WSEAS TRANSACTIONS on Zhaolin Jiang, Jing Wang

Fast Algorithms for Solving RFPrLR Circulant Linear Systems

ZHAOLIN JIANG JING WANG Department of Mathematics Department of Mathematics Linyi University Linyi University Shuangling Road, Linyi city Shuangling Road, Linyi city CHINA CHINA [email protected] [email protected]

Abstract: In this paper, fast algorithms for solving RFPrLR circulant linear systems are presented by the fast algorithm for computing . The unique solution is obtained when the RFPrLR circulant over the complex field C is nonsingular, and the special solution and general solution are obtained when the RFPrLR over the complex field C is singular. The extended algorithms is used to solve the RLPrFL circulant linear systems. Examples show the effectiveness of the algorithms.

Key–Words: RFPrLR circulant matrix, Linear system, Fast algorithm.

1 Introduction meant a square matrix over the complex field C of the form Circulant matrix family have important applications in various disciplines including image processing [1], A = RFPrLRcircfr(a , a , . . . , a − ) communications [2], signal processing [3], encoding 0 1 n 1 [4], and preconditioner [5]. They have been put on   firm basis with the work of P. Davis [6] and Z. L. Jiang a0 a1 . . . an−1   [7].  an−1 a0 + ran−1 . . . an−2   . . . .  The circulant matrices, long a fruitful subject of =  . . .. .  . (1) research [6, 7, 8, 9, 10, 11, 12], have in recent years    a a + ra . . . a  been extended in many directions [24, 25, 26, 27, 28, 2 3 2 1 a a + ra . . . a + ra − 29, 30, 31, 32, 33, 34, 35]. The f(x)-circulant matri- 1 2 1 0 n 1 ces are another natural extension of this well-studied class, and can be found in [13, 14, 15, 16, 17, 18, 19, It can be seen that the matrix over the complex 20, 21, 22, 23]. The f(x)-circulant matrix has a wide field C with an arbitrary first row and the following application, especially on the generalized cyclic codes rule for obtaining any other row from the previous [13]. The properties and structures of the xn −rx−1- one: Get the i+1st row by adding r times the last ele- circulant matrices, which are called RFPrLR circulant ment of the ith row to the first element of the ith row, matrices, are better than those of the general f(x)- and then shifting the elements of the ith row (cycli- circulant matrices, so there are good algorithms for cally) one position to the right. solving the RFPrLR circulant linear system. Note that the RFPrLR circulant matrix is a xn − In this paper, the fast algorithms presented avoid rx − 1 circulant matrix [13], and when r = 0,A be- the problems of error and efficiency produced by comes a circulant matrix [6, 7], when r = 1,A be- computing a great number of triangular functions by comes a RFPLR circulant matrix[17, 18, 19, 20], and means of other general fast algorithms. There is only when r = −1,A becomes a RFMLR circulant matrix error of approximation when the fast algorithm is re- [21, 22, 23]. Thus it is the extension of circulant ma- alized by computers, and only the elements in the first trix, RFPLR circulant matrix and RFMLR circulant row of the RFPrLR circulant matrix and the constant matrix. term r are used by the fast algorithm, so the result of In this paper, let r be a and sat- the computation is accurate in theory. Specially, the n ̸ nn isfy r = − (n−1) . It is easily verified that the poly- result computed by a computer is accurate over the (1 n) nominal g(x) = xn − rx − 1 has no repeated roots in rational field. n rn ≠ n . the complex field if (1−n)(n−1) Definition 1. A row first-plus-rlast right (RFPrLR) We define Θ(1,r) as the basic RFPrLR circulant circulant matrix with the first row (a0, a1, ..., an−1) is matrix, that is,

