Least-Squares Solutions of Generalized Sylvester Equation with Xi Satisfies Different Linear Constraint

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1 Advances n Lnear Algebra & Matrx heory Publshed Onlne June 6 n ScRes. Least-Squares Solutons of Generalzed Sylvester Equaton wth Satsfes Dfferent Lnear Constrant ueln Zhou Dandan Song Qngle Yang Jaofen L * School of Mathematcs and Computng Scence Guangx Colleges and Unverstes Key Laboratory of Data Analyss and Computaton Guln Unversty of Electronc echnology Guln Chna Receved March 6 accepted June 6 publshed June 6 Copyrght 6 by authors and Scentfc Research Publshng Inc. hs work s lcensed under the Creatve Commons Attrbuton Internatonal Lcense (CC BY. Abstract In ths paper an teratve method s constructed to fnd the least-squares solutons of generalzed Sylvester equaton AB B B B C where [ ] s real matrces group and satsfes dfferent lnear constrant. By ths teratve method for any ntal ( ( ( ( matrx group wthn a specal constraned matrx set a least squares soluton group wth satsfyng dfferent lnear constrant can be obtaned wthn fnte teraton steps n the absence of round off errors and the unque least norm least-squares soluton can be obtaned by choosng a specal knd of ntal matrx group. In addton a mnmzaton property of ths teratve method s characterzed. Fnally numercal experments are reported to show the effcency of the proposed method. Keywords Least-Squares Problem Centro-Symmetrc Matrx Bsymmetrc Matrx Iteratve Method. Introducton A a R s sad to be a Centro-symmetrc matrx f aj a n + j n + for all j n. A A matrx ( matrx ( j A a R * Correspondng author. j s sad to be a Bsymmetrc matrx f aj aj a n + j n + for all j n. Let How to cte ths paper: Zhou.L. Song D.D. Yang Q.L. and L J.F. (6 Least-Squares Solutons of Generalzed Sylvester Equaton wth Satsfes Dfferent Lnear Constrant. Advances n Lnear Algebra & Matrx heory

2 m n R SR CSR and BSR denote the set of m n real matrces n n real symmetrc matrces n n real Centro-symmetrc matrces and n n real Bsymmetrc matrces respectvely. Sn ( en en e m n where e denotes th column of n n unt matrx. For a matrx A R we denote ts transpose traced by A tr ( A respectvely. In space R m n we defne nner product as: A B tr ( B A for all AB R then the norm of a matrx A generated by ths nner product s obvously Frobenus norm and denoted by A. Denote n {[ ] } n n n n n n n K R SR CSR BSR. n n n n Obvously K.e. R SR CSR BSR s a lnear subspace of real number feld. In ths paper we manly consder the followng two problems: p n Problem I. Gven matrces A R n q B R ( fnd matrx group K such that Problem II. Denote by that AB B B B C mn AB B B B C. [ ] K ˆ ˆ ˆ ˆ SE such S E the soluton set of Problem I. Fnd matrx group ˆ + ˆ + ˆ + ˆ mn. In fact Problem II s to fnd the least norm soluton of Problem I. here are many valuable efforts on formulatng solutons of varous lnear matrx equatons wth or wthout lnear constrant. For example Baksalary and Kala [] Chu [] [] Peng [] Lao Ba and Le [5] and u We and Zheng [6] consdered the nonsymmetrc soluton of the matrx equaton AB + CYD E ( by usng Moore-Penrose generalzed nverse and the generalzed sngular value decomposton of matrces whle Chang and Wang [7] consdered the symmetrc condtons on the soluton of the matrx equatons Zetak [8] [9] dscussed the Peng [] researched the general lnear matrx equaton AA BYB E A YA C + +. ( l p -soluton and Chebyshev-soluton of the matrx equaton A + YB C. ( A B A B A B C ( l l l wth the bsymmetrc condtons on the solutons. Vec operator and Kronecker product are employed n ths paper so the sze of the matrx s enlarged greatly and the computaton s very expensve n the process of solvng solutons. Iteratve algorthms have been receved much attenton to solve lnear matrx equatons n recent years. For example by extendng the well-known Jacob and Gauss-sedel teratons for Ax b Dng Lu and Dng n [] derved teratve solutons of matrx equatons AB F and generalzed Sylvester matrx equatons AB + CYD F. By absorbng the thought of the conjugate gradent method Peng [] presented an teratve algorthm to solve Equaton (. Peng [] Peng Hu and Zhang [] put forward an teratve method for bsymmetrc soluton of Equaton (. hese matrx-form CG methods are based on short recurrences whch keep work and storage requrement constant at each teraton. However these teraton methods are only defned by the Galerkn condton but lack of a mnmzaton property whch means that the algorthm may exhbt a rather rregular convergence and often results n a very slow convergence. Le and Lao [5] presented that a mnmal resdual algorthm could remedy ths problem and ths algorthm satsfes a mnmzaton property whch ensures that ths method possesses a smoothly convergence. However to our best knowledge the unknown matrx wth dfferent lnear constrant of lnear matrx equatons such as Equatons ((-( has not been consdered yet. No loss of generalty we research the followng case AB B B B C (5 6

