A Laplacian eigenbasis for threshold graphs

Rafael R. Macharete; Renata R. Del-Vecchio; Heber Teixeira; Leonardo de Lima

doi:10.1515/spma-2024-0029

Article Open Access

A Laplacian eigenbasis for threshold graphs

Rafael R. Macharete , Renata R. Del-Vecchio , Heber Teixeira and Leonardo de Lima

Published/Copyright: October 30, 2024

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From the journal Special Matrices Volume 12 Issue 1

Abstract

Let G be a graph on n vertices. In this article, we prove that an eigenbasis of the Laplacian matrix of a star graph of order n is also an eigenbasis of G if and only if G is a threshold graph. As an application of this spectral characterization, we show an infinite family of threshold graphs that are weakly Hadamard diagonalizable.

Keywords: threshold graph; spectral characterization; Laplacian matrix; weakly Hadamard diagonalizable graph

MSC 2010: 05C50; 05C35

1 Introduction

Throughout this article, we consider G = ( V , E ) as a simple graph with vertex set V = { 1 , … , n } and edge set E . The degree of a vertex i of G is the number of incident edges to vertex i and is denoted by d i ( G ) , or d i , when the graph is clear from the context. We write L ( G ) = D ( G ) − A ( G ) for the Laplacian matrix of G , where A ( G ) is the (0, 1)-adjacency matrix and D ( G ) is the diagonal matrix of the vertex degrees of G . The eigenvalues of L ( G ) are arranged by μ 1 ≥ ⋯ ≥ μ n − 1 ≥ μ n = 0 and its corresponding eigenvectors are x 1 , … , x n , where x i j = ( x 1 j , … , x n j ) T for j = 1 , … , n . An eigenbasis of L ( G ) is a basis of R n composed of eigenvectors of L ( G ) . The well-known eigenequation for the Laplacian matrix of G is given by L ( G ) x = μ x and can be rewritten as

(1) ( d i ( G ) − μ ) x i = ∑ { j , i } ∈ E ( G ) x j ,

for each vertex i ∈ V , and the x i s are the entries of an eigenvector x = ( x 1 , … , x n ) T associated with eigenvalue μ . Let S n be a star graph on n vertices. For 1 ≤ k < n , it is easy to see that each x k satisfying

(2) x i k = 1 , if i < k + 1 , − k , if i = k + 1 , 0 , if k + 1 < i ≤ n ,

and x n = 1 n = ( 1 , 1 , … , 1 ) T are the pairwise orthogonal eigenvectors of L ( S n ) associated with eigenvalues 1, n , and 0 of multiplicities n − 2 , 1, and 1, respectively.

There is considerable literature on the Laplacian eigenvalues of a threshold graph [4], but to the best of our knowledge, we have not found articles relating the structure of a threshold graph to its eigenvectors. In this article, we fill this gap by proving that any threshold graph of a given order can be recognized from a specific Laplacian eigenbasis, which is a spectral characterization of a threshold graph. In this sense, we proved that L ( G ) and L ( S n ) share the same eigenvectors when G is a connected threshold graph on n vertices, as stated in Theorem A and proved in Section 3.1. We also extended this result to disconnected threshold graphs in Section 3.2.

Theorem A

Let G be a connected graph on n vertices. Then, G is a threshold graph if and only if { x 1 , … , x n } is an eigenbasis of the Laplacian matrix of G.

Every connected threshold graph G on n vertices has a star S n as a subgraph. Starting from the star that is a subgraph of G , one can obtain any threshold graph on n vertices by adding the appropriate edges. This fact allowed us to study how the eigenvalues and eigenvectors of the star may change during this process of adding edges to obtain G . We used the eigenequation (1), the edge principle of Merris [8], and other tools we developed here. We prove that the eigenvectors of L ( S n ) do not change in this process. The eigenvectors of L ( S n ) may, at most, jump from one eigenspace to another during the process of adding edges to the star. It is worth highlighting that the threshold graphs are determined by the Laplacian spectrum, i.e., no two threshold graphs with n vertices have the same spectrum unless they are isomorphic. Despite this, all threshold graphs of order n share the same eigenvector basis.

This article is organized as follows. In Section 2, we prove structural results, define threshold graphs, and state a useful result available in the literature. In Section 3, we prove the main results, and in Section 4, we apply our results to find an infinite family of weakly Hadamard diagonalizable (WHD) graphs.

2 Structural tools

In this section, we show that an appropriate change in a graph G by adding or removing edges yields a new graph G ′ such that an eigenvector x to L ( G ) is also still an eigenvector to L ( G ′ ) under some conditions. Merris [8] proved that this is true when the entries x i and x j of the eigenvector x corresponding to the edge { i , j } ∈ E are equal.

Theorem 2.1

([8], Edge principle) Let μ be an eigenvalue of L ( G ) associated with eigenvector x . If x i = x j , then μ is an eigenvalue of L ( G ′ ) associated with x , where G ′ is obtained from G by deleting or adding an edge e = { i , j } depending on whether or not it is an edge of G.

The results we obtained in this section show that we can even go beyond the edge principle. We show that it is possible to add or remove an edge { i , j } from G where the corresponding entries x i and x j are not equal, and still guarantee that x is also an eigenvector to the Laplacian matrix of the new graph G ′ . Those results are in Lemmas 2.2 and 2.3, which are important structural tools used to prove our main theorems of Section 3.

