 Open Access
 Total Downloads : 15
 Authors : K. Ananthi , Dr. J. Ravi Sankar , Dr. N. Selvi
 Paper ID : IJERTCONV3IS22025
 Volume & Issue : NCEASE – 2015 (Volume 3 – Issue 22)
 Published (First Online): 24042018
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
On Domination and Energy of Zero Divisor Graphs
K. Ananthi Department of Mathematics VSA Group of Institutions
Salem, India 636 010.
Dr. J. Ravi Sankar School of Advanced Sciences
VIT University Vellore, India 632 014.
Dr. N. Selvi Department of Mathematics
ADM College for Women, Nagapattinam, India611001
Abstract The energy E (G) of a graph G is the sum of the absolute values of the eigenvalues of G. In this paper, we study the characterization of eigenvalues and energy of adjacency matrix corresponding to zerodivisor graphs of finite commutative ring. Also, we study the eigenvalue and energy of
Zn where n=2p, 3p, 5p, pq, p2, 4p where p, q are distinct
prime numbers. Finally, we give the relationship between the domination number, energy, rank and eigenvalues of complete zerodivisor graphs.
Keywords Commutative ring, Zerodivisor graph, Energy graph.
I. INTRODUCTION
The energy E(G), of a graph G is defined to be the sum of the absolute values of its eigenvalues. Hence, if A(G) is the adjacency matrix of G, and { 1 ,2 ,3…n } are the
The concept of energy has been generalized in two different directions. Let A be an mÃ—n matrix and A* denote its adjoint(conjugate transpose of A).The singular values s1(A) s2(A) sm(A) of a matrix A are the square roots of the eigenvalues of AA*.Note that if A is an nÃ—nHermitian matrix(i.e , A = A*) , then the singular values of A are the absolute values of the eigenvalues of A.For any
m
m
AMm,ndefine the energy of A , E(A) = si ( A) .From
i1
the above, we note that the usual energy of a graphG , E(G) = E(A(G)).
Another generalization of energy is defined as follows:Let Mbe a matrix associated with G. Suppose Âµ1,
Âµ2,Âµn are the eigenvalues of M and is the average of Âµ1,
Âµ2,Âµn.The M energy of G is then defined as the absolute
n
n
n
eigenvalues of A(G), then E(G)= i
i1
. The set
deviation EM(G)= i
i1

.If M is the adjacency matrix
{ , ,…,
} is the spectrum of G and denoted by Spec G.
A(G), then =0.Hence, the usual energy E(G) = EA(G).For
1 2 n
The totally disconnected graph
K c has zero energy while the
notation and zerodivisor graph terminology, we in general follow [4,5,6,7].
n
n
complete graph Kn with the maximum possible number of edges (among graphs on n vertices) has energy 2(n1).Graphs for which the energy is greater than 2(n1) are called hyperenergetic graphs. If 2(n1), then G is called nonhyperenergetic.
The energy of G was first defined by I. Gutman in 1978 as the sum of the absolute values of the eigenvalues of A(G) in [2]. Energy of graph originated from theoretical chemistry. Huckel molecular orbital theory is a field of theoretical chemistry where graph eigen values occur. The carbon atoms of a hydrocarbon system correspond to vertices of a graph associated with the molecule. From Huckel theory, the energy of a molecular graph is equal to the – electron energy of a conjugated hydrocarbon [3].
Let R be a commutative ring and let Z(R) be its set of zerodivisors. The zerodivisor graph of a ring is the graph (simple) whose vertex set is the set of nonzero zerodivisors, and an edge is drawn between two distinct vertices if their product is zero. Throughout this paperff, we consider the
commutative ring by R and zerodivisor graph Rby
Zn . The idea of a zerodivisor graph of a commutative
ring was introduced by I.Beck in [1], where he was mainly interested in colourings. The zerodivisor graph is very useful to find the algebraic structures and properties of rings.
Claude Berge introduced the theory of Domination in 1958. The inspiration for this concept was drawn from the classical problem of covering chessboard with minimum number of chess pieces.
The most common definition given of a dominating set is that it is a set of vertices D V in a graph G = (V, E )
Theorem 2: For any graph Z , where p is any prime
3 p
3 p
3 p
3 p
having the property that every vertex v VD is adjacent to
number, then the Eigenvalues are 2p 1,
2p 1
atleast one vertex in D . The domination number (G) is the cardinality of a smallest dominating set of G.
and EZ 2 2( p 1) .
Throughout this paper G will denote a simple (no loops or multiple edges), undirected graph with n vertices and m edges. If {v1, v2, vn} is the set of vertices of G , then the adjacency matrix of G, A(G) = A = [aij] is an n Ã— n matrix, where aij= 1 if vi and vj are adjacent and aij= 0 otherwise. Thus, A is a symmetric (0, 1) matrix with real eigenvalues and zeros on the diagonal. If 1 , 2 ,…n are the eigenvalues of A, then we
have 1 2 … n and 1 2 …n 0 .
Proof: The vertex set of Z is {3,6,9,3(p1),p,2p}. Let u and v be two vertices in Z with maximum degree.
3 p
3 p
3 p
3 p
3 p
3 p
Let u=p and v=2p then there exist any other vertex wp2p in Z such that,is adjacent to both u and v. That is, uw
=vw= 0. But uv=2p2 which is not divided by 3p. Therefore u and v are nonadjacent vertices. Then the vertex set V can be partitioned into two parts V1 and V2 such that V1={u,v}={p,2p} and V2={3,6,9,,3(p1)}. Clearly V1=2 and V2=p1, then
II EIGENVALUES & ENERGY OF ZERO DIVISOR GRAPHS
V=V1+V2=p+1.
Case(i): Let p=5.The eigenvalues of
(Z15 ) are
In this paper, we evaluate the eigenvalues of zerodivisor
graph and find the energy of some generalized zero divisor 8 , 8 .Then the energy of
graphs.
(Z15 )= Sum of the
2 p
2 p
Theorem1: For any graph Z , where p is any prime
absolute values of the eigenvalues= 8
8 2 8 .
number, then the eigenvalues are
p 1,
p 1 and
Case (ii): Let p=7.The Eigenvalues of
(Z21) are
2 p
2 p
EZ
2
12, 
12 
. Then the energy of 
(Z21) = Sum 
of 
the 

