On the distance energy of k-uniform hypergraphs

Kshitij Sharma; Swarup Kumar Panda

doi:10.1515/spma-2023-0188

Artikel Open Access

On the distance energy of k-uniform hypergraphs

Kshitij Sharma und Swarup Kumar Panda

Veröffentlicht/Copyright: 20. Juli 2023

Veröffentlicht von

Veröffentlichen auch Sie bei De Gruyter Brill

Manuskript einreichen Informationen für Autor*innen Erkunden Sie dieses Fachgebiet

Aus der Zeitschrift Special Matrices Band 11 Heft 1

Abstract

In this article, we extend the concept of distance energy for hypergraphs. We first establish a relation between the distance energy and the distance spectral radius. Then, we obtain some bounds for the distance energy in terms of some invariant of hypergraphs such as the determinant of the distance matrix, number of vertices, and Wiener index along with the distance energy of join of k -uniform hypergraphs. Furthermore, it is shown that the determinant of the distance matrix of k -uniform hyperstar on n vertices is ( − 1 ) n − 1 ( n − 1 ) k n − k k − 1 . Later, the distance spectrum of k -uniform hyperstar is obtained, which gives the explicit distance energy of k -uniform hyperstar.

Keywords: hypergraph; distance matrix; distance degree; spectral radius; distance energy; hyperstar

MSC 2010: 05C15; 05C50; 05C65; 15A18

1 Introduction

A hypergraph ℋ is an ordered pair ℋ = ( V ( ℋ ) , E ( ℋ ) ) , where V = V ( ℋ ) is a nonempty finite set called the vertex set of ℋ and E = E ( ℋ ) is a family of subsets of V called the hyperedge set of ℋ . In a hypergraph, two vertices are said to be adjacent if there is a hyperedge that contains both of these vertices. Two hyperedges are said to be adjacent if their intersection is not empty. A hypergraph is said to be simple if all of its hyperedges are distinct and no hyperedge is a subset of another hyperedge. A hypergraph is said to be finite if V has a finite number of elements. The number of vertices in a hyperedge is called the cardinality of the hyperedge, and if all the hyperedges have same cardinality say k , then that hypergraph is known as the k-uniform hypergraph. The degree, d i , of a vertex i ∈ V , which is the number of hyperedges that contain i . Hypergraph theory was introduced by Claude [5]. Hypergraph theory found applications in chemistry [8–10].

For u , v ∈ V ( ℋ ) , a walk of length p from u to v in ℋ is defined to be a sequence of vertices and hyperedges v 0 e 1 v 1 … v p − 1 e p v p with v 0 = u and v p = v such that e i contains the vertices v i − 1 and v i , and v i − 1 ≠ v i for i = 1 , … , p . If all v i and e i are distinct, then we say that this walk is a path. If there is a path from u to v for any u , v ∈ V ( ℋ ) , then we say that ℋ is connected.

Let ℋ be a connected hypergraph. The distance between two vertices u and v in ℋ is the length of a shortest path from u to v , and it is denoted by d ℋ ( u , v ) . The diameter of ℋ is the maximum distance between the pair of vertices, and it is denoted by d ( ℋ ) or d .

Let ℋ be a connected k -uniform hypergraph with vertices n . The distance matrix of ℋ is the n × n matrix D ( ℋ ) = [ ( D ( ℋ ) ) i j ] such that

[ ( D ( ℋ ) ) i j ] = d ℋ ( i , j ) i ≠ j , 0 i = j .

The row sum of the i th row of D ( ℋ ) is known as the distance degree (or vertex transmission) of the i th vertex, and is denoted by D i ( ℋ ) . The second distance degree for the ith vertex of a hypergraph ℋ with n vertices is defined as, T i ( ℋ ) = ∑ j = 1 n d ℋ ( i , j ) D j ( ℋ ) . So D 1 ( ℋ ) , D 2 ( ℋ ) , … , D n ( ℋ ) , and T 1 ( ℋ ) , T 2 ( ℋ ) , … , T n ( ℋ ) are the distance degree sequence and second distance degree sequence of ℋ , respectively. A hypergraph ℋ is said to be r distance regular if D i ( ℋ ) = r for all i = 1 , 2 , … , n , and a hypergraph ℋ is said to be r pseudo distance regular if T i ( ℋ ) D i ( ℋ ) = r for all i = 1 , 2 , … , n .

The eigenvalues of D ( ℋ ) are called the distance eigenvalues of ℋ . Since D ( ℋ ) is real and symmetric, the distance eigenvalues of ℋ are real. Let λ 1 ( ℋ ) ≥ λ 2 ( ℋ ) ≥ ⋯ ≥ λ n ( ℋ ) be the eigenvalue corresponding to the distance matrix of a hypergraph ℋ . The matrix D ( ℋ ) is non-negative and irreducible. By Perron-Frobenius theorem, λ 1 ( ℋ ) is simple and there is a unique unit positive eigenvector corresponding to λ 1 ( ℋ ) . The eigenvalue λ 1 ( ℋ ) is called the distance spectral radius.

We define S ( ℋ ) to be the sum of the squares of the distances between all unordered pairs of vertices and M 1 ( ℋ ) to be the sum of the squares of the distance degree of the vertices. So we have the following:

∑ i = 1 n λ i ( ℋ ) = 0 S = S ( ℋ ) = ∑ u < v d ℋ 2 ( u , v ) 2 S ( ℋ ) = ∑ i = 1 n λ i 2 ( ℋ ) M 1 ( ℋ ) = ∑ i ∈ V D i 2 ( ℋ ) .

The Wiener index W ( ℋ ) of a hypergraph ℋ is defined to be the sum of distances between every unordered pair vertices, that is, W ( ℋ ) = ∑ u < v d ℋ ( u , v ) = 1 2 ∑ i = 1 n D i ( ℋ ) . We have the following:

W ( ℋ ) ≥ ( n − 1 ) n 2 S ( ℋ ) ≥ ( n − 1 ) n 2 .

The distance matrix is a mathematical object, which is being increasingly used in different fields that include chemistry. The origins of the distance matrix go back to the very first article of Cayley [4] in 1841. Therefore, it is of interest to study the distance matrix and its eigenvalues. The distance spectrum of an ordinary graph has been studied extensively (see [1]).

Graphs are used to present hydrocarbon molecules, and it is known that the approximation of π -electron energy may be computed from the eigenvalues of the graph, so keeping that concept, in 1977 Gutman [7] defined graph energy, and later on, Nikiforov [13] extended the concept of graph energy to matrices. Since hypergraphs have been found their application in chemistry, so the same definition of energy can be extended for hypergraphs. The distance energy of a hypergraph ℋ is defined as the sum of the absolute values of all the eigenvalues of the distance matrix. Mathematically, we have,

DE ( ℋ ) = ∑ i = 1 n ∣ λ i ( ℋ ) ∣ .

In Section 2, we give bounds for distance energy in terms of diameter ( d ), number of vertices ( n ), sum of the square of the distance between all the vertices ( S ) and determinant of distance matrix. We also find the distance energy of join of two k -uniform hypergraphs and give the bound for the distance energy of join of k k-uniform hypergraphs. In Section 3, we supply formula for the determinant of k -uniform hyperstar in form of number of vertices n and k . Later, we give the spectrum of the k -uniform hyperstar.

2 Distance energy

The distance energy of a hypergraph ℋ is defined as the sum of the absolute value of all the eigenvalues of the distance matrix. Mathematically, we have

DE ( ℋ ) = ∑ i = 1 n ∣ λ i ( ℋ ) ∣ .

