Research Article Finite Iterative Algorithm for Solving a Complex of Conjugate and Transpose Matrix Equation

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1 indawi Publishing Corporation Discrete Mathematics Volume 013, Article ID 17063, 13 pages Research Article Finite Iterative Algorithm for Solving a Complex of Conjugate and Transpose Matrix Equation Mohamed A. Ramadan, 1 Talaat S. El-Danaf, 1 and Ahmed M. E. Bayoumi 1 Department of Mathematics, Faculty of Science, Menoufia University, Shebeen El-Koom, Egypt Department of Mathematics, Faculty of Education, Ain Shams University, Cairo, Egypt Correspondence should be addressed to Mohamed A. Ramadan; mramadan@eun.eg Received 4 August 01; Accepted 4 November 01 Academic Editor: Franck Petit Copyright 013 Mohamed A. Ramadan et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Weconsideraniterativealgorithmforsolvingacomplexmatrixequationwithconjugateandtransposeoftwounknownsofthe form: A 1 VB 1 C 1 WD 1 A VB C WD A 3 V B 3 C 3 W D 3 A 4 V T B 4 C 4 W T D 4 =E. With the iterative algorithm, the existence of a solution of this matrix equation can be determined automatically. When this matrix equation is consistent, for any initial matrices V 1, W 1 the solutions can be obtained by iterative algorithm within finite iterative steps in the absence of round-off errors. Some lemmas and theorems are stated and proved where the iterative solutions are obtained. A numerical example is given to illustrate the effectiveness of the proposed method and to support the theoretical results of this paper. 1. Introduction Consider the complex matrix equation: A 1 VB 1 C 1 WD 1 A VB C WD A 3 V B 3 C 3 W D 3 A 4 V T B 4 C 4 W T D 4 =E, (1) where A 1, A, C 1, C C m r, B 1, B, D 1, D C s n, A 3, A 4, C 3, C 4 C m s, E C m n and B 3, B 4, D 3, D 4 C r n are given matrices, while V, W C r s are matrices to be determined. In the field of linear algebra, iterative algorithms for solving matrix equations have received much attention. Basedontheiterativesolutionsofmatrixequations,Dingand Chen presented the hierarchical gradient iterative algorithms for general matrix equations [1, and hierarchical least squares iterative algorithms for generalized coupled Sylvester matrix equations and general coupled matrix equations [3, 4. The hierarchical gradient iterative algorithms [1, and hierarchical least squares iterative algorithms [1, 4, 5for solving general (coupled) matrix equations are innovational and computationally efficient numerical ones and were proposed based on the hierarchical identification principle [3, 6which regards the unknown matrix as the system parameter matrix to be identified. Iterative algorithms were proposed for continuous and discrete Lyapunov matrix equations by applying the hierarchical identificationprinciple [7. Recently, the idea of the hierarchical identification was also utilized to solve the so-called extended Sylvester-conjugate matrix equations in [8. From an optimization point of view, a gradient-based iteration was constructed in [9 to solve the general coupled matrix equation. A significant feature of the method in [9 is that a necessary and sufficient condition guaranteeing the convergence of the algorithm can be explicitly obtained. Some complex matrix equations have attracted attention from many researchers since it was shown in [10 thatthe consistence of the matrix equation AX XB = C can be characterized by the consimilarity [11 13 oftwopartitioned matrices related to the coefficient matrices A, B, and C. By consimilarity Jordan decomposition, explicit solutions were obtained in [10, 14. Some explicit expressions of the solution to the matrix equation AX XB = C were established in [15, and it was shown that this matrix equation has a unique solution if and only if AA and BB hav no common eigenvalues. Research on solving linear matrix equations has been actively engaged in for many years. For example, Navarra et al.