Lemma 2.2

Let G = ( V , E ) be a graph on n vertices and x ∈ R n such that x = ( x 1 , x 2 , … , x k , 0 , … , 0 ) is an eigenvector of L ( G ) associated with eigenvalue μ , in a way that ∑ l = 1 k x l = 0 and x l ≠ 0 , for all l ∈ { 1 , … , k } . Take the vertex partition W 1 = { 1 , … , k } and W 2 = { k + 1 , … , n } . If i ∈ W 2 and e l = { i , l } ∉ E , for l = 1 , … , k , then x is an eigenvector of L ( G ′ ) , where G ′ = G + e 1 + ⋯ + e k , associated with eigenvalue μ + 1 .

Proof

Let j ∈ W 1 . Adding − x j + x j on both sides of equation (1) for G , we obtain

(3) ( d j ( G ) + 1 − ( μ + 1 ) ) x j = − x j + x j + ∑ { u , j } ∈ E ( G ) x u .

Since d j ( G ′ ) = d j ( G ) + 1 and x i = 0 , we can rewrite (3) as follows:

(4) ( d j ( G ′ ) − ( μ + 1 ) ) x j = x i + ∑ { u , j } ∈ E ( G ) x u , ( d j ( G ′ ) − ( μ + 1 ) ) x j = ∑ { u , j } ∈ E ( G ′ ) x u .

If j ∈ W 2 and j ≠ i , note that x j = 0 and ∑ { u , j } ∈ E ( G ′ ) x u = 0 . So, we have

(5) ( d j ( G ′ ) − ( μ + 1 ) ) x j = ∑ { u , j } ∈ E ( G ′ ) x u .

Let j = i ∈ W 2 . In this case, x i = x j = 0 and d j ( G ′ ) = d j ( G ) + k . Adding ( k − 1 ) x j on both sides of equation (1) for G , we have

(6) ( d j ( G ) + k − ( μ + 1 ) ) x j = ( k − 1 ) x j + ∑ { u , j } ∈ E ( G ) x u , ( d j ( G ′ ) − ( μ + 1 ) ) x j = ∑ { u , j } ∈ E ( G ) x u .

Since ∑ l = 1 k x l = 0 , we rewrite equation (6) as

(7) ( d j ( G ′ ) − ( μ + 1 ) ) x j = ∑ l = 1 k x l + ∑ { u , j } ∈ E ( G ) x u ( d j ( G ′ ) − ( μ + 1 ) ) x j = ∑ { u , j } ∈ E ( G ′ ) x u .

Therefore, from equations (4), (5), and (7), we obtain that x is an eigenvector of L ( G ′ ) associated with eigenvalue μ + 1 , and the proof is complete.□

Lemma 2.3

Let G = ( V , E ) be a graph on n vertices, and suppose that x = x n − k = ( 1 , 1 , … , 1 , − n + k , 0 , 0 , … , 0 ︸ k − 1 ) is an eigenvector of L ( G ) associated with eigenvalue μ . Let G ′ = G + e 1 + e 2 + ⋯ + e n − k , where e l = { l , n − k + 1 } ∉ E ( G ) , for l = 1 , … , n − k . Then, x is an eigenvector of L ( G ′ ) associated with eigenvalue μ + ( n − k ) + 1 .

Proof

Consider a partition of the vertex set of G given by W 1 = { 1 , … , n − k } , W 2 = { n − k + 1 } , and W 3 = { n − k + 2 , … , n } .

Let i ∈ W 1 . Adding x i − ( n − k + 1 ) x i in both sides of equation (1) for G , we obtain

(8) ( d i ( G ) + 1 − ( μ + n − k + 1 ) ) x i = x i − ( n − k + 1 ) x i + ∑ { u , j } ∈ E ( G ) x u .

In this case, note that d i ( G ′ ) = d i ( G ) + 1 , x i = 1 , and x n − k + 1 = − ( n − k ) . From these facts, equation (8) can be rewritten as

(9) ( d i ( G ′ ) − ( μ + n − k + 1 ) ) x i = x n − k + 1 + ∑ { u , i } ∈ E ( G ) x u ( d i ( G ′ ) − ( μ + n − k + 1 ) ) x i = ∑ { u , i } ∈ E ( G ′ ) x u .

Let i = n − k + 1 ∈ W 2 . Adding ( n − k ) x i − ( n − k + 1 ) x i in both sides of (1) of G and noting that d i ( G ′ ) = d i ( G ) + n − k , we obtain

(10) ( d i ( G ′ ) − ( μ + n − k + 1 ) ) x i = − x i + ∑ { u , i } ∈ E ( G ) x u .

Since x i = − n + k = − ∑ j = 1 n − k x j , from equation (10), we obtain

(11) ( d i ( G ′ ) − ( μ + n − k + 1 ) ) x i = ∑ { u , i } ∈ E ( G ′ ) x u .

Now, if i ∈ W 3 , we have x i = 0 and equality

(12) ( d i ( G ′ ) − ( μ + n − k + 1 ) ) x i = ∑ { u , i } ∈ E ( G ′ ) x u

holds. Therefore, from equations (9), (11), and (12), we have that x is an eigenvector of L ( G ′ ) associated with egeinvalue μ + ( n − k ) + 1 , and the proof is complete.□

Note that the previous results hold for any graph, but we will usually apply them to the family of threshold graphs. There are several different ways to define a threshold graph, as can be seen in [6]. We use the characterization by a binary generating sequence as follows.

