## 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 2.7.

[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.

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