2 Discrete Mathematics studied a representation of the general common solution of the matrix equations A 1 XB 1 = C 1 and A XB = C [16; Van der Woude obtained the existence of a common solution X for the matrix equations A i XB j =C ij [17; Bhimasankaram considered the linear matrix equations AX=B,CX=D, and EXF = G [18. Mitra has provided conditions for the existence of a solution and a representation of the general common solution of the matrix equations AX = C and XB = D and the matrix equations A 1 XB 1 =C 1 and A XB =C [19, 0. Ramadan et al. [1 introduced a complete, general, andexplicitsolutiontotheyakubovichmatrixequationv AVF = BW, and the matrix equation (AXB, GX) = (C, D) has some important results that have been developed. In [, necessary and sufficient conditions for its solvability and the expression of the solution were derived by means of generalized inverse. Moreover, in [ the least-squares solution was also obtained by using the generalized singular value decomposition. While in [3, when this matrix equation is consistent, the minimum-norm solution was given by the use of the canonical correlation decomposition. In [4, based on the projection theorem in ilbert space, an analytical expression of the least-squares solution was given for the matrix equation (AXB, GX) = (C, D) by making use of the generalized singular value decomposition and the canonical correlation decomposition. In [5, by using the matrix rank method a necessary and sufficient condition was derived for the matrix equations AX 1 B = C and GX = D to have a common least square solution. In the aforementioned methods, the coefficient matrices of the considered equations arerequiredtobefirstlytransformedintosomecanonical forms. Recently, an iterative algorithm has presented in [6 to solve the matrix equation (AXB, GX) = (C, D). Different from the above mentioned methods, this algorithm can be implemented by initial coefficient matrices and can provide a solution within finite iteration steps for any initial values. Very recently, in [7anewoperatorofconjugate product for complex polynomial matrices is proposed. It is shown that an arbitrary complex polynomial matrix can be converted into the so-called Smith normal form by elementary transformations in the framework of conjugate product. Meanwhile, the conjugate product and the Sylvester-conjugate sum are also proposed in [8.Basedontheimportantpropertiesof the above new operators, a unified approach to solve a general class of Sylvester-polynomial-conjugate matrix equations is given. The complete solution of the Sylvester-polynomialconjugate matrix equation is obtained. In [9 by using a real inner product in complex matrix spaces, a solution can be obtained within finite iterative steps for any initial values in the absence of round-off errors. In [30 iterative solutions to a class of complex matrix equations are given by applying the hierarchical identification principle. This paper is organized as follows. First, in Section, we introduce some notations, a lemma, and a theorem that will be needed to develop this work. In Section 3, we proposeiterativemethodstoobtainnumericalsolutiontothe complexmatrixequationwithconjugateandtransposeoftwo unknowns of the form: A 1 VB 1 C 1 WD 1 A VB C WD A 3 V B 3 C 3 W D 3 A 4 V T B 4 C 4 W T D 