Conrgnc Thors for Two Itrt Mthods A sttonry trt thod for solng th lnr syst: Ax = b (.) ploys n trton trx B nd constnt ctor c so tht for gn strtng stt x of x for = 2... x Bx c + = +. (.2) For such n trton to conrg to th soluton x t ust b consstnt wth th orgnl lnr syst nd t ust conrg. To b consstnt w sply nd for x to b fxd pont tht s: x = Bx + c. (.3) Snc tht s qulnt to ( I Bx ) = c th consstnc condton cn b sttd ndpndnt of x by syng AI ( B) c= b. (.4) Th sst wy to dlop consstnt sttonry trt thod s to splt th trx A : A = M + N (.5) thn rwrt Ax = b s Th trton wll thn b Mx Mx= Nx+ b. (.6) + = Nx + b. (.7) Rcstng ths n th for bo w h B M = N nd c= M b. It s sy to show tht ths trton s consstnt for ny splttng s long s M s nonsngulr. Obously to b prctcl th trx M ust b slctd so tht th syst My = d s sly sold. Populr chocs for M r dgonl trcs (s n th Jcob thod) lowr trngulr trcs (s n th Guss-Sdl nd SOR thods) nd trdgonl trcs. Conrgnc: Thus constructng consstnt trtons s sy th dffcult ssu s constructng conrgnt consstnt trtons. Howr notc tht f s quton (.3) subtrctd fro quton (.2) w obtn whr s th rror x x. + = B Our frst rsult on conrgnc follows dtly fro ths. Thor : (.8)
Th sttonry trt thod for solng th lnr syst: x + = Bx + c for = 2... conrgs for ny ntl crtor ctor nor x f B < for so trx nor tht s consstnt wth Proof: Lt. b trx nor consstnt wth ctor nor. nd such tht B <. W thn h + = B B nd spl nduct rgunt shows tht n gnrl Snc B < ust conrg to zro (nd thus (.9) B. (.) x conrg to x ) ndpndnt of. Ths thor prods suffcnt condton for conrgnc. Wthout proof w offr ths thor tht prods both ncssry nd suffcnt condtons for conrgnc. It ploys th spctrl rdus of trx: ρ ( A) = th bsolut lu of th lrgst gnlu of A n bsolut lu. Thor 2: Th sttonry trt thod for solng th lnr syst: x + = Bx + c for = 2... conrgs for ny ntl ctor x f nd only f ( B) ρ <. Th sst wy to pro ths uss th Jordn Norl For of th trx B. Notc tht th thor dos not sy tht f ρ( B) th trton wll not conrg. It sys tht f ρ( B) th trton wll not conrg for so ntl ctor x. In prctcl trs though th dffrnc s nor: th only wy to h conrgnc wth ρ( B) s to h n ntl rror hng no coponnt n ny drcton of n gnctor of B corrspondng to n gnlu t lst on n bsolut lu. Ths s probblty zro nt. Th followng thor uss Thor to show th Jcob trton conrgs f th trx s strctly row dgonlly donnt. Rcll tht Jcob trton s + x = ( b x for = 2...n (.) )/ nd tht strct row dgonl donnc sys tht < for = 2... n. (.2)
Th splttng for th Jcob thod s A = D+ ( L+ U) whr DL nd U r th dgonl strct lowr trngl nd strct uppr trngl of th trx rspctly. Thus th trton trx s D ( L+ U). Thor 3: Th Jcob trt thod + x = ( b x for = 2... n )/ for solng th lnr syst Ax = b conrgs for ny ntl ctor strctly row dgonlly donnt. x f th trx A s Proof: Lt. ndct th nfnty ctor nor s wll s ts subordnt trx nor. To pro th thor t suffcs to show D ( L+ U) bsolut lus of th trx D ( L+ U). Ths r <. To tht nd consdr th row sus n but proprty (.2) gur- nts tht ths s strctly lss thn on. Th xu of th row sus n bsolut lu s lso strctly lss thn on so D ( L+ U) < s wll. Th nxt thor uss Thor 2 to show th Guss-Sdl trton lso conrgs f th trx s strctly row dgonlly donnt. Rcll tht Guss-Sdl trton s x b x x = (.3) + = ( + )/ for 2...n < > Th splttng for th Guss-Sdl thod s A= ( L+ D) + U. Thus th trton trx s ( L+ D) U.
Thor 4: Proof: Th Guss-Sdl trt thod x + = ( b x + x )/ for = 2... n < > for solng th lnr syst Ax= b conrgs for ny ntl ctor strctly row dgonlly donnt. Accordng to Thor 2 t suffcs to show gnctor corrspondng to n gnlu λ so = ρ( ( L+ D) U). W shll show x f th trx A s ρ( ( L+ D) U) <. To tht nd lt b ny λ of ( L+ D) U such λ < nd thus ρ ( (L+ D) U) <. W h U = λ( L + D) (.4) ( L+ D) U= λ. (.5) In coponnt fshon ths sys = λ. (.6) > Lt dnot n ndx of corrspondng to th lrgst coponnt n bsolut lu. Tht s = x{ } (.7) so W lso h for row n prtculr. (.8) > > = λ Ddng by th ncssrly post lus nd = λ + < λ < λ w h <
so But snc nd λ λ > > < < It s sy to show tht < t follows tht > < λ > < > = + < > > λ <. <. (.9) (.2) < x so th bound on th spctrl rdus trton trx of th Guss-Sdl thod s strctly lss thn th bound of th nfnty nor of th trton trx of th Jcob thod. Tht dos not gurnt tht th Guss-Sdl trton lwys conrgs fstr thn th Jcob trton. Howr t s oftn obsrd n prctc tht Guss-Sdl trton conrgs bout twc s fst s th Jcob trton. To s ths gn tht. Cll ths quntty 2 θ. W h θ > nd f θ s sll thn > < > < 4θ. Yt ( θ ) + ( θ) = 2θ nd f w gn for x 2 2 thn our bound for th nor of th Jcob trton trx s 2θ whl our bound on th spctrl rdus trton trx of th Guss-Sdl thod s 4θ.
Notc tht f th trton conrgs s to so tolrnc ε rqurs lu of σ for so fctor of bout σ thn to rduc ln ε. If lnσ σ thn ln ε ln ε lnε ln σ ( σ ) so w stt bout stps. Wth Jcob w h ( σ ) σ 2θ ln ε lnε but wth Guss-Sdl w h whch ustfs th cl tht Jcob conrgs twc s σ 4θ fst. Lstly wthout proof w stt nothr thor for conrgnc of th Guss-Sdl trton. Thor 5: Th Guss-Sdl trt thod x + = ( b x + x )/ for = 2... n < > for solng th lnr syst Ax sytrc nd post dfnt. = b conrgs for ny ntl ctor x f th trx A s