3 whch has four unknown matrces and each s requred to satsfy dfferent lnear constrant. We should pont out that the matrces A B C are expermentally occurrng n practces so they may not satsfy solvablty condtons. Hence we should study the least squares solutons.e. Problem I. Notng that t s obvous dffcultes to solve ths problem by conventonal methods such as matrx decomposton and ver operator hence teratve method s consdered. Absorbng the thought of the mnmal resdual algorthm presented by Le and Lao [5] and combng the trat of problem we conduct an teratve method for solvng Problem I. hs method can both mantan the short recurrence and satsfy a mnmzaton property.e. the approxmaton soluton mnmzes the resdual norm of Equaton (5 over a specal affne subspace whch ensures that ths method converges smoothly. he paper s organzed as follows. In Secton we frst conduct an teratve method for solvng Problem I and then descrbe the basc propertes of ths method we also solve Problem II by usng ths teratve method. In Secton we show that the method possesses a mnmzaton property. In Secton we present numercal experments to show the effcency of the proposed method and use some conclusons n Secton 5 to end our paper.. he Iteratve Method for Solvng Problem I and II In ths secton we frstly ntroduce some lemmas whch are requred for solvng Problem I we then conduct an teratve method to obtan the soluton of Problem I. We show that for any ntal matrx group ( ( ( ( ( k ( k ( k ( k K the matrx group sequences generated by the teratve method converge to a soluton of Problem I wthn fnte teraton steps n the absence of roundoff errors. We also show that the unque least norm soluton of Problem I can be obtaned by choosng a specal knd of ntal matrx group. Lemma. [6] [7]. A matrx CSR f and only f SnSn. A matrx BSR f and only f SnSn. Lemma. Suppose that a matrx R then + SnSn CSR. Suppose that a matrx SR then + SnSn BSR. Proof: Its proof s easy to obtan from Lemma. Lemma. Suppose that A R SR Y CSR Z BSR then A A+ A A Y [ A + SnASn] Y A Z A A S ( A A S Z n n. Proof: It s easy to verfy from drect computaton. Lemma. (Projecton heorem [8]. Let be a fnte dmensonal nner product space M be a subspace of and M be the orthogonal complement subspace of M. For a gven x there always exsts an m M such that x m x m m M where. s the norm assocated wth the nner product defned n. Moreover m M s the unque mnmzaton vector n M f and only f ( ( x m M e. x m M. Lemma 5. Suppose R s the resdual of matrx group K correspondng to Equaton (5.e. R C AB A B AB A B f the followng condtons are satsfed smultaneously A RB A RB + BR A A RB (6 + Sn A RB Sn A RB + B R A + S A RB + B R A S ( n n 6

