Two special kinds of least squares solutions for the quaternion matrix equation AXB+CXD=E

Transcription

1 Electronic Journal of Linear Algebra Volume 23 Volume 23 (2012) Article Two special kinds of least squares solutions for the quaternion matrix equation AXB+CXD=E Shi-fang Yuan Qing-wen Wang Follow this and additional works at: Recommended Citation Yuan Shi-fang and Wang Qing-wen (2012) "Two special kinds of least squares solutions for the quaternion matrix equation AXB+CXD=E" Electronic Journal of Linear Algebra Volume 23 DOI: This Article is brought to you for free and open access by Wyoming Scholars Repository It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository For more information please contact

2 Volume 23 pp February TWO SPECIAL KINDS OF LEAST SQUARES SOLUTIONS FOR THE QUATERNION MATRIX EQUATION AXB +CXD = E SHI-FANG YUAN AND QING-WEN WANG Abstract By using the complex representation of quaternion matrices the Moore Penrose generalized inverse and the Kronecker product of matrices the expressions of the least squares η-hermitian solution with the least norm and the expressions of the least squares η-anti-hermitian solution with the least norm are derived for the matrix equation AXB+CXD = E over quaternions Key words Matrix equation Least squares solution η-hermitian matrix η-anti-hermitian matrix Moore Penrose generalized inverse Kronecker product Quaternion matrices AMS subject classifications 65F05 65H10 15A33 1 Introduction ThroughoutthispaperletQR m n SR n n ASR n n C m n and Q m n be the skew field of quaternions the set of all m n real matrices the set of all n n real symmetric matrices the set of all n n real anti-symmetric matrices the set of all m n complex matrices and the set of all m n 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 For A Q m n A A T A H and A + denote the conjugate matrix the transpose matrix the conjugate transpose matrix and the Moore Penrose generalized inverse matrix of matrix A respectively A quaternion a can be uniquely expressed as a = a 0 +a 1 i+a 2 j +a 3 k with real coefficients a 0 a 1 a 2 a 3 and i 2 = j 2 = k 2 = 1ij = ji = k and a can be uniquely expressed as a = c 1 + c 2 j where c 1 and c 2 are complex numbers The following quaternion involutions of a quaternion a = a 0 +a 1 i+a 2 j +a 3 k defined as 4 a i = iai = a 0 +a 1 i a 2 j a 3 k Received by the editorsonjuly Accepted forpublication on January Handling Editor: Bryan L Shader Department of Mathematics Shanghai University Shanghai PR China and School of Mathematics and Computational Science Wuyi University Jiangmen PR China (ysf301@yahoocomcn) Supported by Natural Science Fund of China ( ) Guangdong Natural Science Fund of China ( ) and Program for Guangdong Excellent Talents in University Guangdong Education Ministry China (LYM10128) Department of Mathematics Shanghai University Shanghai PR China (wqw858@yahoocomcn) Supported by grants from the Natural Science Foundation of China ( ) the Natural Science Foundation of Shanghai (11ZR ) the PhD Programs Foundation of Ministry of Education of China ( ) and Shanghai Leading Academic Discipline Project (J50101) 257