4 =Eusing iterative method. In Section 4, numerical example is given to explore the simplicity and the neatness of the presented methods.. Preliminaries The following notations, definitions, lemmas, and theorems will be used to develop the proposed work. We use A T, A, A,tr(A),and A to denote the transpose, conjugate, conjugatetranspose,thetrace,andthefrobeniusnormofa matrix A, respectively. We denote the set of all m ncomplex matrices by C m n,andre(a) denote the real part of number a. Definition 1 (inner product [31). A real inner product space is a vector space V over the real field R together with an inner product. That is, with a map, :V V R. () Satisfying the following three axioms for all vectors x, y, z V and all scalars a R: (1) symmetry: x, y = y, x () linearity in the first argument: ax, y = a x, y, x y, z = x, z y,z (3) positive definiteness: x, x > 0forallx=0, two vectors u, v V are said to be orthogonal if u, v = 0. The following theorem defines a real inner product on space C m n over the field R. Theorem (see [3). In the space C m n over the field R, an inner product can be defined as 3. The Main Result (3) A, B = Re [tr (A B). (4) In this section, we propose an iterative solution to a complex matrix equation with conjugate and transpose of two unknowns: A 1 VB 1 C 1 WD 1 A VB C WD A 3 V B 3 C 3 W D 3 A 4 V T B 4 C 4 W T D 4 =E defined in (1) wherea 1,A,C 1,C C m r, B 1,B,D 1, D C s n, A 3,A 4,C 3,C 4 C m s, E C m n and B 3,B 4, D 3,D 4 C r n are given matrices, while V, W C r s are matrices to be determined. The following finite iterative algorithm is presented to solve it. Algorithm 3. (1) Input A 1, A, C 1, C, B 1, B, D 1, D, A 3, A 4, C 3, C 4, B 3, B 4, D 3, D 4, E; () Chosen arbitrary matrices V 1, W 1 C r s ; (5)

3 Discrete Mathematics 3 (3) Set R 1 =E A 1 V 1 B 1 C 1 W 1 D 1 A V 1 B For i=1,fromalgorithm 3 we have Re {tr [P 1 (V V 1 )Q 1 (W W 1 ) C W 1 D A 3 V 1 B 3 C 3 W 1 D 3 A 4 V T 1 B 4 C 4 W T 1 D 4, P 1 =A 1 R 1B 1 A R 1 B B 3R 1 A 3 B 4 R T 1 A 4, Q 1 =C 1 R 1D 1 C R 1 D D 3R 1 C 3 D 4 R T 1 C 4, k=:1; (4) If R k =0, then stop; else go to Step 5; (5) Set V k1 =V k W k1 =W k P k Q k P k, P k Q k Q k, R k1 =E A 1 V k1 B 1 C 1 W k1 D 1 A V k1 B (6) = Re {tr [(A 1 R 1B 1 A R 1 B B 3 R 1 A 3 B 4 R T 1 A 4 B ) (V V 1 ) (C 1 R 1D 1 C R 1 D D 3 R 1 C 3 D 4 R T 1 C 4 D ) (W W 1 ) = Re {tr [R 1 A 1 (V V 1 )B 1 R 1 A (V V 1 ) B R 1 B 3 (V V 1 )A 3 R 1 B 4 (V V 1 ) A 4 R 1 C 1 (W W 1 )D 1 R 1 C (W W 1 ) D R 1 D 3 (W W 1 )C 3 R 1 D 4 (W W 1 ) C 4. (9) C W k1 D A 3 V k1 B 3 C 3 W k1 D 3 A 4 V T k1 B 4 C 4 W T k1 D 4, P k1 =A 1 R k1b 1 A R k1 B B 3R k1 A 3 B 4 R T k1 A 4 R k1 P k, Q k1 =C 1 R k1d 1 C R k1 D D 3 R k1 C 3 D 4 R T k1 C 4 R k1 Q k; (7) From properties of trace and conjugate Re {tr [P 1 (V V 1 )Q 1 (W W 1 ) = Re {tr [R 1 A 1 (V V 1 )B 1 R 1 A (V V 1 ) B A 3 (V V 1 ) B 3 R 1 A 4(V V 1 ) T B 4 R 1 R 1 C 1 (W W 1 )D 1 R 1 C (W W 1 ) D C 3 (W W 1 ) D 3 R 1 (6) If R k1 =0, then stop; else let k=k1go to Step 5. ToprovetheconvergencepropertyofAlgorithm 3, we first establish the following basic properties. Lemma 4. Suppose that the matrix equation (1) is consistent and V, W are arbitrary solutions of (1).Thenforanyinitial matrices V 1 and W 1,wehave Re {tr [P i (V V i )Q i (W W i ) = R i, (8) where the sequence {V i, {P i, {W i, {Q i,and{r i are generated by Algorithm 3 for i = 1,,.... Proof. We apply mathematical induction to prove the conclusion. C 4 (W W 1 ) T D 4 R 1 = Re {tr [R 1 A 1 (V V 1 )B 1 R 1 A (V V 1 )B R 1 A 3(V V 1 ) B 3 R 1 A 4(V V 1 ) T B 4 R 1 C 1 (W W 1 )D 1 R 1 C (W W 1 )D R 1 C 3(W W 1 ) D 3 R 1 C 4(W W 1 ) T D 4

