Download DETERMINANT OF ADJACENCY MATRIX OF SQUARE CYCLE

International Journal of Pure and Applied Mathematics
Volume 90 No. 4 2014, 413-421
ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version)
url: http://www.ijpam.eu
doi: http://dx.doi.org/10.12732/ijpam.v90i4.3
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DETERMINANT OF ADJACENCY MATRIX
OF SQUARE CYCLE GRAPH
Nitiphoom Adsawatithisakul1 , Decha Samana2 §
1,2 Department
of Mathematics
Faculty of Science
King Mongkut’s Institute of Technology Ladkrabang
Bangkok 10520, THAILAND
Abstract: Square Cycle, Cn2 is a graph that has n vertices and two vertices
u and v are adjacent if and only if distance between u and v not greater than
2. In this paper, we show that the determinant of adjacency matrix of square
cycle Cn2 are as follows

 0, n ≡ 0, 2, 4 mod 6,
2
det(A(Cn )) =
16, n ≡ 3 mod 6,

4, n ≡ 1, 5 mod 6.
AMS Subject Classification: 05C50
Key Words: determinant, square cycle graph, adjacency matrix
1. Introduction
Let G be a simple graph with n vertices. We denote det(A(G)) is the determinant of adjacency matrix of G and E(G; k) is kth eigenvalues of the adjacency
matrix which det(A(G)) and E(G; k) are independent of the choice of vertices
Received:
May 21, 2013
§ Correspondence
author
c 2014 Academic Publications, Ltd.
url: www.acadpubl.eu
414
N. Adsawatithisakul, D. Samana
Figure 1: d-th power of cycle graph
in adjacency matrix and are an invariant of G.
In [2] and [4], they determined the determinant of adjacency matrix of some
graphs, such as Kn , Cn , Pn and Wn . B. Gyurov and J. Cloud [7] has determined
determinant of Pin-wheel graph. Moreover, there are studies of graph which
satisfy some properties of determinant for example, M. Doob [5] construct circulant graph with det(A(G)) = −deg(G), S. Hu [9] and A. Abdollahi [1] have
found that the determinant of graphs with exactly one cycle and exactly two
cycles, respectively.
Cycle power, Cnd is a graph that has n vertices and distance each pair of
vertex is less or equal d. For example,
If d = 2, n ≥ 6, it is called square cycle graph.
Furthermore, there are studies of cycle power such as: C.N. Campos and
C.P. de Mello [3], M. Krivelevich and A. Nachmias [10] studied about the colouring in cycle power, Y. Hoa, C. Woo and P. Chen [8] investigate the sandpile
group in cycle power, D. Li and M. Liu [11] consider cycle power and their
complements which satisfy Hadwiger’s conjecture.
From figure 1 graph C62 and graph C82 , we write to adjacency matrix

0
1

1
2
A(C6 ) = 
0

1
1
1
0
1
1
0
1
1
1
0
1
1
0
0
1
1
0
1
1
1
0
1
1
0
1

1
1

0
,
1

1
0

0
1

1

0
2
A(C8 ) = 
0

0

1
1
1
0
1
1
0
0
0
1
1
1
0
1
1
0
0
0
0
1
1
0
1
1
0
0
0
0
1
1
0
1
1
0
0
0
0
1
1
0
1
1
1
0
0
0
1
1
0
1