3 Volume 23 pp February SF Yuan and QW Wang a j = jaj = a 0 a 1 i+a 2 j a 3 k a k = kak = a 0 a 1 i a 2 j +a 3 k For any A Q m n A can be uniquely expressed as A = A 1 +A 2 j where A 1 A 2 C m n and A H = Re(A 1 ) T Im(A 1 ) T i Re(A 2 ) T j Im(A 2 ) T k The complex representation matrix of A = A 1 +A 2 j Q m n is denoted by A 1 A 2 f(a) = C 2m 2n A 2 A 1 Notice that f(a) is uniquely determined by A For A Q m n B Q n s we have f(ab) = f(a)f(b) (see 39) We define the inner product: AB =tr(b H A) for all AB Q m n Then Q m n is a right Hilbert inner product space and the norm of a matrix generated by this inner product is the quaternion matrix Frobenius norm For A = (a ij ) Q m n B = (b ij ) Q p q the symbol A B = (a ij B) Q mp nq stands for the Kronecker product of A and B Definition A matrix A Q n n is η-hermitian if A ηh = A and a matrix A Q n n is η-anti-hermitian if A ηh = A where A ηh = ηa H η η {i j k} η-hermitian matrices and η-anti -Hermitian matrices are denoted by ηhq n n and ηaq n n respectively Various aspects of the solutions for real complex or quaternion matrix equations such as AXB = C AX + XB = C AXB + CX T D = E X AXB = C and AXB + CYD = E have been widely investigated (see and references cited therein) For the matrix equation (11) AXB +CXD = E if B and C are identity matrices then the matrix equation (11) reduces to the wellknown Sylvester equation 3 If C and D are identity matrices then the matrix equation (11) reduces to the well-known Stein equation There are many important results about their solutions For example Mitra 14 and Tian 16 considered the solvability condition for the complex and real matrix equation (11) respectively Hernández and Gassó 5 obtained the explicit solution of the matrix equation (11) Mansour 13 studied the solvability condition for the matrix equation (11) in the operator algebra For the quaternion matrix equation (11) Huang 7 obtained necessary and sufficient conditions for the existence of a solution or a unique solution using the method of complex representation of quaternion matrices Using the complex representation of quaternion matrices the Moore Penrose generalized inverse and the Kronecker product of matrices Yuan Liao and Lei 38 studied the least squares Hermitian problems for the quaternion matrix equation (AXBCXD) = (EF) η- Hermitian and η-anti-hermitian quaternion matrices are important class of matrices

4 Volume 23 pp February Two Special Kinds of Least Squares Solutions for the Quaternion Matrix Equation 259 applied in widely linear modelling due to the quaternion involution properties (see for details) In this paper we use the results of 38 to consider the least squares η-hermitian and η-anti-hermitian problems for quaternion matrix equation (11) The related problems are described as follows let Problem IGivenA Q m n B Q n s C Q m n D Q n s ande Q m s H L = {X X ηhq m n AXB+CXD E = Find X H H L such that X H = min X H L X min X m n AX 0B+CX 0 D E } 0 ηhq Problem II Given A Q m n B Q n s C Q m n D Q n s and E Q m s let A L = {X X ηaq n n AXB +CXD E = Find X A A L such that X A = min X A L X min X 0 ηaq m n AX 0B +CX 0 D E } ThesolutionX H ofproblemiiscalledthe leastsquaresη-hermitiansolutionwith the least norm and the solution X A of Problem II is called the least squares η-anti- Hermitian solution with the least norm for matrix equation (11) over quaternions Our approach to solving these problems is based on the way of studying vec(abc) mentioned in 38 which can overcome the difficulty from the noncommutative multiplication of quaternions and turns Problems I and II of quaternion matrix equation (11) into a system of real equations respectively This paper is organized as follows In Section 2 we analyze the structure of two special matrix sets: ηhq n n and ηaq n n In Section 3 we derive the explicit expression for the solution of Problem I In Section 4 we derive the explicit expression for the solution of Problem II Finally in Section 5 we give numerical algorithms and numerical examples for Problems I and II respectively 2 The structure of ηhq n n and ηaq n n In this section we analyze the structure of ηhq n n and ηaq n n Definition 21 For matrix A Q n n let a 1 = (a 11 2a 21 2a n1 ) a 2 = (a 22 2a 32 2a n2 ) a n 1 = (a (n 1)(n 1) 2a n(n 1) ) a n = a nn and denote

5 Volume 23 pp February SF Yuan and QW Wang by vec S (A) the following vector: (21) vec S (A) = (a 1 a 2 a n 1 a n ) T Q n(n+1) 2 Definition 22 For matrix B Q n n let b 1 = (b 21 b 31 b n1 ) b 2 = (b 32 b 42 b n2 ) b n 2 = (b (n 1)(n 2) b n(n 2) ) b n 1 = b n(n 1) and denote by vec A (B) the following vector: (22) vec A (B) = 2(b 1 b 2 b n 2 b n 1 ) T Q n(n 1) 2 Lemma Suppose X R n n Then (i) X SR n n vec(x) = K S vec S (X) where vec S (X) is represented as (21) and the matrix K S R n2 n(n+1) 2 is of the following form: K S = 1 2 2e1 e 2 e 3 e n 1 en e e2 e 3 e n 1 en e e e e 2 0 2en 1 en e e 2 0 e n 1 2en (ii) X ASR n n vec(x) = K A vec A (X) where vec A (X) is represented as (22) and the matrix K A R n2 n(n 1) 2 is of the following form: K A = 1 2 e 2 e 3 e n 1 en e e 3 e n 1 en 0 0 e e e e 2 0 en e e 2 e n 1 where e i is the i-th column of I n Obviously KS TK S = In(n+1) and KA TK A = 2 In(n 1) 2 We identify q Qwith acomplexvector q C 2 anddenote suchanidentification by the symbol = that is c 1 +c 2 j = q = q = (c 1 c 2 )