4 4 Discrete Mathematics = Re {tr [R 1 (A 1V B 1 C 1 W D 1 A V B = Re {tr [R k1 A 1 (V V k1 )B 1 C W D A 3 V B 3 C 3 W D 3 A 4 V T B 4 C 4 W T D 4 A 1 V 1 B 1 C 1 W 1 D 1 A V 1 B C W 1 D A 3 V 1 B 3 C 3 W 1 D 3 A 4 V T 1 B 4 C 4 W T 1 D 4). (10) R T k1 A (V V k1 ) B R k1 B 3 (V V k1 )A 3 R k1 B 4 (V V k1 ) A 4 R k1 C 1 (W W k1 )D 1 In view that V and W are solutions of matrix equation (1), with this relation we have Re {tr [P 1 (V V 1 )Q 1 (W W 1 ) R T k1 C (W W k1 ) D R k1 D 3 (W W k1 )C 3 R k1 D 4 (W W k1 ) C 4 = Re {tr [R 1 (E A 1V 1 B 1 C 1 W 1 D 1 A V 1 B C W 1 D = Re {tr [R 1 R 1 = R 1. A 3 V 1 B 3 C 3 W 1 D 3 A 4 V T 1 B 4 C 4 W T 1 D 4) This implies that (8)holds for i=1. Assume that (8)holds for=k.that is, (11) R k1 (P k (V V k1 ) Q k (W W k1 )). (13) From properties of trace and conjugate we get Re {tr [P k1 (V V k1 )Q k1 (W W k1 ) Re {tr [P k (V V k ) Q k (W W k ) =. (1) Thenwehavetoprovethattheconclusionholdsfori=k1; it follows from Algorithm 3 that = Re {tr [R k1 A 1 (V V k1 )B 1 R T k1 A (V V k1 ) B Re {tr [P k1 (V V k1 )Q k1 (W W k1 ) = Re { { tr [ (A 1 R k1b 1 A R k1 B { [ B 3 R k1 A 3 B 4 R T k1 A 4 R k1 P k) (C 1 R k1d 1 C R k1 D (V V k1 ) D 3 R k1 C 3 D 4 R T k1 C 4 R k1 Q k) (W W k1 ) A 3 (V V k1 ) B 3 R k1 A 4 (V V k1 ) T B 4 R k1 R k1 C 1 (W W k1 )D 1 Rk1 T C (W W k1 ) D C 3 (W W k1 ) D 3 R k1 C 4 (W W k1 ) T D 4 R k1 R k1 (P k (V V k1 ) Q k (W W k1 ))

5 Discrete Mathematics 5 = Re {tr [R k1 A 1 (V V k1 ) B 1 R k1 A (V V k1 )B R k1 A 3(V V k1 ) B 3 R k1 A 4(V V k1 ) T B 4 R k1 C 1 (W W k1 )D 1 R k1 C (W W k1 )D R k1 C 3(W W k1 ) D 3 R k1 C 4(W W k1 ) T D 4 R k1 Re {tr [P k (V V k Q k (W W k = Re {tr [R k1 (A 1V B 1 C 1 W D 1 P k Q k P k) A V B C W D P k Q k Q k) A 3 V B 3 C 3 W D 3 A 4 V T B 4 C 4 W T D 4 A 1 V k1 B 1 C 1 W k1 D 1 A V k1 B C W k1 D A 3 V k1 B 3 C 3 W k1 D 3 A 4 V T k1 B 4 C 4 W T k1 D 4) R k1 Re {tr [P k (V V k ) Q k (W W k ) P k Q k (P k P k Q k Q k). (14) In view that V and W are solutions of matrix equation (1), with relation (14)onehas Re {tr [P k1 (V V k1 )Q k1 (W W k1 ) = Re {tr [R k1 (E A 1V k1 B 1 C 1 W k1 D 1 A V k1 B C W k1 D A 3 V k1 B 3 C 3 W k1 D 3 A 4 V T k1 B 4 C 4 W T k1 D 4) R k1 [ P k Q k ( P k Q k ) = Re {tr [R k1 R k1 = R k1. Then relation (8) holds by mathematical induction. (15) Lemma 5. Suppose that the matrix equation (1) is consistent and the sequences {R i, {P i,and{q i are generated by Algorithm 3 with any initial matrices V 1, W 1,suchthat =0 for all i=1,,...,k,andthen R i Re {tr (R j R i)=0, Re {tr (P j P i Q j Q i)=0, i,j=1,,...,k, i=j. Proof. We apply mathematical induction. Step 1. We prove that (16) Re {tr (R i1 R i)=0, (17) Re {tr (P i1 P i Q i1 Q i)=0, i=1,,...,k. (18) FirstfromAlgorithm3we have R k1 =E A 1 V k1 B 1 C 1 W k1 D 1 A V k1 B C W k1 D A 3 V k1 B 3 C 3 W k1 D 3 A 4 V T k1 B 4 C 4 W T k1 D 4, R k1 =E A 1 (V k P k Q k P k)b 1 C 1 (W k P k Q k Q k)d 1 A (V k P k Q k P k)b C (W k P k Q k Q k)d A 3 (V k C 3 (W k A 4 (V k P k Q k P k) B 3 P k Q k Q k) D 3 T P k Q k P k) B 4

