C_G$ and $z_2\in V_2(B)\cap C_G$. We obtain the non-trivial components $C_1$ and $C_2$ of $G^{\prime }$ such that $z_1 \in V_{C_1}$ and $z_2\in V_{C_2}$ where $G^{\prime }$ is the subgraph constructed from G by deleting all the edges of B. As in case 1, it is easily seen that $d(x,z_1)\le m$ for all $x\in V_{C_1}$ and $d(y,z_2)\le m$ for all $y\in V_{C_2}$. Define

$$\begin{aligned} U_1&= \{ x\in V_{C_1}: d(x,z_1)=m\},\quad U_{2} = \{ y\in V_{C_1}:0 \le d(y,z_1)< m\},\\ U_3&= \{ x\in V_{C_2}: d(x,z_2)=m\},\quad U_{4} = \{ y\in V_{C_2}:0 \le d(y,z_2)<m\},\\ U_{5}&= V_1(B)\setminus \{z_1\}~ \text {whenever}|V_1(B)|\ge 2, \quad \text {and}\\ U_{6}&= V_2(B)\setminus \{z_2\}~\text {whenever }|V_2(B)|\ge 2. \end{aligned}$$

Subcase 2.1: Suppose that $|V_1(B)|\ge 2$. Then, by ($P_1$), $|V_2(B)|\ge 2$. We now see that $\{U_1,U_2,\ldots , U_{6}\}$ partitions $V_G$. Let $a,b \in V_G$. It is easy to verify that

$$\begin{aligned} e(a)&= {\left\{ \begin{array}{ll} 2m+1 &{} \text {if } a\in U_1\cup U_3,\\ d(a,z_1)+(m+1) &{} \text {if } a\in U_2,\\ d(a,z_2)+(m+1) &{} \text {if } a\in U_4,\\ m+2 &{} \text {if } a\in U_5\cup U_6. \end{array}\right. } \end{aligned}$$

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

(4.5)

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

Let $\widetilde{E(G)}$ denote the $4 \times 4$ block matrix, which is the leading principal submatrix of E(G) in (4.5).

Subcase 2.2: Let $|V_1(B)|=1$. Then by ($P_1$), $|V_2(B)|= 1$. Note that $\{U_1,U_2,U_3, U_{4}\}$ partitions $V_G$, and $E(G)=\widetilde{E(G)}$. The rest of the proof is similar to case 1.

(ii) Consider the matrix E(G) given in (4.5). Suppose that

$$\begin{aligned} E(G){{\textbf {v}}}=\mu {{\textbf {v}}} ~\text {for some }0\ne \mu \in {\mathbb {R}} \text {and}~ {{\textbf {v}}}= (\mathbf {v_{1}}^{\prime },\mathbf {v_{2}}^{\prime },\ldots ,\mathbf {v_{6}}^{\prime })^{\prime } \in {\mathbb {R}}^n, \end{aligned}$$

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

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

For $m=1$, Theorem 4.2 need not be true. The following example illustrates this.

Example 4.3

Consider the following graph H:

The eccentricity matrix of H is given by

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

Note that ${{\,\textrm{diam}\,}}(H)=3$ and the rows and columns of E(H) are indexed by $\{u_1,u_2,u_3,u_4,u_5\}$. The eigenvalues of E(H) are $2,-2,-4.1394,-0.7849$ and 4.9243. So, the spectrum of E(G) is not symmetric about origin.

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

Let $G\in {\mathscr {B}}$ be such that ${{\,\textrm{diam}\,}}(G)$ is even. In this subsection, we compute the inertia of E(G) and show that the spectrum of 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) where $G\in {\mathscr {B}}$, see Theorem 4.7.

The following lemma is needed to prove Theorem 4.6.