To proceed further, we need the following lemma.

Lemma 2.1

[11] Let ℋ be a connected hypergraph with n vertices, and W be the Wiener index corresponding to ℋ . Then,

λ 1 ( ℋ ) ≥ 2 W ( ℋ ) n ,

with equality holds if and only if D 1 ( ℋ ) = D 2 ( ℋ ) = ⋯ = D n ( ℋ ) .

In this section, we establish a relation between the distance spectral radius and the distance energy of a hypergraph. We also obtain bounds for the distance energy of hypergraphs in terms of hypergraphs invariants.

The following theorem gives a relation between the distance energy and the distance spectral radius of a hypergraph. The proof is trivial, so we omit it.

Theorem 2.2

Let ℋ be a connected hypergraph with n ≥ 3 vertices, then, DE ( ℋ ) ≥ 2 λ 1 ( ℋ ) . With equality if and only if ℋ has only one positive eigenvalue.

The following result supplies a lower bound for the distance energy in terms of the number of vertices and the Wiener index.

Lemma 2.3

Let ℋ be a connected hypergraph with n vertices, and W be the Wiener index corresponding to ℋ . Then,

DE ( ℋ ) ≥ 4 W ( ℋ ) n ,

with equality holds if and only if ℋ has only one positive eigenvalue and D 1 ( ℋ ) = D 2 ( ℋ ) = ⋯ = D n ( ℋ ) .

Proof

Using Theorem 2.2, we have DE ( ℋ ) ≥ 2 λ 1 ( ℋ ) . By Theorem 2.1, we have λ 1 ( ℋ ) ≥ 2 W ( ℋ ) n . Hence, DE ( ℋ ) ≥ 4 W ( ℋ ) n . Note that DE ( ℋ ) = 4 W ( ℋ ) n if and only if DE ( ℋ ) = 2 λ 1 ( ℋ ) and λ 1 ( ℋ ) = 2 W ( ℋ ) n . Using Theorems 2.1 and 2.2, we have DE ( ℋ ) = 4 W ( ℋ ) n if and only if ℋ has only one positive eigenvalue and D 1 ( ℋ ) = D 2 ( ℋ ) = ⋯ = D n ( ℋ ) . □

The following theorem supplies a lower bound for the distance energy in terms of k .

Theorem 2.4

Let ℋ be a connected k-uniform hypergraph with n vertices, and m be the number of hyperedges. Then,

DE ( ℋ ) ≥ 2 ( k − 1 ) .

Equality holds if and only if k = n .

Proof

Using Lemma 2.3 and the proof of Theorem 2.1, we have DE ( ℋ ) ≥ 4 W n and 2 W ≥ ( k − 1 ) n . Hence, DE ( ℋ ) ≥ 2 ( k − 1 ) . The proof is complete.□

Lemma 2.5

Let ℋ be a connected k-uniform hypergraph with n vertices and diameter 2. Then,

∑ i = 1 n λ i 2 ( ℋ ) ≤ n ( 4 n − 3 k − 1 ) .

Proof

Since we know that,

∑ i = 1 n λ i 2 ( ℋ ) = ∑ i = 1 n ( D 2 ( ℋ ) ) i i = ∑ i = 1 n ∑ j = 1 n d ℋ ( i , j ) d ℋ ( j , i ) = ∑ i = 1 n ∑ j = 1 n d ℋ ( i , j ) 2 .

Since ℋ is a k -uniform hypergraph, each row of D has at least k − 1 entries, which are equal to 1. So there are at least ( k − 1 ) n entries in D ( ℋ ) , which are equal to 1. Then,

∑ i = 1 n λ i 2 ( ℋ ) ≤ ( k − 1 ) n + 2 2 ( n ( n − 1 ) − ( k − 1 ) n ) .

Thus, we have

□ ∑ i = 1 n λ i 2 ( ℋ ) ≤ ( 4 n − 3 k − 1 ) n .

The following result supplies an upper bound for the distance energy under certain condition.

Theorem 2.6

Let ℋ be a connected k-uniform hypergraph with n vertices and diameter 2. If 4 n 2 − ( 3 k + 1 ) n − λ 1 2 ( ℋ ) > 0 , then

DE ( ℋ ) ≤ λ 1 ( ℋ ) + ( n − 1 ) ( 4 n 2 − ( 3 k + 1 ) n − λ 1 2 ( ℋ ) ) .

Proof

After applying Cauchy-Schwartz inequality on ( ∣ λ 2 ( ℋ ) ∣ , ∣ λ 3 ( ℋ ) ∣ , … , ∣ λ n ( ℋ ) ∣ ) and ( 1 , 1 , … , 1 ) , we have,

∑ i = 2 n ∣ λ i ( ℋ ) ∣ 2 ≤ ( n − 1 ) ∑ i = 2 n ∣ λ i 2 ( ℋ ) ∣ ∑ i = 1 n ∣ λ i ( ℋ ) ∣ − λ 1 ( ℋ ) 2 ≤ ( n − 1 ) ∑ i = 1 n ∣ λ i 2 ( ℋ ) ∣ − λ 1 2 ( ℋ ) .

Using Lemma 2.5, we have

□ ( DE ( ℋ ) − λ 1 ( ℋ ) ) 2 ≤ ( n − 1 ) ( ( 4 n − 3 k − 1 ) n − λ 1 2 ( ℋ ) ) DE ( ℋ ) − λ 1 ( ℋ ) ≤ ( n − 1 ) ( 4 n 2 − ( 3 k + 1 ) n − λ 1 2 ( ℋ ) ) DE ( ℋ ) ≤ λ 1 ( ℋ ) + ( n − 1 ) ( 4 n 2 − ( 3 k + 1 ) n − λ 1 2 ( ℋ ) ) .

The following theorem supplies an upper bound for the distance energy in terms of n , k , diameter d , and minimum degree δ .

Theorem 2.7

Let ℋ be a connected k-uniform hypergraph with n vertices, minimum degree δ , and diameter d. Then,

DE ( ℋ ) ≤ n n d 2 − d ( d − 1 ) ( 4 d + 1 ) 6 − ( k + δ − 2 ) ( d 2 − 1 ) − 1 .

Proof

Let i be a vertex of ℋ and N be the cardinality of the set { t ∈ V ( ℋ ) : t ∼ i } . Let δ i be the degree of the vertex i . Therefore, there are δ i hyperedges containing the vertex i and each hyperedge contains at most k − 1 vertices that are adjacent to i . Also, there is at least one hyperedge that contains k − 1 vertices adjacent to i . Then, N ≥ k − 1 + δ i − 1 ≥ k − 2 + δ , as δ i ≥ δ .

We know that DE ( ℋ ) ≤ 2 n S . Then, we find a bound for 2 S including δ , as we know that

2 S ( ℋ ) = ∑ u , v ∈ V d ℋ ( u , v ) 2 2 S ≤ n ( 1 2 N i + 2 2 + ⋯ + ( d − 1 ) 2 + d 2 ( n − 1 − ( d − 2 ) − N i ) ) 2 S ≤ n ( d − 1 ) d ( 2 d − 1 ) 6 + N i − 1 + ( n − 1 − ( d − 2 ) − N i ) d 2 2 S ( ℋ ) ≤ n n d 2 − d ( d − 1 ) ( 4 d + 1 ) 6 − N i ( d 2 − 1 ) − 1 .