6 Volume 23 pp February Two Special Kinds of Least Squares Solutions for the Quaternion Matrix Equation 261 For A = A 1 +A 2 j Q m n we have A = Φ A = (A 1 A 2 ) vec(a 1 )+vec(a 2 )j = vec(a) vec(a1 ) = vec(φ A ) = vec(a 2 ) so that vec(a1 ) vec(a) = vec(φ A ) = vec(a 2 ) We denote A = (Re(A 1 )Im(A 1 )Re(A 2 )Im(A 2 )) vec(re(a 1 )) vec( vec(im(a 1 )) A) = vec(re(a 2 )) vec(im(a 2 )) Notice that vec(φ A ) = vec( A) In particular for A = A 1 + A 2 i C m n with A 1 A 2 R m n we have A = A = (A 1 A 2 ) and vec(a 1 )+vec(a 2 )i = vec(a) = vec( vec(a1 ) A) = vec(a 2 ) Addition of two quaternion matrices A = A 1 +A 2 j and B = B 1 +B 2 j satisfies (A 1 +B 1 )+(A 2 +B 2 )j = (A+B) = Φ A+B = (A 1 +B 1 A 2 +B 2 ) whereas multiplication satisfies AB = (A 1 +A 2 j)(b 1 +B 2 j) = (A 1 B 1 A 2 B 2 )+(A 1 B 2 +A 2 B 1 )j So AB = Φ AB moreover Φ AB can be expressed as Φ AB = (A 1 B 1 A 2 B 2 A 1 B 2 +A 2 B 1 ) B1 B 2 = (A 1 A 2 ) B 2 B 1 = Φ A f(b) For X = X 1 +X 2 j ηhq n n by Definition 11 we have X H η = ηx (Re(X 1 ) T Im(X 1 ) T i Re(X 2 ) T j Im(X 2 ) T k)η

7 Volume 23 pp February SF Yuan and QW Wang = η(re(x 1 )+Im(X 1 )i+re(x 2 )j +Im(X 2 )k) Re(X 1 ) T = Re(X 1 ) Im(X 1 ) T = { Im(X1 ) η = i Im(X 1 ) η i Re(X 2 ) T = { Re(X2 ) η = j Re(X 2 ) η j Im(X 2 ) T = { Im(X2 ) η = k Im(X 2 ) η k Theorem 24 If X Q n n then X ηhq n n vec( X) = K ηh vec ηh ( X) where K S vec S (Re(A 1 )) 0 K A 0 0 K ih = vec ih ( vec A (Im(A 1 )) A) = 0 0 K S 0 vec S (Re(A 2 )) K S vec S (Im(A 2 )) K jh = K S K S K A K S vec jh ( A) = vec S (Re(A 1 )) vec S (Im(A 1 )) vec A (Re(A 2 )) vec S (Im(A 2 )) K kh = K S K S K S K A vec kh ( A) = vec S (Re(A 1 )) vec S (Im(A 1 )) vec S (Re(A 2 )) vec A (Im(A 2 )) Proof We prove it only the case of η = j similar arguments handle the cases that η = i and η = k For X Q n n we have X jhq n n X H j = jx Re(X 1 ) T = Re(X 1 ) Im(X 1 ) T = Im(X 1 ) Re(X 2 ) T = Re(X 2 ) Im(X 1 ) T = Im(X 1 )