Lemma 4.4

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

Proof

Let P(u, v) be a diametrical path in G where $u,v \in V_G$. We first prove the result for the case ${{\,\textrm{diam}\,}}(G)$ is even. Assume that ${{\,\textrm{diam}\,}}(G)=2\,m$ for some $m\ge 2$. Then by Lemmas 3.7 and 3.12, we have ${{\,\textrm{diam}\,}}(T_G)=2m$ and $|C(T_G)|=1$. Let $C(T_G)=\{z\}$. Since $C(T_G)\subseteq C(G)$, we have $z\in C(G)$. By Lemmas 3.8 and 3.12, $e_G(z)=e_{T_G}(z)=m=e_G(x)$ for all $x \in C(G)$. Note that there exists a vertex a in the path P(u, v) such that $d(u,a)=m=d(a,v)$. Therefore, $e_G(a)\ge m$. We prove the result by showing that $a \in C(G)$. To do this, we claim that $d(a,y)\le m$ for all $y\in V_G$. On the contrary, assume that $d(a,y_0)>m$ for some $y_0\in V_G$. Then a and $y_0$ do not belong to the same block, and $y_0$ does not lie on the path P(u, v). Clearly, either a lies in ${P(u,y_0)}$ for some shortest path $P(u,y_0)$ or a lies in ${P(v,y_0)}$ for some shortest path $P(v,y_0)$, otherwise $(P_3)$ fails. So, $d(u,y_0)=d(u,a)+d(a,y_0)>d(u,a)+m=2m$ or $d(v,y_0)>2m$, which is a contradiction. Hence, the result follows in this case. The proof is similar when ${{\,\textrm{diam}\,}}(G)$ is odd. $\square$

The notion of diametrically distinguished vertex is introduced in [14] 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 4.5

Let $G \in {\mathscr {B}}$ and let $u\in V_G$. Then u is said to be diametrically distinguished if there exists a diametrical path containing the vertex u and u is adjacent to z for some $z\in C(G)$.

Theorem 4.6

Let $G \in {\mathscr {B}}$ be such that ${{\,\textrm{diam}\,}}(G)=2\,m$ with $m\ge 2$. Let the center of the tree $T_G$, associated with G, be $\{z\}$ and k be the number of elements in the center C(G). Then the following hold:

• (i) If $z \not \in C_G$ then ${{\,\textrm{In}\,}}(E(G))={\left\{ \begin{array}{ll} (3,k+1,n-k-4) &{} \text {when }m=2, \\ (2,2,n-4) &{} \text {otherwise}, \end{array}\right. }$ where n is the number of vertices of G.

• (ii) If $z \in C_G$, then ${{\,\textrm{In}\,}}(E(G))=(r,r,n-2r)$ where r is the number of distinct blocks of G having a diametrically distinguished vertex.

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

Proof

By Lemma 3.11, we have $z \in C(G)$, and by Lemma 3.12, $e(z)=m$.

(i) Suppose that $z\not \in C_G$. Let $z \in V_1(B)$ for some block B of G. Then by Remark 3.10 and Theorem 3.14, B is a bridge block such that $w_1,w_2\in V_2(B) \cap C_G$ and $C(G)=V_1(B)$. Without loss of generality, assume that $e(w_1)\le e(w_2)$. By Lemma 3.9, $e(z)=e(w_2)-1$ which implies $e(w_2)=m+1$. Since $w_1 \not \in C(G)$ and $e(z)\le e(w_1)\le e(w_2)$, we get $e(w_1)=m+1$. Construct a subgraph $G^{\prime }$ from G by deleting all the edges of the block B. Then by $(P_3)$ and $(P_4)$, $G^{\prime }$ has two non-trivial components $C_1$ and $C_2$ containing the vertices $w_1$ and $w_2$, respectively, and the remaining $|V_B|-2$ components of $G^{\prime }$ are complete graphs of order one. We claim that $d(x,w_1)\le m-1$ for all $x \in V_{C_1}$. Suppose there exists $x_0 \in V_{C_1}$ such that $d(x_0,w_1)> m-1$. Then by ($P_3$) and Remark 3.4 that $w_1 \in {\bar{P}}(x_0,z)$ for all paths ${\bar{P}}(x_0,z)$. This implies $d(x_0,z)=d(x_0,w_1)+d(w_1,z)>m$ which yields $e(z)>m$, a contradiction. Hence, $d(x,w_1)\le m-1$ for all $x \in V_{C_1}$. Similarly, $d(y,w_2)\le m-1$ for all $y \in V_{C_2}$. For $i=1,2$, define

$$\begin{aligned} U_i&= \{ x\in V_{C_i}: d(x,w_i)=m-1\},\\ U_{2+i}&= \{ y\in V_{C_i}:0 \le d(y,w_i)<m-1\},\\ U_{5}&= V_1(B)~ \text {and}\\ U_{6}&= V_2(B)\setminus \{w_1,w_2\}~\text {whenever }|V_2(B)|\ge 3. \end{aligned}$$

