Research Article Least Squares Pure Imaginary Solution and Real Solution of the Quaternion Matrix Equation AXB + CXD = E with the Least Norm

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1 Applied Mathematics, Article ID , 9 pages Research Article Least Squares Pure Imaginary Solution and Real Solution of the Quaternion Matrix Equation AXB + CXD = E with the Least Norm Shi-Fang Yuan School of Mathematics and Computational Science, Wuyi University, Jiangmen 5900, China Correspondence should be addressed to Shi-Fang Yuan; ysf301@aliyun.com Received 5 December 013; Accepted 3 February 014; Published 15 April 014 Academic Editor: Qing-Wen Wang Copyright 014 Shi-Fang Yuan. 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. Using the Kronecker product of matrices, the Moore-Penrose generalized inverse, and the complex representation of quaternion matrices, we derive the expressions of least squares solution with the least norm, least squares pure imaginary solution with the least norm, and least squares real solution with the least norm of the quaternion matrix equation AXB + CXD = E, respectively. 1. Introduction Quaternions were introduced by Irish mathematician Sir William Rowan Hamilton in The family of quaternions is a skew field or noncommutative division algebra, since the characteristic property of quaternions is their noncommutativity under multiplication. A quaternion q can be uniquely expressed as q=q 0 +q 1 i+q j+q 3 k with real coefficients q 0, q 1, q, q 3, i = j = k = 1, ij = ji = k, and q can be uniquely expressed as q = c 1 +c j,where c 1 and c are complex numbers. Thus, every pure imaginary quaternion q can be uniquely expressed as q =q 1 i+q j+ q 3 k. Throughout this paper, let Q, R m n,c m n,q m n,and IQ m n betheskewfieldofquaternions,thesetofallm n real matrices, the set of all m ncomplex matrices, the set of all m n quaternion matrices, and the set of all m npure imaginary quaternion matrices, respectively. For A C m n,re(a) and Im(A) denote the real part and the imaginary part of matrix A, respectively.fora Q m n, A, A T, A H,andA + denote the conjugate matrix, the transpose matrix, the conjugate transpose matrix, and the Moore-Penrose generalized inverse matrix of matrix A, respectively. For any A Q m n, A canbeuniquelyexpressedasa= A 1 +A j,wherea 1,A C m n. The complex representation matrix of A=A 1 +A j Q m n is denoted by f (A) =[ A 1 A A A 1 ] C m n. (1) Notice that f(a) is uniquely determined by A.ForA Q m n, B Q n s,wehavef(ab) = f(a)f(b) (see [). Denote the trace of a square matrix A=(a ij ) Q n n by tr(a) = a 11 +a + +a nn. We define the inner product A, B = tr(b H A) for all A, B Q m n.thenq m n is a Hilbert inner product space and the norm of a matrix generated by this inner product is the quaternion matrix Frobenius norm.the-normof the vector x is denoted by x. Various aspects of the solutions of matrix equations such as AXB = C, AX + XB = C, AXB + CYD = E, (AXB, CXD) = (E, F) have been investigated. See, for example, [ 35].For the matrix equation AXB + CXD = E, () if B and C are identity matrices, then the matrix equation () reduces to the well-known Sylvester equation [5, 36]. If C and D are identity matrices, then the matrix equation ()