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Obviously, the RLPrFL circulant matrix over the   complex field C is the extension of left circulant 0 1 . .. 0 0   matrix[6, 7, 36].  0 0 ... 0 0  For the convenience of application, we give the    ......  obvious results in following lemmas. Θ(1,r) =  . . . . .  (2)   0 0 ... 0 1 Lemma 3. Let A = RFPrLRcircfr(a0, a1,..., 1 r . . . 0 0 C n×n. an−1) be a RFPrLR circulant matrix over and let B = RLPrFLcircfr(an−1, an−2, . . . , a1, a0) be a It is easily verified that the g(x) = RLPrFL circulant matrix over C. Then xn − rx − 1 is both the minimal polynomial and ˆ the characteristic polynomial of the matrix Θ(1,r), i.e., BIn = A Θ is nonsingular nonderogatory. In addition, (1,r) or n B = AIˆn, Θ(1,r) = In + rΘ(1,r). where In view of the structure of the powers of the basic   RFPrLR circulant matrix Θ , it is clear that 0 0 ... 0 1 (1,r)    0 0 ... 1 0  Iˆn = . (5) A = RFPrLRcircfr(a0, a1, . . . , an−1)  ......  n∑−1 1 0 ... 0 0 i = aiΘ(1,r). (3) i=0 Lemma 4 ([16]). Let C[x] be the polynomial over a field C, and let f(x), g(x) ∈ C[x]. Suppose Thus, A is a RFPrLR circulant matrix if and only that the polynomial matrix if A = f(Θ(1,r)) for some polynomial f(x). The ( ) n∑−1 i f(x) 1 0 polynomial f(x) = aix will be called the rep- i=0 g(x) 0 1 resenter of the RFPrLR circulant matrix A. In addition, the product of two RFPrLR circulant is changed into the polynomial matrix matrices is a RFPrLR circulant matrix and the inverse ( ) d(x) u(x) v(x) of a nonsingular RFPrLR circulant matrix is also a RFPrLR circulant matrix. Furthermore, a RFPrLR 0 s(x) t(x) circulant matrices commute under multiplication. by a series of elementary row operations, then Definition 2. A row last-plus-rfirst left (RLPrFL) (f(x), g(x)) = d(x) circulant matrix with the first row (a0, a1, . . . , an−1) C is meant a square matrix over the complex field of and the form f(x)u(x) + g(x)v(x) = d(x).

n ∼ A = RLPrFLcircfr(a0, a1, . . . , an−1) Lemma 5. C[x]/⟨x − rx − 1⟩ = C[Θ(1,r)].   C a0 a1 . . . an−1 Proof. Consider the following -algebra homomor-   phism  a1 . . . ra0 + an−1 a0    φ : C[x] → C[Θ ]  . .. . .  (1,r) =  . . . .  . (4)   f(x) → A = f(Θ(1,r)) an−2 . . . ran−3 + an−4 an−3 ra0 + an−1 . . . ran−2 + an−3 an−2 for f(x) ∈ C[x]. It is clear that φ is an C-algebra epimorphism. So we have It can be seen that the matrix over the complex C ∼ C field C with an arbitrary first row and the following [x]/kerφ = [Θ(1,r)]. rule for obtaining any other row from the previous C one: Get the i+1st row by adding r times the first ele- Since [x] is a principal ideal integral domain, there ∈ C ment of the ith row to the last element of the ith row, is a monic polynomial p(x) [x] such that kerφ = ⟨ ⟩ n − − and then shifting the elements of the ith row (cycli- p(x) . Since x rx 1 is the minimal polynomial n − − cally) one position to the left. of Θ(1,r), then p(x) = x rx 1.