8 Volume 23 pp February Two Special Kinds of Least Squares Solutions for the Quaternion Matrix Equation 263 vec(re(a 1 )) K S vec S (Re(A 1 )) vec( vec(im(a 1 )) 0 K S 0 0 vec S (Im(A 1 )) X) = = vec(re(a 2 )) 0 0 K A 0 vec A (Re(A 2 )) vec(im(a 2 )) K S vec S (Im(A 2 )) = K jh vec jh ( X) Similar to the discussions mentioned above we have Theorem 25 If X Q n n then X ηaq n n vec( X) = K ηa vec ηa ( X) where K A vec A (Re(A 1 )) 0 K S 0 0 K ia = vec ia ( vec S (Im(A 1 )) A) = 0 0 K A 0 vec A (Re(A 2 )) K A vec A (Im(A 2 )) K ja = K A K A K S K A vec ja ( A) = vec A (Re(A 1 )) vec A (Im(A 1 )) vec S (Re(A 2 )) vec A (Im(A 2 )) K ka = K A K A K A K S vec ka ( A) = vec A (Re(A 1 )) vec A (Im(A 1 )) vec A (Re(A 2 )) vec S (Im(A 2 )) 3 The solution of Problem I It is wellknownthat fora C m n B C n s and C C s t (31) vec(abc) = (C T A)vec(B) However in the setting of quaternion matrices if A Q m n B Q n s and C Q s t (31) does not hold because Q is not commutative under multiplication

9 Volume 23 pp February SF Yuan and QW Wang Nevertheless by applying the method in 38 to study vec(abc) we can turn Problem I for quaternion matrix equation (11) into the least squares problem with the least norm for a system of real equations and solve Problem I Lemma Let A = A 1 + A 2 j Q m n B = B 1 + B 2 j Q n s and C = C 1 +C 2 j Q s t be given Then vec(φ vec(φ ABC ) = (f(c) T A 1 f(cj) H B ) A 2 ) vec( Φ jbj ) Lemma 32 For X = X 1 +X 2 j ηhq n n let I n 2 ii n I n 2 ii n 2 W = I n 2 ii n I n 2 ii n 2 Then vec(φ X ) vec( Φ jxj ) = WK ηh vec ηh ( X) Proof For X = X 1 +X 2 j ηhq n n by Theorem 24 we have vec(x 1 ) vec(φ X ) vec(x 2 ) = vec( Φ jxj ) vec(x 1 ) vec(x 2 ) I n 2 ii n vec(re(a 1 )) 0 0 I n 2 ii n 2 vec(im(a 1 )) = I n 2 ii n vec(re(a 2 )) 0 0 I n 2 ii n 2 vec(im(a 2 )) = WK ηh vec ηh ( X) By Theorem 24 and Lemmas 31 and 32 we have Lemma 33 If A = A 1 + A 2 j Q m n X = X 1 + X 2 j ηhq n n and B = B 1 +B 2 j Q n s then vec(φ AXB ) = (f(b) T A 1 f(bj) H A 2 )WK ηh vec ηh ( X)

10 Volume 23 pp February Two Special Kinds of Least Squares Solutions for the Quaternion Matrix Equation 265 Lemma 34 1 The matrix equation Ax = b with A R m n and b R n has a solution x 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 x = A + b+(i A + A)y Lemma 35 1 The least squares solutions of the matrix equation Ax = b with A R m n and b R n can be represented as x = A + b+(i A + A)y 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 Based on our earlier discussions we now turn our attention to Problem I The following notation is necessary for deriving the solutions of Problem I For A = A 1 + A 2 j Q m n B Q n s C = C 1 +C 2 j Q m n D Q n s E Q m s set (32) P = (f(b) T A 1 f(bj) H A 2 )+(f(d) T C 1 f(dj) H C 2 )WK ηh P 1 = Re(P) = Im(P) e = vec(re(φ E)) vec(im(φ E )) and R = (I P + 1 P 1)P T 2 Z = (I +(I R + R) P + 1 P+T 1 P T 2 (I R+ R)) 1 H = R + +(I R + R)Z P + 1 P+T 1 (I P T 2 R + ) S 11 = I P 1 P + 1 +P+T 1 P T 2 Z(I R+ R) P + 1 S 12 = P +T 1 P T 2 (I R + R)Z S 22 = (I R + R)Z From the results in 12 we have + = (P 1 + HT P 1 + HT ) I + = S11 S 12 S T 12 S 22 + = P + 1 P 1 +RR +