2 Applied Mathematics reduces to the well-known Stein equation[37]. There are many important results about the matrix equation (). For example, Hernández and Gassó [38] obtained the explicit solution of the matrix equation (). Mansour [1 considered the solvability condition of the matrix equation () inthe operator algebra. Mitra [39]studied the solvabilityconditions of matrix equation (). For the quaternion matrix equation (). Huang [40] obtained necessary and sufficient conditions for the existence of a solution or a unique solution using the method of complex representation of quaternion matrices. Note that some authors have investigated the real and pure imaginary solutions to the quaternion matrix equations. For example, Au-Yeung and Cheng [] considered the pure imaginary quaternionic solutions of the Hurwitz matrix equations. Wang et al. [4 studiedthequaternion matrix equation AXB = C and obtained necessary and sufficient conditions for the existence of a real solution or pure imaginary solution of the quaternion matrix equation AXB = C. Using the complex representation of quaternion matrices and the Moore-Penrose generalized inverse, Yuan et al. [4] derived the expressions of the least squares solution with the least norm, the least squares pure imaginary solution with the least norm, and the least squares real solution with the least norm for the quaternion matrix equation AX = B, respectively. Motivated by the work mentioned above, in this paper, we will consider the related problem of quaternion matrix equation (). Problem 1. Given A Q m n, B Q k s, C Q m n, D Q k s, E Q m s,let H L = {X X Q n k, AXB + CXD E Find X H H L such that = min X 0 Q n k AX 0B+CX 0 D E }. (3) X H = min X H L X. (4) Problem. Given A Q m n, B Q k s, C Q m n, D Q k s, E Q m s,let J L = {X X IQ n k, AXB + CXD E Find X J J L such that = min X 0 IQ n k AX 0B+CX 0 D E }. (5) X J = min X J L X. (6) Problem 3. Given A Q m n, B Q k s, C Q m n, D Q k s, E Q m s,let A L = {X X R n k, AXB + CXD E Find X A A L such that = min X 0 R n k AX 0B+CX 0 D E }. (7) X A = min X A L X. (8) The solution X H of Problem 1 is called the least squares solution with the least norm; the solution X J of Problem is called the least squares pure imaginary solution with the least norm; and the solution X A of Problem 3 is called the least squares real solution with the least norm for matrix equation () over the skew field of quaternions. This paper is organized as follows. In Section, we derive the explicit expression for the solution of Problem 1. In Section 3, we derive the explicit expression for the solution of Problem. In Section 4, we derive the explicit expression for the solution of Problem 3.Finally,inSection 5,wereport numerical algorithms and numerical examples to illustrate our results.. The Solution of Problem 1 To study Problem 1, we begin with the following lemmas. Lemma 4 (see [43]). The matrix equation Ax=b,withA R m n and b R n,hasasolutionx R n if and only if in this case it has the general solution where y R n is an arbitrary vector. AA + b=b; (9) x=a + b+(i A + A) y, (10) Lemma 5 (see [43]). The least squares solutions of the matrix equation Ax=b,withA R m n and b R n,canberepresented as x=a + b+(i A + A) y, (11) where y R n is an arbitrary vector, and the least squares solution of the matrix equation Ax = b with the least norm is x= A + b. We identify q Q with a complex vector q C and denote such an identification by the symbol,thatis; c 1 +c j=q q=(c 1,c ). (1) For A=A 1 +A j Q m n,wehavea Φ A =(A 1,A ) and A = Φ A (13) = Re A 1 + Im A 1 + Re A + Im A.