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By Lemma 5, we have the following lemma. where A is given in Equation (1), T Lemma 6. Let A = RFPrLRcircfr(a0, a1,..., X = (x1, x2, . . . , xn) , an−1) be a RFPrLR-circulant matrix. Then A is non- T singular if and only if b = (bn−1, . . . , b1, b0) . (f(x), g(x)) = 1, If A is nonsingular, then Equation (6) has a unique solution X = A−1b. where A n∑−1 If is singular and Equation (6) has a solution, i then the general solution of Equation (6) is f(x) = aix i=0 {1,2} {1,2} X = A b + (In − A A)Y, is the representer of A and where Y is an arbitrary n-dimension column vector. n − − g(x) = x rx 1. The key of the problem is how to find A−1b, {1,2} {1,2} Proof. A is nonsingular if and only if f(x) + ⟨xn − A b and A A. For this purpose, we at first rx−1⟩ is an invertible element in C[x]/⟨xn −rx−1⟩. prove the following results. By Lemma 5, if and only if there exists Theorem 7. Let A = RFPrLRcircfr(a0,a1,··· , u(x) + ⟨xn − rx − 1⟩ ∈ C[x]/⟨xn − rx − 1⟩ an−1) be a nonsingular RFPrLR circulant matrix of T order n over C and b = (bn−1, . . . , b1, b0) . Then such that there exists a unique RFPrLR circulant matrix f(x)u(x) + ⟨xn − rx − 1⟩ ≡ 1 + ⟨xn − rx − 1⟩, C = RFPrLRcircfr(c0, c1, . . . , cn−1) u(x), v(x) ∈ C[x] if and only if there exist such that of order n over C such that the unique solution of f(x)u(x) + (xn − rx − 1)v(x) = 1 Equation (6) is the last column of C, i.e. n T if and only if (f(x), x − rx − 1) = 1. X = (cn−1, . . . , c1, c0 + rcn−1) .

In [37] for an m × n matrix A, any solution to the Proof. Since matrix A = RFPrLRcircfr(a0, a1,..., equation AXA = A is called a generalized inverse an−1) is nonsingular, then by Lemma 6 we have n∑−1 of A. In addition, if X satisfies X = XAX, then A i and X are said to be semi-inverses A{1,2}. (f(x), g(x)) = 1, where f(x) = aix is the rep- i=0 In this paper, we only consider square matrices resenter of A and g(x) = xn − rx − 1. A. k In [38] the smallest positive integer for which Let B = RFPrLRcircfr(b0 − rbn−1, b1,..., Ak+1 Ak index A. ( )= rank( ) holds is called the of bn−1) be the RFPrLR circulant matrix of order n con- T If A has index 1, the generalized inverse X of A is structed by b = (b − , . . . , b , b ) . Then the repre- # n 1 1 0 called the group inverse A of A. Clearly A and X senter of B is are group inverses if and only if they are semi-inverses n∑−1 and AX = XA. − i In [39] and [40] a semi-inverse X of A was con- b(x) = (b0 rbn−1) + bix . sidered in which the nonzero eigenvalues of X are the i=1 reciprocals of the nonzero eigenvalue of A. These ma- Therefore, we can change the polynomial matrix trices were called spectral inverses. It was shown in   A . . [40] that a nonzero matrix has a unique spectral in-  f(x) . 1 0 . b(x)  verse, As, if and only if A has index 1: when As is the . . group inverse A# of A. g(x) . 0 1 . 0 into the polynomial matrix 2 Fast algorithms for solving the   . . RFPrLR circulant linear system  1 . u(x) v(x) . c(x)  . . and the RLPrFL circulant linear 0 . s(x) t(x) . c1(x) system by a series of elementary row operations. By Lemma 4, we have Consider the RFPrLR circulant linear system ( )( ) ( ) u(x) v(x) f(x) 1 = , AX = b, (6) s(x) t(x) g(x) 0

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( )( ) ( ) u(x) v(x) b(x) c(x) = , Proof. Since A = RFPrLRcircfr(a0, a1, . . . , an−1), n∑−1 s(x) t(x) 0 c1(x) i then the representer of A is f(x) = aix and the i.e. i=0 characteristic polynomial of Θ(1,r) is f(x)u(x) + g(x)v(x) = 1, u(x)b(x) = c(x). g(x) = xn − rx − 1. Substituting x by Θ(1,r) in the above two equa- ( ) tions respectively, we have f(x) We can change the polynomial matrix g(x) f(Θ )u(Θ ) + g(Θ )v(Θ ) = I , ( ) (1,r) (1,r) (1,r) (1,r) n d(x) into the polynomial matrix by a series of u(Θ(1,r))b(Θ(1,r)) = c(Θ(1,r)). 0 elementary row operations. Since A is singular, by Since d(x) f(Θ ) = A, g(Θ ) = 0 Lemma 4 and Lemma 6, we know that is the (1,r) (1,r) largest common factor, which is not equal to 1, of and f(x) and g(x). Let b(Θ(1,r)) = B, f(x) = d(x)f (x) then 1 Au(Θ(1,r)) = In, (7) and u(Θ(1,r))B = c(Θ(1,r)). (8) g(x) = d(x)g1(x),