11 Volume 23 pp February SF Yuan and QW Wang Theorem 36 Let A Q m n B Q k s C Q m n D Q k s and E Q m s Let P 1 and e be as in (32) Then { H L = X vec( } (33) X) = K ηh (P 1 + HT P 1 + HT )e+(i P 1 + P 1 RR + )y where y R 2n2 +n is an arbitrary vector Proof By Lemma 33 AXB +CXD E 2 = Φ AXB +Φ CXD Φ E 2 By Lemma 35 it follows that and thus vec ηh ( X) = = vec(φ AXB )+vec(φ CXD ) vec(φ E ) 2 = Qvec ηh( X) vec(re(φ 2 E)) vec(im(φ E )) ( ) = vec ηh ( 2 X) e + e+ I + (P 1 ) y vec( X) = K ηh (P + 1 HT P + 1 HT )e+k ηh (I P + 1 P 1 RR + )y By Lemma 34 and Theorem 36 we get the following conclusion Corollary 37 The quaternion matrix equation (11) has a solution X ηhq m n if and only if S11 S 12 (34) e = 0 S12 T S 22 In this case denote by H E the solution set of (11) Then { H E = X vec( } X) = K ηh (P 1 + HT P 1 + HT )e+(i P 1 + P 1 RR + )y where y R 2n2 +n is an arbitrary vector Furthermore if (34) holds then the quaternion matrix equation (11) has a unique solution X H E if and only if (35) rank = 2n 2 +n

12 Volume 23 pp February Two Special Kinds of Least Squares Solutions for the Quaternion Matrix Equation 267 In this case (36) { H E = X vec( } X) = K ηh (P 1 + HT P 1 + HT )e Theorem 38 Problem I has a unique solution X H H L This solution satisfies (37) vec( XH ) = K ηh (P 1 + HT P 1 + HT )e Proof From (33) it is easy to verify that the solution set H L is nonempty and is a closed convex set Hence Problem I has a unique solution X H H L We now prove that the solution X H can be expressed as (37) From (33) we have by Lemma 35 and (33) min X H L X = min X H L vec( X) Thus vec( XH ) = K ηh + e vec( XH ) = K ηh (P + 1 HT P + 1 HT )e Corollary 39 The least norm problem X H = min X H E X has a unique solution X H H E and X H can be expressed as (37) 4 The solution of Problem II We now discuss the solution of Problem II Similar to Lemmas 32 and 33 we have the following conclusions Lemma 41 For X = X 1 +X 2 j ηaq n n one gets vec(φ X ) = WK ηa vec ηa ( X) vec( Φ jxj ) Lemma 42 For A = A 1 + A 2 j Q m n X = X 1 + X 2 j ηaq n n and B = B 1 +B 2 j Q n s one gets vec(φ AXB) = (f(b) T A 1f(Bj) H A 2)WK ηavec ηa( X)

13 Volume 23 pp February SF Yuan and QW Wang For A = A 1 +A 2 j Q m n B Q n s C = C 1 +C 2 j Q m n D Q n s and E Q m s set Q = (f(b) T A 1 f(bj) H A 2 )+(f(d) T C 1 f(dj) H C 2 )WK ηa (41) Q 1 = Re(Q) Q 2 = Im(Q) e = vec(re(φ E)) vec(im(φ E )) and R 1 = (I Q + 1 Q 1)Q T 2 Z 1 = (I +(I R + 1 R 1)Q 2 Q + 1 Q+T 1 Q T 2 (I R+ 1 R 1)) 1 H 1 = R + 1 +(I R+ 1 R 1)Z 1 Q 2 Q + 1 Q+T 1 (I Q T 2R + 1 ) 11 = I Q 1 Q + 1 +Q+T 1 Q T 2 Z 1(I R + 1 R 1)Q 2 Q = Q +T 1 Q T 2(I R + 1 R 1)Z 1 22 = (I R + 1 R 1)Z 1 From the results in 12 we have Q1 I Q 2 + Q1 = (Q + 1 HT 1 Q 2Q + 1 HT 1 ) Q 2 + Q = T Q1 Q 2 Q 2 + Q1 Q 2 = Q + 1 Q 1 +R 1 R + 1 Theorem 43 Let A Q m n B Q k s C Q m n D Q k s and E Q m s Let Q 1 Q 2 and e be as in (41) Then { A L = X vec( } X) = K ηa (Q + 1 HT 1 Q 2Q + 1 HT 1 )e+(i Q+ 1 Q 1 R 1 R 1 + )y where y R 2n2 n is an arbitrary vector By Lemma 34 and Theorem 43 we get the following conclusion Corollary 44 The quaternion matrix equation (11) has a solution X ηhq m n if and only if (42) e = 0 T 12 22