3 Applied Mathematics 3 We denote A=(Re A 1, Im A 1, Re A, Im A ), vec ( A) = [ vec (Re A 1 ) vec (Im A 1 ) [ vec (Re A )]. (14) [ vec (Im A )] Notice that Φ A = A. Inparticular,forA=A 1 +A i C m n with A 1,A R m n,wehavea A=(A 1,A ),and vec (A 1 )+vec (A )i=vec (A) vec ( A) = [ vec (A 1) vec (A ) ]. (15) Addition of two quaternion matrices A=A 1 +A j and B=B 1 +B j is defined by (A 1 +B 1 )+(A +B )j= (A+B) whereas multiplication is defined as AB = (A 1 +A j) (B 1 +B j) Φ A+B =(A 1 +B 1,A +B ), (16) = (A 1 B 1 A B )+(A 1 B +A B 1 )j. (17) So Φ A+B = Φ A +Φ B, AB Φ AB ;moreover,φ AB can be expressed as Φ AB = (A 1 B 1 A B,A 1 B +A B 1 ) = (A 1,A )[ B 1 B B B 1 ] =Φ A f (B). (18) Lemma 6 (see [34]). Let A=A 1 +A j Q m n, B=B 1 + B j Q n s,andc=c 1 +C j Q s t be given. Then vec (Φ ABC )=(f(c) T A 1, f(cj) H A )[ vec (Φ B) vec ( Φ jbj ) ]. Lemma 7. For X=X 1 +X j Q n k,let (19) Proof. For X=X 1 +X j Q n k,wehave vec (X 1 ) [ vec (Φ [ X) [[[[ vec ( Φ jxj ) ]= vec (X ) vec (X 1 ) ] [ vec (X )] = [ I nk ii nk 0 0 vec (Re (X 1 )) 0 0 I nk ii nk vec (Im (X 1 )) [ I nk ii nk 0 0 ] [ vec (Re (X ))] [ 0 0 I nk ii nk ] [ vec (Im (X ))] =Kvec ( X). () By Lemmas 6 and 7,wehavethefollowing. Lemma 8. If A=A 1 +A j Q m n, X=X 1 +X j Q n k, and B=B 1 +B j Q k s,then vec (Φ AXB )=(f(b) T A 1, f(bj) H A )Kvec ( X). (3) Based on our earlier discussions, we now turn our attention to Problem 1. The following notations are necessary for deriving the solutions of Problem 1. ForA = A 1 +A j Q m n, B Q k s, C=C 1 +C j Q m n, D Q k s, E Q m s, set P=(f(B) T A 1 +f(d) T C 1, f(bj) H A + f(dj) H C )K, P 1 = Re (P), P = Im (P), e=[ vec (Re (Φ E)) vec (Im (Φ E )) ], R= (I 4nk P + 1 P 1)P T, H=R + +(I ms R + R) ZP P + 1 P+T 1 (I 4nk P T R+ ), (4) Z=(I ms +(I ms R + R) P P + 1 P+T 1 P T (I ms R + R)) 1, S 11 =I ms P 1 P + 1 +P+T 1 P T Z(I ms R + R) P P + 1, S 1 = P +T 1 P T (I ms R + R) Z, S = (I ms R + R) Z. (5) Then K= [ I nk ii nk I nk ii nk [ I nk ii nk 0 0 ]. (0) [ 0 0 I nk ii nk ] From the results in [33], one has [ P + =(P + P 1 HT P P + 1,HT ), [ P + [ P =P + P P 1 P 1 +RR +, (6) [ vec (Φ X) vec ( Φ jxj ) ]=Kvec ( X). (1) I 4nk [ P [ P + P P =[ S 11 S 1 S T 1 S ].

4 4 Applied Mathematics Theorem 9. Let A Q m n, B Q k s, C Q m n, D Q k s, and E Q m s,andletp 1, P, e be as in (4).Then H L = {X vec ( X) = (P + 1 HT P P + 1,HT )e (7) +(I P + 1 P 1 RR + )y}, Proof. By Lemmas 5 and 8,wecanget AXB + CXD E = Φ AXB+CXD E = vec (Φ AXB+CXD E) = vec (Φ AXB +Φ CXD Φ E ) = vec (Φ AXB)+vec (Φ CXD ) vec (Φ E ) = P vec ( X) vec (Φ E ) = (P 1 +ip ) vec ( X) = [vec (Re (Φ E )) + i vec (Im (Φ E ))] [P vec ( X) e. P (8) By Lemma 5,itfollowsthat vec ( X) = [ P + e+[i P 4nk [ P + [ P ] y; (9) P P thus vec ( X) = (P + 1 HT P P + 1,HT )e+(i 4nk P + 1 P 1 RR + )y. (30) The proof is completed. By Lemma 4 and Theorem 9,wegetthefollowingconclusion. Corollary 10. The quaternion matrix