By Equation (7), we know that u(Θ(1,r)) is a unique then − inverse A 1 of A. According to Equation (8) and the (f1(x), g1(x)) = 1. characters of the RFPrLR circulant matrix, we know Since (d(x), g1(x)) = 1, we have that the last column of C is

T −1 (f(x), g1(x)) = (d(x)f1(x), g1(x)) = 1. (cn−1, . . . , c1, c0 + rcn−1) = A b. Since AA−1b = b, then Since (d(x), g1(x)) = 1 and (f(x), g1(x)) = 1, we have −1 T A b = (cn−1, . . . , c1, c0 + rcn−1) (f(x)d(x), g1(x)) = 1. − − is the solution of Equation (6). Since both A 1 and B Let B = RFPrLRcircfr(b0 rbn−1, b1,..., are unique, then A−1B is also unique. So bn−1) be the RFPrLR circulant matrix of order n con- T structed by b = (bn−1, . . . , b1, b0) . Then the repre- T −1 X = (cn−1, . . . , c1, c0 + rcn−1) = A b senter of B is is unique. n∑−1 i b(x) = (b0 − rbn−1) + bix . Theorem 8. Let A = RFPrLRcircfr(a0, a1,..., i=1 an−1) be a singular RFPrLR circulant matrix T of order n over C and b = (bn−1, . . . , b1, b0) . Therefore, we can change the polynomial matrix If the solution of Equation (6) exists, then   . . . there exist a unique RFPrLR circulant matrix  f(x)d(x) . 1 0 . d(x)b(x) . d(x)f(x)  C = RFPrLRcircfr(c0, c1, . . . , cn−1) and a unique . . . RFPrLR circulant matrix g1(x) . 0 1 . 0 . 0 into the polynomial matrix E = RFPrLRcircfr(e0, e1, . . . , en−1)   C . . . of order n over such that  1 . u(x) v(x) . c(x) . e(x)  T . . . X1 = (cn−1, . . . , c1, c0 + rcn−1) 0 . s(x) t(x) . c1(x) . e1(x) is a unique special solution of Equation (6) and by a series of elementary row operations. Then, by Lemma 4, we have X2 = X1 + (In − E)Y f(x)d(x)u(x) + g1(x)v(x) = 1, (9) is a general solution of Equation (6), where Y is an arbitrary n-dimension column vector. d(x)u(x)b(x) = c(x), (10)

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d(x)u(x)f(x) = e(x). (11) Since both A and T are RFPrLR circulant matri- ces, then AT = TA. If there exists another RFPrLR Therefore circulant matrix T1 such that f(x)d(x)u(x)f(x) + g(x)f1(x)v(x) = f(x). (12) AT1A = A, T1AT1 = T1,T1A = AT1.

Substituting x by Θ(1,r) in Equations (10), (11) and (12), respectively, we have Let AT = TA = H and AT1 = T1A = F . Clearly H2 = H and F 2 = F . Thus we have

d(Θ(1,r))u(Θ(1,r))b(Θ(1,r)) = c(Θ(1,r)), H = AT = AT1AT = FH, d(Θ(1,r))u(Θ(1,r))f(Θ(1,r)) = e(Θ(1,r)), F = T1A = T1AT A = FH. f(Θ(1,r))d(Θ(1,r))u(Θ(1,r))f(Θ(1,r))+ So H = F. Hence g(Θ(1,r))f1(Θ(1,r))v(Θ(1,r)) = f(Θ(1,r)). T = T AT = HT = FT = T1AT Since f(Θ(1,r)) = A, g(Θ(1,r)) = 0 and b(Θ(1,r)) = B, then = T1H = T1F = T1AT1 = T1.