14 Volume 23 pp February Two Special Kinds of Least Squares Solutions for the Quaternion Matrix Equation 269 In this case denote by A E the solution set of (11) Then { A E = X vec( } X) = K ηa (Q + 1 HT 1 Q 2Q + 1 HT 1 )e+(i Q+ 1 Q 1 R 1 R 1 + )y where y R 2n2 n is an arbitrary vector Furthermore if (42) holds then the quaternion matrix equation (11) has a unique solution X A E if and only if (43) rank Q1 Q 2 = 2n 2 n In this case (44) { A E = X vec( } X) = K ηa (Q + 1 HT 1 Q 2Q + 1 HT 1 )e Theorem 45 Problem II has a unique solution X A A L This solution satisfies (45) vec( X A ) = K ηa (Q + 1 HT 1 Q 2Q + 1 HT 1 )e Corollary 46 The least norm problem X A = min X A E X has a unique solution X A A E and X A can be expressed as (45) 5 Numerical verification Based on the discussions in Sections 2 3 and 4 we report two numerical algorithms and three numerical examples to find the solutions of Problems I II in this section Algorithms 51 and 52 provide the methods to find the solutions of Problems I and II When the consistent conditions for matrix equation (11) hold Examples 53 and 54 consider the numerical solutions of Problems I and II for X jhq n n and X kaq n n respectively In Example 55 we use a matrix to perturb the matrix E of Example 54 obtain the inconsistent matrix equation (11) thus we can analyze the least squares solution with the least norm for matrix equation (11) in Problem II For demonstration purpose and avoiding the matrices with large norm to interrupt the solutions of Problems I and II we only consider the cases of small n = 5 and take the coefficient matrices in Examples 53 and 54

15 Volume 23 pp February SF Yuan and QW Wang Algorithm 51 (for Problem I) (1) Input A B C D and E (A Q m n B Q n s C Q m n D Q n s and E Q m s ) (2) Compute P 1 R H S 11 S 12 S 22 and e (3) If (34) and (35) hold then calculate X H (X H H E ) according to (36) (4) If (34) holds then calculate X H (X H H E ) according to (37) otherwise go to next step (5) Calculate X H (X H H L ) according to (37) Algorithm 52 (for Problem II) (1) Input A B C D and E (A Q m n B Q n s C Q m n D Q n s and E Q m s ) (2) Compute Q 1 Q 2 R 1 H and e (3) If (42) and (43) hold then calculate X A (X A A E ) according to (44) (4) If (42) holds then calculate X A (X A A E ) according to (44) otherwise go to next step (5) Calculate X A (X A A L ) according to (45) Example 53 Taking A = A 1 +A 2 j B = B 1 +B 2 j C = C 1 +C 2 j D = D 1 +D 2 j X = X 1 +X 2 j E = AXB +CXD where A 1 = I 5 ones(3 5) i A 2 = I C 1 = I C 2 = I B 1 = (I ) B 2 = ( I )i D 1 = ones(56)i D 2 = ones(56) X 1 = i i i i i i i i i i i i i i i i i i i i i i i i i X 2 = i i i i i i i i i i i i i i i i i i i i i i i i i Obviously X jhq 5 5 Let Φ A = (A 1 A 2 ) Φ B = (B 1 B 2 ) Φ C = (C 1 C 2 ) Φ D = (D 1 D 2 )

16 Volume 23 pp February Two Special Kinds of Least Squares Solutions for the Quaternion Matrix Equation 271 Φ X = (X 1 X 2 ) Φ E = Φ A f(x)f(b)+φ C f(x)f(d) By using Matlab 77 and Algorithm 51 we obtain S11 S 12 rank = 55 e = S12 T S 22 According to Algorithm 51 (3) we can see the matrix equation AXB +CXD = E has a unique j-hermitian solution and a unique Hermitian solution with the least norm X H H E and we can get X H X = Example 54 Suppose ABCD are the same as in Example 53 X = X 1 + X 2 j kaq n n and E = AXB +CXD where X 1 = i i i i i i i i i i i i i i i i i i i i 0 X 2 = i i i i i i i i i i i i i i i i i i i i i i i i i Let Φ A = (A 1 A 2 ) Φ B = (B 1 B 2 ) Φ C = (C 1 C 2 ) Φ D = (D 1 D 2 ) Φ X = (X 1 X 2 ) Φ E = Φ A f(x)f(b)+φ C f(x)f(d) By using Matlab 77 and Algorithm 52 we obtain S11 S 12 rank = 45 e = S12 T S 22 According to Algorithm 52 (4) we can see the matrix equation AXB +CXD = E has infinite k-anti-hermitian solutions and a unique k-anti-hermitian solution with the least norm X A A E for Problem II and we can get X A X = Example 55 Suppose A B C D X Φ A Φ B Φ C Φ D Φ X