equation () has a solution X Q n k if and only if [ S 11 S 1 S T 1 S ]e=0. (31) In this case, denote by H E the solution set of ().Then H E = {X vec ( X) = (P + 1 HT P P + 1,HT )e (3) +(I P + 1 P 1 RR + )y}, Furthermore, if (31) holds, then the quaternion matrix equation () has a unique solution X H E ifandonlyif rank [ P 1 P ]=4nk. (33) In this case, H E ={X vec ( X) = (P + 1 HT P P + 1,HT )e}. (34) Theorem 11. Problem 1 has a unique solution X H H L.This solution satisfies vec ( X H )=(P + 1 HT P P + 1,HT )e. (35) Proof. From (7), it is easy to verify that the solution set H L is nonempty and is a closed convex set. Hence, Problem 1 has auniquesolutionx H H L. We now prove that the solution X H canbeexpressedas (35). From (7), we have by Lemma 5 and (7), Thus, min X = min X H L X H L vec ( X) ; (36) vec ( X) = [ P + e. (37) P vec ( X) = (P + 1 HT P P + 1,HT )e. (38) Thus we have completed the proof. Corollary 1. The least norm problem X H = min X H E X (39) has a unique solution X H H E and X H can be expressed as (35). 3. The Solution of Problem We now discuss the solution of Problem.ForX=X 1 +X j IQ n k,wehavere(x 1 )=0. For A=A 1 +A j Q m n, B Q k s, C=C 1 +C j Q m n, D Q k s, E Q m s,set Q= (f(b) T A 1 +f(d) T C 1, f(bj) H A + f(dj) H C ) [ ii nk I nk ii nk [ ii nk 0 0 ], [ 0 I nk ii nk ] (40)

5 Applied Mathematics 5 Thus we have Q 1 = Re (Q), Q = Im (Q), e=[ vec (Re (Φ E)) vec (Im (Φ E )) ], (41) R 1 =(I 3nk Q + 1 Q 1)Q T, H 1 =R + 1 +(I ms R + 1 R 1) Z 1 Q Q + 1 Q+T 1 (I 3nk Q T R+ 1 ), Z 1 = (I ms +(I R + 1 R 1) Q Q + 1 Q+T 1 QT (I ms R + 1 R 1)) 1, Δ 11 =I ms Q 1 Q + 1 +Q+T 1 QT Z 1 (I ms R + 1 R 1)Q Q + 1, Δ 1 = Q +T 1 QT (I ms R + 1 R 1)Z 1, Δ =(I ms R + 1 R 1)Z 1. [ Q + =(Q + Q 1 HT 1 Q Q + 1,HT 1 ), (4) In this case, denote by J E the pure imaginary solution set of (). Then J E = { vec (Im (X 1 )) { X [ vec (Re (X ))] { [ vec (Im (X ))] +(I 3nk Q + 1 Q 1 R 1 R + 1 )y} }, } =(Q + 1 HT 1 Q Q + 1,HT 1 )e (46) Furthermore, if (45) holds, then the quaternion matrix equation () has a unique solution X J E if and only if In this case, J E = { vec (Im (X 1 )) { X [ vec (Re (X ))] { [ vec (Im (X ))] rank [ Q 1 Q ]=3nk. (47) =(Q + 1 HT 1 Q Q + 1,HT 1 )e} }. } (48) [ Q + [ Q =Q + Q Q 1 Q 1 +R 1 R + 1, (43) Theorem 15. Problem has a unique solution X J = ImX J1 i+ ReX J j+imx J k J L. This solution satisfies I 3nk [ Q [ Q + Q Q =[ Δ 11 Δ 1 Δ T 1 Δ ]. We now study Problem. Since the methods are the same as in Section, we only describe the following results using Lemmas 4 and 5 and Theorem 9 and omit their detailed proofs. Theorem 13. Let A Q m n, B Q k s, C Q m n, D Q k s, and E Q m s ;letq 1, Q, E be as in (41). ThenthesetJ L of Problem can be expressed as J L = { vec (Im (X 1 )) { X [ vec (Re (X ))] =(Q + 1 HT 1 Q Q + 1,HT 1 )e { [ vec (Im (X ))] +(I 3nk Q + 1 Q 1 R 1 R + 1 )y} }, } (44) Corollary 14. The quaternion matrix equation () has a solution X IQ m n if and only if [ Δ 11 Δ 1 Δ T 1 Δ ]e=0. (45) vec (Im (X J1 )) [ vec (Re (X J ))] [ vec (Im (X J ))] Corollary 16. The least norm problem =(Q + 1 HT 1 Q Q + 1,HT 1 )e. (49) X J = min X J E X (50) has a unique solution X J = ImX J1 i+re X J j+imx J k J E and X J can be expressed as (49). 