6 6 Discrete Mathematics C 4 (W k T P k Q k Q k) D 4 =E A 1 V k B 1 C 1 W k D 1 A V k B C W k D = R 1 R 1 P 1 Q 1 Re {tr [P 1 A 1 R 1B 1 Q 1 C 1 R 1D 1 A 3 V k B 3 C 3 W k D 3 A 4 V T k B 4 C 4 W T k D 4 P k Q k (A 1 P k B 1 C 1 Q k D 1 A P k B C Q k D A 3 P k B 3 C 3 Q k D 3 A 4 P T k B 4 C 4 Q T k D 4), R k1 =R k P k Q k (A 1 P k B 1 C 1 Q k D 1 A P k B C Q k D A 3 P k B 3 C 3 Q k D 3 A 4 P T k B 4 C 4 Q T k D 4). (19) For i=1,itfollowsfrom(19)that Re {tr (R R 1) = Re { { tr [(R 1 { [ R 1 P 1 Q 1 (A 1 P 1 B 1 C 1 Q 1 D 1 = R 1 R 1 P 1 Q 1 P 1 A R 1 B Q 1 C R 1 D B 3 R 1 A 3P 1 D 3R 1 C 3Q 1 B 4 R T 1 A 4P 1 D 4R T 1 C 4Q 1 Re {tr [P 1 (A 1 R 1B 1 A R 1 B B 3 R 1 A 3 B 4 R T 1 A 4) Q 1 (C 1 R 1D 1 C R 1 D D 3 R 1 C 3 D 4 R T 1 C 4 ) = R 1 R 1 P 1 Q 1 Re {tr (P 1 P 1 Q 1 Q 1), Re {tr (R R 1) = R 1 R 1 P 1 Q 1 ( P 1 Q 1 )=0. (0) A P 1 B C Q 1 D A 3 P 1 B 3 C 3 Q 1 D 3 = Re {tr (R 1 R 1) R 1 P 1 Q 1 A 4 P T 1 B 4 C 4 Q T 1 D 4)) tr (P 1 A 1 R 1B 1 Q 1 C 1 R 1D 1 P 1 A R 1B Q 1 C R 1D P 1 A 3 R 1B 3 Q 1C 3 R 1D 3 P 1 A 4 R 1B 4 Q 1C 4 R 1D 4 ) R 1 This implies that (17) is satisfied for i=1. From Algorithm 3 we have Re {tr (P P 1 Q Q 1) = Re { { tr [(A 1 R B 1 A R B B 3R A 3 { [ B 4 R T A 4 R R 1 P 1) P 1 (C 1 R D 1 C R D D 3R C 3 D 4 R T C 4 R R 1 Q 1) Q 1

7 Discrete Mathematics 7 = Re {tr [R A 1P 1 B 1 R A P 1 B R B 3 P 1A 3 R B 4 P1 A 4 R R 1 P 1 P 1 R C 1Q 1 D 1 R C Q 1 D R D 3 Q 1C 3 R D 4 Q1 C 4 R R 1 Q 1 Q 1 = Re {tr [R A 1P 1 B 1 R A P 1 B A 3 P 1 B 3R A 4 P T 1 B 4R R C 1Q 1 D 1 R C Q 1 D C 3 Q 1 D 3R C 4 Q T 1 D 4R R R 1 (P 1 P 1 Q 1 Q 1) = Re {tr [R A 1P 1 B 1 R A P 1 B R A 3P 1 B 3 R A 4P T 1 B 4 R C 1Q 1 D 1 R C Q 1 D R C 3Q 1 D 3 R C 4Q T 1 D 4 R R 1 (P 1 P 1 Q 1 Q 1) = Re {tr [R (A 1P 1 B 1 C 1 Q 1 D 1 A P 1 B C Q 1 D A 3 P 1 B 3 C 3 Q 1 D 3 A 4 P T 1 B 4 C 4 Q 1 D 4) Assume that (17) and(18) hold for i = k 1,from Algorithm3we have Re {tr (R k1 R k) = Re { { tr [(R k { [ P k Q k Re {tr (R k1 R k) = Re {tr (R k R k) (A 1 P k B 1 C 1 Q k D 1 A P k B C Q k D A 3 P k B 3 C 3 Q k D 3 A 4 P T k B 4 C 4 Q T k D 4)) P k Q k tr (P k A 1 R kb 1 Q k C 1 R kd 1 = P k Q k P k A R kb Q k C R kd P k A 3 R kb 3 Q kc 3 R kd 3 P k A 4 R kb 4 Q kc 4 R kd 4 ) Re {tr (P k A 1 R kb 1 Q k C 1 R kd 1 R k, R R 1 ( P 1 Q 1 ) = P 1 Q 1 R 1 Re {tr [R (R 1 R ) R R 1 ( P 1 Q 1 ) = P 1 Q 1 R 1 ( R ) R R 1 ( P 1 Q 1 )=0. This implies that (18) is satisfied for i =1. (1) P k A R k B Q k C R k D B 3 R k A 3P k = R 1 R 1 P k Q k D 3R k C 3Q k B 4 R T k A 4P k D 4R T k C 4Q k ) Re {tr [P k (A 1 R kb 1 A R k B B 3 R k A 3 B 4 R T k A 4) Q k (C 1 R kd 1 C R k D D 3 R k C 3 D 4 R T k C 4)

8 8 Discrete Mathematics = P k Q k Re {tr [P k (P k R k 1 P k 1) Q k (Q k R k 1 Q k 1) = P k Q k ( P k Q k )=0. Thus (17)holds fori=k. Also, fromalgorithm3we have Re {tr (P k1 P k Q k1 Q k) = Re { { tr [(A 1 R k1b 1 A R k1 B { [ B 3 R k1 A 3 B 4 R T k1 A 4 R k1 P k) P k (C 1 R k1d 1 C R k1 D D 3 R k1 C 3 D 4 R T k1 C 4 R k1 Q k) Q k () R k1 P k P k R k1 C 1Q k D 1 R k1 C Q k D C 3 Q k D 3R k1 C 4 Q T k D 4R k1 R k1 = Re {tr [R k1 (A 1P k B 1 C 1 Q k D 1 Q k Q k A P k B C Q k D A 3 P k B 3 C 3 Q k D 3 A 4 P T k B 4 C 4 Q T k D 4) R k1 (P k P k Q k Q k) = P k Q k Re {tr (R k1 (R k R k1 )) R k1 ( P k Q k ) = P k Q k ( R k1 ) R k1 ( P k Q k )=0. (3) This implies that (17)and(18)hold fori=k. Then