d(Θ(1,r))u(Θ(1,r))B = c(Θ(1,r)), (13) So T is unique. Hence TB = C, TA = E and T b are also unique. d(Θ(1,r))u(Θ(1,r))A = e(Θ(1,r)), (14) By Theorem 7 and Theorem 8, we have the fol- Ad(Θ )u(Θ )A = A. (15) (1,r) (1,r) lowing fast algorithm for solving the RFPrLR circu- In the same way, we have lant linear system (6): Step 1. From the RFPrLR circulant linear system (6), d(Θ(1,r))u(Θ(1,r))Ad(Θ(1,r))u(Θ(1,r)) we get the polynomial = d(Θ(1,r))u(Θ(1,r)). (16) n∑−1 i n By Equation (15) and (16), we know T = f(x) = aix , g(x) = x − rx − 1 {1,2} d(Θ(1,r))u(Θ(1,r)) is a semi-inverses A of A. i=0 Let and n∑−1 C = TB = c(Θ(1,r)) i b(x) = (b0 − rbn−1) + bix ; = RFPrLRcircfr(c0, c1, . . . , cn−1). i=1 and let (Step 2. )Change the polynomial matrix f(x) b(x) into the polynomial matrix E = TA = e(Θ(1,r)) ( g(x) 0 ) = RFPrLRcircfr(e , e , . . . , e − ). d(x) c(x) 0 1 n 1 by a series of elementary row 0 c1(x) According to Equation (13) and the characters of the operations; RFPrLR circulant matrix, we know that the last col- umn of C is Step 3. If d(x) = 1, then the RFPrLR circulant lin- ear system (6) has a unique solution. Substituting x T (cn−1, . . . , c1, c0 + rcn−1) = T b. by Θ(1,r) in polynomial c(x), we obtain a RFPrLR circulant matrix Since AX = b has a solution, then AT b = b, i.e., T b is a solution of Equation (6). We have C = c(Θ(1,r)) = RFPrLRcircfr(c0, c1, . . . , cn−1).

A[T b + (In − E)Y ] = AT b + A(In − E)Y So the unique solution of AX = b is − = b + AY AEY T (cn−1, . . . , c1, c0 + rcn−1) ; = b + AY − AT AY = b + AY − AY = b, Step 4. If d(x) ≠ 1, d(x) dividing g(x), we get the quotient g1(x) and change the polynomial matrix so X2 = X1 + (In − E)Y is the general solution ( ) of Equation (6), where Y is an arbitrary n-dimension f(x)d(x) d(x)b(x) d(x)f(x) d(x)f(x)b(x) column vector and X1 = T b. g1(x) 0 0 0

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into the polynomial matrix and a unique RFPrLR circulant matrix ( ) 1 c(x) e(x) r(x) E = RFPrLRcircfr(e0, e1, . . . , en−1) 0 c1(x) e1(x) r1(x) of order n over C such that by a series of elementary row operations; T X1 = (c0 + rcn−1, c1, . . . , cn−2, cn−1) Step 5. Substituting x by Θ(1,r) in polynomial r(x), we get a RFPrLR circulant matrix is the unique special solution of AX = b and