4. The Solution of Problem 3 We now discuss the solution of Problem 3. ForX = X 1 + X j R n k,wehaveim(x 1 )=Re(X )=Im(X )=0,and X=Re(X 1 ).Thus,wehavethefollowinglemmas. For A=A 1 +A j Q m n, B Q k s, C=C 1 +C j Q m n, D Q k s, E Q m s,set T=(f(B) T A 1 +f(d) T C 1, I nk f(bj) H A + f(dj) H C ) [ 0 [ I nk ], [ 0 ] (51)

6 6 Applied Mathematics T 1 = Re (T), T = Im (T), e = [ vec (Re (Φ E)) vec (Im (Φ E )) ], (5) Theorem 19. Problem 3 has a unique solution X A A L.This solution satisfies R = (I nk T + 1 T 1)T T, H =R + +(I ms R + R )Z T T + 1 T+T 1 (I nk T T R+ ), Z =(I ms +(I ms R + R ) T T + 1 T+T 1 TT (I ms R + R )) 1, (53) vec (Re X A )=(T + 1 HT T T + 1,HT )e. (60) Corollary 0. The least norm problem X A = min X A E X (61) has a unique solution X A A E and X A can be expressed as (60). Λ 11 =I ms T 1 T + 1 +T+T 1 TT Z (I ms R + 1 R 1)T T + 1, Λ 1 = T +T 1 TT (I ms R + R )Z, Λ = (I ms R + R )Z. We have [ T + =(T + T 1 HT T T + 1,HT ), [ T + [ T =T + T T 1 T 1 +R R +, I nk [ T [ T + T T =[ Λ 11 Λ 1 Λ T 1 Λ ]. (54) By Lemma 5, we can easily get the following results for Problem 3. Theorem 17. Let A Q m n, B Q k s, C Q m n, D Q k s,ande Q m s ;lett 1, T, e be as in (5). Thentheset A L of Problem 3 can be expressed as A L = {X vec (Re (X 1 )) = (T + 1 HT T T + 1,HT )e +(I nk T + 1 T 1 R R + )y}, (55) Corollary 18. The quaternion matrix equation () has a solution X R m n ifandonlyif [ Λ 11 Λ 1 Λ T 1 Λ ]e=0. (56) In this case, denote by A E the real solution set of ().Then A E = {X vec (Re (X 1 )) = (T + 1 HT T T + 1,HT )e +(I nk T + 1 T 1 R R + )y}, (57) Furthermore, if (56) holds, then the quaternion matrix equation () has a unique solution X A E if and only if rank [ T =nk. (58) T In this case, A E ={X vec (Re (X 1 )) = (T + 1 HT T T + 1,HT )e}. (59) 5. Numerical Verification Based on the discussions in Sections, 3, and4, wereport numerical tests in this section. We give three numerical algorithms and four numerical examples to find the solutions of Problems 1,,and3. Algorithms 1,,and3 provide the methods to find the solutionsofproblems 1,, and 3. If the consistent conditions for matrix equation ()hold,examples 4 and 5 consider the numerical solutions of Problem 1 for X Q n k. In Examples 6 and 7, if the consistent conditions for matrix equation () arenotsatisfied,wecancomputetheleastsquaressolution with the least norm in Problems and 3 by Algorithms and 3, respectively. For demonstration purpose and avoiding the matrices with large norm to interrupt the solutions of Problems 1 and, we only consider the coefficient matrices of