The Symmetric and Antipersymmetric Solutions of the Matrix Equation A 1 X 1 B 1 + A 2 X 2 B A l X l B l = C and Its Optimal Approximation

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1 The Symmetric Antipersymmetric Soutions of the Matrix Equation A 1 X 1 B 1 + A 2 X 2 B A X B C Its Optima Approximation Ying Zhang Member IAENG Abstract A matrix A (a ij) R n n is said to be symmetric antipersymmetric matrix if a ij a ji a n j+1n i+1 for a 1 i j n Peng gave the bisymmetric soutions of the matrix equation A 1X 1B 1+A 2X 2B 2+ +A X B C where [X 1 X 2 X ] is a rea matrices group Based on this work an adjusted iterative method is proposed to find the symmetric antipersymmetric soutions of the above matrix equation When the matrix equation is consistent for any initia symmetric antipersymmetric matrix group [X 1 X 2 X ] the east norm symmetric antipersymmetric soution group can be obtained In addition for a given symmetric antipersymmetric matrix group [ X 1 X 2 X ] the optima approximation symmetric antipersymmetric soution group can be obtained Given numerica exampes show that the iterative method is efficient Index Terms Iterative method Matrix equation Symmetric antipersymmetric Least-norm soution group Optima approximation soution I INTRODUCTION AMatrix A (a ij ) R n n is said to be symmetric antipersymmetric matrix if a ij a ji a n j+1n i+1 for a 1 i j n Let R m n SR n n SA n denote the set of m n rea matrices n n rea symmetric matrices n n rea symmetric antipersymmetric matrices respectivey S n (S n (e n e n 1 e 1 )) denotes the n n reverse unit matrix (e i denotes ith coumn of n n unit matrix) The superscripts T represents the transpose of a matrix In space R m n we define inner products as: < A B > trace(b T A) for a A B R m n Then the norm of A generated by this inner product is Frobenius norm denoted by A Probem I (See [1]) Given A i R p ni B i R ni q (i 1 2 ) C R p q find matrix group [X 1 X 2 X ] with X i SA ni i 1 2 such that A 1 X 1 B 1 + A 2 X 2 B A X B C (1) Probem II (See [2]) When Probem I is consistent et S E denote its soution group set for given matrix group [ X 1 X 2 X ] with X i SA ni (i 1 2 ) find Manuscript received 19 Oct 2016; revised 25 Nov 2016 This work was supported by the Nationa Natura Science Foundation of China under Grant Nos Liaoning Province Nature Science Foundation under Grant Nos the Fundamenta Research Funds for the Centra Universities under Grant Nos Y Zhang is with the Department of Mathematics Daian Maritime University Daian Liaoning China e-mai: zhgyg77@sinacom [ X 1 X 2 X ] S E with X i SA ni such that X 1 X X 2 X X X 2 min [X 1X 2X ] S E [ X 1 X X 2 X X X 2 ] (2) In [3] an iterative method is constructed to find the bisymmetric soutions of matrix equation A 1 X 1 B 1 + A 2 X 2 B A X B C where [X 1 X 2 X ] is rea matrices group In this paper by adjusting the agorithm in [3] we can find symmetric antipersymmetric soutions of the above matrix equation In eectricity contro theory processing of digita signas symmetric antipersymmetric matrices have wide appication II ITERATIVE METHODS FOR SOLVING PROBLEM I AND II Firsty we introduce some emmas used to sove Probem I Lemma 21(see [4]) A matrix X SA n if