International Journal of Algebra, Vol. 1, 2007, no. 11, 541 - 546

Inverse Matrix of Symmetric Circulant Matrix on Skew Field

Jun-qing Wang and Chang-zhou Dong1

College of Mathematics Science Tianjin Polytechnic University Tianjin 300160, P. R. China

Abstract

Introduced the notion of symmetric circulant matrix on skew field, an easy method is given to determine the inverse of symmetric circu- lant matrix on skew field , with this method , we derived the formula of determine inverse about several special type of symmetric circulant matrix.

Mathematics Subject Classification: 05A19, 15A15

Keywords: skew field; symmetric circulant matrix; inverse matrix

1 Introduction

Circulant matrix have application in code theory, applied mathematics, engineering mathematics, physics, mathematical statistics and so on, espe- cially, circulant matrix take up an important status in error correcting code theory, reference [1] gave a deep discuss on circulant matrix theory, in this paper, we generalized this notion to skew field, then, given the notion and the property of solution of symmetric circulant matrix on skew field, and then, given the formula of determine inverse about several special type of symmetric circulant matrix.

[email protected] 542 Jun-qing Wang and Chang-zhou Dong

2 Preparation Knowledge

Let K is a skew field, Mn(k) represent matrix of order n on K, set In represent identity matrix of order n, suppose A ∈ Mn(k), matrix as ⎡ ⎤ a0 a1 a2 ... an−2 an−1 ⎢ ⎥ ⎢ ⎥ ⎢ a1 a2 a3 ... an−1 a0 ⎥ A ⎢ ⎥ , = ⎢ ...... ⎥ ⎣ ...... ⎦ an−1 a0 a1 ... an−3 an−2 we call it symmetric circulant matrix on K, notation is  A = Sc a0 a1 a2 ... an−2 an−1 .

Especially, we call  D = Sc 010... 00 as basic symmetric circulant matrix. It is easy to know that the property of solution of symmetric circulant matrix are as follow: 1. D2k−1 = D; D2k = I. 2. The product of two symmetric circulant matrix is a circulant matrix. 3. Suppose symmetric circulant matrix A is invertible, then A−1,kA(k ∈ K) is symmetric circulant matrix also. Lemma Let  A = Sc a0 a1 a2 ... an−2 an−1 , suppose A is invertible , then ,  −1 A = Sc b0 b1 b2 ... bn−2 bn−1 .

Where b0,b1,...,bn−1 is the unique solution of right system of linear equa- tions ⎡ ⎤ ⎡ ⎤ x0 1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ x1 ⎥ ⎢ 0 ⎥ A ⎢ ⎥ ⎢ ⎥ ⎢ . ⎥ = ⎢ . ⎥ (1) ⎣ . ⎦ ⎣ . ⎦ xn−1 0 Proof. Inverse matrix of symmetric circulant matrix 543

By AA−1 = I, We can get:

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ a a ... a b b ... b 10... 0 0 1 n−1 0 1 n−1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ a a ... a ⎥ ⎢ b b ... b ⎥ ⎢ 01... 0 ⎥ ⎢ 1 2 0 ⎥ ⎢ 1 2 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ . . . . ⎥ ⎣ ...... ⎦ ⎣ ...... ⎦ ⎣ . . .. . ⎦ an−1 a0 ... an−2 bn−1 b0 ... bn−2 00... 1

So ⎡ ⎤ ⎡ ⎤ x0 1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ x1 ⎥ ⎢ 0 ⎥ A ⎢ ⎥ ⎢ ⎥ ⎢ . ⎥ = ⎢ . ⎥ ⎣ . ⎦ ⎣ . ⎦ xn−1 0 According to A is invertible. We can get: system of linear equations (1) have unique solutionb0,b1,...,bn−1.

3 Inverse matrix of two special type symmet- ric circulant matrix.

Theorem 1 Suppose  A = Sc a, d, . . . , d , and a, d is the unite in K. Then,  −1 A = Sc b0,b1, ..., bn−1 .

Where −1 −1 −1 b0 =[−ad a + n(d − a)+a] [−ad − (n − 1)] −1 b1 = b2 = ...= bn−1 =(d − a) + b0 =(d − a)−1 +[−ad−1a + n(d − a)+a]−1[−ad−1 − (n − 1)] Proof.