absolute 
values 
of 
the 
12, 
12 
. Then the energy of 
(Z21) = Sum 
of 
the 

absolute 
values 
of 
the 
p 1 .
Proof: The vertex set of Z 2 p is {2, 4, 6 2(p1), p}. Let
Eigenvalues= 12 12 12 .
u=4 and v=p then 2p must divides uv. That is 2p divides 4p.
Clearly, u and v are adjacent vertices. Similarly, any vertices
Case(iii):Let p>7, is a prime number.
2 p
2 p
u in V Z and v=p then 2p must divides uv. It seems
In general, the eigenvalues of
Z3 p are
2 p
2 p
that p is adjacent to all the vertices in V Z
.Let u=4p
2( p 1), 2( p 1). Then the energy of
Z =
3 p
3 p
and
Sum of the absolute values of the
2 p
2 p
2 p
2 p
5 p
5 p
w=6p in V Z such that uw 0 It means that 2p does
eigenvalues= 2( p 1) 2( p 1) 2 2( p 1).
not divide uv=24. Clearly, no two vertices in
Z
are
Theorem 3: For any graph Z ,where p is any prime
adjacent, except p.
Case(i): Let p=3.The eigenvalues of
Z6
number, then the eigenvalues are
are 2 , 2 .
4p 1,
4p 1
Then the energy of Z6
=Sum of the absolute values of
and EZ5 p 2 4( p 1) .
the eigenvalues= 2
2 2 2.
Proof: The vertex set of
(Z5 p )
is {5,10, . . .5(p1), p,
Case(ii): Let p=5.The eigenvalues of
Z
are 4 , 4 .
2p,3p,4p}. Clearly V ((Z5 p )) p 3 .Let u and v be any
Then the energy of Z10
10
=Sum of the absolute values of
two vertices in (Z5 p ) with maximum and minimum
the eigenvalues= 4
4 4.
degree, respectively. Let u = p and v=10 then 5p must divide uv which implies that u and v are adjacent.
Case (iii): Let p>5, is a prime number.
In general, the eigenvalues of
2 p
2 p
p 1, p 1 . Then the energy of Z
Z are
2 p
2 p
=Sum of the
Let u=p and v=2p then 5p does not divide uv=2p2, which implies that u and v are non adjacent vertices. Then the vertex set V can be partitioned into two parts V1 and V2, where V1={p,2p,3p,4p} and V2= {5,10,,5(p1)}.Clearly
absolute values of the
any vertices u and v in V1are nonadjacent as same as in V2.
eigenvalues=
p 1
p 1 2
p 1 .
Let u=p in V1 and v=10 are in V2 then 5p divides uv=10p. Finally, we note that, every vertex in V1 is adjacent to all the
vertices
V Z5 p
in V2.Moreover
V1 V2 and V1 V2 .
Case(i):Let p=7.The eigenvalues of (Z35 ) are
( p 1)(q 1), ( p 1)(q 1) and
24, 24 .Then the energy of (Z35 ) = Sum of the
EZ pq 2 ( p 1)(q 1) .
absolute values of the eigenvalues= 24 24 2 24 .
Proof: The proof is by the method of induction on p and q. The vertex set of Z is {p,2p,3p,,p(q
pq
pq
Case(ii):Let p=11.The eigenvalues of (Z55 ) are
1),q,2q,3q,,(p1)q}.
2q
2q
Case(i): Let p=2,q is any prime>2,Using theorem (2.1), for
40 , 40 .Then the energy of (Z55 ) = Sum of the
any graph Z , where q>2 is any prime number, then the
absolute values of the
eigenvalues= 40 40 2 40 .
eigenvalues are q 1 , q 1 .Without loss of generality, the eigenvalues can be written as
