1 Introduction

Abstract

The Kite graph, denoted by is obtained by appending a complete graph to a pendant vertex of a path . In this paper, firstly we show that no two non-isomorphic kite graphs are cospectral w.r.t adjacency matrix. Let be a graph which is cospectral with and the clique number of is denoted by . Then, it is shown that . Also, we prove that graphs are determined by their adjacency spectrum.


Key Words: Kite graph, cospectral graphs, clique number, determined by adjacency spectrum

2010 Mathematics Subject Classification: 05C50, 05C75

On the spectral characterization of Kite graphs


Sezer Sorgun  , Hatice Topcu



Department of Mathematics,

Nevþehir Hacı Bektaş Veli University,

Nevþehir 50300, Turkey.

e-mail: srgnrzs@gmail.com, haticekamittopcu@gmail.com

[3mm]


(Received June 4, 2015)

1 Introduction

All of the graphs considered here are simple and undirected. Let be a graph with vertex set and edge set .For a given graph , if does not contain as a subgraph, then is called . A complete subgraph of is called a clique of G. The clique number of G is the number of vertices in the largest clique of and it is denoted by . Let be the (0,1)-adjacency matrix of G and the degree of the vertex . The polynomial is the characteristic polynomial of G where is the identity matrix. Eigenvalues of the matrix are called adjacency eigenvalues. Since is real and symmetric matrix, adjacency eigenvalues are all real numbers and will be ordered as . Adjacency spectrum of the graph G consists the adjacency eigenvalues with their multiplicities. The largest eigenvalue of a graph is known as its spectral radius.

Two graphs and are said to be cospectral if they have same spectrum (i.e. same characteristic polynomial). A graph is determined by adjacency spectrum, shortly DAS, if every graph cospectral with is isomorphic to . It has been conjectured by the first author in [6] that almost all graphs are determined by their spectrum, DS for short. But it is difficult to show that a given graph is DS. Up to now, only few graphs are proved to be DS [2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 15]. Recently, some papers have been appeared that focus on some special graphs (oftenly under some conditions) and prove that these special graphs are DS or non-DS [2, 3, 4, 7, 9, 10, 11, 12, 13, 15]. For a recent widely survey, one can see [6].

The Kite graph, denoted by , is obtained by appending a complete graph with vertices to a pendant vertex of a path graph with vertices . If , it is called short kite graph.

In this paper, firstly we obtain the characteristic polynomial of kite graphs and show that no two non-isomorphic kite graphs are cospectral w.r.t adjacency matrix. Then for a given graph which is cospectral with , the clique number of is . Also we prove that graphs are DAS for all .

2 Preliminaries

First, we give some lemmas that will be used in the next sections of this paper.

Lemma 2.1.

[3] Let be a pendant vertex of a graph and be the vertex which is adjacent to . Let be the induced subgraph obtained from by deleting the vertex . If and are deleted, the induced subgraph is obtained. Then,

Lemma 2.2.

[5] For matrices and , followings are equivalent :

(i) and are cospectral

(ii) and have the same characteristic polynomial

(iii) for

Lemma 2.3.

[5] For the adjacency matrix of a graph , the following parameters can be deduced from the spectrum;

(i) the number of vertices

(ii) the number of edges

(iii) the number of closed walks of any fixed length.

Let be the number of subgraphs of a graph which are isomorphic to and let be the number of closed walks of length in .

Lemma 2.4.

[12] The number of closed walks of length 2 and 3 of a graph are given in the following, where m is number of edges of .

(i) and .

In the rest of the paper, we denote the number of subgraphs of a graph which are isomorphic to complete graph with .

Theorem 2.5.

[1] For any integers and , if we denote the spectral radius of with then

Theorem 2.6.

[14] Let be a graph with n vertices, m edges and spectral radius . If is , then

Theorem 2.7.

[4] Let denote the graph obtained by attaching m pendant edges to a vertex of complete graph . The graph and its complement are determined by their adjacency spectrum.

3 Characteristic Polynomials of Kite Graphs

We use similar method with [3] to obtain the general form of characteristic polynomials of graphs. Obviously, if we delete the vertex with one degree from short kite graph, the induced subgraph will be the complete graph . Then, by deleting the vertex with one degree and its adjacent vertex, we obtain complete graph with vertices, . From Lemma 2.1, we get

Similarly, for induced subgraphs will be and respectively. By Lemma 2.1, we get

By using these polynomials, let us calculate the characteristic polynomial of where . Again, by Lemma 2.1 we have

Coefficients of above equation are , . Simultaneously, we get

and coefficients of above equation are , . Then for , we have

and coefficients of above equation are . In the following steps, for , . From this difference equation, we get

Now, let and . Then, we have

and by calculation the characteristic polynomial of any kite graph, , where , is

Theorem 3.1.