{1,2} ˆ − R = r(Θ(1,r)) = AA B. X2 = X1 + In(In E)Y If the last column of R isn’t b, then AX = b has no is the general solution of AX = b, where Y is an solution. Otherwise, the RFPrLR circulant linear sys- arbitrary n-dimension column vector and Iˆn is given tem AX = b has a solution. Substituting x by Θ(1,r) in Equation (5). in polynomial c(x) and e(x), we have two RFPrLR circulant matrices By Theorem 9 and Theorem 10, we can get the fast algorithm for solving the RLPrFL circulant linear {1,2} C = c(Θ(1,r)) = A B system AX = b, where and A = RLPrFLcircfr(an−1, . . . , a1, a0), E = e(Θ ) = A{1,2}A. (1,r) T X = (x1, x2, . . . , xn) , So the unique special solution of AX = b is T b = (bn−1, . . . , b1, b0) . T X = (c − , . . . , c , c + rc − ) 1 n 1 1 0 n 1 Step 1. From the RLPrFL circulant linear system and the general solution of AX = b is AX = b, we get the polynomial − n∑−1 X2 = X1 + (In E)Y, i n f(x) = aix , g(x) = x − rx − 1 where Y is an arbitrary n-dimension column vector. i=0 The advantage of the above algorithm is that it and can solve AX = b whether the coefficient matrix of n∑−1 i AX = b is singular or nonsingular. b(x) = (b0 − rbn−1) + bix ; By Lemma 3 and Theorem 7, we have the follow- i=1 ing theorem. (Step 2. )Change the polynomial matrix f(x) b(x) Theorem 9. Let A = RLPrFLcircfr(an−1, . . . , a1, into the polynomial matrix a0) be a nonsingular RLPrFL circulant matrix of or- ( g(x) 0 ) T der n over C and b = (b − , . . . , b , b ) . Then there d(x) c(x) n 1 1 0 by a series of elementary row exists a unique RFPrLR circulant matrix 0 c1(x) operations; C = RFPrLRcircfr(c0, c1, . . . , cn−1) Step 3. If d(x) = 1, then the RLPrFL circulant linear of order n over C such that the unique solution of system AX = b has a unique solution. Substituting AX = b is x by Θ(1,r) in polynomial c(x), we have a RFPrLR T circulant matrix X = (c0 + rcn−1, c1, . . . , cn−2, cn−1) . C = c(Θ(1,r)) = RFPrLRcircfr(c0, c1, . . . , cn−1). By Lemma 3 and Theorem 8, we have the follow- ing theorem. So the unique solution of AX = b is

− T Theorem 10. Let A = RLPrFLcircfr(an 1, . . . , a1, (c0 + rcn−1, c1, . . . , cn−2, cn−1) ; a0) be a singular RLPrFL circulant matrix of order T n over C and b = (bn−1, . . . , b1, b0) . If the solution Step 4. If d(x) ≠ 1, dividing g(x) by d(x) , we get of AX = b exists, then there exists a unique RFPrLR the quotient g1(x) and change the polynomial matrix circulant matrix ( ) f(x)d(x) d(x)b(x) d(x)f(x) d(x)f(x)b(x) C = RFPrLRcircfr(c0, c1, . . . , cn−1) g1(x) 0 0 0