Definition 2.4

Given a sequence { b i } of 0 ′ s and 1 ′ s with n elements, the threshold graph associated with the binary sequence { b i } is the graph on n vertices constructed recursively, starting with an empty graph, and for i = 1 , … , n

Add the isolated vertex i if b i = 0 ,
Add the vertex i adjacent to all vertices with label less than i if b i = 1 .

A vertex i is an isolated vertex if b i = 0 and a dominating vertex if b i = 1 .

3 Main results

In this section, we determine an eigenbasis of the Laplacian matrix of any threshold graph G . We split this section into two parts. In the first part, we consider that G is connected and state our main results supported by the lemmas of Section 2. In the second part, we assume that G is disconnected, and using some of the previous results, we prove that our main result also holds for disconnected threshold graphs.

3.1 Connected threshold graphs

Recall that x 1 , … , x n are the eigenvectors of L ( S n ) . We are going to prove that x 1 , x 2 , … , x n are also the eigenvectors of any connected threshold graph G on n vertices. Our proof considers Theorem 2.1 and Lemmas 2.2 and 2.3 applied to the star graph S n . Once S n is a subgraph of every connected threshold of n vertices, we prove that it is possible to add edges in S n in such a way that the new graph G has the same eigenvectors of L ( S n ) . However, note that the eigenspaces associated with the eigenvalues of G can be different from the eigenspaces of S n .

Proposition 3.1

Let G be a connected threshold on n vertices. Then, { x 1 , … , x n } is an eigenbasis of L ( G ) .

Proof

If G is isomorphic to S n , the result is straightforward. Suppose that G is connected and not isomorphic to S n with binary sequence b = ( 0 , b 2 , b 3 , … , b n − 1 , 1 ) . Since the star S n is a subgraph of G , we start from it to obtain to G by adding the appropriate edges and studying their effects under the eigenvectors of the resulting graphs. Let B = { x 1 , … , x n } . Take i 1 such that b i 1 is the first entry equal to one in b , i.e., b i 1 = 1 , and b l = 0 for all l < i 1 . Let G 1 be the resulting graph obtained from S n by adding the edges e t = { t , i 1 } with t = 1 , 2 , … , i 1 − 1 , i.e., G 1 = S n + e 1 + e 2 + ⋯ + e i 1 − 1 . Note that G 1 has binary sequence given by b ( 1 ) = ( 0 , … , 0 , 1 ⏟ i 1 , 0 , … , 0 , 1 ) , and is consequently threshold. Also, G 1 is a subgraph of G . Now, we want to prove that x k is still an eigenvector of L ( G 1 ) for each k considering the following three possible cases: k < i 1 , k = i 1 , and k > i 1 .

Case 1: Take x k such that k > i 1 , i.e., take the following eigenvectors of L ( S n ) :

x i 1 + 1 = ( 1 , 1 , … , 1 , 1 ⏟ i 1 , − ( i 1 + 1 ) , 0 , … , 0 ) , x i 1 + 2 = ( 1 , 1 , … , 1 , 1 ⏟ i 1 , 1 , − ( i 1 + 2 ) , 0 , … , 0 ) , ⋮ x n − 1 = ( 1 , 1 , … , 1 , 1 ⏟ i 1 , 1 , … , 1 , − ( n − 1 ) ) .

Since the entries x l k are all equal to 1 for l < i 1 , by Theorem 2.1, the resulting graph G 1 has the vectors x k as eigenvectors of L ( G 1 ) for k > i 1 .

Case 2: Let k = i 1 and take the eigenvector x i 1 = ( 1 , 1 , … , 1 , − i 1 ⏟ i 1 , 0 , … , 0 ) ∈ B . From Lemma 2.3, it follows that x i 1 is also an eigenvector of L ( G 1 ) .

Case 3: Let k < i 1 and take the vectors x k ∈ B such that k < i 1 , i.e.,

x 1 = ( 1 , − 1 , 0 , … , 0 ⏟ i 1 , … , 0 ) , x 2 = ( 1 , 1 , − 2 , 0 , … , 0 ⏟ i 1 , … , 0 ) , ⋮ x i 1 − 1 = ( 1 , 1 , … , 1 , − ( i 1 − 1 ) , 0 ⏟ i 1 , … , 0 ) .

Let W 1 = { 1 , 2 , … , i 1 − 1 } and W 2 = { i 1 , i 1 + 1 , … , n } be a partition of the vertices of G 1 . From Lemma 2.2, these vectors are the eigenvectors of L ( G 1 ) . Hence, the eigenvectors of L ( G 1 ) are the vectors of B . Now, suppose that there exists b i 2 = 1 for i 1 < i 2 < n , where b i 2 is the next entry after b i 1 that is equal to 1. It means that in the way of obtaining G from G 1 , we obtain a new graph G 2 by adding the edges e t ′ = { t , i 2 } for t = 1 , 2 , … , i 2 − 1 , i.e., G 2 = G 1 + e 1 ′ + e 2 ′ + ⋯ + e i 2 − 1 ′ . Note that G 2 has binary sequence b ( 2 ) = ( b 1 ( 2 ) , … , b n ( 2 ) ) with b j ( 2 ) = 1 if j ∈ { i 1 , i 2 } and b j ( 2 ) = 0 elsewhere and therefore is threshold. Again, G 2 is a subgraph of G . We are going to prove that L ( G 1 ) and L ( G 2 ) share the same eigenvectors by analyzing again the cases k ≤ i 2 , k = i 2 , and k > i 2 .