4 then the matrx group s a soluton of Problem I. Proof: Let obvously Z s a lnear subspace of { [ ] } L Z Z AB B B B K K denote p q R. For matrx group C AB B B B then C L. Applyng to Lemma we know that s a soluton of Problem I f and only f.e. for all [ ] K ( C C L R AB B B B. By Lemma t s easy to verfy that f the equatons of (6 are satsfed smultaneously the expresson above holds whch means s a soluton of Problem I. Lemma 6. Suppose that matrx group s a soluton of Problem I then arbtrary matrx group [ ] SE can be express as where matrx group K satsfes AB B B B. (7 Proof: Assume that matrx group s a soluton of Problem I. If S then by Lemma 5 and ts proof process we have E AB B B B C ( ( ( ( A + B + A + B + A + B + A + B C AB B B B R AB B B B + R whch mples matrx group Conversely f matrx group [ ] satsfes (7. Y Y Y Y K where matrx group K satsfes (7 then AYB + AYB + AYB + AYB C AB B B B R AB B B B C whch means matrx group [ ] Y Y Y Y S E. satsfes dfferent lnear con- Next we develop teratve algorthm for the least-squares solutons wth strant of matrx equaton p n where A R n q B R the unknown matrces group to be solved. AB B B B C ( and C are gven constant matrces and [ ] K s 6

5 Algorthm. For an arbtrary ntal matrx group Step. ( r r r r R C A B P ARB r r ( ( ( ( P ( ARB + BR A P ( ARB + Sn ARBS n P ARB + BR A + S ARB + BR A S Q P r. ( n ( n K compute Step. If Pkr then stop else k: k r + and compute Step. M AQ B k r kr r r P + Q r kr r ( k+ ( k k r r k kr M k R R M k+ k k k P AR B P AMB k+ k+ k k k k Pk+ ( A Rk+ B + BRk+ A Pk A MkB + B Mk A Pk+ ( AR k+ B + Sn AR k+ BS n k Pk AM k B + Sn AMBS k n Pk+ ( AR k+ B + BR k+ A + Sn ( AR k+ B + BR k+ A Sn k Pk AM kb + BM k A + Sn ( AM kb + BM k A Sn r kr Pk+ r r βk Q k+ r Pk+ r + βkqk r r P Step. Go to step. Remark. Obvously matrces sequence ( r r r ( P Q r generated by Algorthm satsfes P R P SR P CSR P BSR n n n n n n n n ( n n ( n n ( n n ( n n Q R Q SR Q CSR Q BSR R SR CSR BSR R s the resdual of Equaton (5 when Algorthm mples that f Pkr ( r then the correspondng matrx group ( k ( k ( k ( k s the soluton of Problem I. 6

6 In the next part we wll show the basc propertes of teraton method by nducton. Frst for convenence of dscusso the later context we ntroduce the followng conclusons from Algorthm. For all jt ( j r tr r r j r tr r r R R A P B A R R B P A ( R Rj B Pt + A ( R Rj B + B( R Rj A Pt + A ( R Rj B + Sn A ( R Rj B Sn P t + A ( R R B + B ( R R A + S A ( R R B + B ( R R A S P r P r j j n j j n t P + P P. tr jr tr r Lemma 7. For matrces P r Q r (r and M generated by Algorthm f there exst a postve nteger k such that Pr r and for all k then we have r jr r P P j k j M M j k j j r P Q j < k. r jr Proof: For k t follows that r Assume that the conclusons hold for all j s ( s k P P P P AMB + P P AMB + BM A ( n n ( r r + P P AMB + S AMBS + P P AM B + BM A + S ( AM B + BM A S n n P r AP r rbr M P r r r ( β r M M. r r + r r r r r + β r r M M M A P Q B M A P B M + +. R R AP r rbr β M P r β M r r P Q P P. r r r r r r < < then P P M M P Q jr sr j s sr jr r r 6