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into the polynomial matrix We transform the polynomial matrix A(x) by a series ( ) of elementary row operations as follows: 1 c(x) e(x) r(x) ( ) 2 + x + x3 1 + 2x + x2 0 c (x) e (x) r (x) A(x) = 1 1 1 −1 − 3x + x4 0 w w by a series of elementary row operations; w  (2) − x(1) Step 5. Substituting x by Θ(1,r) in polynomial r(x), ( ) we get a RFPrLR circulant matrix 2 + x + x3 1 + 2x + x2 2 −1 − 5x − x p3(x) { } w R = r(Θ ) = AA 1,2 B. w (1,r) w  (1) + x(2) If the last column of R isn’t b, then AX = b has no ( ) solution. Otherwise, the RLPrFL circulant linear sys- 2 − 5x2 1 + 2x − 2x3 − x4 tem AX = b has a solution. Substituting x by Θ −1 − 5x − x2 −x − 2x2 − x3 (1,r) w in polynomial c(x) and e(x), we have two RFPrLR w w circulant matrices  (1) − 5(2) ( ) {1,2} C = c(Θ(1,r)) = A B 7 + 25x p1(x) −1 − 5x − x2 −x − 2x2 − x3 w and w {1,2} w 1 E = e(Θ(1,r)) = A A.  (2) + 25 x(1) ( ) So the unique special solution of AX = b is 7 + 25x p1(x) 118 T −1 − x p2(x) X = (c + rc − , c , . . . , c − , c − ) 25 w 1 0 n 1 1 n 2 n 1 w w 118(1) + 625(2) and  ( ) X = X + Iˆ (I − E)Y 2 1 n n 201 p3(x) −1 − 118 x p (x) is the general solution of AX = b, where Y is an 25 w 2 arbitrary n-dimension column vector. w w 1  201 (1) ( ) 3 Examples 1 p4(x) − − 118 . 1 25 x p2(x) Example 1. Solve the RFP3LR circulant linear sys- tem AX = b, where where p (x) = 1 + 7x + 10x2 + 3x3 − x4, A = RFP3LRcircfr(2, 1, 0, 1) 1 − 24 − 43 2 − 15 3 3 4 − 1 5 p2(x) = 25 x 25 x 25 x + 25 x 25 x , 2 − 3 − 4 − 5 and p3(x) = 118+226x+105x 21x 43x 25x , T 118 226 35 2 − 7 3 − 43 4 − 25 5 b = (0, 1, 2, 1) . p4(x) = 201 + 201 x+ 67 x 67 x 201 x 201 x . From A = RFP3LRcircfr(2, 1, 0, 1) and b = Since d(x) = 1, then the RFP3LR circulant linear sys- (0, 1, 2, 1)T , we get the polynomial tem AX = b has a unique solution. On the other hand, 118 226 35 7 43 25 3 − − 4 c(x) = + x+ x2− x3− x4− x5. f(x) = 2 + x + x , g(x) = 1 3x + x 201 201 67 67 201 201 and Substituting x by Θ(1,3) in polynomial c(x), we have b(x) = 1 + 2x + x2. the RFP3LR circulant matrix 25 24 10 7 Then ( ) C = c(Θ(1,3)) = RFP3LRcircfr( , , , − ). f(x) b(x) 67 67 67 67 A(x) = g(x) 0 So the unique solution of AX = b is the last column ( ) of C, i.e., 3 2 2 + x + x 1 + 2x + x 7 10 24 4 = − − 4 . X = (− , , , )T . 1 3x + x 0 67 67 67 67

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√ √ d(x)f(x)b(x) = 19 − 6 10 + 3( 10 − 3)x Example 2. Solve the RFP6LR circulant linear sys- √ tem + 3x2 + (3 + 10)x3, AX = b, where Thus we structure matrix B(x) and transform the √ polynomial matrix B(x) by a series of elementary row A = RFP6LRcircfr(1, 3 + 10) operations as follows: B(x) = and √ ( ) b = (1, 3 + 10)T . f(x)d(x) d(x)b(x) d(x)f(x) d(x)f(x)b(x) √ g1(x) 0 0 0 From√ A = RFP6LRcircfr(1, 3 + 10) and b = ( ) T r (x) s (x) s (x) s (x) (1, 3 + 10) , we get the polynomial = √ 1 1 2 3 − 10 − 3 + x 0 0 0 √ w − − 2 w √ f(x) = 1 + (3 + 10)x, g(x) = 1 6x + x w  (1) − (3 + 10)x(2) and √ ( ) b(x) = 10 − 3 + x. √ r2(x) s1(x) s2(x) s3(x) − 10 − 3 + x 0 0 0 w Then ( ) w √ f(x) b(x) w A(x) =  (1) − (21 + 6 10)(2) g(x) 0 ( √ ) √ √ ( ) 40 10 + 120 s1(x) s2(x) s3(x) 1 + (3 + 10) 10 − 3 + x √ −3 − 10 + x 0 0 0 = − − 2 . w 1 6x + x 0 w √ w 1 − We transform the polynomial matrix A(x) by a series  40 ( 10 3)(1) of elementary row operations as follows: ( ) 1 s (x) s (x) s (x) ( √ √ ) √ 4 5 6 . 1 + (3 + 10)x 10 − 3 + x −3 − 10 + x 0 0 0 A(x) = −1 − 6x + x2 0 w w √ w  (2) − (( 10 − 3)x − 1)(1) where ( √ √ ) √ √ − − 2 1 + (3 + 10)x 10 3 + x r1(x) = √10 3 + 2x + ( √10 + 3)x , 0 q (x) − w 1 r2(x) = 10 √3 + (21 +√ 6 10)x, w √ − − 2 w s1(x) = 19√ 6 10 + (2 √10 6)x + x ,  ( 10 − 3)(1) − 2 s2(x) = 10 √3 + 2x √+ ( 10 + 3)x , √ ( √ ) 2 3 s3(x) = 19−6 10+3( 10−3)x+3x +(3+ 10)x , 10 − 3 + x q1(x) √ √ √ , s (x) = 37 10−117 + 19−6 10 x + 10−3 x2, 0 q2(x) 4 40√ √ 20 40 19−6 10 10−3 1 2 s5(x) = √40 + 20 x√+ 40 x , 37 10−117 57−18 10 where s6(x) =√ 40 + 40 x 3 10−9 2 1 3 √ √ + 40 x + 40 x . q1(x) = 19 − 6 10 + ( 10 − 3)x, √ √ √ Substituting x by Θ(1,6) in polynomial 2 q2(x) = −3 + 10 + (6 10 − 18)x − ( 10 − 3)x . √ √ √ 37 10 − 117 57 − 18 10 − ̸ r(x) = + x It is obvious that d(x) = 10 3 + x = 1, it 40 40 denoted that A is singular. √ 3 10 − 9 1 Then + x2 + x3, √ 40 40 − − g1(x) = g(x)/d(x) = 10 3 + x, we get the RFP6LR circulant matrix √ √ 2 d(x)f(x) = 10 − 3 + 2x + ( 10 + 3)x , R = r(Θ ) = AA{1,2}B √ √ (1,6) √ d(x)b(x) = 19 − 6 10 + (2 10 − 6)x + x2, = RFP6LRcircfr( 10 − 3, 1).