small sizes in numerical experiments. Algorithm 1 (for Problem 1). We have the following. (1) Input A, B, C, D, ande (A Q m n, B Q k s, C Q m n, D Q k s,ande Q m s ). () Compute P 1, P, R, H, Z, S 11, S 1, S, e. (3) If (31) and(33) hold,thencalculatex H (X H H E ) according to (34). (4) If (31)holds,thencalculateX H (X H H E ) according to (35). Otherwise go to next step. (5) Calculate X H (X H H L ) according to (35). Algorithm (for Problem ). We have the following. (1) Input A, B, C, D, ande (A Q m n, B Q k s, C Q m n, D Q k s,ande Q m s ). () Compute Q 1, Q, R 1, Z 1, H 1, Δ 11, Δ 1, Δ, e. (3) If (45) and(47) hold,thencalculatex J (X A J E ) according to (48). (4) If (45) holds,thencalculatex J (X J J E ) according to (49). Otherwise go to next step. (5) Calculate X A (X J J L ) according to (49). Algorithm 3 (for Problem 3). We have the following. (1) Input A, B, C, D, ande (A Q m n, B Q k s, C Q m n, D Q k s,ande Q m s ).

7 Applied Mathematics 7 () Compute T 1, T, R, Z, H, Λ 11, Λ 1, Λ, e. (3) If (56) and(58) hold,thencalculatex A (X A A E ) according to (59). (4) If (56)holds,thencalculateX A (X A A E ) according to (60). Otherwise go to next step. (5) Calculate X A (X A A L ) according to (60). Example 4. Let m=6, n=6, k=5, s=5, A=A 1 +A j, B = B 1 +B j, C=C 1 +C j, D = D 1 +D j, X=X 1 +X j, (6) E=AXB+CXD,where A 1 = eye (m) + ones (n) i, A = magic (m) + ones (n) i, B 1 = ones (k) + eye (k) i, B = eye (s) + magic (s) i, Example 5. Let m=6, n=6, k=5, s=5, A=A 1 +A j, B = B 1 +B j, C=C 1 +C j, D = D 1 +D j, X=X 1 +X j, E=AXB+CXD,where A 1 = ones (n) i, A = zeros (n), B 1 = ones (k), B = eye (s) i, C 1 = magic (n), C = eye (n) i, D 1 = zeros (k), D = ones (s) i, X 1 = rand (n, k) + randn (n, k) i, X = randn (n, k) + rand (n, k) i. Let (66) (67) C 1 = eye (m) + magic (n) i, C = magic (m) + eye (n) i, D 1 = ones (k) + ones (k) i, D = eye (s) + ones (s) i, X 1 = rand (n, k) + randn (n, k) i, X = randn (n, k) + rand (n, k) i. (63) Φ A =(A 1,A ), Φ B =(B 1,B ), Φ C =(C 1,C ), Φ D =(D 1,D ), Φ X =(X 1,X ), Φ E =Φ A f(x) f(b) +Φ C f(x) f(d). (68) By using matlab 7.7 and Algorithm 1,weobtain Let Φ A =(A 1,A ), Φ B =(B 1,B ), rank [ P 1 P ] = 40, [S 11 S 1 S T 1 S ]e = (69) Φ C =(C 1,C ), Φ D =(D 1,D ), Φ X =(X 1,X ), Φ E =Φ A f(x) f(b) +Φ C f (X) f(d). (64) By using matlab 7.7 and Algorithm 1,weobtain rank [ P 1 P ] = 10, [S 11 S 1 S T 1 S ]e = (65) According to Algorithm 1 (3), we can see the matrix equation AXB + CXD = E has a unique solution which is a unique solution with the least norm X H H E.Wecanget X H X = According to Algorithm 1 (3), wecanseethematrixequa- tion AXB + CXD = E has infinite solution and a unique solution with the least norm X H H E,andwecanget X H X = Example 6. Suppose A, B, C, D, X, Φ A, Φ B, Φ C, Φ D, Φ X are the same as in Example 5; byusingmatlab7.7and Algorithm,weobtain rank [ P 1 P ] = 30, [S 11 S 1 S T 1 S ]e = (70) According to Algorithm (4), we can see the matrix equation AXB+CXD = E has infinite pure imaginary least squares solutionsandauniquepureimaginarysolutionwiththeleast