1
1

0

0
.
0

1

1
0
We see adjacency matrix of C62 and C82 is a circulant matrix because a main
diagonal of matrix is equal to zero and entries in first row satisfy a1j = a1,(n−j+2)
DETERMINANT OF ADJACENCY MATRIX...
415
for j = 2, ..., n and aij = ai+1,j+1 , then a square cycle graph is a circulant graph.
It is interesting to study determinant of adjacency matrix of square cycle graph.
Proposition 1. (see [2]) Suppose that [0, a2 , ..., an ] is the first row of the
adjacency matrix of a circulant graph G. Then the eigenvalues of graph G is
denoted E(G; k),
E(G; k) =
n
X
aj z j−1
j=1
where z = e
2kπi
n
, k = 1, 2, ..., n.
Square cycle graph is a circulant graph then eigenvalues of square cycle
graph is
E(Cn2 ; k) = z + z 2 + z n−2 + z n−1 .
(1)
Determinant of a square matrix can be find by eigenvalue its as below
Theorem 2. (see [6]) Let λ1 , ..., λn be a eigenvalues of a square matrix A.
Then
det(A) = λ1 λ2 ...λn .
Next, we present lemma that will be used in the proof of determinant of
adjacency matrix of square cycle graph.
2. Main Results
Lemma 3. Let q be a positive number. Then
2q
Y
(cos
k=1
4q+1
Y
3kπ
3kπ
kπ
kπ
(cos
cos
)
cos
)
6q + 3
6q + 3
6q + 3
6q + 3
k=2q+2
6q+2
Y
(cos
k=4q+3
3kπ
kπ
cos
) = 2−12q . (2)
6q + 3
6q + 3
Proof. The left hand side of (2) is
2q
6kπ
2kπ
Y
sin 6q+3
sin 6q+3
k=1
kπ
3kπ
2 sin 6q+3
2 sin 6q+3
4q+1
Y
k=2q+2
6kπ
2kπ
sin 6q+3
sin 6q+3
kπ
3kπ
2 sin 6q+3
2 sin 6q+3
6q+2
Y
k=4q+3
6kπ
2kπ
sin 6q+3
sin 6q+3
kπ
3kπ
2 sin 6q+3
2 sin 6q+3
416
N. Adsawatithisakul, D. Samana

 Q
2q
6kπ
2kπ
sin
sin
1
k=q+1
6q+3
6q+3 
= 12q  Qq
(6k−3)π
(2k−1)π
2
k=1 sin 6q+3 sin 6q+3
!
Q4q+1
6kπ
2kπ
k=2q+2 sin 6q+3 sin 6q+3
Q4q+1
kπ
3kπ
k=2q+2 sin 6q+3 sin 6q+3
!
Q6q+2
6kπ
2kπ
k=4q+3 sin 6q+3 sin 6q+3
Q6q+2
3kπ
kπ
k=4q+3 sin 6q+3 sin 6q+3
Q

q
(6k−3)π
(2k−1)π
sin
sin
1
k=1
6q+3
6q+3 
= 12q  Qq
(6k−3)π
(2k−1)π
2
k=1 sin 6q+3 sin 6q+3
Q

(6k−(6q+3))π
(2k−(2q+1))π
3q+1
sin
sin
6q+3
6q+3
 Qk=2q+2

(6k−(6q+3))π
(2k−(2q+1))π
3q+1
sin
sin
k=2q+2
6q+3
6q+3

Q
(2k−(4q+3))π
5q+2
(6k−(12q+9))π
sin
sin
6q+3
6q+3

 Qk=4q+3
5q+2
(6k−(12q+9))π
(2k−(4q+3))π
sin
sin
k=4q+3
6q+3
6q+3
= 2−12q .
Lemma 4. Let q be a positive integer. Then
6q
Y
(cos
k=1
3kπ
kπ
cos
) = 2−12q .
6q + 1
6q + 1
Proof. It can be proved by
6q
6kπ
2kπ
Y
sin 6q+1
sin 6q+1
kπ
3kπ
cos
)=
(cos
3kπ
kπ
6q + 1
6q + 1
2 sin 6q+1
2 sin 6q+1
k=1
k=1
 Q

6q
6kπ
2kπ
1  k=3q+1 sin 6q+1 sin 6q+1 
= 12q Q3q
(6k−3)π
(2k−1)π
2
k=1 sin 6q+1 sin 6q+1

Q
(2k−1)π
3q
(6k−3)π
1
k=1 sin 6q+1 sin 6q+1 
= 12q  Q3q
(6k−3)π
(2k−1)π
2
k=1 sin 6q+1 sin 6q+1
6q
Y
= 2−12q .
Lemma 5. Let q be a positive integer. Then
DETERMINANT OF ADJACENCY MATRIX...
6q+4
Y
(cos
k=1
417
3kπ
kπ
cos
) = 2−2(6q+4) .
6q + 5
6q + 5
Proof. It can be proved by
6q+4
Y
k=1
(cos
6q+4
6kπ
2kπ
Y sin 6q+5
sin 6q+5
kπ
3kπ
cos
)=
3kπ
kπ
6q + 5
6q + 5
2 sin 6q+5
2 sin 6q+5
k=1
 Q