After rearranging, we have

(1) 2 S ( ℋ ) ≤ n n d 2 − d ( d − 1 ) ( 4 d + 1 ) 6 − ( k + δ − 2 ) ( d 2 − 1 ) − 1 .

Thus, we have

□ DE ( ℋ ) ≤ n n d 2 − d ( d − 1 ) ( 4 d + 1 ) 6 − ( k + δ − 2 ) ( d 2 − 1 ) − 1 .

Remark 2.8

Since S ( ℋ ) = ∑ u < v d ℋ 2 ( u , v ) and W = ∑ u < v d ℋ ( u , v ) , clearly, we have S ≤ W .

Proposition 2.9

Let ℋ be a connected hypergraph. Then, λ 1 ( ℋ ) ≤ 2 S .

Proof

Note that λ 1 2 ( ℋ ) , … , λ n 2 ( ℋ ) are the eigenvalues of the matrix D 2 ( ℋ ) . Therefore, ∑ i = 1 n λ i 2 ( ℋ ) = ∑ i = 1 n ∑ j = 1 n d ℋ 2 ( i , j ) . Then, λ 1 2 ( ℋ ) ≤ ∑ i = 1 n ∑ j = 1 n d ℋ 2 ( i , j ) = 2 S . Hence, λ 1 ( ℋ ) ≤ 2 S . □

Theorem 2.10

Let ℋ be a connected non-singular hypergraph with n vertices, then,

DE ( ℋ ) ≤ ( 3 W ) n 2 ∣ det ( D ( ℋ ) ) ∣ n − 1 .

Proof

Consider the matrix

x 11 x 12 … x 1 n x 21 x 22 … x 2 n ⋮ ⋮ ⋱ ⋮ x n 1 x n 2 … x n n = 1 1 ∣ λ 1 ( ℋ ) ∣ … 1 ∣ λ 1 ( ℋ ) ∣ 1 ∣ λ 2 ( ℋ ) ∣ 1 … 1 ∣ λ 2 ( ℋ ) ∣ ⋮ ⋮ ⋱ ⋮ 1 ∣ λ n ( ℋ ) ∣ 1 ∣ λ n ( ℋ ) ∣ … 1 .

Now applying Holder’s inequality for x i j , we have

∑ j = 1 n ∏ i = 1 n x i j 1 ∕ n ≤ ∏ i = 1 n ∑ j = 1 n x i j 1 ∕ n .

For the right-hand side (RHS), we have

1 + n − 1 ∣ λ 1 ( ℋ ) ∣ 1 ∕ n 1 + n − 1 ∣ λ 2 ( ℋ ) ∣ 1 ∕ n ⋯ 1 + n − 1 ∣ λ n ( ℋ ) ∣ 1 ∕ n ≤ 1 + n − 1 ∣ λ 1 ( ℋ ) ∣ 1 + n − 1 ∣ λ 2 ( ℋ ) ∣ ⋯ 1 + n − 1 ∣ λ n ( ℋ ) ∣ .

Since, W ≥ n − 1 , we obtain

1 + n − 1 ∣ λ 1 ( ℋ ) ∣ 1 ∕ n 1 + n − 1 ∣ λ 2 ( ℋ ) ∣ 1 ∕ n ⋯ 1 + n − 1 ∣ λ n ( ℋ ) ∣ 1 ∕ n ≤ 1 + W ∣ λ 1 ( ℋ ) ∣ 1 + W ∣ λ 2 ( ℋ ) ∣ ⋯ 1 + W ∣ λ n ( ℋ ) ∣ .

Using Proposition 2.9, we have ∣ λ i ( ℋ ) ∣ ≤ ∣ λ 1 ( ℋ ) ∣ ≤ 2 S ≤ 2 W , for all i . Thus, 1 ≤ 2 W ∣ λ i ( ℋ ) ∣ . Thus, we have

RHS ≤ 2 W ∣ λ 1 ( ℋ ) ∣ + W ∣ λ 1 ( ℋ ) ∣ 2 W ∣ λ 2 ( ℋ ) ∣ + W ∣ λ 2 ( ℋ ) ∣ ⋯ 2 W ∣ λ n ( ℋ ) ∣ + W ∣ λ n ( ℋ ) ∣ . RHS ≤ ( 3 W ) n ∣ λ 1 ( ℋ ) λ 2 ( ℋ ) … λ n ( ℋ ) ∣ = ( 3 W ) n ∣ det ( D ( ℋ ) ) ∣ .

Now consider the left-hand side (LHS).

LHS = 1 ∣ λ 2 ( ℋ ) ∣ 1 ∕ n ∣ λ 3 ( ℋ ) ∣ 1 ∕ n … ∣ λ n ( ℋ ) ∣ 1 ∕ n + 1 ∣ λ 1 ( ℋ ) ∣ 1 ∕ n ∣ λ 3 ( ℋ ) ∣ 1 ∕ n … ∣ λ n ( ℋ ) ∣ 1 ∕ n + ⋯ + 1 ∣ λ 1 ( ℋ ) ∣ 1 ∕ n ∣ λ 2 ( ℋ ) ∣ 1 ∕ n … ∣ λ n − 1 ( ℋ ) ∣ 1 ∕ n LHS = 1 ∣ det ( D ( ℋ ) ) ∣ 1 ∕ n ∑ i = 1 n ∣ λ i ( ℋ ) ∣ 1 ∕ n .

Since LHS ≤ RHS, therefore

1 ∣ det ( D ( ℋ ) ) ∣ 1 ∕ n ∑ i = 1 n ∣ λ i ( ℋ ) ∣ 1 ∕ n ≤ ( 3 W ) n ∣ det ( D ( ℋ ) ) ∣ ∑ i = 1 n ∣ λ i ( ℋ ) ∣ 1 ∕ n ≤ ( 3 W ) n ∣ det ( D ( ℋ ) ) ∣ n − 1 n

Note that ∑ i = 1 n ∣ λ i ( ℋ ) ∣ 1 ∕ n ≥ ∑ i = 1 n ∣ λ i ( ℋ ) ∣ 1 ∕ n . Then,

∑ i = 1 n ∣ λ i ( ℋ ) ∣ 1 ∕ n ≤ ( 3 W ) n ∣ det ( D ( ℋ ) ) ∣ n − 1 n .

Hence, DE ( ℋ ) ≤ ( 3 W ) n 2 ∣ det ( D ( ℋ ) ) ∣ n − 1 . □

The following result supplies an upper bound for the distance energy in terms of M 1 ( ℋ ) = ∑ i ∈ V D i 2 ( ℋ ) .

Theorem 2.11

Let ℋ be a connected hypergraph with n vertices. Then,

DE ( ℋ ) < M 1 ( ℋ ) + 2 ( n − 1 ) n S ,

where M 1 ( ℋ ) = ∑ i ∈ V D i 2 ( ℋ ) .

Proof

We know that, DE ( ℋ ) = ∑ i = 1 n ∣ λ i ( ℋ ) ∣ . After squaring both sides, we have

DE 2 ( ℋ ) = ∑ i = 1 n ∣ λ i ( ℋ ) ∣ 2 = ∑ i = 1 n λ i 2 ( ℋ ) + ∑ i ≠ j ∣ λ i ( ℋ ) ∣ ∣ λ j ( ℋ ) ∣ .

Note that λ 1 ( ℋ ) ≥ λ i ( ℋ ) for all i ≠ 1 , and there is at least one eigenvalue λ p ( ℋ ) , p ≠ 1 such that λ 1 ( ℋ ) > λ p ( ℋ ) . Otherwise, λ 1 ( ℋ ) = λ 2 ( ℋ ) = ⋯ = λ n ( ℋ ) . Then, the distance matrix of ℋ is scalar multiple of the identity matrix, which is impossible.