17 Volume 23 pp February SF Yuan and QW Wang are the same as in Example 54 Φ G = ones(812) and let Φ E = Φ A f(x)f(b) + Φ C f(x)f(d)+φ G By using Matlab 77 and Algorithm 52 we obtain S11 S 12 rank = 45 e = S12 T S 22 According to Algorithm 52 (5) we can see the matrix equation AXB +CXD = E has infinite least squares k-anti-hermitian solutions and a unique least squares k- anti-hermitian solution with the least norm X A A L for Problem II and we can get X A X = and X A = X A1 +X A2 j where X A1 = i i i i i i i i i i i i i i i i i i i i 0 X A2 = i i i i i i i i i i i i i i i i i i i i i i i i i In addition the related numerical results are also verified and listed in Table 1 and Table 2 where + S11 N(P 1 ) = I S 12 S12 T S 22 and N(Q 1 Q 2 ) = I Q1 Q 2 Q1 Q T Table 1 Numerical results for Example 53 S 11 S 12 S 22 H Z N(P 1 ) E e 014 Table 2 Numerical results for Example 54 and Example H 1 Z 1 N(Q 1 Q 2 ) E54 E e 014

18 Volume 23 pp February Two Special Kinds of Least Squares Solutions for the Quaternion Matrix Equation 273 Examples53 54 and 55are used to showthe feasibility ofalgorithms 51 and 52 Acknowledgment The authors would like to thank the referees very much for their valuable suggestions and comments which resulted in a great improvement of the original manuscript REFERENCES 1 B-I Adi and TNE Greville Generalized Inverses: Theory and Applications John Wiley and Sons New York MDehghan and MHajarian On the reflexive solutions of the matrix equation AXB+CYD = E Bull Korean Math Soc 46: JD Gardiner AJ Laub JJ Amato and CB Moler Solution of Sylvester matrix equation AXB T +CXD T = E ACM Trans Math Software 18: T Ell and SJ Sangwine Quaternion involutions and anti-involutions Comput Math Appl 53: V Hernández and M Gassó Explicit solution of the matrix equation AXB CXD = E Linear Algebra Appl 121: RA Horn and FZ Zhang A generalization of the complex Autonne-Takagi factorization to quaternion matrices Linear Multilinear Algebra DOI: / LP Huang The matrix equation AXB GXD = E over the quaternion field Linear Algebra Appl 234: TS Jiang and MS Wei On solutions of the matrix equations X AXB = C and X AXB = C Linear Algebra Appl 367: TS Jiang and MS Wei On a solution of the quaternion matrix equation X A XB = C and its application Acta Math Sin 21: YT Li and WJ Wu Symmetric and skew-antisymmetric solutions to systems of real quaternion matrix equations Comput Math Appl 55: AP Liao ZZ Bai and Y Lei Best approximate solution of matrix equation AXB+CYD = E SIAM J Matrix Anal Appl 27: JR Magnus L-structured matrices and linear matrix equations Linear Multilinear Algebra 14: A Mansour Solvability of AXB CXD = E in the operators algebra B(H) Lobachevskii J Math 31: SK Mitra The matrix equation AXB+CXD = E SIAM J Appl Math 32: MA Ramadan MA Abdel Naby and AME Bayoumi On the explicit solutions of forms of the Sylvester and the Yakubovich matrix equations Math Comput Modelling 50: YG Tian The solvability of two linear matrix equations Linear Multilinear Algebra 48: CC Took and DP Mandic Augmented second-order statistics of quaternion random signals Signal Process 91: CC Took DP Mandic and FZ Zhang On the unitary diagonalisation of a special class of quaternion matrices Appl Math Lett 24: MH Wang XH Chen and MS Wei Iterative algorithms for solving the matrix equation AXB +CX T D = E Appl Math Comput 187: QW Wang A system of matrix equations and a linear matrix equation over arbitrary regular rings with identity Linear Algebra Appl 384:

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