6q+4
6kπ
2kπ
sin
sin
1
k=3q+3
6q+5
6q+5 
= 2(6q+4)  Q3q+2
(6k−3)π
(2k−1)π
2
k=1 sin 6q+5 sin 6q+5

Q
(2k−1)π
(6k−3)π
3q+2
sin
sin
1
k=1
6q+5
6q+5 
= 2(6q+4)  Q3q+2
(2k−1)π
(6k−3)π
2
k=1 sin 6q+5 sin 6q+5
= 2−2(6q+4) .
Next, we use Lemma 3, 4 and 5 to find determinant of adjacency matrix of
square cycle graph.
Theorem 6. Let Cn2 be a square
positive integer. Then

 0,
2
det(A(Cn )) =
16,

4,
cycle graph with n vertices and n be a
n ≡ 0, 2, 4 mod 6,
n ≡ 3 mod 6,
.
n ≡ 1, 5 mod 6.
Proof. Let E(Cn2 ; k) be a kth eigenvalue of adjacency matrix of square cycle
graph Cn2 . From (1), We get
E(Cn2 ; k) = e
=e
2kπi
n
2kπi
n
+e
+e
4kπi
n
4kπi
n
+e
+e
2k(n−2)πi
n
2knπi
n
·e
+e
2k(n−1)πi
n
−4kπi
n
+e
2knπi
n
·e
−2kπi
n
.
By Euler’s formula, we obtain
2kπ
2kπ
4kπ
4kπ
+ i sin
) + (cos
+ i sin
)+
n
n
n
n
−4kπ
−2kπ
−2kπ
−4kπ
(cos
+ i sin
) + (cos
+ i sin
)
n
n
n
n
4kπ
2kπ
+ 2 cos
.
= 2 cos
n
n
E(Cn2 ; k) = (cos
418
N. Adsawatithisakul, D. Samana
We can rewrite
E(Cn2 ; k) = 4(cos
3kπ
kπ
cos
).
n
n
(3)
From (3) We have
det(A(Cn2 ))
=
=
n
Y
k=1
n
Y
E(Cn2 ; k)
4(cos
k=1
kπ
3kπ
cos
).
n
n
(4)
Consider n as follows
Case I. n ≡ 0, 2, 4 mod 6.
Since n is even and 1 ≤ k ≤ n, consider (3) when k = n2 . Then
n
π
3nπ
n
E(Cn2 ; ) = 4(cos 2 cos 2 )
2
n
n
= 0.
From (4), we obtain
det(A(Cn2 )) =
n
Y
E(Cn2 ; k)
k=1
= 0.
Therefore, det(A(Cn2 )) = 0 when n ≡ 0, 2, 4 mod 6.
Case II. n ≡ 3 mod 6 Then n = 6q + 3, ∃q ∈ Z+ .
From (4), we obtain
det(A(Cn2 )) =
=
n
Y
E(Cn2 ; k)
k=1
6q+3
Y
4(cos
k=1
kπ
3kπ
cos
)
n
n
3(2q + 1)π
(2q + 1)π
3(4q + 2)π
(4q + 2)π
cos
)4(cos
cos
)
6q + 3
6q + 3
6q + 3
6q + 3
2q
3kπ
(6q + 3)π 12q Y
kπ
3(6q + 3)π
4(cos
cos
)2
cos
)
4(cos
6q + 3
6q + 3
6q + 3
6q + 3
= 4(cos
k=1
DETERMINANT OF ADJACENCY MATRIX...
4q+1
Y
4(cos
6q+2
Y
3kπ
3kπ
kπ
kπ
4(cos
cos
)
cos
)
6q + 3
6q + 3
6q + 3
6q + 3
k=4q+3
k=2q+2
= (−2)(−2)(4)212q
2q
Y
4(cos
k=1
4q+1
Y
3kπ
kπ
cos
)
6q + 3
6q + 3
4(cos
kπ
3kπ
cos
)
6q + 3
6q + 3
4(cos
3kπ
kπ
cos
).
6q + 3
6q + 3
k=2q+2
6q+2
Y
k=4q+3
Using Lemma 3, we have
det(A(Cn2 )) = (−2)(−2)(4)(212q )(2−12q )
= 16.
Therefore det(A(Cn2 )) = 16 when n ≡ 3 mod 6.
Case III. n ≡ 1 mod 6 and n ≡ 5 mod 6. We consider 2 subcases.
Subcase 3.1, n ≡ 1 mod 6, by (4), we obtain
det(A(Cn2 )) =
=
n
Y
E(Cn2 ; k)
k=1
6q+1
Y
22 (cos
k=1
= 4(cos
3kπ
kπ
cos
)
6q + 1
6q + 1
6q
3kπ
3(6q + 1)π
(6q + 1)π 12q Y
kπ
4(cos
cos
)2 (
cos
).
6q + 1
6q + 1
6q + 1
6q + 1
k=1
Using Lemma 4, we have
det(A(Cn2 )) = 4(212q )(2−12q )
= 4.
Subcase 3.2, n ≡ 5 mod 6, by (4), we obtain
det(A(Cn2 )) =
=
n
Y
E(Cn2 ; k)
k=1
6q+5
Y
k=1
22 (cos
419
3kπ
kπ
cos
)
6q + 5
6q + 5
420
N. Adsawatithisakul, D. Samana
6q+4
3kπ
(6q + 5)π 2(6q+4) Y
kπ
3(6q + 5)π
4(cos
cos
)2
(
cos
)).
= 4(cos
6q + 5
6q + 5
6q + 5
6q + 5
k=1
Using Lemma 5, we have
det(A(Cn2 )) = 4(22(6q+4) )(2−2(6q+4) )
= 4.
From subcase 3.1 and 3.2, we obtain
det(A(Cn2 )) = 4 for n ≡ 1, 5 mod 6.
From case I, II and III,

