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).
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
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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...
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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
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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
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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
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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.
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[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.
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