We have

DE 2 ( ℋ ) < M 1 ( ℋ ) + ( n − 1 ) n λ 1 2 ( ℋ ) .

Using Proposition 2.9, we have

□ DE 2 ( ℋ ) < M 1 ( ℋ ) + 2 ( n − 1 ) n S . DE ( ℋ ) < M 1 ( ℋ ) + 2 ( n − 1 ) n S .

Next, we supply an upper bound for the Nordhaus-Gaddum-type inequality for the distance energy.

Theorem 2.12

Let ℋ be a connected k-uniform hypergraph with n vertices and m hyperedges. Let m ¯ be the number of hyperedges of ℋ ¯ , where ℋ ¯ is the complement of ℋ . If m ≤ n k and m ¯ ≤ n k , then,

DE ( ℋ ) + DE ( ℋ ¯ ) ≥ 2 n n k − 4 .

Proof

Using the proof of Theorem 2.1, we have 2 W ( ℋ ) ≥ ( k − 1 ) n . Since m ≤ n k , we have 2 W ( ℋ ) ≥ m − n . Using Lemma 2.3, we have, DE ( ℋ ) ≥ 2 ( m − n ) n . In our case, m ≤ n k and m ¯ ≤ n k for ℋ and ℋ ¯ , then

DE ( ℋ ) ≥ 2 ( m − n ) n and DE ( ℋ ¯ ) ≥ 2 ( m ¯ − n ) n .

Adding both of them, we have

DE ( ℋ ) + DE ( ℋ ¯ ) ≥ 2 ( m − n ) n + 2 ( m ¯ − n ) n = 2 ( m + m ¯ ) n − 4 .

It is known that m + m ¯ = n k . Thus, DE ( ℋ ) + DE ( ℋ ¯ ) ≥ 2 n n k − 4 . □

There are some bounds for the distance energy of a connected graph in terms of its determinant, which are available in [16]. We note that those bounds are true for the distance energy of a connected non-singular hypergraphs and the proofs are same. Next, we state those results without giving their proofs.

Theorem 2.13

Let ℋ be a connected hypergraph with n vertices, and is non-singular, then,

DE ( ℋ ) ≤ 2 n S ∣ det ( D ( ℋ ) ) ∣ 1 ∕ n .

Proof

The proof is same as [16, Lemma 2.4].□

Theorem 2.14

Let ℋ be a connected non-singular hypergraph with n vertices, then

DE ( ℋ ) ≤ 2 S + 2 ( n − 1 ) S ∣ det ( D ( ℋ ) ) ∣ 1 ∕ n .

Proof

The proof is same as [16, Theorem 2.8].□

Theorem 2.15

Let ℋ be a connected non-singular hypergraph with n vertices, then

n n − 1 n ∣ det ( D ( ℋ ) ) ∣ 1 ∕ n ≤ DE ( ℋ ) ≤ ( 4 S ) n 2 ∣ det ( D ( ℋ ) ) ∣ n − 1 .

Proof

The proof is same as [16, Theorem 2.5].□

Next, we give the distance energy of join of k -uniform hypergraphs.

Definition 2.16

[14] Let ℋ 1 ( V 1 , E 1 ) and ℋ 2 ( V 2 , E 2 ) be two k -uniform hypergraphs. The join ℋ ( V , E ) = ℋ 1 ⊕ ℋ 2 of ℋ 1 and ℋ 2 is the hypergraph with the vertex set V = V 1 ⋃ V 2 and edge set E = ⋃ i = 0 2 E i , where E 0 = { e ⊆ V : ∣ e ∣ = k , e ∩ V i ≠ ϕ ∀ i = 1 , 2 } .

Theorem 2.17

Let ℋ 1 and ℋ 2 be two k-uniform hypergraphs with V 1 and V 2 as vertex set having cardinality n 1 and n 2 , respectively, and let E 1 and E 2 be the edge set. Then, the join of ℋ 1 and ℋ 2 will have distance energy 2 ( n − 1 ) for k ≥ 3 , where n = n 1 + n 2 .

Proof

Let ℋ 1 ( V 1 , E 1 ) and ℋ 2 ( V 2 , E 2 ) be two k -uniform hypergraphs with the vertex set V 1 and V 2 and edge set E 1 and E 2 . Then, we have the join as ℋ = ℋ 1 ⊕ ℋ 2 , with the vertex set V = V 1 ⋃ V 2 and edge set E = ⋃ i = 0 2 E i , where E 0 = { e ⊆ V : ∣ e ∣ = k , e ∩ V i ≠ ϕ ∀ i = 1 , 2 } . For e ∈ E 0 , the edge e must contain at least one vertex from V 1 and at least one vertex from V 2 , and the rest of the k − 2 vertices can be from any of the vertex sets. Therefore, for any two vertices in V 1 , we can have an edge in E 0 containing them, i.e., for u i , u j be any two arbitrary vertices in V 1 , and v 1 , v 2 , … , v k − 2 are vertices in V 2 , then the set { u i , u j , v 1 , v 2 , … , v k − 2 } is in E 0 . Thus, d ℋ ( u i , u k ) = 1 ∀ i , j = 1 , 2 , … , n 1 and i ≠ j . Similarly, for any two vertices of V 2 we have d ℋ ( v i , v j ) = 1 for all v i , v j ∈ V 2 ∀ i , j = 1 , 2 , … , n 2 and i ≠ j . Now, let u i ∈ V 1 and v j ∈ V 2 , so we have an edge e = { u i , v j , u 1 , u 2 , … , u p , v 1 , v 2 , … , v l } , where u s ∈ V 1 , v t ∈ V 2 , i ≠ s , j ≠ t for s = 1 , 2 , … , p and t = 1 , 2 , … , l such that ∣ e ∣ = k . Thus, e ∈ E 0 ; hence, d ℋ ( u i , v j ) = 1 for all u i ∈ V 1 and v j ∈ V 2 , for i = 1 , 2 , … , n 1 and j = 1 , 2 , … , n 2 . So we have for u i , u j ∈ V ,

d ℋ ( u i , u j ) = 1 i ≠ j , 0 i = j .

Thus, D ( ℋ ) = J − I , where J is the matrix of all 1’s and I is the identity matrix both of order n . Eigenvalues corresponding to D ( ℋ ) are λ 1 ( ℋ ) = n − 1 and λ i ( ℋ ) = − 1 for i = 2 , 3 , … , n . Hence, DE ( ℋ ) = 2 ( n − 1 ) . □

Definition 2.18

Let ℋ 1 , ℋ 2 , … , ℋ l , be l , k -uniform hypergraphs with the vertex set V 1 , V 2 , … , V l and edge set E 1 , E 2 , … , E l , respectively. Then, the join of these l hypergraphs is ℋ = ℋ 1 ⊕ ℋ 2 ⊕ ⋯ ⊕ ℋ l with the vertex set V = ⋃ i = 1 l V i and edge set E = ⋃ i = 0 l E i , where E 0 = { e ⊆ V : ∣ e ∣ = k , e ∩ V i ≠ ϕ ∀ i = 1 , 2 , … , l } .

Remark 2.19

We cannot join more than k , k -uniform hypergraphs simultaneously.