No two non-isomorphic kite graphs have the same adjacency spectrum.

Proof.

Assume that there are two cospectral kite graphs with number of vertices respectively, and . Since they are cospectral, they must have same number of vertices and same characteristic polynomials. Hence, and we get

i.e.

i.e.

Let . It follows that . Then, we have

By using the fact that and , we get

Since , the derivation of th of equals to zero again. Thus, we have

i.e.

i.e.

since . This is a contradiction with our assumption that . For , we get the similar contradiction. So must be equal to . Hence and these graphs are isomorphic. ∎

4 Spectral Determination of Graphs

Lemma 4.1.

Let be a graph which is cospectral with . Then we get

.

Proof.

Since is cospectral with , from Lemma 2.3, has the same number of vertices, same number of edges and same spectrum with . So, if has vertices and edges, and . Also, . From Theorem 2.6, we say that if then isn’t . It means that, contains as a subgraph. Now, we claim that for , . By Theorem 2.5, we’ve already known that . Hence, we need to show that, when , . Indeed,

By the help of Mathematica, for we can see

i.e.

i.e.

Since and , we get

Thus, we proved our claim and so contains as a subgraph such that . Consequently, .

Theorem 4.2.

graphs are determined by their adjacency spectrum for all .

Proof.

If or , graphs are actually the path graphs or . Also if , then we obtain the lollipop graph . As is known, these graphs are already DAS [3]. Hence we will continue our proof for . For a given graph with vertices and edges, assume that is cospectral with . Then by Lemma 2.3 and Lemma 2.4, , and . From Lemma 3.2.1, . When , . It’s well-known that complete graph is DS. So . If , then contains at least one clique with size . It means that the edge number of is greater than or equal to . But it is a contradiction since . Hence, . Because of these, . Let us investigate the three cases, respectively, , , .

CASE 1 : Let . Then . So, contains at least one clique with size . This clique is denoted by . Let us label the five vertices, respectively, with which are not in the clique and call the set of these five vertices with . We demonstrate this case by the following figure.

Figure 1:

For , denotes the number of adjacent vertices of in . By the fact that , for all we say

(1)

Also, denotes the number of common adjacent vertices in of and such that and . Similarly, if then

(2)

Moreover, denotes the number of edges between the vertices of and denotes the number of cliques with size 3 which are composed by vertices of .

First of all, since the number of edges of is equal to ,

. It follows that

(3)

Similarly, by using , we get

. Hence, we have

(4)

If , then . Clearly, this is contradiction. Also if , then which implies . Again this is a contradiction. For this reason, we will continue for .

Clearly, . So, we will investigate the cases of .

Subcase 1

Let . Then, and from (3), we have

(5)

Hence, we get

Clearly,

Since, the spectrum of does not contain zero, has not an isolated vertex. So, from this fact and (1), we get for all . Hence, by (5), we get

(6)

From (1) and (5), which implies . Where ,

(7)

This means that, . But this result contradicts with (4).

Subcase 2

Let . Then and from (3) we get

(8)

Since and by (2), . From here and (1), we have

(9)

By using (8) and (1), we obtain which implies . Since is an integer, . Where ,

(10)

This means that, . But this result contradicts with (4).

Subcase 3

Let . Then and by (3), we get

(11)

By using similar way with last subcase, we obtain

(12)

and . If , we have

(13)

By (12) and (13), we get . This result contradicts with (4) as in Subcase 2.

Subcase 4

Let . Then and

By using similar way again, we obtain

(14)

Since , we have

(15)

By (14) and (15), we have . We get same contradiction with (4).

Subcase 5

Let .Then and . Similarly, we obtain

(16)

Since , we have

(17)

By (16) and (17), we get same contradiction with (4).

Subcase 6

Let .Then and . Similarly, we obtain

(18)

Since , we get

(19)

By (18) and (19), we get same contradiction with (4).

Subcase 7

Let .Then and . Similarly, we obtain

(20)

Since , we have

(21)

By (20) and (21), we get same contradiction with (4).

Subcase 8

Let .Then and . Also here,

Hence, in the same way as former subcases, we obtain

(22)

Since , we get

(23)

So, by (22) and (23), we get same contradiction with (4).

Subcase 9

Let . Then and . Such as in the last subcase, we get

Hence, we obtain

(24)

Since , we get