ony if X X T S n XS n Lemma 22 Suppose that a matrix X SR n n then X S n XS n SA n Lemma 23 Suppose that A R n n X SA n then Proof < A X > 1 4 [A + AT S n (A + A T )S n ] X 1 [A + 4 AT S n(a + A T )S n] X 1 [ A X + 4 AT X S nas n X S na T S n X ] 1 [ A X + 4 AT X T A S nxs n A T S nxs n ] A X (3) Next by adjusting the method in [3] an iterative method to obtain the symmetric antipersymmetric soution groups of Probem I Agorithm 21 Step 1 Input matrices A i R p ni B i R ni q X i SA ni ni (i 1 2 ) C R p q ; Step 2 Cacuate R 0 C i1 A ix i B i ; Y 0i A T i R 0Bi T i 1 2 ; P 0i 1 4 [Y 0i + Y0i T S n i Y 0i S ni S ni Y0i T S n i ]; i 1 2 ; k:0; Step 3 If R k 0 then stop; ese k : k + 1; Step 4 Cacuate X (k) i X (k 1) i + R k 1 2 P k 1j 2 P k 1i i 1 ;

2 R k C i1 A ix (k) i B i R R k 1 k 1 2 ( P k 1j 2 i1 A ip k 1i B i ); Y ki A T i R kbi T i 1 2 ; P ki 1 4 [Y ki + Yki T S n i Y ki S ni S ni Yki T S n i ] + R k 2 R k 1 P 2 k 1i i 1 2 ; Step 5 goto step 3 Simiar with the discussions in [3] we need verify whether Agorithm 21 has the foowing basic properties Lemma 24 Suppose that the sequences {R i } {P ir }(R i 0 i k r 1 2 ) are generated by Agorithm 21 then R i R j 0 P ir P jr 0 (i j k i j) Proof We verify the concusion by induction with the method of [3] Step 1 Show that < R 0 R 1 > 0 0 when k 1 < R 0 R 1 >< R 0 R 0 R 0 2 R 0 2 R 0 2 P 0r 2 P 0r 2 < 1 4 [Y0r + Y T 0r S nr (Y 0r + Y0r)S T nr ] P 0r > 0 < P 0r P 1r > < P 0r 1 4 [Y1r + Y 1r T S nr Y 1rS nr (4) < P 0r P 1r > A rp 0rB r > S nr Y1rS T nr ] + R 1 2 R 0 P 2 0r > < P 0r 1 4 [Y1r + Y 1r T S nr Y 1rS nr S nr Y1rS T nr ] > + R 1 2 R 0 P 0r Step 2 Suppose that (4) hods when k s then when k s + 1 < R s R s+1 > < R s R s R s 2 P sr 2 P sr 2 A rp srb r > < 1 4 [Ysr + Y T sr S nr (Y sr + Ysr)S T nr ] P sr > 0 < P sr P s+1r > < P sr 1 4 [Y s+1r + Y T s+1r S nr Y s+1r S nr S nr Ys+1rS T nr ] + Rs+1 2 R s P 2 sr > < P sr 1 4 [Y s+1r + Ys+1r T S nr Y s+1r S nr S nr Ys+1rS T nr ] > + Rs+1 2 R s P 2 sr 2 0 (5) (6) (7) (8) For j 1 2 s 1 we have that < R j R s+1 > < R j R s P sr 2 P sr 2 A rp srb r > < 1 4 [Yjr + Y T jr S nr (Y jr + Yjr)S T nr ] P sr > 0 < P jr P s+1r > < P jr 1 4 [Ys+1r + Y s+1r T S nr Y s+1rs nr S nr Ys+1rS T nr ] > + R s+1 2 R s 2 < P jr P sr > 0 (9) (10) From steps 1 2 the concusion < R i R j > 0 < P ir P jr > 0 hod for a i j k(i j) by the principe of induction Next we verify that if Probem I is consistent for any initia symmetric antipersymmetric matrix group [X 1 X 2 X ] the matrix group sequences [X (k) 1 X (k) 2 X (k) ] generated by the iterative method converge to a symmetric antipersymmetric soution group within at most pq iterative steps in the absence of roundoff errors Lemma 25 Suppose that Probem I is consistent [X1 X2 X ] is a symmetric antipersymmetric soution group For any initia symmetric antipersymmetric matrix group [X 1 X 2 X ] the sequences {X (i) j }{R i } {P ij }(j 1 2 ) generated by Agorithm 21 satisfy < P ij Xj X (i) j > R i 2 (i ) (11) Proof We prove the concusion by induction with the method in [3] When i 0 < P 0j Xj X j > < 1 4 [Y0j + Y T 0j S nj Y 0jS nj S nj Y T 0jS nj ] Xj X j > R 0 2 (12) Suppose that the concusion hods for i s(s 0) that is < P sj X j X (s) j > R s 2 (13) then when i s + 1 < P s+1j Xj X(s+1) j > < 1 4 [Y s+1j + Ys+1j T S n j Y s+1j S nj S nj Ys+1j T S n j ] + Rs+1 2 R s P 2 sj Xj X(s+1) j > R s+1 2 By the principe of induction the concusion P ij X j X(i) j > R i 2 hods for a i (14) <