Let  −1 B = A = Sc b0,b1, ..., bn−1 , 544 Jun-qing Wang and Chang-zhou Dong the right system of linear equations, ⎡ ⎤ ⎡ ⎤ x0 1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ x1 ⎥ ⎢ 0 ⎥ A ⎢ ⎥ ⎢ ⎥ ⎢ . ⎥ = ⎢ . ⎥ ⎣ . ⎦ ⎣ . ⎦ xn−1 0

A is its coefficient. Do elementary line transformation on the augmented matrix A¯ of A, ⎡ ⎤ ⎡ ⎤ a d d ... d d a d d ... d d ⎢ 1 ⎥ ⎢ 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ddd...da0 ⎥ ⎢ d − a 00... 0 a − d −1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ A¯ = ⎢ ddd...ad0 ⎥ → ⎢ 000... a− dd− a 0 ⎥ . ⎢ ⎥ ⎢ ⎥ ⎢ ...... ⎥ ⎢ ...... ⎥ ⎣ ...... ⎦ ⎣ ...... ⎦ dad...dd0 0 a − d 0 ... 0 d − a 0 ⎧ ⎪ ⎨⎪ ab0 + d(b1 + b2 + ...+ bn−1)=1, d − a b a − d b − , i , ,...,n− So, ⎪ ( ) 0 +( ) n−1 = 1 ( =1 2 2) ⎩⎪ (a − d)bi +(d − a)bn−1 =0, Solve it we can get:

−1 −1 −1 b0 =[−ad a + n(d − a)+a] [−ad − (n − 1)]

−1 b1 = b2 = ...= bn−1 =(d − a) + b0 =(d − a)−1 +[−ad−1a + n(d − a)+a]−1[−ad−1 − (n − 1)]

So  −1 A = Sc b0,b1, ..., bn−1 , Theorem 2 Let  A = Sc a, a + d, a +2d, ..., a+(n − 1)d , if A is invertible ,then  −1 A = Sc b0,b1, ..., bn−1 .

Where −1 −1 −1 b0 =[−ad a + n(d − a)+a] [−ad − (n − 1)], −1 n−1 −1 b1 = b2 = ...= bn−2 = n (a + id) , i=0 Inverse matrix of symmetric circulant matrix 545

−1 bn−1 =(d − a) + b0.

Let ⎡ ⎤ ⎡ ⎤ x0 1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ x1 ⎥ ⎢ 0 ⎥ A ⎢ ⎥ ⎢ ⎥ ⎢ . ⎥ = ⎢ . ⎥ ⎣ . ⎦ ⎣ . ⎦ xn−1 0

A is its coefficient. Do elementary line transformation on the augmented matrix A¯ of A,

⎡ ⎤ aa+ da+2d ... a+(n − 1)d 1 ⎢ a + da+2da+3d ... a 0 ⎥ ¯ ⎢ ⎥ A = ⎢ ...... ⎥ ⎣ ...... ⎦ a +(n − 1)da a+ d ... a+(n − 2)d 0 ⎡ ⎤ n−1 −1 ⎢ ... a id ⎥ ⎢ 111 11( + ) ⎥ ⎢ i=0 ⎥ ⎢ a dada d ... a n − da ⎥ → ⎢ + +2 +3 +( 1) 0 ⎥ ⎢ ...... ⎥ ⎣ ...... ⎦ a +(n − 1)da a+ d ... a+(n − 3)da+(n − 2)d 0 ⎡ ⎤ n−1 −1 ⎢ 111... 11 (a + id) ⎥ ⎢ ⎥ ⎢ i=0 ⎥ ⎢ n−1 −1 ⎥ ⎢ −1 ⎥ ⎢ 012... n− 2 −1 −(1 + d a) (a + id) ⎥ ⎢ i=0 ⎥ ⎢ −1 ⎥ → ⎢ n−1 ⎥ ⎢ 000... −n 0 − (a + id) ⎥ ⎢ ⎥ ⎢ i=0 ⎥ ⎢ ...... ⎥ ⎢ ...... ⎥ ⎢ . . . . . . ⎥ ⎣ n−1 −1 ⎦ 0 −n 0 ... 00 − (a + id) i=0

⎧So, the solution of equation can be decided by those equations as follows: −1 ⎪ n−1 ⎪ b b ... b a id , ⎪ 0 + 1 + + n−1 = ( + ) ⎪ i=0 ⎨ n−1 −1 −1 b1 +2b2 + ...+(n − 2)bn−2 − bn−1 = −(1 + d a) (a + id) , ⎪ ⎪ i=0 ⎪ n−1 −1 ⎪ ⎩ −nbj = (a + id) , i=0 (j =1, 2,...,n− 2) Solve it we can get:

−1 −1 −1 b0 =[−ad a + n(d − a)+a] [−ad − (n − 1)], 546 Jun-qing Wang and Chang-zhou Dong

−1 n−1 −1 b1 = b2 = ...= bn−2 = n (a + id) , i=0 −1 bn−1 =(d − a) + b0.

So,  −1 A = Sc b0,b1, ..., bn−1 .

References

[1] Brualdi R.A., Introductory Combinatorics, North-Holland, New York. (1977)

[2] Ke Zh., Wei W. D., Combinatorial Theory, Science Publisher, Beijing, (1981).

[3] Tongji University, Advanced Mathematics, Advanced Education Publisher, Shanghai (1993).

[4] Lu K. C., Lu K. M., Combinatorial Mathematics, Qinghua University Pub- lisher, Beijing (1991).

Received: August 27, 2007