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Since the last column of R is b, the RFP6LR cir- [4] T. A. Gulliver and M. Harada, New nonbinary culant linear system AX = b has a solution. self-dual codes, IEEE Trans. Inform. Theory. 54, Substituting x by Θ(1,6) in polynomial 2008, pp. 415–417. √ √ √ [5] X. Q. Jin, V. K. Sin and L. L. Song, Circulant- 37 10 − 117 19 − 6 10 10 − 3 c(x) = + x + x2 block preconditioners for solving ordinary dif- 40 20 40 ferential equations, Appl. Math. Comput. 140 (2- and 3), 2003, pp. 409–418. √ √ [6] P. Davis, Circulant Matrices, Wiley, New York, 19 − 6 10 10 − 3 1 e(x) = + x + x2, 1979. 40 20 40 [7] Z. L. Jiang and Z. X. Zhou, Circulant Matri- we have the RFP6LR circulant matrices ces, Chengdu Technology University Publishing Company, Chengdu, 1999. (in Chinese) C = c(Θ ) = A{1,2}B (1,6) √ √ [8] J. Gutirrez-Gutirrez, Positive integer powers of 19 10 − 60 10 − 3 10 complex symmetric circulant matrices, Appl. = RFP6LRcircfr( , ) 20 20 Math. Comput. 202 (2), 2008, pp. 877–881. and [9] J. Rimas, On computing of arbitrary positive in- teger powers for one type of even order sym- {1,2} E = e(Θ(1,6)) = A A metric circulant matrices-II, Appl. Math. Com- √ √ put. 174 (1), 2006, pp. 511–523. 10 − 3 10 10 = RFP6LRcircfr( , ). [10] J. Rimas, On computing of arbitrary positive in- 20 20 teger powers for one type of even order sym- So the unique special solution of AX = b is the last metric circulant matrices-I, Appl. Math. Comput. column of C, i.e., 172 (1), 2006, pp. 86–90. √ √ − [11] J. Rimas, On computing of arbitrary positive in- 10 3 10 10 T X1 = ( , ) teger powers for one type of odd order symmet- 20 20 ric circulant matrices-II, Appl. Math. Comput. and a general solution of AX = b is 169 (2), 2005, pp. 1016–1027.

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