relations (17) and(18) holds by mathematical induction. Step. We want to show that = Re {tr [R k1 A 1P k B 1 R k1 A P k B R k1 B 3 P ka 3 R k1 B 4 Pk A 4 R k1 P k P k R k1 C 1Q k D 1 R k1 C Q k D R k1 D 3 Q kc 3 R k1 D 4 Qk C 4 R k1 Q k Q k Re (tr (R il R i)) = 0, Re (tr (P il P i Q il Q i)) = 0 (4) holds for l 1. We will prove this conclusion by induction. The case of l=1hasbeenproveninstep1.nowweassume that (4)holdsforl s, s 1.Theaimistoshowthat Re (tr (R is1 R i)) = 0, Re (tr (P is1 P i Q is1 Q i)) = 0. First we prove the following: (5) = Re {tr [R k1 A 1P k B 1 R k1 A P k B A 3 P k B 3R k1 A 4P T k B 4R k1 Re (tr (R s1 R 0)) = 0, Re (tr (P s1 P 0 Q s1 Q 0)) = 0. (6)

9 Discrete Mathematics 9 By using Algorithm 3,from(19) and induction we have Re {tr (R s1 R 0) = Re { { tr [ (R s R s { [ P s Q s (A 1 P s B 1 C 1 Q s D 1 A P s B Re {tr (P s1 P 0 Q s1 Q 0) = Re {tr [(A 1 R s1b 1 A R s1 B B 3 R s1 A 3 B 4 R T s1 A 4 R s1 R s P s) P 0 C Q s D A 3 P s B 3 C 3 Q s D 3 (C 1 R s1d 1 C R s1 D = Re { tr (R s R 0) A 4 P T s B 4 C 4 Q T s D 4)) R 0 R s P s Q s D 3 R s1 C 3 D 4 R T k1 C 4 R s1 R s Q s) Q 0 = Re {tr [R s1 A 1P 0 B 1 R s1 A P 0 B Re {tr (R s1 R 0) tr (P s A 1 R 0B 1 Q s C 1 R 0D 1 P s A R 0B Q s C R 0D P s A 3 R 0B 3 Q sc 3 R 0D 3 = Re {tr (R s R 0) R s P s Q s P s A 4 R 0B 4 Q sc 4 R 0D 4 ), Re {tr (P s A 1 R 0B 1 Q s C 1 R 0D 1 R s1 B 3 P 0A 3 R s1 B 4 P0 A 4 R s1 R s P s P 0 R s1 C 1Q 0 D 1 R s1 C Q 0 D R s1 D 3 Q 0C 3 R s1 D 4 Q0 C 4 R s1 = Re {tr [R s1 A 1P 0 B 1 R s1 A P 0 B A 3 P 0 B 3R s1 A 4P T 0 B 4R s1 R s1 R s P s P 0 R s1 C 1Q 0 D 1 R s Q s Q 0 = R s P s Q s = P s A R 0 B Q s C R 0 D B 3 R 0 A 3P s D 3 R 0 C 3Q s B 4 R T 0 A 4P s D 4 R T 0 C 4Q s ) Re {tr (P s (A 1 R 0B 1 A R 0 B R s B 3 R 0 A 3 B 4 R T 0 A 4) Q s (C 1 R 0D 1 C R 0 D D 3 R 0 C 3 D 4 R T 0 C 4)) P s Q s Re {tr (P s P 0 Q s Q 0)=0, R s1 C Q 0 D C 3 Q 0 D 3R s1 C 4 Q T 0 D 4R s1 R s1 R s Q s Q 0 = Re {tr [R s1 (A 1P 0 B 1 C 1 Q 0 D 1 A P 0 B C Q 0 D A 3 P 0 B 3 C 3 Q 0 D 3 A 4 P T 0 B 4 C 4 Q T 0 D 4) R s1 R s (P s P 0 Q s Q 0) = P 0 Q 0 R 0 Re {tr (R s1 (R 0 R 1 ))= 0. Then (6)isholds. (7)

10 10 Discrete Mathematics From Algorithm 3 we have Re {tr (P is1 P i Q is1 Q i) = Re {tr [(A 1 R is1b 1 A R is1 B B 3 R is1 A 3 B 4 R T is1 A 4 R is1 R is P is) P i (C 1 R is1d 1 C R is1 D D 3 R is1 C 3 D 4 R T is1 C 4 R is1 R is Q is) = Re {tr [R is1 A 1P i B 1 R is1 A P i B Q i = P i Q i R i Re {tr (R is1 (R i R i1 )) = P i Q i R i Re {tr (R is1 R i). Also from (19)wehave Re {tr (R is1 R i) = Re { { tr [(R is R is { [ P is Q is (A 1 P is B 1 C 1 Q is D 1 A P is B C Q is D A 3 P is B 3 C 3 Q is D 3 (8) R is1 B 3 P ia 3 R is1 B 4 Pi A 4 R is1 R is P is P i R is1 C 1Q i D 1 R is1 C Q i D R is1 D 3 Q ic 3 R is1 D 4 Qi C 4 R is1 = Re {tr [R is1 A 1P i B 1 R is1 A P i B R is Q is Q i A 3 P i B 3 R is1 A 4P T i B 4 R is1 R is1 R is P is P i R is1 C 1Q i D 1 R is1 C Q i D C 3 Q i D 3 R is1 C 4 Q T k D 4R is1 R is1 = Re {tr [R is1 (A 1P i B 1 C 1 Q i D 1 R is Q is Q i A 4 P T is B 4 C 4 Q T is D 4)) R i = Re {tr (R is R i) R is P is Q is Re {tr (R is1 R i) = R is P is Q is tr (P is A 1 R ib 1 Q is C 1 R id 1 P is A R ib Q is C R id P is A 3 R ib 3 Q isc 3 R id 3 P is A 4 R ib 4 Q isc 4 R id 4 ), Re {tr (P is A 1 R ib 1 Q is C 1 R id 1 A P i B C Q i D A 3 P i B 3 C 3 P i D 3 A 4 P T i B 4 C 4 Q T i D 4) R is1 R is (P is P i Q is Q i) P is A R i B Q is C R i D B 3 R i A 3 P is D 3R i C 3 Q is B 4 R T i A 4P is D 4R T i C 4Q is )