8 8 Applied Mathematics norm X J J L for Problem and we can get X J X = ,andX J = Im X J1 i+re X J j+im X J k,where Im X J1 = , [ [ ] Re X J = , [ [ ] Im X J = [ ] [ ] (71) Example 7. Suppose A, B, C, D, X, Φ A, Φ B, Φ C, Φ D, Φ X are the same as in Example 5. Byusingmatlab7.7and Algorithm,weobtain rank [ P 1 P ] = 10, [S 11 S 1 S T 1 S ]e = 1.75e (7) According to Algorithm 3 (5), we can see the matrix equation AXB + CXD = E has infinite least squares real solutions and a unique least squares real solution with the least norm X A A L for Problem 3 and we can get X A X = e + 017,and X A =10 16 [ [ [ ]. ] (73) Examples 4, 5, 6,and7 areusedtoshowthefeasibilityof Algorithms 1,,and3. Conflict of Interests The author declares that there is no conflict of interests regarding the publication of this paper. Acknowledgments This work is supported by Natural Science Foundation of China (no ), Guangdong Natural Science Fund of China (no ), and Science and Technology ProjectofJiangmenCity,China. References [ F. Z. Zhang, Quaternions and matrices of quaternions, Linear Algebra and Its Applications,vol.51,pp.1 57,1997. [] Y.-H. Au-Yeung and C.-M. Cheng, On the pure imaginary quaternionic solutions of the Hurwitz matrix equations, Linear Algebra and Its Applications,vol.419,no.-3,pp ,006. [3] 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 , [4] M. Dehghan and M. Hajarian, Finite iterative algorithms for the reflexive and anti-reflexive solutions of the matrix equation A 1 X 1 B 1 +A X B =C, Mathematical and Computer Modelling,vol.49,no.9-10,pp ,009. [5] J.D.Gardiner,A.J.Laub,J.J.Amato,andC.B.Moler, Solution of the Sylvester matrix equation AXB T +CXD T = E, ACM Transactions on Mathematical Software, vol.18,no.,pp.3 31, 199. [6] Z.-H. He and Q.-W. Wang, A real quaternion matrix equation with applications, Linear and Multilinear Algebra, vol. 61,no. 6, pp , 013. [7] N. Li and Q.-W. Wang, Iterative algorithm for solving a class of quaternion matrix equation over the generalized (P, Q)- reflexive matrices, Abstract and Applied Analysis, vol. 013, Article ID , 15 pages, 013. [8] N.Li,Q.-W.Wang,andJ.Jiang, Anefficientalgorithmforthe reflexive solution of the quaternion matrix equation AXB + CX H D=F, Applied Mathematics, vol.013,article ID17540,14pages,013. [9] Y.-T. Li and W.-J. Wu, Symmetric and skew-antisymmetric solutions to systems of real quaternion matrix equations, Computers & Mathematics with Applications, vol.55,no.6,pp , 008. [10] A.-P. Liao and Y. Lei, Least-squares solution with the minimum-norm for the matrix equation (AXB, GXH) = (C, D), Computers & Mathematics with Applications,vol.50,no. 3-4, pp , 005.

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