Theorem 2.20

Let ℋ 1 , ℋ 2 , … , ℋ k , be k, k-uniform hypergraphs with at least one of the hypergraphs having diameter greater than 2. Then, the join of these hypergraphs have the diameter 2 and

DE ( ℋ ) ≥ 2 2 n + k − 4 − ( 2 k − 3 ) m k n .

Proof

Let ℋ 1 , ℋ 2 , … , ℋ k , be k , k -uniform hypergraphs with the vertex set V 1 , V 2 , … , V k and edge set E 1 , E 2 , … , E k , respectively. Let ℋ be the join of these k hypergraphs with m edges, then we have the vertex set V = ⋃ i = 1 k V i and edge set E = ⋃ i = 0 k E i , where E 0 = { e ⊆ V : ∣ e ∣ = k , e ∩ V i ≠ ϕ ∀ i = 1 , 2 , … , k } . Any edge e ∈ E 0 will be of the form e = { u 1 , u 2 , … , u k } where u i ∈ V i , i = 1 , 2 , … , k ; thus, e ∩ V i = u i for any e ∈ E 0 and u i ∈ V i .

Consider u i , u j ∈ V 1 , such that d ℋ 1 ( u i , u j ) ≠ 1 . Let e 1 , e 2 ∈ E 0 , such that u i ∈ e 1 and u j ∈ e 2 and e 1 ∩ e 2 ≠ ϕ . Let v l be that common vertex between e 1 and e 2 ; thus, we have a path of length 2 from u i to u j , as u i e 1 v l e 2 u j in ℋ . The distance d ℋ ( u i , u j ) cannot be 1, as if it is 1, then there exists an edge e such that both u i , u j ∈ e , which is not possible as u i , u j in ℋ 1 , d ℋ 1 ( u i , u j ) ≠ 1 and in ℋ cannot contain e by definition of ℋ . Hence, d ℋ ( u i , u j ) = 2 for i ≠ j and u i , u j ∈ V 1 . Similarly, for any two vertices v i , v j ∈ V t for hypergraph ℋ t , such that d ℋ t ( v i , v j ) ≠ 1 , we have d ℋ ( v i , v j ) = 2 . Now, let u ∈ V i and v ∈ V j such that i ≠ j , be any two vertices for hypergraphs ℋ i and ℋ j , respectively, we have e ∈ E o such that e = { u , v , a 1 , a 2 , … , a k − 2 } , where a s ∈ V s such that s ≠ i , j . Hence, d ℋ ( u , v ) = 1 . Thus, the diameter of the hypergraph ℋ is 2.

Let N i be the cardinality of the set of the neighbours of the i th vertex, i.e., the vertices that are adjacent to i . Since the adjacent vertices are at the distance 1, so we have N i vertices at distance 1, and as the diameter is 2, we have ( n − 1 − N i ) vertices at the distance 2, where n = ∑ i = 1 k ∣ V i ∣ . Let δ i denote the degree of the i th vertex. Each set contains at most k − 1 vertices that are adjacent to i . Also, there is at least one hyperedge that contains k − 1 vertices that are adjacent to i . Then, N i ≥ k − 1 + δ i − 1 = k − 2 + δ i . Also, N i ≤ ( k − 1 ) δ i . Along ∑ i = 1 n δ i = k m [17, p. 142].

Let x = [ 1 , 1 , … , 1 ] T be all one vectors. By Rayleigh principle,

λ 1 ( ℋ ) ≥ x T D ( ℋ ) x x T x λ 1 ( ℋ ) ≥ 1 n ∑ i = 1 n ( N i + 2 ( n − 1 − N i ) ) λ 1 ( ℋ ) ≥ 1 n ∑ i = 1 n ( ( k − 2 + δ i ) + 2 ( n − 1 − ( k − 1 ) δ i ) ) λ 1 ( ℋ ) ≥ 1 n ( ( k − 2 ) n + k m + 2 ( n ( n − 1 ) − k m ( k − 1 ) ) ) .

Thus, we have

λ 1 ( ℋ ) ≥ 2 n + k − 4 − ( 2 k − 3 ) m k n .

Also, DE ( ℋ ) ≥ 2 λ 1 ( ℋ ) . Hence,

□ DE ( ℋ ) ≥ 2 2 n + k − 4 − ( 2 k − 3 ) m k n .

Lemma 2.21

[12] Suppose both A and B are non-negative irreducible matrices and B ≤ A (namely, B i j ≤ A i j for each pair of i, j). Then, ρ ( B ) ≤ ρ ( A ) with equality if and only if B = A .

Theorem 2.22

Let ℋ be a hypergraph with n vertices, then

DE ( ℋ ) ≥ 2 ( n + λ n ( ℋ ) ) ,

where λ n ( ℋ ) is the smallest eigenvalue and equality holds if and only if D ( ℋ ) = J − I , where J is the matrix of all 1’s and I is the identity matrix.

Proof

For any hypergraph ℋ , we have D ( ℋ ) ≥ J − I , where J is the matrix of all 1’s and I is the identity matrix. Then, from Lemma 2.21 λ 1 ( ℋ ) ≥ n − 1 .

Now, we show that λ n ( ℋ ) ≤ − 1 . Consider it is not, then we have λ n ( ℋ ) > − 1 . Thus, we have λ i ( ℋ ) > − 1 for i = 2 , 3 , … , n . Hence, we have, ∑ i = 2 n λ i ( ℋ ) > − ( n − 1 ) . This implies Trace ( D ( ℋ ) ) = ∑ i = 1 n λ i ( ℋ ) > 0 . A contradiction to the fact that Trace ( D ( ℋ ) ) = 0 . Thus, λ n ( ℋ ) ≤ − 1 . Therefore, we have, λ 1 ( ℋ ) − λ n ( ℋ ) ≥ n − 1 + 1 = n . Hence, DE ( ℋ ) ≥ 2 ( n + λ n ( ℋ ) ) as DE ( ℋ ) ≥ 2 λ 1 ( ℋ ) .

Now, for the equality part, consider all the inequalities as equalities; thus, we have λ 1 ( ℋ ) = n − 1 and λ n ( ℋ ) = − 1 for D ( ℋ ) = J − I . Conversely, we have D ( ℋ ) = J − I . Thus, DE ( ℋ ) = 2 ( n − 1 ) .□

3 Distance energy of hyperstar

In [6], Graham and Pollack showed that if T is a tree on n vertices with distance matrix D , then the determinant of T is ( − 1 ) n − 1 ( n − 1 ) 2 n − 2 , and this is a function of only the number of vertices. In [3], Bapat et al. showed that the determinant of the q -distance matrix of a tree T with n vertices is ( − 1 ) n − 1 ( n − 1 ) ( 1 + q ) n − 2 . In [2], the authors showed that the determinant of the bipartite distance matrix of a non-singular tree on 2 p vertices is a multiple of 2 p − 1 . In [15], Sivasubramanian showed that the determinant of the distance matrix of a 3-uniform hypertree T on n = 2 k + 1 vertices is 3 k − 1 2 k . It is natural to ask to calculate the determinant for other uniform hypergraphs. In this section, we show that the determinant of the distance matrix of a connected k -uniform hyperstar on n vertices is a function of only the number of vertices n and k .

The following is the definition of k -uniform hyperstar.