3 Theorem 21 Suppose that Probem I is consistent then for any initia symmetric an antipersymmetric matrix group [X 1 X 2 X ] a symmetric an antipersymmetric soution group can be obtained within at most pq iteration steps for Agorithm 21 Theorem 22 Probem I is consistent if ony if there exists a nonnegative integer number k such that R k 0 or P kj 0 for some j {1 2 } The proofs of the two theorems are simiar with the proofs of Theorem in [3] so they are omitted From Lemma 25 if there exists a nonnegative integer number k such that P kj 0 for a j {1 2 } but R k 0 then Probem I is inconsistent Hence the sovabiity of Probem I can be judged automaticay by Agorithm 21 Lemma 26 Probem I has a symmetric antipersymmetric soution groups if ony if the foowing inear matrix equations is consistent A 1X 1B 1 + A 2X 2B A X B C B1 T X 1A T 1 + B2 T X 2A T B T X A T C T (15) A 1S n1 X 1S n1 B A S n X S n B C B1 T S n1 X 1S n1 A T B T S n X S n A T C T Proof Suppose that Probem I has a symmetric antipersymmetric soution group [Y 1 Y 2 Y ] then Y i Yi T S ni Y i S ni (i 1 2 ) A 1 Y 1 B 1 + A 2 Y 2 B A Y B C B1 T Y 1 A T 1 + B2 T Y 2 A T B T Y A T (A 1 Y1 T B 1 + A 2 Y2 T B A Y T B ) T C T A 1 S n1 Y 1 S n1 B A S n Y S n B (A 1 Y 1 B A Y B ) C B1 T S n1 Y 1 S n1 A T B T S n Y S n A T (B1 T Y 1 A T B T Y A T ) CT (16) Hence the symmetric antipersymmetric soution group [Y 1 Y 2 Y ] is a soution group of the inear matrix equations (15) namey the inear matrix equations (15) is consistent Conversey suppose that the inear matrix equations (15) is consistent then there exists a matrix group [Ȳ1 Ȳ2 Ȳ](Ȳi R ni ni i 1 2 ) such that A 1 Ȳ 1B 1 + A 2 Ȳ 2B A Ȳ B C B1 T Ȳ1AT 1 + B2 T Ȳ2AT B T Ȳ A T C T (17) A 1S n1 Ȳ 1S n1 B A S n Ȳ S n B C B1 T S n1 Ȳ 1S n1 A T B T S n Ȳ S n A T C T Let Y i Ȳi+Ȳ T i Sn i (Ȳi+Ȳ T i )Sn i 4 then Y i SA n n A 1Y 1B 1 + A 2Y 2B A Y B A 1 Ȳ1+Ȳ 1 T Sn 1 (Ȳ1+Ȳ 1 T )Sn 1 B A 2 Ȳ2+Ȳ 2 T Sn 2 (Ȳ2+Ȳ 2 T (18) )Sn 2 B A Ȳ+Ȳ T Sn (Ȳ+Ȳ T )Sn B 4 C Therefore [Y 1 Y 2 Y ] is a symmetric antipersymmetric soution group of Probem I Remark From the proof process of Lemma 26 any symmetric antipersymmetric soution groups of Probem I must be the soution groups of the inear matrix equations (15) Let S E denote the soution group set of the inear matrix equations (15) then S E S E where S E is the soution group set of Probem I Simiar with the discussions in [3] checking [X1 X2 X ] being the east Frobenius norm symmetric antipersymmetric soution group of probem I is equivaent to verify that [X1 X2 X ] is the east Frobenius norm symmetric antipersymmetric soution group of the inear matrix equation (15) Lemma 27 (see [3]) Suppose that the consistent systems of the inear equations Ax b has a soution x R(A T ) the x is an unique east Frobenius norm soution of the systems of inear equations Next we verify that if et the initia symmetric antipersymmetric matrices be X j A T j HBT j +B jh T A j S nj (A T j HBT j + B jh T A j )S nj j 1 2 where H is