Let $x_0 \in V_G$ such that $e(w_1)=d(x_0,w_1)$. This implies $x_0 \not \in V_{C_1}$ as $e(w_1)=m+1$. Since $m\ge 2$ and $w_1 \in V_2(B)$, $x_0 \not \in V_B$. Therefore, $x_0 \in V_{C_2}$. By ($P_3$), $w_2 \in {\bar{P}}(x_0,w_1)$ for all paths ${\bar{P}}(x_0,w_1)$. Hence, $d(x_0,w_1)=d(x_0,w_2)+d(w_2,w_1)=d(x_0,w_2)+2$ which yields $d(x_0,w_2)=m-1$. Therefore, $x_0 \in U_2$. Similarly, using the fact that $e(w_2)=m+1$, we see that $U_1\ne \emptyset$. Assume that $|V_2(B)|\ge 3$. Now, it is clear that $\{U_i:1\le i\le 6\}$ partitions $V_G$.

Since $C(G)=V_1(B)$, $e(a)=e(z)=m$ for all $a \in U_5$. By Lemma 3.9, $e(b)=m+1$ for all $b \in U_6$. By ($P_3$), we see that $w_1,w_2 \in P_G(x,y)$ for all $x \in V_{C_1}$ and $y \in V_{C_2}$. Fix $a_0 \in U_1$ and $b_0 \in U_2$. Similar to subcase 1.1 of item (i) in Theorem 4.2, we can compute the eccentricity of the remaining vertices of $V_G$, using the shortest paths $P(a_0,w_1)$ in $C_1$ and $P(b_0,w_2)$ in $C_2$, which are given below:

$$\begin{aligned} e(a)&= {\left\{ \begin{array}{ll} 2m &{} \text {if } a\in U_1\cup U_2,\\ d(a,w_1)+(m+1) &{} \text {if } a\in U_3,\\ d(a,w_2)+(m+1) &{} \text {if } a\in U_4. \end{array}\right. } \end{aligned}$$

The eccentricity matrix E(G) is given by

(4.6)

where $P_{a*}=P_{b*}$ for all $a,b\in U_1$, $Q_{u*}=Q_{v*}$ for all $u,v\in U_2$ and $R={\left\{ \begin{array}{ll} 2(J_k-I_k) &{} \text {if }m=2, \\ O &{} \text {if }m\ge 3. \end{array}\right. }$ The structures of P and Q yield that ${{\,\textrm{rank}\,}}(E(G)\left( [U_i\mid V_G]\right) )=1$ for all $i=1,2,3,4$.

Case 1: Assume that $m\ge 3$. Then the principal submatrix $R=O$. For each $a \in U_{5}\cup U_{6}$, observe that $E(G)_{a*}=\alpha E(G)_{w_1*}+\beta E(G)_{w_2*}$ for some real numbers $\alpha$ and $\beta$. Hence, ${{\,\textrm{rank}\,}}(E(G)) \le 4$. The rest of the proof is similar to that of Theorem 4.2 by considering the principal submatrix

$$\begin{aligned} \begin{pmatrix} 0 &{} 2m &{} 0 &{} 2m-1\\ 2m &{} 0 &{} 2m-1 &{} 0 \\ 0 &{} 2m-1 &{} 0 &{} 0 \\ 2m-1 &{} 0 &{} 0 &{}0 \end{pmatrix}. \end{aligned}$$

If $\vert V_2(B)|=2$, then $U_6=\emptyset$. Therefore, E(G) is a $5 \times 5$ block matrix. The result in this subcase can be verified similarly.