Case 1 ′ : Take the vectors x k for k > i 2 , i.e., consider the following eigenvectors of L ( S n ) :

x i 2 + 1 = ( 1 , 1 , … , 1 , 1 ⏟ i 2 , − ( i 2 + 1 ) ︸ k , 0 , … , 0 ) , x i 2 + 2 = ( 1 , 1 , … , 1 , 1 ⏟ i 2 , 1 , − ( i 2 + 2 ) ︸ k , 0 , … , 0 ) , ⋮ x n − 1 = ( 1 , 1 , … , 1 , 1 ⏟ i 2 , 1 , … , 1 , − ( n − 1 ) ︸ k ) .

Similar to what we have done for graph G 1 in Case 1, since all entries x l k are equal to 1 for l < i 2 , from Theorem 2.1, the vectors x k with k > i 2 are the eigenvectors of L ( G 2 ) as well.

Case 2 ′ : Consider x i 2 = ( 1 , 1 , … , 1 , − i 2 ⏟ i 2 , 0 , … , 0 ) ∈ B . As G 2 = G 1 + e 1 ′ + e 2 ′ + ⋯ + e i 2 − 1 ′ , where e t ′ = { t , i 2 } for t = 1 , 2 , … , i 2 − 1 , from Lemma 2.3, it follows that x i 2 is also an eigenvector of L ( G 2 ) .

Case 3 ′ : Consider x k ∈ B for k < i 2 , i.e., take the vectors:

x 1 = ( 1 , − 1 , 0 , … , 0 ⏟ i 2 , … , 0 ) , x 2 = ( 1 , 1 , − 2 , 0 , … , 0 ⏟ i 2 , … , 0 ) , ⋮ x i 2 − 1 = ( 1 , 1 , … , 1 , − ( i 2 − 1 ) , 0 ⏟ i 2 , … , 0 ) .

Consider the vertex partition of G 2 as W 1 = { 1 , 2 , … , i 2 − 1 } and W 2 = { i 2 , … , n } . Since x k with k < i 2 are the eigenvectors of L ( G 1 ) , from Lemma 2.2, it follows that they are also the eigenvectors of L ( G 2 ) .

So, the vectors x k are also the eigenvectors of L ( G 2 ) with k = 1 , 2 , … , n − 1 . Therefore, taking b i p + 1 in an inductive way, as the next entry equal to 1 after b i p , we can note that the threshold graph G p + 1 = G p + e 1 + e 2 + ⋯ + e i p + 1 − 1 obtained from the threshold G p by adding the edges e t = { t , i p + 1 } such that t = 1 , 2 , … , i p + 1 − 1 , for some p ∈ N , has eigenvectors x k with k = 1 , 2 , … , n − 1 , and those vectors are also the eigenvectors of L ( G p ) . Note that the binary sequence of G p + 1 is given by b ( p + 1 ) = ( b 1 ( p + 1 ) , … , b n ( p + 1 ) ) , where b j ( p + 1 ) = 1 for all j ∈ { i 1 , i 2 , … , i p + 1 } and b j ( p + 1 ) = 0 elsewhere, which implies that G p + 1 is threshold. Then, we can make as many steps as needed in the binary sequence of G for all i such that b i = 1 , and we will have that x k with k = 1 , 2 , … , n − 1 are the eigenvectors of L ( G ) . Moreover, x n is an eigenvector of L ( G ) associated with eigenvalue 0 since G is connected. Hence, since G and its order n are arbitrary, it follows that all connected threshold graphs on n vertices have B as an eigenbasis of L ( G ) . □

$Figure 1 Graph G 1 {G}_{1} .$

Figure 1

Graph G 1 .

Example 3.2

Let G 1 be the threshold graph of Figure 1. Its binary sequence is given by b 1 = ( 0 , 0 , 0 , 1 , 0 , 1 ) . Note that G 1 has the graph S 6 as a subgraph. In other words, we have that G 1 = S 6 + { 1 , 4 } + { 2 , 4 } + { 3 , 4 } . The eigenvalues of L ( S 6 ) are 1, 6, and 0 of multiplicities 4, 1, and 1, respectively. An eigenbasis of L ( S 6 ) is given by B = { x 1 , x 2 , x 3 , x 4 , x 5 , x 6 } , where x 1 = ( 1 , − 1 , 0 , 0 , 0 , 0 ) , x 2 = ( 1 , 1 , − 2 , 0 , 0 , 0 ) , x 3 = ( 1 , 1 , 1 , − 3 , 0 , 0 ) , x 4 = ( 1 , 1 , 1 , 1 , − 4 , 0 ) , x 5 = ( 1 , 1 , 1 , 1 , 1 , − 5 ) , and x 6 = ( 1 , 1 , 1 , 1 , 1 , 1 ) . Moreover, the vectors x 1 , x 2 , x 3 , and x 4 are associated with eigenvalue 1, the eigenvector x 5 is associated with eigenvalue 6, and the eigenvector x 6 is associated with eigenvalue 0. Note that b 4 is the first entry equal to one in b 1 . Then, by Lemma 2.2, we add one unit to the eigenvalue 1 for 1 ≤ k ≤ i − 2 = 2 , and we obtain that ( 2 , x 1 ) and ( 2 , x 2 ) are the eigenpair for L ( G 1 ) . When k = 3 , i.e., k = i − 1 , by Lemma 2.3, we add 1 + k to the eigenvalue 1, and hence, we have that ( 5 , x 3 ) is the eigenpair for L ( G 1 ) . Finally, by Theorem 2.1 (edge principle), we obtain that the eigenpairs ( 1 , x 4 ) , ( 6 , x 5 ) , and ( 0 , x 6 ) are preserved in the graph G 1 . Therefore, the eigenbasis B of L ( S 6 ) is also an eigenbasis of L ( G 1 ) . Table 1 summarizes the result.