arbitrary then the symmetric antipersymmetric soution group [X1 X2 X ] obtained by the iterative method is the east Frobenius norm symmetric antipersymmetric soution group Theorem 23 Suppose that Probem I is consistent If we choose the initia symmetric antipersymmetric matrices X j A T j HBT j + B j H T A j S nj (A T j HBT j + B j H T A j )S nj j 1 2 H is arbitrary Let X 1 0 X 2 0 X 0 then the symmetric antipersymmetric soution group [X1 X2 X ] obtained by Agorithm 21 is the unique east Frobenius norm symmetric antipersymmetric soution group of Probem I Proof With the proof method in [3] in term of Theorem 21 if we take X i A T i HBT i +B ih T A i S ni (A T i HBT i + B i H T A i )S ni (i 1 2 ) We can obtain the symmetric antipersymmetric soution group [X1 X2 X ] of probem I in finite iteration steps Xi (i 1 2 ) can be expressed as X i A T i Y B T i +B iy T A i S ni (A T i Y B T i +B iy T A i)s ni (19) where Y R p q Next we wi verify that [X1 X2 X ] is the unique east Frobenius norm symmetric antipersymmetric soution group of Probem I From the Remark we ony need show that [X1 X2 X ] is the east Frobenius norm symmetric antipersymmetric soution group of the inear matrix equation (15) For matrix A R m n et vec(a) denote the foowing mn vector containing a the entries of matrix A: vec(a) (A(: 1) T A(: 2) T A(: n) T ) R mn (20) where A(: i) denote ith coumn of matrix A A B denote the Kronecker product of matrices A B Then inear matrix equation (15) is equivaent to the system of inear matrix equations B T A A B T (B T Sn ) (A S n ) (A S n ) (B T Sn ) Noting that vec(x 1 ) vec(x 2 ) vec(x ) B T 1 A1 BT 2 A2 A 1 B T 1 A 2 B T 2 (B T 1 Sn 1 ) (A1Sn 1 ) (BT 2 Sn 2 ) (A2Sn 2 ) (21) (A 1S n1 ) (B T 1 Sn 1 ) (A2Sn 2 ) (BT 2 Sn 2 ) vec(x 1) vec(x 2) vec(x ) vec(c) vec(c T ) vec(c) vec(c T ) vec(a T 1 Y BT 1 + B1Y T A 1 S n1 (A T 1 Y BT 1 + B1Y T A 1)S n1 ) vec(a T 2 Y BT 2 + B2Y T A 2 S n2 (A T 2 Y BT 2 + B2Y T A 2)S n2 ) vec(a T Y BT + B Y T A S n (A T Y BT + B Y T A )S n )

4 B 1 A T 1 A T 1 B1 (Sn 1 B1) (Sn 1 AT 1 ) B 2 A T 2 A T 2 B2 (Sn 2 B2) (Sn 2 AT 2 ) B A T A T B (S n B ) (S n A T ) (S n1 A T 1 ) (Sn 1 B1) (S n2 A T 2 ) (Sn 2 B2) (S n A T ) (Sn B ) vec(y ) vec(y T ) vec(y ) vec(y T ) B T 1 A1 BT 2 A2 A 1 B T 1 A 2 B T 2 (B T 1 Sn 1 ) (A1Sn 1 ) (BT 2 Sn 2 ) (A2Sn 2 ) (A 1S n1 ) (B T 1 Sn 1 ) (A2Sn 2 ) (BT 2 Sn 2 ) T B T A A B T (B T Sn ) (A S n ) (A S n ) (B T Sn ) R vec(y ) vec(y T ) vec(y ) vec(y T ) B T 1 A1 BT 2 A2 A 1 B T 1 A 2 B T 2 (B T 1 Sn 1 ) (A1Sn 1 ) (BT 2 Sn 2 ) (A2Sn 2 ) (A 1S n1 ) (B T 1 Sn 1 ) (A2Sn 2 ) (BT 2 Sn 2 ) T B T A A B T (B T Sn ) (A S n ) (A S n ) (B T Sn ) Hence from Lemma 27 [vec(x1 ) vec(x2 ) vec(x )] is the unique east Frobenius norm symmetric antipersymmetric soution group of the matrix equation (21) Since vec operator is isomorphic [X1 X2 X ] is the unique east Frobenius norm symmetric antipersymmetric soution group of the inear marix equation (15) thus it is aso the unique east Frobenius norm symmetric antiperwymmetric soution group of Probem I When Probem I is consistent its symmetric antipersymmetric soution group set S E is nonempty then Let A 1 X 1 B 1 + A 2 X 2 B A X B C A 1 (X 1 X 1 )B A (X X )B C A 1 X1 B 1 A 2 X2 B 