Case 2: Let $m=2$. Then $U_3=\{w_1\}$ and $U_4=\{w_2\}$. Note that the principal submatrix R of E(G) in (4.6) is $2(J_k-I_k)$ where $k=|C(G)|=|U_5|$. Let $C=\begin{pmatrix} 2J_{k\times |U_1|}&2J_{k\times |U_2|}&{\textbf{0}}&{\textbf{0}} \end{pmatrix}$ and $D=2(J_k-I_k)$. Then $D^{-1}=\frac{1}{2(k-1)}J_k-\frac{1}{2}I_k$. Let A denote the $4 \times 4$ block leading principal submatrix of E(G) in (4.6) and let $M=A-C^{\prime }D^{-1}C$. We have

$$\begin{aligned} M=\begin{pmatrix} \mu J_{|U_1|}&{} (\mu +4)J_{|U_1|\times |U_2|} &{}{\textbf{0}} &{} 3{\textbf{e}} \\ (\mu +4)J^{\prime }&{}\mu J_{|U_2|} &{} 3{\textbf{e}} &{} {\textbf{0}} \\ {\textbf{0}}^{\prime }&{} 3{\textbf{e}}^{\prime } &{} 0&{}0\\ 3{\textbf{e}}^{\prime } &{} {\textbf{0}}^{\prime }&{} 0 &{} 0\\ \end{pmatrix}~\text {where}~\mu =\frac{-2k}{k-1}. \end{aligned}$$

Consider the principal submatrix $N=\left( {\begin{matrix} \mu &{} \mu +4 &{} 0&{}3 \\ \mu +4 &{} \mu &{} 3&{}0 \\ 0 &{} 3 &{} 0 &{}0&{}\\ 3&{}0&{}0&{}0 \end{matrix}}\right)$ of M. Apply Theorem 2.1 to N, by taking A as $2\times 2$ leading principal submatrix of N, we get ${{\,\textrm{In}\,}}(N)=(2,2,0)$. Therefore, by Theorem 2.3, $n_+(M)\ge 2$ and $n_-(M)\ge 2$.

Subcase 2.1: Suppose that $|V_2(B)|=2$. Then $U_6=\emptyset$ and so E(G) is a $5\times 5$ block matrix. Since ${{\,\textrm{rank}\,}}(M)=4$, we have ${{\,\textrm{In}\,}}(M)=(2,2,n-k-4)$. Using Theorem 2.1 to E(G), we get ${{\,\textrm{In}\,}}(E(G))={{\,\textrm{In}\,}}(2J_k-2I_k)+{{\,\textrm{In}\,}}(M)=(1,k-1,0)+(2,2,n-k-4) =(3,k+1,n-k-4)$.

Subcase 2.2: If $|V_2(B)|\ge 3$ then $U_6\ne \emptyset$. By subcase 2.1, we have ${{\,\textrm{rank}\,}}(E(G))\ge k+4$. Since each column in $U_6$ is a linear combination of columns in $U_3$ and $U_4$, we have ${{\,\textrm{rank}\,}}(E(G))=k+4$. Therefore, the result follows using ${{\,\textrm{In}\,}}(X)$ and Theorem 2.3 where X is a $5 \times 5$ block leading principal submatrix of E(G).

(ii) Assume that $z\in C_G$. Then by Theorem 3.14, $C(G)=\{z\}$. Let $\deg (z)=p$ where $\deg (z)$ denotes the degree of the vertex z. Then by $(P_4)$, $p\ge 2$. Let $w_1,w_2, \ldots , w_p$ be the vertices in G that are adjacent to z. Now obtain the subgraph $G^{\prime }$ from G by deleting all the edges that are incident with z. By the construction of $G^{\prime }$, it is clear that $V_G=V_{G^{\prime }}$, and $G^{\prime }$ has a component $K_1$ with the vertex set $\{z\}$. Let the components of $G^{\prime }$ be $C_1,C_2, \ldots ,C_q$. Among these, let $C_1,C_2, \ldots ,C_r$ be the components containing the vertices $x_1,x_2, \ldots , x_r$, respectively, such that $d(x_i,z)=m$ for all $i=1,2,\ldots ,r$. Since $e(z)=m$, existence of $C_i$ containing such $x_i$ is guaranteed. We claim that $r \ge 2$. Let $P_G(a,b)$ be a diametrical path in G where $a,b\in V_G$. Then, by Lemma 4.4, $z \in P_G(a,b)$, and hence $d(a,z)=m=d(b,z)$. Therefore, $a,b \in \cup _{i=1}^r V_{C_i}$. If $a\in V_{C_i}$ and $b\in V_{C_j}$ with $C_i\ne C_j$, then the claim follows. Using $(P_3)$, we observe that any path $P^{\prime }_G(a,b)$ between a and b in G contains the central vertex z as $z \in P_G(a,b)$. This implies a and b do not belong to a single component $C_i$ and hence $r\ge 2$.