Definition 3.1

A hypertree is a connected hypergraph with no hypercycles. For a k uniform hypertree ℋ with vertices n if there is a disjoint partition, V ( ℋ ) = { u } ∪ V 1 ∪ V 2 ⋯ ∪ V m of the vertex set V ( ℋ ) such that ∣ V i ∣ = k − 1 for all i = 1 , 2 , … , m and we define hyperedge set, E ( ℋ ) = { { u } ∪ V i : 1 ≤ i ≤ m } , then such a hypergraph is called a hyperstar centered at u . Note that a k-uniform hyperstar with n vertices always has n − 1 k − 1 hyperedges.

The following is our main theorem in this section.

Theorem 3.2

Let ℋ be a connected k-uniform hyperstar with n vertices. Then,

det ( D ( ℋ ) ) = ( − 1 ) n − 1 ( n − 1 ) k n − k k − 1 .

Proof

The hypergraph ℋ has n vertices u 1 , u 2 , … , u n and it is a k -uniform hyperstar; then, there is a central vertex that is contained in all the hyperedges. Let u 1 be the central vertex. No other vertex is contained in more than one hyperedges. We have to allocate n − 1 vertices in mutually exclusive sets of cardinality k − 1 , so we have a total of m = n − 1 k − 1 number of sets, say A 1 = { u 2 , u 3 , … , u k } , A 2 = { u k + 1 , u k + 2 , … , u 2 k − 1 } , … , A m = { u n − k + 2 , u n − k + 3 , … , u n } . Since u 1 is contained in all the hyperedges, thus the distance of all the vertices from u 1 is 1, that is, d ℋ ( u 1 , u i ) = 1 for all u i ∈ V ( ℋ ) \ { u 1 } .

It is clear that A i and A j are mutually disjoint for i , j = 1 , … , n and i ≠ j . Let u i ∈ A i and u j ∈ A j and let E i and E j be the hyperedges containing u i and u j , respectively. If we want to traverse from the vertex u i ∈ A i to the vertex u j ∈ A j , then we have to traverse along the path u i E i u 1 E j u j and this is the only path from u i to u j . Hence, the distance between them is 2. So we have, d ℋ ( u i , u j ) = 2 , for u i ∈ A i and u j ∈ A j , and clearly, if u l ∈ A i , then d ℋ ( u i , u l ) = 1 . Thus, the distance matrix of ℋ is

D ( ℋ ) = 0 1 T 1 T … 1 T 1 C D … D 1 D C … D ⋮ ⋮ ⋮ ⋱ ⋮ 1 D D … C ,

where C = 0 1 … 1 1 0 … 1 ⋮ ⋮ ⋱ ⋮ 1 1 … 0 k − 1 × k − 1 , D = 2 2 … 2 2 2 … 2 ⋮ ⋮ ⋱ ⋮ 2 2 … 2 k − 1 × k − 1 and 1 = 1 1 ⋮ 1 k − 1 × 1 .

We now make D ( ℋ ) into a block lower triangular matrix by applying the elementary row operation R i = R i + c R j . We use R i to denote the ith row of D ( ℋ ) , and r i to denote the ith row of 1 , C and D . There are m ( m + 1 ) blocks in D ( ℋ ) below the first row. We use ( s , t ) to denote the position for those m ( m + 1 ) blocks. The block at ( s , 1 ) is 1 for s = 1 , … , m ; the block at ( s , s + 1 ) is C for s = 1 , … , m , and rest of the blocks are D .

We can write the 1st row R 1 of D ( ℋ ) as m + 1 partition R 1 ′ , R 2 ′ , … , R m + 1 ′ , the size of R 1 ′ is 1 and the size of R i ′ is k − 1 for i = 2 , … , m + 1 . Therefore, R 1 = [ R 1 ′ , R 2 ′ , … , R m + 1 ′ ] . It is clear that [ ( r i ) ( s , 1 ) , … , ( r i ) ( s , m + 1 ) ] is one of the row of D ( ℋ ) for i = 1 , … , k − 1 and s = 1 , … , m .

We are now applying the following row operations.

( r i ) ( s , t ) → ( r i ) ( s , t ) − ( r 1 ) ( s , t ) for all 2 ≤ i ≤ k − 1 , 1 ≤ s ≤ m , and 1 ≤ t ≤ m + 1 .

Then, we have

D ′ ( ℋ ) = 0 1 T 1 T … 1 T 1 ′ C ′ D ′ … D ′ 1 ′ D ′ C ′ … D ′ ⋮ ⋮ ⋮ ⋱ ⋮ 1 ′ D ′ D ′ … C ′ ,

where C ′ = 0 1 1 … 1 1 − 1 0 … 0 1 0 − 1 … 0 ⋮ ⋮ ⋮ ⋱ ⋮ 1 0 0 … − 1 , D ′ = 2 2 2 … 2 0 0 0 … 0 0 0 0 … 0 ⋮ ⋮ ⋮ ⋱ ⋮ 0 0 0 … 0 and 1 ′ = [ 1 , 0 , … , 0 ] T .

Since D ( ℋ ) and D ′ ( ℋ ) are row equivalent matrices and we are using only R i → R i + c R j type of row operation, we have det ( D ( ℋ ) ) = det ( D ′ ( ℋ ) ) .

We now apply the following row operations:

R 1 → R 1 + ∑ s = 1 m ∑ t = 1 m + 1 ( U ) ( s , t ) , ( r 1 ) ( s , t ) → ( r 1 ) ( s , t ) + ( U ) ( s , t ) 1 ≤ s ≤ m and 1 ≤ t ≤ m + 1 ,

where ( U ) ( s , t ) = ∑ i = 2 k − 1 ( r i ) ( s , t ) . We have,

D ″ ( ℋ ) = 0 1 ″ T 1 ″ T … 1 ″ T 1 ′ C ″ D ′ … D ′ 1 ′ D ′ C ″ … D ′ ⋮ ⋮ ⋮ ⋱ ⋮ 1 ′ D ′ D ′ … C ″ ,

where

C ″ = k − 2 0 0 … 0 1 − 1 0 … 0 1 0 − 1 … 0 ⋮ ⋮ ⋮ ⋱ ⋮ 1 0 0 … − 1 , D ′ = 2 2 2 … 2 0 0 0 … 0 0 0 0 … 0 ⋮ ⋮ ⋮ ⋱ ⋮ 0 0 0 … 0 , 1 ″ = [ k − 1 , 0 , … , 0 ] and 1 ′ = [ 1 , 0 , … , 0 ] T .

We now apply the following operation:

( r 1 ) ( s , t ) → ( r 1 ) ( s , t ) + 2 ∑ i = 1 i ≠ s m ∑ j = 1 j ≠ t m + 1 ( U ) ( i , j ) .

Thus, our matrix reduces to

D ‴ ( ℋ ) = 0 1 ″ T 1 ″ T … 1 ″ T 1 ′ C ″ D ″ … D ″ 1 ′ D ″ C ″ … D ″ ⋮ ⋮ ⋮ ⋱ ⋮ 1 ′ D ″ D ″ … C ″ ,

where

C ″ = k − 2 0 0 … 0 1 − 1 0 … 0 1 0 − 1 … 0 ⋮ ⋮ ⋮ ⋱ ⋮ 1 0 0 … − 1 , D ″ = 2 ( k − 1 ) 0 0 … 0 0 0 0 … 0 0 0 0 … 0 ⋮ ⋮ ⋮ ⋱ ⋮ 0 0 0 … 0 , 1 ″ = [ k − 1 , 0 , … , 0 ] and 1 ′ = [ 1 , 0 , … , 0 ] T .

Here, R 1 ′ = 0 and R i ′ = [ k − 1 , 0 , … , 0 ] for i = 2 , … , m . We now apply the following operation:

( r 1 ) ( s , t ) → ( r 1 ) ( s , t ) − 2 R t ′ .

