1 Introduction

# On the spectral characterization of Kite graphs

## 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]

## 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,

 PA(G)(λ)=λPA(G1)(λ)−PA(G2)(λ)
###### 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

 p−1+1p2+1p3<ρ(Kitep,q)
###### Theorem 2.6.

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

 μ≤√2m(r−1r)
###### 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

 PA(Kitep,1)(λ) = λPA(Kp)(λ)−PA(Kp−1)(λ) = λ(λ−p+1)(λ+1)p−1−[(λ−p+2)(λ+1)p−2] = (λ+1)p−2[(λ2−λp+λ)(λ+1)−λ+p−2] = (λ+1)p−2[λ3−(p−2)λ2−λp+p−2]

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

 PA(Kitep,2)(λ) = λPA(Kitep,1)(λ)−PA(Kp))(λ) = λ(λPA(Kp)(λ)−PA(Kp−1)(λ))−PA(Kp))(λ) = (λ2−1)PA(Kp)(λ)−λPA(Kp−1)(λ)

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

 PA(Kitep,1)(λ) = λPA(Kp)(λ)−PA(Kp−1)(λ)

Coefficients of above equation are , . Simultaneously, we get

 PA(Kitep,2)(λ) = (λ2−1)PA(Kp)(λ)−λPA(Kp−1)(λ)

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

 PA(Kitep,3)(λ) = λPA(Kitep,2)(λ)−PA(Kitep,1))(λ) = (λ(λ2−1)−λ)PA(Kp)(λ)−((λ2−1)PA(Kp−1)(λ))

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

 an=n∑k=0(λ+√λ2−42)k(λ−√λ2−42)n−k

Now, let and . Then, we have

 an=n∑k=0u2k−n=u−n(1−u2n+2)1−u2

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

 PA(\emphKitep,q)(u+u−1) = an−pPA(Kp)(u+u−1)−an−p−1PA(Kp−1)(u+u−1) = u−n+p(1−u2n−2p+2)1−u2.((u+u−1−p+1).(u+u−1+1)p−1) −u−n+p+1(1−u2n−2p+4)1−u2.((u+u−1−p+2).(u+u−1+1)p−2) = u−n+p(1+u−u−1)p−21−u2[(2−p).(1+u−1−u2n−2p+2−u2n−2p+3) +(u−2−u2n−2p+4)] = u−q(1+u−u−1)p−21−u2[(2−p).(1+u−1−u2q+2−u2q+3) +(u−2−u2q+4)]
###### 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

 PA(Kitep1,q1)(u+u−1)=PA(Kitep2,q2)(u+u−1)

i.e.

 u−n+p1(1+u−u−1)p1−21−u2[(2−p1).(1+u−1−u2n−2p1+2−u2n−2p1+3) +(u−2−u2n−2p1+4)] = u−n+p2(1+u−u−1)p2−21−u2[(2−p2).(1+u−1−u2n−2p2+2−u2n−2p2+3) +(u−2−u2n−2p2+4])

i.e.

 up1.(1+u−u−1)p1.[(2−p1).(1+u−1−u2n−2p1+2−u2n−2p1+3) +(u−2−u2n−2p1+4)] = up2.(1+u−u−1)p2.[(2−p2).(1+u−1−u2n−2p2+2−u2n−2p2+3) +(u−2−u2n−2p2+4)]

Let . It follows that . Then, we have

 up1−p2.(1+u−u−1)p1−p2{[(2−p1).(1+u−1−u2n−2p1+2−u2n−2p1+3) +(u−2−u2n−2p1+4)]−[(2−p2).(1+u−1−u2n−2p2+2−u2n−2p2+3) +(u−2−u2n−2p2+4)]}=0

By using the fact that and , we get

 f(u) = [(2−p1).(1+u−1−u2n−2p1+2−u2n−2p1+3)+(u−2−u2n−2p1+4)] −[(2−p2).(1+u−1−u2n−2p2+2−u2n−2p2+3)+(u−2−u2n−2p2+4)] = 0

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

 [(p1−2)(2n−2p2+4)!(u−2n+2p2−6)]−[(p2−2).(2n−2p2+4)!(u−2n+2p2−6)]=0

i.e.

 [(p1−2)−(p2−2)].(u−2n+2p2−6)=0

i.e.

 p1=p2

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 Kitep,2 Graphs

###### Lemma 4.1.

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

 w(G)≥p−2q+1

.