We now find the possible components of $G^{\prime }$ containing a diametrically distinguished vertex. Let S be the set of all diametrically distinguished vertices of G and let $w\in S$. Then w is adjacent to z and there exists a diametrical path $P_G(a_0,b_0)$ containing w. By Lemma 4.4, $z \in P_G(a_0,b_0)$. Then either $d(a_0,w)=m-1$ or $d(b_0,w)=m-1$. Without loss of generality, assume that $d(a_0,w)=m-1$. That is, the path $P_G(a_0,b_0)$ yields a path $P_G(a_0,w)$ with $z\not \in P_G(a_0,w)$, and $d(a_0,z)=m$. This implies $a_0\in V_{C_i}$ for some $i=1,2,\ldots ,r$. By the maximality of $C_i$, the path $P_G(a_0,w)$ is completely contained in $C_i$. That is, $w\in V_{C_i}$. Hence, diametrically distinguished vertices are necessarily belong to $\cup _{i=1}^rC_i$.

We claim that there are exactly r distinct blocks of G that contain a diametrically distinguished vertex. For each $i \in \{1,2,\ldots ,r\}$, if we prove $V_{C_i}\cap S \ne \emptyset$ and $V_{C_i}\cap S \subseteq V_{B_i}$ for some block $B_i$ in G with $B_1, B_2, \ldots , B_r$ are pairwise distinct then the claim follows. Let $1\le i \le r$. By the definition of $C_i$, there exists $x_i \in V_{C_i}$ such that $d(x_i,z)=m$. Let $P_G(x_i,z)$ be a shortest path between $x_i$ and z. Let $w_i$ be the vertex in the path $P_G(x_i,z)$ such that $w_i$ is adjacent to z in G. Then we see that $d(x_i,w_i)=m-1$ which follows from the subpath $P_G(x_i,w_i)$ obtained from $P_G(x_i,z)$. Since $x_i\in V_{C_i}$, and $z \not \in P_G(x_i,w_i)$, by the maximality of $C_i$, $P_G(x_i,w_i)$ lies in $C_i$. Choose $j \in \{1,2,\ldots ,r\}$ such that $C_j \ne C_i$. Then there exist $x_j$ and $w_j$ in $V_{C_j}$ such that $d(x_j,w_j)=m-1$. Also, the subpath $P_G(x_j,w_j)$ obtained from a shortest path $P_G(x_j,z)$, completely lies in $C_j$. Since the components $C_i$ and $C_j$ are disjoint, we have $P_G(x_i,w_i)\cap P_G(x_j,w_j)=\emptyset$. Note that, by ($P_3$), $z\in {\bar{P}}_G(x_i,x_j)$ for all paths ${\bar{P}}_G(x_i,x_j)$. This implies $d(x_i,x_j)=d(x_i,z)+d(z,x_j)=m+m={{\,\textrm{diam}\,}}(G)$. That is, any shortest path between $x_i$ and $x_j$ in G is a diametrical path. In particular, the shortest paths $P_G(x_i,z)$ and $P_G(x_j,z)$ induce a diametrical path containing $w_i$. This implies $w_i\in S$ and hence $V_{C_i}\cap S \ne \emptyset$.