We have

D ″ ″ ( ℋ ) = 0 0 T 0 T … 0 T 1 ′ C ‴ 0 k − 1 × k − 1 … 0 k − 1 × k − 1 1 ′ 0 k − 1 × k − 1 C ‴ … 0 k − 1 × k − 1 ⋮ ⋮ ⋮ ⋱ ⋮ 1 ′ 0 k − 1 × k − 1 0 k − 1 × k − 1 … C ‴ ,

where

C ‴ = − k 0 0 … 0 1 − 1 0 … 0 1 0 − 1 … 0 ⋮ ⋮ ⋮ ⋱ ⋮ 1 0 0 … − 1 , 0 = [ 0 , 0 , … , 0 ] and 1 ′ = [ 1 , 0 , … , 0 ] T .

We now apply the operation R 1 → R 1 + k − 1 k ∑ s = 1 m ∑ t = 1 m + 1 ( r 1 ) ( s , t ) .

We have

D ′′′′′ ( ℋ ) = n − 1 k 0 T 0 T … 0 T 1 ′ C ‴ 0 k − 1 × k − 1 … 0 k − 1 × k − 1 1 ′ 0 k − 1 × k − 1 C ‴ … 0 k − 1 × k − 1 ⋮ ⋮ ⋮ ⋱ ⋮ 1 ′ 0 k − 1 × k − 1 0 k − 1 × k − 1 … C ‴ .

Then, det ( D ′′′′′ ( ℋ ) ) = det ( D ( ℋ ) ) = n − 1 k ( det ( C ‴ ) ) m .

Note that det ( C ‴ ) = ( − 1 ) k − 1 k .

Therefore, det ( D ( ℋ ) ) = n − 1 k ( ( − 1 ) k − 1 k ) m = ( ( − 1 ) k − 1 ) m ( n − 1 ) k m − 1 .

Since m = n − 1 k − 1 , we have

□ det ( D ( ℋ ) ) = ( − 1 ) n − 1 ( n − 1 ) k n − k k − 1 .

Now, we give the distance energy of the hyperstar.

Theorem 3.3

Let ℋ be a k-uniform hyperstar, with n vertices. Then, the distance spectrum of ℋ is − 1 , − k , 2 ( n − 1 ) − k − ( 2 ( n − 1 ) − k ) 2 + 4 ( n − 1 ) 2 and 2 ( n − 1 ) − k + ( 2 ( n − 1 ) − k ) 2 + 4 ( n − 1 ) 2 with multiplicity ( k − 2 ) m , m − 1 , 1 , and 1, respectively, where m = n − 1 k − 1 , and

DE ( ℋ ) = 2 ( n − 1 ) − k + ( 2 ( n − 1 ) − k ) 2 + 4 ( n − 1 ) .

Proof

The block form of the distance matrix of ℋ is

D ( ℋ ) = 0 1 T 1 T … 1 T 1 C D … D 1 D C … D ⋮ ⋮ ⋮ ⋱ ⋮ 1 D D … C ,

where C = 0 1 … 1 1 0 … 1 ⋮ ⋮ ⋱ ⋮ 1 1 … 0 k − 1 × k − 1 , D = 2 2 … 2 2 2 … 2 ⋮ ⋮ ⋱ ⋮ 2 2 … 2 k − 1 × k − 1 , and 1 = 1 1 ⋮ 1 k − 1 × 1 .

We use R i to denote the ith row of D ( ℋ ) and r i to denote the ith row of 1 , C and D . There are m ( m + 1 ) blocks in D ( ℋ ) below the first row.

First, we show that − 1 is an eigenvalue of D ( ℋ ) , so for that, we show det ( D ( ℋ ) − ( − 1 ) I ) = 0 , where I is the identity matrix of order n .

D ( ℋ ) + I = 1 1 T 1 T … 1 T 1 C ′ D … D 1 D C ′ … D ⋮ ⋮ ⋮ ⋱ ⋮ 1 D D … C ′ ,

where C ′ = 1 1 … 1 1 1 … 1 ⋮ ⋮ ⋱ ⋮ 1 1 … 1 k − 1 × k − 1 , D = 2 2 … 2 2 2 … 2 ⋮ ⋮ ⋱ ⋮ 2 2 … 2 k − 1 × k − 1 , and 1 = 1 1 ⋮ 1 k − 1 × 1 .

Clearly, the rows R 2 and R 3 are identical; hence, the determinant is 0. Thus, − 1 is an eigenvalue of D ( ℋ ) . Since C ′ = J and D = 2 J , where J is the matrix of all 1’s having rank 1. So the second row of block distance matrix [ 1 C ′ D … D ] will have rank 1. Similarly, all the other rows will have rank 1. All the first rows of each block row, i.e., first row of [ 1 C ′ D … D ] , [ 1 D C ′ … D ] , … [ 1 D D … C ′ ] , are linearly independent in D ( ℋ ) + I ; thus, the rank of the D ( ℋ ) + I is m + 1 . Hence, the nullity will be n − m − 1 = ( k − 2 ) m . Hence, the multiplicity of − 1 is ( k − 2 ) m .

Now, we show that − k is an eigenvalue of D ( ℋ ) , for that, we show det ( D ( ℋ ) − ( − k ) I ) = 0 , where I is the identity matrix of order n .

D ( ℋ ) + k I = k 1 T 1 T … 1 T 1 C ″ D … D 1 D C ″ … D ⋮ ⋮ ⋮ ⋱ ⋮ 1 D D … C ″ ,

where C ″ = k 1 … 1 1 k … 1 ⋮ ⋮ ⋱ ⋮ 1 1 … k k − 1 × k − 1 , D = 2 2 … 2 2 2 … 2 ⋮ ⋮ ⋱ ⋮ 2 2 … 2 k − 1 × k − 1 , and 1 = 1 1 ⋮ 1 k − 1 × 1 .

Now, we apply the row operations on the D ( ℋ ) + k I ,

( r i ) ( s , t ) → ( r i ) ( s , t ) − R 1 for all 1 ≤ i ≤ k − 1 , 1 ≤ s ≤ m , and 1 ≤ t ≤ m + 1 .

We have

D ( ℋ ) + k I = k 1 T 1 T … 1 T 1 ′ C ‴ D ′ … D ′ 1 ′ D ′ C ‴ … D ′ ⋮ ⋮ ⋮ ⋱ ⋮ 1 ′ D ′ D ′ … C ‴ ,

where

C ‴ = k − 1 0 … 0 0 k − 1 … 0 ⋮ ⋮ ⋱ ⋮ 0 0 … k − 1 k − 1 × k − 1 , D ′ = 1 1 … 1 1 1 … 1 ⋮ ⋮ ⋱ ⋮ 1 1 … 1 k − 1 × k − 1 , and 1 ′ = − ( k − 1 ) − ( k − 1 ) ⋮ − ( k − 1 ) k − 1 × 1 .

For all s and t we apply

( r 1 ) ( s , t ) → ( r 1 ) ( s , t ) + ∑ i = 2 k − 1 ( r i ) ( s , t ) .