Table 1

Eigenvectors and eigenvalues for the Laplacian matrix of the connected threshold graphs S 6 and G 1

Eigenvector of L	Eigenvalue of L
	For L ( S 6 )	For L ( G 1 )
x 1	1	2
x 2	1	2
x 3	1	5
x 4	1	1
x 5	6	6
x 6	0	0

Proposition 3.3

Let G = ( V , E ) be a connected graph on n vertices such that { x 1 , … , x n } is an eigenbasis of L ( G ) . Then, G is threshold.

Proof

Since { x 1 , … , x n } are the eigenvectors of L ( G ) , consider the i th row of the eigenequation

[ L ( G ) ] i x k = μ k x i k ,

for i < n . It implies that

(13) l i , 1 x 1 k + l i , 2 x 2 k + … + l i , i − 1 x i − 1 k + l i , i x i k + … + l i , n x n k = μ k x i k .

From (13), for each k ∈ { 1 , … , i − 1 } , we obtain

(14) l i , 1 − l i , 2 = 0 l i , 1 + l i , 2 − 2 l 1 , 3 = 0 ⋮ l i , 1 + l i , 2 + ⋯ + l i , i − 2 − ( i − 2 ) l i , i − 1 = 0 l i , 1 + l i , 2 + l i , 3 + ⋯ + l i , i − 1 − ( i − 1 ) l i , i = − μ i − 1 ( i − 1 ) .

Using the first i − 2 equations of (14), we obtain

l i , 1 = l i , 2 = ⋯ = l i , i − 1 .

For j < i , each entry l i , j of L ( G ) is equal to 0 or − 1 . It means that vertex i is either connected to all previous vertices with labels 1 , … , i − 1 or vertex i is connected to none of them. So, we can represent G by a binary sequence b i ∈ { 0 , 1 } for all i = 1 , … , n − 1 .

Now, we need to prove that b n = 1 . Take the n th row of L ( G ) . For 1 ≤ k ≤ n − 1 and using the n − 2 equation of (13), we obtain l n , 1 = l n , 2 = ⋯ = l n , n − 1 . Using the last equation given by

l n , 1 + l n , 2 + l n , 3 + ⋯ + l n , n − 1 − ( n − 1 ) l n , n = − μ n − 1 ( n − 1 ) ,

we have that l n , n − l n , 1 = μ n − 1 . Note that if l n , 1 = 0 , then l n , n = ∑ j = 1 n − 1 l n , j = 0 . In this case, l n , n = μ n − 1 = 0 , which is a contradiction since G is connected. So, l n , j = − 1 for all 1 ≤ j ≤ n − 1 , which implies that vertex n is a dominanting vertex in G . This completes the proof.□

Now, we state the main result of this article.

Theorem A

Let G be a connected graph on n vertices. Then, G is a threshold graph if and only if { x 1 , … , x n } is an eigenbasis of the Laplacian matrix of G.

Proof

The proof is a straightforward consequence of Propositions 3.1 and 3.3.□

3.2 Disconnected threshold graphs

Consider now that G is a disconnected threshold graph. We are going to prove that the eigenvectors of L ( G ) are the same as those of L ( S n − p ⊔ p K 1 ) , where p ≥ 1 is the number of isolated vertices. We show that n − p eigenvectors of L ( G ) can be obtained from the eigenvectors of L ( S n − p ) by adding p entries equal to 0. The remaining p eigenvectors are the standard vectors e j for a convenient j . Let G = H ⊔ p K 1 , where H is a connected threshold graph with n − p vertices. For each 1 ≤ k ≤ n − p , let y k be the n -vector with entries:

(15) y j k = x j k , if 1 ≤ j ≤ n − p , 0 , otherwise ,

where each x j k is given by (2). For 1 ≤ l ≤ p , write

(16) y j n − p + l = 1 , if j = n − p + l , 0 , otherwise .

We will prove that { y 1 , y 2 , … , y n } is an eigenbasis of L ( G ) . The next propositions are important to prove the main result of this section.

Proposition 3.4

Let G be a disconnected threshold graph on n vertices having p of them as isolated vertices. Then, { y 1 , y 2 , … , y n } is an eigenbasis of L ( G ) .