Case (iii): Let p>11 , is a prime number.In general, the
( p 1)(q 1) ,
( p 1)(q 1) , where p=2.
eigenvalues of
Z
are
5 p
5 p
4( p 1),
4( p 1) . Then
Then, EZ 2q 2
(q 1) 2
(2 1)(q 1) 2
( p 1)(q 1),
the energy of
Z5 p =Sum of the absolute values of the
where p=2.
Case(ii): Let p=3,q is any prime>3,Using theorem (2.2), for
p
p
Eigenvalues= 4( p 1) 4( p 1) 2
4( p 1) .
any graph Z , where q>3 is any prime number, then the
Theorem 4: For any graph Z
2 , where p>2 is any prime
3q
eigenvalues are 2(q 1), 2(q 1) . Without loss of
number, then the eigenvalues are
1,1,… 1, ( p 2) and
generality, the eigenvalues can be written as
( p2)times
( p 1)(q 1) , ( p 1)(q 1) ,where p=3.
E Z 2 2( p 2) .
Then,
p
E Z3q
2 2(q 1) 2
(3 1)(q 1) 2
( p 1)(q 1),
Proof: If p is any prime, then
p
p
V Z 2 ={p,2p,3p,4p,,(p1)p}. Clearly p is adjacent to
where p=3.
5q
5q
p
p
all the vertices in Z 2
. Also note that, any two vertices
p
p
Case(iii):Let p =5,q is any prime>5,Using theorem (2.3), for any graph Z , where q>5 is any prime number, then the
p
p
in Z 2
is adjacent and hence
Z 2
is a complete
eigenvalues are
4(q 1),
4(q 1) . Without loss of
graph, namely K p1 .
generality, the eigenvalues can be written as
3
3
Case(i): Let p=3.The eigenvalues of Z 2 are 1,1. Then
( p 1)(q 1), ( p 1)(q 1) ,where p=4. Then,
3
3
the energy of Z 2 =Sum of the absolute values of the
eigenvalues=2.
EZ5q 2 4(q 1) 2 (5 1)(q 1) 2 ( p 1)(q 1),
5
5
Case(ii): Let p=5.The eigenvalues of
Z 2 are 1,1,1,3.
where p=5.
Case(iv): Let p<q,The vertex set of
Z , is
pq
pq
5
5
Then the energy of the eigenvalues=6.
Z 2 =Sum of the absolute values of
pq
pq
pq
pq
{p,2p,3p,,p(q1),q,2q,3q,,(p1)q}.Using theabove cases, the eigenvalues of
7
7
Case(iii): Let p=7.The eigenvalues of Z 2 are 1,1,1,1
1,5. Then the energy of Z 2 =Sum of the absolute values
Z
are
( p 1)(q 1),
( p 1)(q 1)
where p and q
7 are distinct primes with p<q. Then the energy of
Z pq is
of the eigenvalues=10.
Case(iv): Let p>7, is a prime number.The eigenvalues of
EZ pq
( p
1)(q
1)
( p
1)(q
1) 2
( p
1)(q
1)
p
p
Z 2
are 1,1,…,1, ( p 2) .
( p2) times
.
Theorem6:If p > 4 is any prime, thenthe eigenvalues are
Then the energy of
Z
2 = Sum of the absolute values of
in the form of x, 2.4142x,where x and 2.4142x are less
p
p
the
than p and E((z4 p )) 2(x 2.4142x) .
eigenvalues=
eigenvalues=
1,… 1 ( p 2)times
1,… 1 ( p 2)times
p 2 2 p 2 .
p 2 2 p 2 .
pq
pq
Theorem5: In Z
, if p and q are distinct prime numbers
Proof: The vertex of
(Z4 p )
is {2,4, , 2(2p1),p,2p,3p}
with p<q, then the eigenvalues are
with
V ((Z4P
2 p 1.Let v = 2p be a vertex and let w
be any vertex except p and 3p such that 4p divides vw.
G 
((Zn ) 
E((Zn ) 
((Zn ) 
Eigenvalues 
(Z6 ) 
1 
2 
2 
1,1 
(Z25 ) 
1 
6 
4 
1,1,1,3 
(Z49 ) 
1 
10 
6 
1,1,1,1, 1,5 
(Z121) 
1 
18 
10 
1,1,1,1, 11,1,1, 1,9, 
.. 