###### 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,

 (√2m(r−1r))2−(p−1+1p2+1p3)2 = (p2−p+2q)(r−1)−r(p−1+1p2+1p3)2 = (p2−p+2q)(r−1)− (r(p2+p3)p5)(2(p−1)+(p2+p3)p5) = (pr−p2+p+(2q−1)r−2q)− (r(p2+p3)p5)(2(p−1)+(p2+p3)p5)

By the help of Mathematica, for we can see

 (pr−p2+p+(2q−1)r−2q)−(r(p2+p3)p5)(2(p−1)+(p2+p3)p5)<0

i.e.

 (√2m(r−1r))2−(p−1+1p2+1p3)2<0

i.e.

 (√2m(r−1r))2<(p−1+1p2+1p3)2

Since and , we get

 √2m(r−1r)

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

 xi≤p−4 (1)

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

 xi∧j≤p−5 (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

 5∑i=1xi+d=(p2)+2−(p−32)=3p−4 (3)

Similarly, by using , we get

. Hence, we have

 5∑i=1(xi2)+∑i∼jxi∧j+α=(p3)−(p−33)=3p22−15p2+10 (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∑i=1xi=3p−4 (5)

Hence, we get

 5∑i=1(xi2)+∑i∼jxi∧j+α=5∑i=1(xi2)

Clearly,

 5∑i=1(xi2)≤max{5∑i=1(xi2)}

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

 5∑i=1(xi2) ≤ max{5∑i=1(xi2)} (6) ≤ 3(p−42)+(72) = 3p22−27p2+51

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

 3p22−27p2+51<3p22−15p2+10 (7)

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

Subcase 2

Let . Then and from (3) we get

 5∑i=1xi=3p−5 (8)

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

 5∑i=1(xi2)+∑i∼jxi∧j+α ≤ max{5∑i=1(xi2)}+p−5 (9) ≤ 3(p−42)+(72)+p−5 = 3p22−25p2+46

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

 3p22−25p2+46<3p22−15p2+10 (10)

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

Subcase 3

Let . Then and by (3), we get

 5∑i=1xi=3p−6 (11)

By using similar way with last subcase, we obtain

 5∑i=1(xi2)+∑i∼jxi∧j+α ≤ 3(p−42)+(62)+2(p−5) (12) = 3p22−23p2+35

and . If , we have

 3p22−23p2+35<3p22−15p2+10 (13)

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

Subcase 4

Let . Then and

 5∑i=1xi=3p−7

By using similar way again, we obtain

 5∑i=1(xi2)+∑i∼jxi∧j+α ≤ 3(p−42)+(52)+3(p−5)+1 (14) = 3p22−21p2+26

Since , we have

 3p22−21p2+26<3p22−15p2+10 (15)

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

Subcase 5

Let .Then and . Similarly, we obtain

 5∑i=1(xi2)+∑i∼jxi∧j+α ≤ 3(p−42)+(42)+4(p−5)+1 (16) = 3p22−19p2+17

Since , we have

 3p22−19p2+17<3p22−15p2+10 (17)

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

Subcase 6

Let .Then and . Similarly, we obtain

 5∑i=1(xi2)+∑i∼jxi∧j+α ≤ 3(p−42)+(32)+5(p−5)+2 (18) = 3p22−17p2+10

Since , we get

 3p22−17p2+10<3p22−15p2+10 (19)

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

Subcase 7

Let .Then and . Similarly, we obtain

 5∑i=1(xi2)+∑i∼jxi∧j+α ≤ 3(p−42)+(22)+6(p−5)+4 (20) = 3p22−15p2+5

Since , we have

 3p22−15p2+5<3p22−15p2+10 (21)

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

Subcase 8

Let .Then and . Also here,

 ∑i∼jxi∧j≤5∑i=1xi+2(p−5)=5p−21

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

 5∑i=1(xi2)+∑i∼jxi∧j+α ≤ 3(p−42)+5p−21+4 (22) = 3p22−17p2+13

Since , we get

 3p22−17p2+13<3p22−15p2+10 (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

 ∑i∼jxi∧j≤5∑i=1xi+3(p−5)=6p−27

Hence, we obtain

 5∑i=1(xi2)+∑i∼jxi∧j+α ≤ 3(p−42)+6p−27+5 (24) = 3p22−15p2+8

Since , we get

 3p22−15p2+8<