Proof

Let G = H ⊔ p K 1 , where H is a connected threshold with n − p vertices and p isolated vertices. From Proposition 3.1, { x 1 , x 1 , … , x n − p } is an eigenbasis of L ( H ) . Since

L ( G ) = L ( H ) 0 ( n − p ) × p 0 ( n − p ) × p T 0 p × p ,

we have that L ( G ) y k = μ k y k , for 1 ≤ k ≤ n − p , where μ k is an eigenvalue of L ( H ) , and for n − p + 1 ≤ k ≤ n , we have that L ( G ) y k = 0 y k = 0 . Therefore, { y 1 , y 2 , … , y n } is an eigenbasis of L ( G ) , and the proof is complete.□

Proposition 3.5

Let G = ( V , E ) be a disconnected graph on n vertices such that { y 1 , y 2 , … , y n } is an eigenbasis of L ( G ) . Then, G is threshold.

Proof

Let G = H ⊔ p K 1 , where H is a connected threshold with n − p vertices and p isolated vertices. Take i th row of L ( G ) such that i < n − p . Since { y 1 , y 2 , … , y n } is an eigenbasis of L ( G ) , we have

[ L ( G ) ] i y k = μ k y i k ,

for 1 ≤ k < i − 1 . Since y j k = x j k for all j = 1 , … , n − p , by the same arguments of the ones used in Proposition 3.3, we obtain that H is threshold. Therefore, G is threshold.□

Theorem 3.6

Let G be a disconnected graph on n vertices. Then, G is threshold if and only if { y 1 , y 2 , … , y n } is an eigenbasis of L ( G ) .

Proof

The proof is a straightforward consequence of Propositions 3.4 and 3.5.□

4 Application to WHD matrix

Recently, Adm et al. [1] introduced the weak Hadamard matrices. A matrix W ∈ R n × n is called weak Hadamard if W has all entries from the set { − 1 , 0 , 1 } and W T W is tridiagonal. Also, Adm et al. [1] defined that a graph G is WHD if L ( G ) is diagonalizable by a weak Hadamard matrix W , i.e.,

L ( G ) = W Λ W − 1 ,

where Λ is the diagonal matrix of the eigenvalues of L ( G ) .

In this section, we build an infinite family of threshold graphs that are WHD. Our motivations are related to (i) showing the relevance of our results from Section 3 by obtaining a new infinite family of threshold graphs, which are also WHD since those graphs are not easy to find; (ii) the fact that it could be useful to create new graphs with Laplacian perfect state transfer among those WHD threshold graphs since the eigenvectors are known for a given order and the eigenvalues are easily obtained from the degree sequence. A related work on this subject is the recent work of McLaren et al. [7], and Johnston et al. [5]; (iii) an interesting problem raised by Adm et al. [1] of determining all cographs that are WHD. While completely solving this problem seems difficult, we found some new threshold graphs that are WHD and partially answer the question.

Note that Adm et al. [1] proved that the threshold graphs K k c ∨ K n are WHD under certain conditions on n and k (Lemma 4.6 of [1]). Next, we present an alternative and more complete proof of this result since we show that the matrix of the eigenvectors is WHD, in such a way that it is clearer here than in the proof provided by Adm et al. [1]. Moreover, we extend their result and prove that the graph G ′ obtained as G ′ = K k c ∨ K n + e , where e is an edge, preserves the properties of being threshold and WHD. Hence, we obtain a new infinite family of thresholds that are WHD using the methods developed in Sections 2 and 3. To this end, we denote the eigenspace associated with an eigenvalue μ of L ( G ) by ℰ L ( μ ) = { x ∈ R n , x ≠ 0 : L ( G ) x = μ x } , and to prove these results, we build an eigenbasis of the Laplacian matrix of the desired graphs from the eigenbasis of L ( S n ) described in (2).

Proposition 4.1

Let G = K k c ∨ K n , with k ≥ 3 . If n − k ∈ { 0 , 1 , 2 } , then G is a WHD graph.

Proof

Note that G = K k c ∨ K n is a threshold graph on n + k vertices and binary sequence b = ( 0 , 0 , … , 0 ︸ k , 1 , 1 , … , 1 ︸ n ) . The eigenvalues of L ( G ) are n , n + k , and 0 with multiplicities k − 1 , n , and 1, respectively. We arrange the vertices of G so that the first k vertices are the vertices of K k c . From Proposition 3.1, L ( G ) and L ( S n + k ) share the same eigenvectors, and we obtain that the eigenspaces associated with eigenvalues n , n + k , and 0 of L ( G ) are given by

ℰ L ( n ) = span { x 1 , x 2 , … , x k − 1 } , ℰ L ( n + k ) = span { x k , … , x n + k − 1 } , ℰ L ( 0 ) = span { x n + k } .

Now, we are going to obtain linear combinations of those vectors to obtain vectors with entries in the set { − 1 , 0 , 1 } . Consider a convenient linear combination given by

u 1 = x 1 and u i = x i − x i − 1 i , for i = 2 , … , k − 1 ,

i.e.,

u i = e i − e i + 1 , for i = 1 , … , k − 1 .

Hence, we obtain

ℰ L ( n ) = span { u 1 , u 2 , … , u k − 1 } ,

and each entry of u i , for i = 1 , … , k − 1 is equal to − 1 , 0, or 1. Similarly, consider a convenient linear combination given by

(17) v i = x i − x i − 1 i , for i = k + 1 , … , n + k − 1 ,

i.e.,

v i = e i − e i + 1 , for i = k + 1 , … , n + k − 1 .