7 r s Pjr Ps+ r Pj Ps samb s + Pj Ps ( AMB s + BM sa s + Pj Ps ( AMB s + Sn AMBS s n + P P AMB + BM A + S ( AMB + BM A S s j s s s n s s n s Pjr AMB r s r s Ar Qjr r r ( Q β M β M M. s j j j s B M j j r r s ( β M M M A P + Q B M A P B r j s+ j r s+ r s s r r j r s+ r r r r R R AP B P P P P j Pj+ r Ps+ r. j j+ r s+ r r j r s+ r j+ r s+ r r j r r j r s Qjr Ps+ r Qj Ps samb s + Qj Ps AMB s + BMsA s + Qj Ps AMB s Sn AMBS s n + + Q P AMB + BM A + S ( AMB + BM A S By the assumpton of Equaton ( we have hen for j s r k j s s s n s s n AQ B M M M. s r jr r s s j s r Qsr Psr Psr + βs Qs r Psr Psr r r r. P P P P AMB + P P AMB + BM A ( s s s s n s n ( s sr s+ r s s s s s s s s + P P AMB + S AMBS + P P AMB + BM A + S ( AMB + BM A S s s s s s n s s n P AP B M s s r sr r s r r Ps s Ar ( Qsr βs Qs r Br Ms r r r s s Ms P. ( β s s+ s r s+ r + s s r r βs s + s r s+ r r r r M M M A P Q B M M A P B β +. s Ms Rs Rs+ AP r s+ rbr βs Ms Ps+ r Ps+ r s r s r 65

8 r Q P Q P AMB + Q P AMB + BM A ( s s s s n s n ( s sr s+ r s s s s s s s s + Q P AMB + S AMBS + Q P AMB + BM A + S ( AMB + BM A S s s s s s n s s n Qsr Psr s Qsr AMB r s r r r r sr s Ms P. P sr P s r r + and the assumpton jr sr show that hen the concluson P P r M j M s + for all j s. By the prncpal of nducton we know that Eq.( holds for all j < k and Equaton ( and Equaton ( hold for all j k j due to the fact that AB BA m n holds for all matrces A and B n R. Lemma 7. shows that the matrx sequence P P P P P P P P r nr r nr generated by Algorthm are orthogonal each other n the fnte dmenson matrx space R. Hence the teratve method wll be termnated at most n r r steps n the absence of roundoff errors. It s worth to note that the conclusons of Lemma 7 may not be true wthout the assumptons and. Hence t s necessary to consder the case that or. If whch mples P r r t follows that P r r. If whch mples M then we have AQ r r Br makng nner product wth R r by both sde yelds AQ r r Br R Qr Ar RB r Qr Pr Pr r r r r. So the dscussons above show that f there exst a postve nteger such that the coeffcent or ( ( ( ( then the correspondng matrx group K s just the soluton of Problem I. ogether wth Lemma 7 and the dscusson about the coeffcent we can conclude the followng theorem. ( ( ( ( heorem. For an arbtrary ntal matrx group K the matrx group sequence ( k ( k ( k ( k K generated by Algorthm wll converge to a soluton of Problem I at nfnte teraton steps n exact arthmetc. By choosng a specal knd of ntal matrx group we can obtan the unque least norm of Problem I. o ths end we frst defne a matrx set as follows A HB ( A HB + BH A S [ ] ( A HB + S n A HB S n ( A HB + BH A + Sn ( A HB + BH A Sn p q where H R. Evdently S s a lnear subspace of K. ( ( ( ( heorem. If we choose the ntal matrx group S especally let 66