11 Discrete Mathematics 11 = R is P is Q is Re {tr [P is (A 1 R ib 1 A R i B = R is P is Q is = Re {tr [P is (P i B 3 R i A 3 B 4 R T i A 4) Q is (C 1 R id 1 C R i D D 3 R i C 3 D 4 R T i C 4) R i R is R i P is Q is R i 1 R i 1 P i 1) Q is (Q i R i R i 1 Q i 1) Re {tr (P is P i 1 Q is Q i 1). (9) Repeating (8) and(9), one can easily obtain for certain α and β tr (P is1 P i Q is1 Q i) = α[tr (P s1 P 1 Q s1 Q 1), tr (R is1 R i)=β[tr (R s1 R 1). (30) Combining these two relations with (6) impliesthat(4) holds for l=s1. From Steps 1 and the conclusion holds by the principle of induction. With the above two lemmas, we have the following theorem. Theorem 6 (see [3). If the matrix equation (1) is consistent, then a solution can be obtained within finite iteration steps by using Algorithm 3 for any initial matrices V 1, W Numerical Example In this section, we present numerical example to illustrate the application of our proposed methods. Example 7. In this example we illustrate our theoretical results of Algorithm 3 for solving the system of matrix equation: Because of the influence of the error of calculation, the residual R(k) is usually unequal to zero in this process of theiteration.weregardthematrixr(k) as a zero matrix if R(k) < Given 3i i 1i A 1 =[ 5 1i 3, 3i i 1i A =[ 5 1i 3, 0 i i A 3 =[ 1 3i 0, A 0 1 3i 1i 4 =[ 0 4i 3i, 1i 3 i 4 C 1 =[ i i 3, C 3i 0 1i =[ 0 4i 1 i, 1 3i i 3i C 3 =[ 1 3i 4i, C 4 =[ 1 i 0 3 i 1i 1, 4i i 0 i B 1 = [ 0 1 i, B = [ 1i 0, [ 4i i [ 1 i 3i 0 1 3i 1 i B 3 = [ 3i 4 i, B 4 =[ 0 i, [ 5 1i [ 1 i i D 1 = [ 1 3i i, D = [ 1i 0, [ i 3i [ 1 i 3i 0 1 3i i D 3 = [ 3i 4 i, D 4 = [ 0 i, [ 5 1i [ i 4i 4 55i 115 5i E=[ 38 i 13 44i. (3) Taking V 1 =[ and W 00 1 =[ we apply Algorithm 3 to compute V k, W k. And iterating 4 steps we get V i i i =[ i i i, [ i i i W i i i =[ i i i [ i i i (33) which satisfy the matrix equation: A 1 VB 1 C 1 WD 1 A VB C WD A 3 V B 3 C 3 W D 3 A 4 V T B 4 C 4 W T D 4 =E. (31) A 1 VB 1 C 1 WD 1 A VB C WD A 3 V B 3 C 3 W D 3 A 4 V T B 4 C 4 W T D 4 =E. (34)

12 1 Discrete Mathematics k (iteration number) log 10 r k Figure 1: The relation between the number of iterations and residual for the example. With the corresponding residual R 4 = E A 1V 4 B 1 C 1 W 4 D 1 A V 4 B C W 4 D A 3 V 4 B 3 C 3 W 4 D 3 A 4 V T 4 B 4 C 4 W T 4 D 4 = Conclusions (35) The above Figure1 shows the convergence curve for the residual function R(k). Inthispaper,aniterativealgorithm constructed to solve a complex matrix equation with conjugate and transpose of two unknowns of the form: A 1 VB 1 C 1 WD 1 A VB C WD A 3 V B 3 C 3 W D 3 A 4 V T B 4 C 4 W T D 4 = E is presented. We proved that the iterative algorithms always converge to the solution for any initial matrices.westatedandprovedsomelemmasandtheorems wherethesolutionsareobtained.theproposedmethod is illustrated by numerical example where the obtained numerical results show that our technique is very neat and efficient. References [1F.Ding,P.X.Liu,andJ.Ding, Iterativesolutionsofthe generalized Sylvester matrix equations by using the hierarchical identification principle, Applied Mathematics and Computation,vol.197,no.1,pp.41 50,008. [ F. Ding and T. Chen, Gradient