Hence, we obtain n − 1 vectors that belong to ℰ L ( n + k ) , and each entry of v i is equal to − 1 , 0, or 1. Since dim ( ℰ L ( n + k ) ) = n , we need another vector v to complete the eigenspace related to the eigenvalue n + k of L ( G ) . Since

L ( G ) = n 0 0 … 0 − 1 − 1 − 1 … − 1 0 n 0 … 0 − 1 − 1 − 1 … − 1 ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ ⋮ ⋮ ⋱ ⋮ 0 0 0 … n − 1 − 1 − 1 … − 1 − 1 − 1 − 1 … − 1 n + k − 1 − 1 − 1 … − 1 − 1 − 1 − 1 … − 1 − 1 n + k − 1 − 1 … − 1 ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ ⋮ ⋮ ⋱ ⋮ − 1 − 1 − 1 … − 1 − 1 − 1 − 1 … n + k − 1 ,

note that v = ( 1 , 1 , … , 1 ︸ k , − 1 , − 1 , … , − 1 ︸ n − 1 , n − k − 1 ︸ 1 ) is an eigenvector of L ( G ) associated with eigenvalue n + k . Moreover, if n − k ∈ { 0 , 1 , 2 } , then the eigenvector v of L ( G ) has entries from { − 1 , 0 , 1 } . Besides, v is not a linear combination of the set { v k + 1 , … , v n − 1 } since the first k entries of each v i are equal to zero. Then, we obtain that

ℰ L ( n + k ) = span { v k + 1 , … , v n − 1 , v }

and each entry of v i , for i = k + 1 , … , n + k − 1 and v are equal to − 1 , 0, or 1.

Let

(18) β = { u 1 , u 2 , … , u k − 1 , v k + 1 , v k + 2 , … , v n − 1 , v , x n + k }

be an eigenbasis of L ( G ) . Note that u 1 is orthogonal to all vectors in β , except for the vector u 2 ; u i , for i = 2 , 3 , … , k − 2 , is orthogonal to all vectors in β , except for the vectors u i − 1 and u i + 1 ; u k − 1 is orthogonal to all vectors in β , except for the vector u k − 2 ; v k + 1 is orthogonal to all vectors in β , except for the vector v k + 2 ; v i , for i = k + 2 , k + 3 , … , n − 2 , is orthogonal to all vectors in β , except for the vectors v i − 1 and v i + 1 ; v n − 1 is orthogonal to all vectors in β , except for the vector v n − 2 ; v and x n + k are orthogonal to all vectors in β .

In this case, if we form the matrix W where the first k − 1 columns are the eigenvectors of ℰ L ( n ) , the next n columns are the eigenvectors of ℰ L ( n + k ) , and the last column is given by the eigenvector of ℰ L ( 0 ) , then we have

W = [ ( u 1 ) T , … , ( u k − 1 ) T ∣ ( v k + 1 ) T , … , ( v n + k − 1 ) T , ( v ) T ∣ x n + k ] ,

and therefore, it follows that W T W is tridiagonal, and the proof is complete.□

Next, we use our previous results to prove that the graphs obtained from K k c ∨ K n plus an edge are also threshold WHD graphs.

Proposition 4.2

Let G = K k c ∨ K n such that n − k ∈ { 0 , 1 , 2 } and k ≥ 4 . Then, G ′ = G + e is a WHD threshold graph.

Proof

From Proposition 4.1, G = K k c ∨ K n is WHD. Let G ′ = G + e . Without loss of generality, take e = { 1 , 2 } . Hence, G ′ can be represented by the binary sequence b = ( 0 , 1 , 0 , 0 , … , 0 ︸ k − 2 , 1 , 1 , … , 1 ︸ n ) . So, G ′ is a threshold graph on n + k vertices. The eigenvalues of L ( G ′ ) are n + 2 , n , n + k and 0 with multiplicities 1, k − 2 , n , and 1, respectively. From Proposition 3.1, L ( G ′ ) and L ( S n + k ) share the same eigenvectors. Also,

ℰ L ( n + 2 ) = span { x 1 } , ℰ L ( n ) = span { x 2 , … , x k − 1 } , ℰ L ( n + k ) = span { x k , … , x n + k − 1 } , ℰ L ( 0 ) = span { x n + k } .

We have that ℰ L ( n + k ) = span { v k + 1 , … , v n + k − 1 , v } , where the vectors v k + 1 , … , v n + k − 1 are given in (17) and v = ( 1 , 1 , … , 1 ︸ k , − 1 , − 1 , … , − 1 ︸ n − 1 , n − k − 1 ︸ 1 ) .

If k is even, we can make the appropriate linear combination of the vectors x 2 , … , x k − 1 to obtain the vectors w j and z j , for j = 1 , … , k 2 − 1 , where

(19) w j = − 1 2 j − 1 x 2 j − 2 − ( j − 1 ) j ( 2 j − 1 ) x 2 j − 1 + ( j + 1 ) j ( 2 j + 1 ) x 2 j + 1 2 j + 1 x 2 j + 1 z j = 1 2 j + 1 ( x 2 j + 1 − x 2 j ) ,

i.e.,

( w 1 ) T = 1 1 − 1 − 1 0 0 0 0 ⋮ 0 0 , ( w 2 ) T = 0 0 1 1 − 1 − 1 0 0 ⋮ 0 0 , … , ( w j ) T = 0 0 0 0 ⋮ 0 0 1 1 − 1 − 1

and

( z 1 ) T = 0 0 1 − 1 0 0 0 0 ⋮ 0 0 , ( z 2 ) T = 0 0 0 0 1 − 1 0 0 ⋮ 0 0 , … , ( z j ) T = 0 0 0 0 ⋮ 0 0 0 0 1 − 1 .