We next prove that $V_{C_i}\cap S \subseteq V_{B_i}$ for some block $B_i$ in G. Let $w \in V_{C_i}\cap S$. Assume that $w_i\in V_{B_i}$ for some block $B_i$ in G. Suppose that $|V_{C_i}\cap S|\ge 2$. Let $w^{\prime } \in V_{C_i}\cap S$ with $w^{\prime }\ne w$. Then $w^{\prime }\in V_{B_i}$, otherwise ($P_3$) fails. Hence, $V_{C_i}\cap S \subseteq V_{B_i}$. Given a block B in G, there exists $l\in \{1,2,\ldots ,q\}$ such that $V_B\setminus \{z\}\subseteq V_{C_l}$, which is clear from the construction of $G^{\prime }$. Since components of $G^{\prime }$ are disjoint, the existed l for the block B is unique. This implies $B_i \ne B_j$ for all $i,j \in \{1,2,\ldots ,r\}$ with $i\ne j$. Thus, there are exactly r distinct blocks of G which have a diametrically distinguished vertex.

We next show that ${{\,\textrm{rank}\,}}\left( E(G)\right) =2r$ by explicitly finding the matrix E(G). By suitably relabeling the vertices $w_1,w_2, \ldots , w_p$, it is assumed that $w_i\in V_{C_i}$ for all $1\le i \le r$. For each $i \in \{1,2,\ldots ,r\}$, define

$$\begin{aligned} U_i&= \{ x\in V_{C_i}: d(x,z)=m\},\\ U_{r+i}&= \{ y\in V_{C_i}:0< d(y,z)<m\},~~~\text { and}\\ U_{2r+1}&= V_{G}\setminus \cup _{i=1}^r\left( U_i\cup U_{r+i}\right) . \end{aligned}$$

It is clear that $x_i\in U_i$, $w_i\in U_{r+i}$, $z \in U_{2r+1}$ and $U_l \cap U_k=\emptyset$ for $l\ne k$. Thus, $\{U_1,U_2,\ldots , U_{2r+1}\}$ partitions $V_G$. Let $a\in V_G$. We claim that $e(a)=d(a,z)+m$. Note that $d(a,x)\le d(a,z)+d(z,x)\le d(a,z)+m$ for all $x\in V_G$, which yields $e(a)\le d(a,z)+m$. Choose an element y such that y and a are in different components of $G^{\prime }$ and $d(y,z)=m$. Since any path between a and y passes through z, we have $d(a,y)=d(a,z)+d(z,y)=d(a,z)+m$ and hence the claim. This implies $e(a)=2\,m$ for all $a\in U_i$.

Next we find the entries of E(G) case-by-case. Let $a,b\in V_G$ and $1\le i,j\le r$ with $i\ne j$.

• If $a\in U_i$ and $b\in U_j$, then $E(G)_{a,b}=2m$ as

  $$\begin{aligned} d(a,b)= d(a,z)+d(z,b)=m+m=2m= \min \{e(a),e(b)\}. \end{aligned}$$

• Note that $E(G)_{a,b}=0$ whenever $a,b\in U_i$ because

  $$\begin{aligned} d(a,b)\le d(a,w_i)+d(w_i,b)=(m-1)+(m-1)< \min \{e(a),e(b)\}. \end{aligned}$$

• Let $a\in U_i$ and $b\in U_{r+j}\cup U_{2r+1}$. Then $e(a)=2m$ and

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

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

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

(4.7)

Since the matrix obtained in (4.7) is of the form $\varepsilon (T)$ given in [14, Theorem 3.2], the remainder of the proof is similar to that of Theorem 3.2 in [14].

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

$$\begin{aligned} \chi _x(E(G))={x}^{n-2l}({x}^{2l}-\delta _1{x}^{2l-1}+\delta _2{x}^{2l-2}- \cdots +(-1)^{2l-1}\delta _{2l-1}x+(-1)^{2l}\delta _{2l}), \end{aligned}$$

where $\delta _{2\,l}\ne 0$ and $\delta _i$ is the sum of all principal minors of order i, for all $i=1,2,\ldots , 2l$. To prove the spectrum of E(G) is not symmetric with respect to the origin, by Lemma 2.4, it enough to find some $i\in \{1,2,\ldots ,2\,l-1\}$ such that $\delta _{i}\ne 0$ and $\delta _{i+1}\ne 0$. It is clear from (4.6) and (4.7) that every $2\times 2$ and $3 \times 3$ principal submatrices $Q_1$ and $Q_2$ of E(G) are, respectively, in the following form:

$$\begin{aligned} Q_1=\begin{pmatrix} 0 &{} \alpha \\ \alpha &{} 0 \end{pmatrix} ~\text {and}~ Q_2= \begin{pmatrix} 0 &{} \alpha &{} \beta \\ \alpha &{} 0 &{} \gamma \\ \beta &{} \gamma &{} 0 \end{pmatrix} ~\text {where }\alpha ,\beta \text { and }\gamma \text {are some non-negative integers.} \end{aligned}$$