Thus, we have

D ( ℋ ) + k I = k 1 T 1 T … 1 T 1 ″ C ″ ″ D ″ … D ″ 1 ″ D ″ C ″ ″ … D ″ ⋮ ⋮ ⋮ ⋱ ⋮ 1 ″ D ″ D ″ … C ″ ″ ,

where

C ″ ″ = k − 1 k − 1 … k − 1 0 k − 1 … 0 ⋮ ⋮ ⋱ ⋮ 0 0 … k − 1 k − 1 × k − 1 , D ″ = k − 1 k − 1 … k − 1 1 1 … 1 ⋮ ⋮ ⋱ ⋮ 1 1 … 1 k − 1 × k − 1 , and 1 ″ = − ( k − 1 ) 2 − ( k − 1 ) ⋮ − ( k − 1 ) k − 1 × 1 .

Here, the first row of each block is same, i.e., the first row of [ 1 ″ C ″ ″ D ″ ⋯ D ″ ] , [ 1 ″ D ″ C ″ ″ ⋯ D ″ ] , … , [ 1 ″ D ″ D ″ ⋯ C ″ ″ ] , as there are two identical rows in D ( ℋ ) + k I ; thus, the determinant is 0. Hence, − k is an eigenvalue of D ( ℋ ) . Except the first row of [ 1 ″ C ″ ″ D ″ ⋯ D ″ ] , [ 1 ″ D ″ C ″ ″ ⋯ D ″ ] , … , [ 1 ″ D ″ D ″ ⋯ C ″ ″ ] the other rows cannot be expressed in the linear combinations of any of the rows; hence, they have rank k − 1 , k − 2 , …, k − 2 , respectively. Thus, the rank of the D ( ℋ ) + k I is ( k − 2 ) m + 2 . Hence, the nullity of the D ( ℋ ) + k I is m − 1 .

Now, two eigenvalues remain, as − 1 and − k add up to n − 2 eigenvalues. Let the eigenvalues be a and b . Since the sum of all the eigenvalues is equal to trace and the product of all the eigenvalues is equal to determinant, so from Theorem 3.2, we have

∑ i = 1 n λ i = 0 and ∏ i = 1 n λ i = ( − 1 ) n − 1 ( n − 1 ) k n − k k − 1 ,

which gives us,

a + b = 2 ( n − 1 ) − k and a b = − ( n − 1 ) .

On solving, we have

a 2 − ( 2 ( n − 1 ) − k ) a − ( n − 1 ) = 0 .

Obtaining us,

a = 2 ( n − 1 ) − k ± ( 2 ( n − 1 ) − k ) 2 + 4 ( n − 1 ) 2 .

Hence, the two eigenvalues are

a = 2 ( n − 1 ) − k + ( 2 ( n − 1 ) − k ) 2 + 4 ( n − 1 ) 2 b = 2 ( n − 1 ) − k − ( 2 ( n − 1 ) − k ) 2 + 4 ( n − 1 ) 2 .

Now, the distance energy of the hyperstar ℋ is

DE ( ℋ ) = ∑ i = 1 n ∣ λ i ∣ ,

DE ( ℋ ) = 2 ( n − 1 ) − k + ( 2 ( n − 1 ) − k ) 2 + 4 ( n − 1 ) 2 + − 2 ( n − 1 ) + k + ( 2 ( n − 1 ) − k ) 2 + 4 ( n − 1 ) 2 + ( m − 1 ) k + ( k − 2 ) m .

Putting m = n − 1 k − 1 , we have

□ DE ( ℋ ) = 2 ( n − 1 ) − k + ( 2 ( n − 1 ) − k ) 2 + 4 ( n − 1 ) .

Funding information: The first author research was supported by Prime Minister Research Fellowship by Government of India (2402790). The second author research was supported by the project MTR/2022/000194 (SERB, India).
Conflict of interest: There is no conflict of interest.
Data availability statement: Our manuscript has no associated data.

References

[1] M. Aouchiche and P. Hansen, Distance spectra of graphs: A survey, Linear Algebra Appl. 458 (2014), 301–386. 10.1016/j.laa.2014.06.010Suche in Google Scholar

[2] R. B. Bapat, R. Jana, and S. Pati, The bipartite distance matrix of a non-singular tree, Linear Algebra Appl. 631 (2021), 254–281. 10.1016/j.laa.2021.09.001Suche in Google Scholar

[3] R. B. Bapat, A. K. Lal, and S. Pati, A q-analogue of the distance matrix of a tree, Linear Algebra Appl. 416 (2006), 799–814. 10.1016/j.laa.2005.12.023Suche in Google Scholar

[4] A. Cayley, A theorem in the geometry of position, Cambridge Math. J. 2 (1841), 267–271. Suche in Google Scholar

[5] B. Claude, Graphs and Hypergraphs, North-Holland Publishing Company, North-Holland, 1973. Suche in Google Scholar

[6] R. L. Graham and H. O. Pollack, On the addressing problem for loop switching, Bell System Tech. J. 50 (1971), 2495–2519. 10.1002/j.1538-7305.1971.tb02618.xSuche in Google Scholar

[7] I. Gutman, The energy of a graph, Ber. Math.-Statist. Sekt. Forschungsz. Graz. 103 (1978), 1–22. Suche in Google Scholar

[8] I. Gutman, E. V. Konstantinova, and V. A. Skorobogatov, Molecular hypergraphs and clar structural formulas of benzenoid hydrocarbons, ACH Models Chem. 136 (1999), 539–548. Suche in Google Scholar

[9] E. V. Konstantinova and V. A. Skorobogatov, Graph and hypergraph models of molecular structure: A comparative analysis of indices, J. Struct. Chem. 39 (1998), 958–966. 10.1007/BF02903615Suche in Google Scholar

[10] E. V. Konstantinova and V. A. Skorobogatov, Application of hypergraph theory in chemistry, Discrete Math. 235 (2001), 365–383. 10.1016/S0012-365X(00)00290-9Suche in Google Scholar

[11] X. Liu and L. Wang, Distance spectral radii of k-uniform bicyclic hypergraphs, Linear Multilinear Algebra, 70 (2022), no. 21, 6190–6210.10.1080/03081087.2021.1948495Suche in Google Scholar

[12] H. Minc, Nonnegative Matrices, John Wiley and Sons, New York, 1988. Suche in Google Scholar

[13] V. Nikiforov, The energy of graphs and matrices, J. Math. Anal. Appl. 326 (2007), 1472–1475. 10.1016/j.jmaa.2006.03.072Suche in Google Scholar

[14] A. Sarkar and A. Banerjee, Joins of hypergraphs and their spectra, Linear Algebra Appl. 603 (2020), 101–129. 10.1016/j.laa.2020.05.029Suche in Google Scholar

[15] S. Sivasubramanian, q-Analogs of distance matrices of 3-hypertrees, Linear Algebra Appl. 431 (2009), 1234–1248. 10.1016/j.laa.2009.04.020Suche in Google Scholar

[16] G. Sridhara, M. R. Rajesh Kanna, and H. L. Parashivmurty, New bounds for distance energy of a graph, J. Indones. Math. Soc. 26 (2020), no. 2, 213–223. 10.22342/jims.26.2.789.213-223Suche in Google Scholar

[17] V. I. Voloshin, Introduction to Graph and Hypergraph Theory, NOVA Science Publishers, Inc., New York, 2009. Suche in Google Scholar

Received: 2023-03-22

Revised: 2023-06-29

Accepted: 2023-06-29

Published Online: 2023-07-20

This work is licensed under the Creative Commons Attribution 4.0 International License.

Artikel in diesem Heft

https://doi.org/10.1515/spma-2023-0188

Schlagwörter für diesen Artikel

hypergraph; distance matrix; distance degree; spectral radius; distance energy; hyperstar

Creative Commons

BY 4.0