9 ( ( we can obtan the least norm soluton of Problem I. ( ( ( ( Proof: By the Algorthm and heorem f we choosng ntal matrx group S we can obtan the soluton ˆ ˆ ˆ ˆ of Problem I wth fnte teraton steps and there exst a matrx p q H R such that the soluton ˆ ˆ ˆ ˆ can be represented that ˆ A HB ˆ ( A HB + BH A ˆ ( A HB + Sn A HB Sn ˆ ( A HB + BH A + Sn ( A HB + BH A Sn. By Lemma 6 we know that arbtrary soluton of Problem I can be express as ˆ ˆ ˆ ˆ where matrx group K satsfes (7. hen So we have r ˆ r r A HB + A HB + BH A + AHB + Sn A HB Sn + A HB + B H A + S ( A HB + B H A S n n H A B. r r r r ˆ + ˆ + r r r r r r r whch mples that matrx group ˆ ˆ ˆ ˆ s the least norm soluton of Problem I. Remark. Snce the soluton of Problem I s no empty so the S s a closed convex lnear subspace hence t s certan that the least norm soluton group ˆ ˆ ˆ ˆ of Problem I s unque and ˆ ˆ ˆ ˆ S. If matrx group [ ] S s a soluton of Problem I then t just be the ˆ. unque least norm soluton of Problem I.e. (. he Mnmzaton Property of Iteratve Method In ths secton the mnmzaton property of Algorthm s characterzed whch ensures the Algorthm converges smoothly. ( ( ( ( heorem. For an arbtrary ntal matrx group K the matrx group ( k ( k ( k ( k generated by Algorthm at the kth teraton step satsfes the followng mnmzaton problem ( k ( k mn [ ] F A B + B C A B + B C where F denote a affne subspace whch has the followng form E 67

10 ( ( { k k } F + span Q Q Q Q Q Q. k Proof: For arbtrary matrx group [ ] F there exst a set of real number { t } Denote ( ( [ ] k + t Q Q. k k ( ( g( t tk A + tq B + + A + tq B C by the concluson Equaton ( n Lemma 7 we have ( k ( ( ( g t t A B + + A B C + t AQ B + + AQ B k k k ( ( ( ( R + t AQ B + + AQ B t AQ B + + AQ B R where R s the correspondng resdual of ntal matrx group the matrx R can be express as Because g( t t k ( R R + A Q rb + A Q rb + + A Q rb. r r r r r r r r r s a contnuous and dfferentable functon wth respect to the k varable t k we easly know that f and only f It follows from the concluso Lemma 7 that By the fact that ( g t t t mn k ( g t t t k t. AQ B R AQ B R Q P P t. r r r r r r r r r r r r r AQ B AQ B AQ B AQ B r r r r r r r r r r r r r r r r ( mn t t tk mn A B + B C. t [ ] F such that. Algorthm show that We complete the proof. ( k ( k ( k ( k heorem shows that the approxmaton soluton mnmzes the resdual norm n the affne subspace F for all ntal matrx group wthn K. Furthermore by the fact ( k ( k ( k ( k F then we have whch shows that the sequence ( k ( k ( k ( k A B + B C A B + B C ( ( ( ( A B + B C A B + B C s monotoncally decreasng. he descent property of the resdual norm of Equaton (5 ensures that the Algorthm possesses fast and smoothly convergence. 68

11 . Numercal Examples In ths secton we present numercal examples to llustrate the effcency of the proposed teraton method. All the tests are performed usng Matlab 7. whch has a machne precson of around 6. Because of the error of calculaton the teraton wll not stop wthn fnte steps. Hence we regard the approxmaton soluton group ( k ( k ( k ( k as the soluton of Problem I f the correspondng P kr ( r satsfes Pkr e. r Example. Gven matrces A B A B A B A B and C as follows: A 6 B A B B A A

12 Choose the ntal matrces B C ( ( where denotes zero matrx n approprate dmenson. Usng Algorthm and teratng 7 steps we have the unque least norm soluton follows: ( 7 ( 7 ( 7 ( ( 7 ( 7 ( 7 ( 7 K as 7