2 A X B X 1 X 1 X 1 X 2 X 2 X 2 X X X C C A 1 X1 B 1 A 2 X2 B 2 A X B (22) then Probem II is equivaent to find the east Frobenius norm symmetric antipersymmetric soution group of the inear matrix equation A 1 X1 B 1 + A 2 X2 B A X B C (23) III EXAMPLE Let n 3 A 1 A 2 B 1 B 2 A 3 B 3 (see [5]) C as foows: B A B 2 A B A C (24) (25) (26) (27) Let X 1 0 X 2 0 X 0 with the precision of Using the Agorithm 21 we can obtain the unique east Frobenius norm symmetric antipersymmetric soution group in 40 iteration steps as foows: X (43) (28) By using the Agorithm 21 et initia matrices X j A T j HBT j + B jh T A j S nj (A T j HBT j + B jh T A j )S nj j 1 2 where H is arbitrary et X1 0 X2 0 X 0 we can obtain the unique east Frobenius norm symmetric antipersymmetric soution group [ X 1 X2 X ] of the inear matrix equation (23) Once above symmetric antipersymmetric matrix group [ X 1 X2 X ] is obtained the unique symmetric antipersymmetric soution group [ ˆX 1 ˆX 2 ˆX ] of probem II can be obtained So ˆX 1 ˆX 2 ˆX can be denoted as ˆX 1 X 1 + X1 ˆX 2 X 2 + X2 ˆX X + X X (43) (29)

5 X (43) (30) R 43 C A 1 X (43) 1 B 1 A 2 X (43) 2 B 2 A 3 X (43) 3 B (31) Let X 3 X 1 X (32) (33) (34) Using the Agorithm 21 we can obtain the unique east Frobenius norm symmetric antipersymmetric soution group of the inear matrix equation (23) in 43 iteration steps as foows: X 1 X (35) (36) X (37) R 49 C A 1 X1 B1 A 2 X2 B2 A 3 X3 B So the symmetric antipersymmetric soution group of Probem II can be obtained as foows: ˆX 1 ˆX 2 (40) ˆX (38) (39) (40) min [ X 1 X X 2 X X 3 X 3 2 ] [X 1 X 2 X 3 ] S E ˆX 1 X1 2 + ˆX 2 X2 2 + ˆX 3 X e + 04 (41)

6 IV CONCLUSION In this paper an iterative method is constructed to find the symmetric antisymmetric soutions of matrix equation A 1 X 1 B 1 + A 2 X 2 B A X B C When the matrix equation is consistent for any initia symmetric antipersymmetric matrix group [X 1 X 2 X ] a symmetric antipersymmetric soution group can be obtained in finite iteration steps the east norm symmetric antipersymmetric soution group can be obtained by choosing a specia kind of initia symmetric antipersymmetric matrix group In addition for a given symmetric antipersymmetric group [ X 1 X 2 X ] the optima approximation symmetric antipersymmetric soution group can be obtained by finding the east norm symmetric antipersymmetric soution group of matrix equation A 1 X1 B 1 + A 2 X2 B A X B C where C C A 1 X1 B 1 A 2 X2 B 2 A X B Given numerica exampes show that the iterative method is efficient REFERENCES [1] Zhen-yun Peng The nearest bisymmetric soutions of inear matrix equations J Comput Math 22(6) 2004 pp [2] K T Jeseph Inverse eigenvaue probem in structura design AIAA J pp [3] Zhuo-hua Peng Xi-yan Hu Lei Zhang The bisymmetric soutions of the matrix equation A 1 X 1 B 1 + A 2 X 2 B A X B C its optima approximation Linear Agebra its Appications pp [4] Dongxiu Xie Yanping Sheng The east-squares soutions of inconsistent matrix equation over symmetric antipersymmetric matrices Appid Mathematics Letters pp [5] Zhuo-hua Peng Xi-yan Hu Lei Zhang The centrosymmetric soutions of matrix equation A 1 X 1 B 1 + A 2 X 2 B A X B C its optima approximation Acta Mathematica Scientia 29A(1) 2009 pp