Since $\det (Q_1)=-\alpha ^2$ and $\det (Q_2)=2\alpha \beta \gamma$, we have $\delta _2\le 0$ and $\delta _3\ge 0$. Also, the following principal submatrices have nonzero determinant:

where $x_i \in U_i$ for $i=1,2$, and $z \in C(T_G)$. Hence, $\delta _2< 0$ and $\delta _3>0$.

Case 2: Suppose that ${{\,\textrm{rank}\,}}(E(G))=k+4$. Since $k=|V_1(B)|$ with cut-vertices $w_1,w_2$ in $V_2(B)$, by ($P_1$), we have $k\ge 2$. If $k=2$, then the proof follows similar to the previous case. If $k\ge 3$, then by item (ii), $n_-\left( E(G)\right) \ge 4>n_+\left( E(G)\right)$, and hence, the proof is complete. $\square$

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

Theorem 4.7

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

Proof

It follows from Theorems 4.2 and 4.6. $\square$

The following example shows that Theorem 4.2 need not be true for a general bi-block graph.

Example 4.8

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

It is easy to see that 0 is not an eigenvalue of the principal submatrix

of E(G) and so $n_0(A)=0$. Using the first conclusion of Theorem 2.3, one can verify that $n_+(A)\ge 3$ and $n_-(A)\ge 3$ by considering the $3 \times 3$ leading principal submatrix of A. Hence, ${{\,\textrm{In}\,}}(A)=(3,3,0)$. Applying the second conclusion of Theorem 2.3 to the matrices E(G) and A, we get $n_+(E(G))\ge 3$ and $n_-(E(G))\ge 3$. Note that the fourth row of E(G) is a scalar multiple of the third row of E(G). Also, the sixth and seventh rows of E(G) are scalar multiples of the first row of E(G). So, the number of linearly independent rows in E(G) is at most 6 and hence ${{\,\textrm{rank}\,}}(E(G))\le 6$. Since ${{\,\textrm{rank}\,}}(E(G))= n_+(E(G))+n_-(E(G))\ge 6$, we get ${{\,\textrm{rank}\,}}(E(G))=6$. This implies that $n_0(E(G))=3$ and hence ${{\,\textrm{In}\,}}(E(G))=(3,3,3)$. Therefore, item (i) of Theorem 4.2 fails in this case. The nonzero eigenvalues of E(G) are $-9.4967$, $-4.3784,~ -2.9329,~ 1.4150,~ 5.2920$ and 10.1010, which are computed using SAGEMATH. Hence, the eigenvalues of E(G) are not symmetric about the origin.

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 [12, 16, 17, 23, 26]. In the next result, 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 4.9

(see [15]) Let $M=(m_{ij})$ be an $n \times n$ nonnegative symmetric matrix and G(M) be the graph on n vertices such that there is an edge between the vertices i and j in G(M) if and only if $m_{ij}\ne 0$. Then M is irreducible if and only if G(M) is connected.

Theorem 4.10

Let $G\in {\mathscr {B}}$ with ${{\,\textrm{diam}\,}}(G)\ge 4$. Then the eccentricity matrix E(G) is irreducible.

Proof

Let $G\in {\mathscr {B}}$. Using Lemma 4.9 and the matrices given in (4.4), (4.5), (4.6) and (4.7), it is direct to see that E(G) is irreducible. $\square$

5 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 $T_G$ for each $G\in {\mathscr {B}}$. A characterization for the spectrum of the eccentricity matrix E(G) is symmetric about the origin being given in terms of the diameter of G. Also, it is proved that the eccentricity matrix of $G\in {\mathscr {B}}$ with ${{\,\textrm{diam}\,}}(G)\ge 4$ is irreducible.