13 wth And ( 7 P7 r e r r.778 and.68. mn AB B B B C [ ] K If we let the ntal matrx ( I ( 5 I ( I ( I wthn K but not wthn S then we have wth And ( 8 ( 8 ( 8 ( 8 notng that [ I 5 I I I ] ( 8 ( 7 P 8 r e r r r 5.97 and.58 >

14 mn AB B B B C [ ] K Example. Suppose that the matrces A B A B A B A B are the same as Example let C AB B B B where I 5I I I that s to say 6 Equaton (5 s consstent over set K. hen smlarly Algorthm. n Peng 8 [] we 7 can conduct 8 another teraton algorthm as follows: ( ( ( ( Algorthm. For an arbtrary ntal matrx group K compute Step. ( r r r r R C A B P ARB r r P ( ARB + BR A P ( ARB + Sn ARBS n P ARB + BR A + S ARB + BR A S Q P r. ( n ( n Step. If R k then stop else k: k+ and compute Step. M AQ B k r kr r r Rk ( k+ ( k k r r k kr + Q r Q kr r R R M k+ k k k P AR B P AMB k+ k+ k k k k Pk+ ( A Rk+ B + BRk+ A Pk A MkB + B Mk A Pk+ ( AR k+ B + Sn AR k+ BS n k Pk AM kb Sn AMBS + k n Pk+ ( AR k+ B + BR k+ A + Sn ( AR k+ B + BR k+ A Sn k Pk AM kb + BM k A + Sn ( AM kb + BM k A Sn Rk + k k+ r k+ r βk k r Rk β Q P + Q r Step. Go to step. he man dfferences of Algorthm and Algorthm are: n Algorthm the selecton of coeffcent make R k + mn and β k such that of k such that R k+ Rk and β k such that AQ r k rbr AQ r r r k rb + r but n Algorthm the choosng Q k r Qkr r + k. Notng that Algorthm satsfes the Galerkn condton but lacks of mnmzaton property. Choosng the ntal matrx where denotes zero matrx n approprate dmenson by makng use of Algorthm and Algorthm we can ( ( 7

15 Fgure. he comparson of resdual norm between these two algorthm. obtan the same least norm soluton group and we also obtan the convergence curves of resdual norm show Fgure. he results n ths fgure show clearly that the resdual norm of Algorthm s monotoncally decreasng whch s n accordance wth the theory establshed n ths paper and the convergence curve s more smooth than that n Algorthm. Acknowledgements We thank the Edtor and the referee for ther comments. Research supported by the Natonal Natural Scence Foundaton of Chna ( References [] Baksalary J.K. and Kala R. (98 he Matrx Equaton AB + CYD E. Lnear Algebra and Its Applcatons [] Chu K.E. (987 Sngular Value and Generlzed Value Decompostons and the Soluton of Lnear Matrx Equatons. Lnear Algebra and Its Applcatons [] Chu K.E. (989 Symmetrc Solutons of Lnear Matrx Equaton by Matrx Decompostons. Lnear Algebra and Its Applcatons [] Peng Z.Y. ( he Solutons of Matrx AC + BYD E and Its Optmal Approxmaton. Mathematcs: heory & Applcatons 99-. [5] Lao A.P. Ba Z.Z. and Le Y. (5 Best Approxmaton Soluton of Matrx Equaton AB + CYD E. SIAM Journal on Matrx Analyss and Applcatons [6] u G. We M. and Zheng D. (998 On the Solutons of Matrx Equaton AB + CYD F. Lnear Algebra and Its Applcatons [7] Chang.W. and Wang J.S. (99 he Symmetrc Soluton of the Matrx Equatons A + YA C AA + BYB C and ( A A B B ( C D. Lnear Algebra and Its Applcatons [8] Zetak K. (98 he lp -Soluton of the Lnear Matrx Equaton A + YB C. Computng

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