based iterative algorithms forsolvingaclassofmatrixequations, IEEE Transactions on Automatic Control,vol.50,no.8,pp ,005. [3 F. Ding and T. Chen, ierarchical gradient-based identification of multivariable discrete-time systems, Automatica,vol.41,no., pp , 005. [4 F. Ding and T. Chen, Iterative least-squares solutions of coupled Sylvester matrix equations, Systems and Control Letters, vol. 54, no., pp , 005. [5 F. Ding and T. Chen, On iterative solutions of general coupled matrix equations, SIAM Journal on Control and Optimization, vol. 44, no. 6, pp , 006. [6 F. Ding and T. Chen, ierarchical least squares identification methods for multivariable systems, IEEE Transactions on Automatic Control,vol.50,no.3,pp ,005. [7 F. Ding and T. Chen, Performance analysis of multi-innovation gradient type identification methods, Automatica, vol. 43, no. 1, pp. 1 14, 007. [8 A.G.Wu,X.Zeng,G.R.Duan,andW.J.Wu, Iterativesolutions to the extended Sylvester-conjugate matrix equations, Applied Mathematics and Computation,vol.17,no.1,pp ,010. [9 B. Zhou, G. R. Duan, and Z. Y. Li, Gradient based iterative algorithm for solving coupled matrix equations, Systems and Control Letters,vol.58,no.5,pp ,009. [10 J.. Bevis, F. J. all, and R. E. artwing, Consimilarity and the matrix equation AX XB = C, in Current Trends in Matrix Theory, pp , North-olland, New York, NY, USA, [11 R. A. orn and C. R. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, UK, [1. Liping, Consimilarity of quaternion matrices and complex matrices, Linear Algebra and Its Applications, vol. 331, no. 1 3, pp.1 30,001. [13 T.Jiang,X.Cheng,andL.Chen, Analgebraicrelationbetween consimilarity and similarity of complex matrices and its applications, JournalofPhysicsA,vol.39,no.9,pp.915 9, 006. [14 J.. Bevis, F. J. all, and R. E. artwig, The matrix equation AX XB = Canditsspecialcases, SIAM Journal on Matrix Analysis and Applications, vol. 9, no. 3, pp , [15A.G.Wu,G.R.Duan,and..Yu, Onsolutionsofthe matrix equations XF AX = C and XF AX =C, Applied Mathematics and Computation, vol.183,no.,pp , 006. [16 A. Navarra, P. L. Odell, and D. M. Young, Representation of the general common solution to the matrix equations A 1 XB 1 =C 1,A XB =C with applications, Computers and Mathematics with Applications, vol.41,no.7-8,pp , 001. [17 J. W. Van der Woude, On the existence of a common solution X to the matrix equations A i XB j =C(i, j) E I, Linear Algebra and Its Applications,vol.375,no.1 3,pp ,003. [18 P. Bhimasankaram, Common solutions to the linear matrix equations AX=C,XB=D,andEXF= G, Sankhya Series A, vol. 38, pp , [19 S. Kumar Mitra, The matrix equations AX = C, XB = D, Linear Algebra and Its Applications,vol.59,pp ,1984. [0 S. K. Mitra, A pair of simultaneous linear matrix equations A 1 XB 1 = C 1,A XB = C and a matrix programming problem, Linear Algebra and Its Applications, vol.131,pp , [1 M. A. Ramadan, M. A. Abdel Naby, and A. M. E. Bayoumi, On the explicit solutions of forms of the Sylvester and the Yakubovich matrix equations, Mathematical and Computer Modelling,vol.50,no.9-10,pp ,009. [ K. W. E. Chu, Singular value and generalized singular value decompositions and the solution of linear matrix equations, Linear Algebra and Its Applications, vol , pp , 1987.

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