(Z 2 ) P 
1 
2p 4 
p1 
1,1,… 1,(P 2) ( p2) times 
G 
((Zn ) 
E((Zn ) 
((Zn ) 
Eigenvalues 
(Z6 ) 
1 
2 
2 
1,1 
(Z25 ) 
1 
6 
4 
1,1,1,3 
(Z49 ) 
1 
10 
6 
1,1,1,1, 1,5 
(Z121) 
1 
18 
10 
1,1,1,1, 11,1,1, 1,9, 
.. 

(Z 2 ) P 
1 
2p 4 
p1 
1,1,… 1,(P 2) ( p2) times 
Clearly, v is adjacent to all the vertices in
V ((Z4P )) expect p and 3p. Let P, S,N(S) be the pendant
set, minimum degree set, neighborhood of S, respectively. Since v has maximum degree then vN(S).
Case(i): Let p = 5.The vertex set of
(Z
20 ) is {2,4,2(10
1),5,10,15} with
V ((Z
20 11. Let v =2p = 10 be a
vertex and letw be any vertex except 5 and 15 such that 20 divides vw. Clearly, v = 10 is adjacent to all the vertices in V ((Z20 )) except 5 and 15 then 10N(S).Let x = 2 and y
= 14 then 28 is not divisible by 20 which implies x and y are non adjacent vertices. But xv = yv = 0.
The eigenvalues of (Z20 ) are 3.6955, 1.5307, 0, 1.5307,
3.6955. Then the energy of
(Z20 ) = Sum of the absolute
values of the eigenvalues = 2(1.5307+3.6955) = 2(1.5307+2.4142(1.5307)) = 10.4524.
Case(ii): Let p = 7.The vertex set of (Z28 ) is {2,4,2(14 1), 7,14,21} with V ( (Z28 2 p 1 15.Let v = 2p
=14 be a vertex and let w be any vertex except 7 and 21 such
that 28 divides vw. Clearly v =14 is adjacent to all the vertices in V ((Z28 )) except 7 and 21 then 14 N(S). The eigenvalues of V ((Z28 )) are 4.5261, 1.8748, 0, 1.8748, 4.5261. Then the energy of V ((Z28 )) = Sum of the absolute values of the eigenvalues = 2(1.8748+4.5261) = 2(1.8748+2.4142(1.8748)) = 12.8018
Case(iii): Let p>7is a prime numberThe vertex set of
V ((Z4 p )) is{2,4,2(2p1),p,2p,3p}with
III CONCLUSION
In this paper, we study the eigenvalues and energy of zero divisor graph over finite commutative rings. Graphs are the most ubiquitous models of both natural and human made structures. In computer science, zero divisor graphs are used to represent networks of communication, network flow, clique problems. For any graph (Z4P ) , where p > 3 is any prime number and 4 <x< 4+x, then the eigenvalues are x , 2.4142x, approximately, where x is a position of consecutive prime number from 5 to so on.
If an Eigen value of a zero divisor graph is a rational number, then it is an integer. Also we note that the energy of a zero divisor graph cannot be an odd integer. The energy of a zero divisor graph cannot be the square root of an oddinteger.The energy of a zero divisor graph cannot be the square root of the double of an odd integer.
V ( (Z4 p )) 2 p 1 .Let v =2p be a vertex and let w be
any other vertex except p and 3p such that 4p divides vw. Clearly , v is adjacent to all the vertices V ((Z4P )) except p and 3p and v = 2p N(S).
REFERENCES

I.Beck, Colouring of Commutative Rings, J. Algebra, 116,(1988),208 226
The eigenvalues of
V ((Z4P ))
are x, 2.4142x, 0,

I.Gutman,The Energy of a Graph,Ber.Math Statist.Sekt.Forschungsz.Graz,103,(1978), 122.
2.4142x, x. Then the energy of V ((Z28 )) = Sum of the absolute values of the eigenvalues =2(x+2.4142x)=6.8284x.
III Relationship between domination, energy, rank and eigenvalues of (Zn ) In this section, we find the bounds which relate the domination number of (Zn ) , energy of
(Zn ), rank of (Zn ) and eigenvalues of (Zn ) ,for n = p2, where p >2 is any prime number.

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