Therefore,

ℰ L ( n ) = span { w 1 , … , w j , z 1 , … , z j } .

Note that each entry of w j and z j , for j = 1 , … , k 2 − 1 , is equal to -1, 0, or 1. Moreover, { z 1 , z 2 , … , z j } forms a mutually orthogonal set and w i is orthogonal to { w i + 2 , w i + 3 , … , w j } , for i = 1 , … , j − 2 . Finally, observe that w i is orthogonal to z k for any i , k . Hence, if we form the matrix

W = [ ( x 1 ) T ∣ ( w 1 ) T , … , ( w j ) T , ( z 1 ) T , … , ( z j ) T ∣ ( v k + 1 ) T , … , ( v n + k − 1 ) T , ( v ) T ∣ ( x n + k ) T ] ,

then it follows that W T W is a tridiagonal matrix.

If k is odd, consider the appropriate linear combination of the vectors x 2 , … , x k − 1 to obtain the vectors w j , z j , and z , for j = 1 , … , k − 1 2 − 1 , where

(20) w j = − 1 2 j − 1 x 2 j − 2 − ( j − 1 ) j ( 2 j − 1 ) x 2 j − 1 + ( j + 1 ) j ( 2 j + 1 ) x 2 j + 1 2 j + 1 x 2 j + 1 , z j = 1 2 j + 1 ( x 2 j + 1 − x 2 j ) , z = 1 n − 1 ( x n − 1 − x n − 2 ) ,

i.e.,

( w 1 ) T = 1 1 − 1 − 1 0 0 0 0 ⋮ 0 0 0 , ( w 2 ) T = 0 0 1 1 − 1 − 1 0 0 ⋮ 0 0 0 , … , ( w j ) T = 0 0 0 0 ⋮ 0 0 1 1 − 1 − 1 0

and

( z 1 ) T = 0 0 1 − 1 0 0 0 0 ⋮ 0 0 0 , ( z 2 ) T = 0 0 0 0 1 − 1 0 0 ⋮ 0 0 0 , … , ( z j ) T = 0 0 0 0 ⋮ 0 0 0 0 1 − 1 0 and ( z ) T = 0 0 0 0 ⋮ 0 0 0 0 0 1 − 1 .

Therefore,

ℰ L ( n ) = span { w 1 , … , w j , z 1 , … , z j , z } .

Observe that each entry of w j , z j , for j = 1 , … , k − 1 2 − 1 , and z is equal to − 1 , 0, or 1. Moreover, { z 1 , z 2 , … , z j } forms a mutually orthogonal set, w i is orthogonal to { w i + 2 , w i + 3 , … , w j } , for i = 1 , … , j − 2 , and w i is orthogonal to z k for any i , k . Finally, note that z is orthogonal to { w 1 , w 2 , … , w j − 1 , z 1 , z 2 , … , z j − 1 } , but z is not orthogonal to neither w j nor z j . In this case, we form the matrix

W = [ ( x 1 ) T ∣ ( w 1 ) T , … , ( w j − 1 ) T ∣ ( w j ) T , ( z ) T , ( z j ) T ∣ ( z 1 ) T , … , ( z j − 1 ) T ∣ ( v k + 1 ) T , … , ( v n + k − 1 ) T , ( v ) T ∣ ( x n + k ) T ] ,

and it follows that W T W is a tridiagonal matrix. So G ′ is WHD, and the proof is complete.□

5 Concluding remarks and future work

In the study of McLaren et al. [7], they build an infinite family of threshold graphs that are WHD of order 2 l for l ≥ 1 . We build here an infinite family of threshold graphs that are WHD and are not completely described by their work. For instance, the graph ( K 2 ⊔ K 6 c ) ∨ K 8 belongs to the family we build but does not belong to the families defined in [7]. In addition, we are able to generate threshold WHD graphs for any number of vertices, and we are not restricted to threshold graphs of order n = 2 l for l ≥ 1 . These facts make the graphs obtained from Proposition 4.2 interesting.

We believe that it is possible to even add more than two edges to the graph K k c ∨ K n − k so that the resulting graph is still threshold WHD. The main results of [2] and [3] can be useful in generating the desired graphs. We will explore this possibility in future work, aiming to describe all threshold graphs that are WHD.

Acknowledgements

The authors would like to thank the anonymous referees for their careful review and comments, which improved the manuscript.

Funding information: The research of Renata Del-Vecchio was supported by CNPq Grant Nos 404788/2023-8 and 308159/2022-5. The research of Leonardo de Lima was supported by CNPq Grant Nos 315739/2021-5 and 403963/2021-4.
Author contributions: All authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results, and approved the final version of the manuscript.
Conflict of interest: The authors state no conflict of interest.
Authorization for the use of human subjects: This is not applicable.
Informed consent: This is not applicable.
Authorization for the use of experimental animals: This is not applicable.
Data availability statement: All data generated or analyzed during this study are included in this published article.

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Received: 2024-03-11

Revised: 2024-09-16

Accepted: 2024-10-14

Published Online: 2024-10-30

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

Articles in the same Issue

https://doi.org/10.1515/spma-2024-0029

Keywords for this article

threshold graph; spectral characterization; Laplacian matrix; weakly Hadamard diagonalizable graph

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