 0, n ≡ 0, 2, 4 mod 6,
16, n ≡ 3 mod 6,
det(A(Cn2 )) =

4, n ≡ 1, 5 mod 6.
References
[1] A. Abdollahi, Determinant of adjacency matrices
http://arxiv.org/abs/0908.3324, Cornell University (2009).
of
graph,
[2] N. Biggs, Algebraic Graph Theory, Cambridge University Press, Cambridge
(1974).
[3] C.N. Campos and C.P. De Mello, A result on the total colouring of powers
of cycles, Electronic Notes in Discrete Mathematics, 18 (2004), 47-52.
[4] D.M. Cvetkovic,M. Doob and H. Sachs, Spectra of Graphs: Theory and
Application, Academic Press, New York, USA (1980).
[5] M. Doob, Circulant graphs with det(−A(G)) = −deg(G) :codeterminants
with Kn , Linear Algebra and its Applications, 340 (2002), 87-96.
[6] L. Goldberg, Matrix Theory with Applications. McGraw - Hill International
Editions, Mathematics and Statistics Series (1991).
[7] B. Gyurov and J. Cloud, On the algebraic properties of Pin-Wheel graphs
and Applications, 73rd annual meeting of the Oklahoma Arkansas section
2011, University of central Oklahoma, USA (2011).
[8] Y. Hoa, C. Woo and P.Chen, On the sandpile group of the square cycle
Cn2 , Linear Algebra and its Applications, 418 (2006), 457-467.
DETERMINANT OF ADJACENCY MATRIX...
421
[9] S. Hu, The Classification and maximum determinants of the adjacency
matrices of graphs with one cycle, J. Math. Study, 36 (2003), 102-104.
[10] M. Krivelevich and A. Nachmias, Colouring powers of cycles from random
lists, European Journal of Combinatorics, 25 (2004), 961-968.
[11] D.Li and M.Liu, Hadwiger’s conjecture for powers of cycles and their complements,European Journal of Combinatorics, 28 (2007), 1152-1155.
422