INTRODUCTION ( ) 1. Errors

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1 INTRODUCTION Numercl lyss volves the study, developmet d lyss of lgorthms for obtg umercl solutos to vrous mthemtcl problems. Frequetly umercl lyss s clled the mthemtcs of scetfc computg. Numercl lyss s the developmet d study of procedures for solvg problems wth computer. The rt d scece of preprg d solvg scetfc d egeerg problems hve udergoe cosderble chges due to the vlble dgtl computg systems. Dgtl computers re the prcpl mes of clculto umercl lyss d cosequetly t s very mportt to uderstd how they operte. A computer hs fte word legth d so oly fed umber of dgts re stored d used durg computto.. Errors Eve storg ect decml umber ts coverted form the computer memory, error s troduced. Ths error s mche depedet. Also t the ed of computto of prtculr problem, the fl result the computer should be coverted to form uderstdble to the user. Therefore ddtol error s commtted t ths stge too. Ths error s clled locl roud off error. Thus we defe Error = True Vlue Computed Vlue wys. I order to determe the ccurcy of ppromte soluto, errors re mesured dfferet Defto : Absolute error = error Defto 3 : Reltve error = Error True Vlue Defto 4 : Roud Off Error : s the qutty R whch must be dded to the fte represetto of computed umber order to me t the true represetto of tht umber. Whe umber N s wrtte flotg pot form wth t dgts, sy, bse s, N ddd 3 d t =... e, d

2 We sy tht the umber N hs t sgfct dgts. For emple,.3 grees wth 3 to oe sgfct dgt. The roud off error for ths represetto wll be.3 3. All the errors defed bove re mche errors d c be mmzed by usg computg ds of hgher precso. () Methemtclly, umercl lyss we usully come cross two types of errors. Iheret Errors It s tht qutty of error whch s preset the sttemet of the problem tself, before fdg ts soluto. It rses due to the smplfed ssumptos mde the mthemtcl modellg of problem. It c lso rse whe the dt s obted from cert physcl mesuremets of the prmeters of the problem. () Tructo Errors These re errors cused by usg ppromte formule computtos. e.g. whe fucto f ( ) s evluted from fte seres, f we use oly frst few terms of the seres to compute vlue of fucto f ( ), we get ppromte swer. Here, the error s due to tructg the seres. Suppose f ( ) = cos. The 4 6 f ( ) = ( ) +...! 4! 6! ( )! If we ret the frst terms, the tructo error (TE) s TE = ( ) + ( ) + ( )... (+ )! (+ 4)! (+ 6)! The study of ths type of error s ssocted wth the problem of covergece. Some specl termology s used to descrbe the rpdty wth whch sequece coverges. Bg O d Lttle o Notto : Let { } d { } The equto o( α ) α be two dfferet sequeces. lm = α = (we sy s lttle oh of α )

3 To vod dvso by zero, we sy tht = o α f ε α d ε s. We wrte o( α ) such tht Cα whe. = (we sy s bg oh of α ) f there s costt C d umber These two ottos gve corse method of comprg two sequeces. They re frequetly used whe both sequeces coverge to. If, to t lest s rpdly s more rpdly th α does. α does. If, α d o( α ) α d o( α ) =, the coverges = the coverges to Defto : The tructo error s the qutty T whch must be dded to the true represetto of the qutty order tht the result s ectly equl to the qutty we re seeg to geerte.. Stblty Numercl Alyss A umber of mthemtcl problems hve solutos tht re qute sestve to smll computtol errors, for emple roud off error. To del wth ths pheomeo, we troduce the cocept of stblty. A umercl method for solvg mthemtcl problem s cosdered stble f the sestvty of the umercl swer to the dt s o greter th the orgl mthemtcl problem. A umercl method s sd to be stble f the effect of y sgle fed roud off error s bouded. 3. Problem Solvg Usg Computers I order to solve gve problem usg computer the mjor steps volved re - () () () (v) Choosg pproprte umercl method Desgg lgorthm Progrmmg Computer Eecuto. I Ut to 4 we dscuss vrous umercl methods (d ther lyss) for solvg trscedetl d polyoml equtos, system of ler equtos, dfferetl equtos, umercl methods vlble to terpolte d ppromte fuctos, tegrto d evluto of ege vlues d ege vectors of symmetrc mtrces. 3

4 UNIT - I TRANSCENDENTAL AND POLYNOMIAL EQUATIONS Itroducto : Oe of the bsc problems scece d egeerg s the computto of roots of equto f ( ) =. The equto f ( ) = s clled lgebrc or polyoml equto f t s purely polyoml. It s clled trscedetl equto f f ( ) cots trgometrc, epoetl or logrthmc fuctos. Defto.. : the -s. A umber ξ s soluto of f ( ) = f f ( ξ ) =. Such ξ s clled root or zero of f ( ) =. Geometrclly, root of f ( ) = s the vlue of t whch the grph of y = f( ) tersects Defto.. : If we c wrte ( ξ ) m f ( ) = g( ) where g( ) s bouded d g ( ξ ) the ξ s ( m ) clled multple root of multplcty m. I ths cse f ( ξ) f ( ξ) f ( ξ) f ( ξ) m d f ( ξ ) for m = the root ξ s clled smple root. () The followg re the bsc propertes of polyoml equto. = ' = " =... = =. Every polyoml equto of th degree, where s postve teger hs ectly roots. () Comple roots occur prs.e. f + b s root of f ( ) =, so s b. () If = s root of f ( ) =, polyoml of degree the f ( ) = ( ) g ( ) where g s polyoml of degree ( ). (v) Descrtes Rule of Sgs : The umber of postve roots of polyoml equto f ( ) = wth rel coeffcets cot eceed the umber of chges sg of the coeffcets the polyoml f ( ) =. Smlrly, the umber of egtve roots of f ( ) = cot eceed the umber of chges the sg of the coeffcets 4

5 3 of f ( ) =. For emple, cosder f ( ) = =. The coeffcets of ths equto re (, 3, 4, 5). As there re three chges sg, the gve equto wll hve t the most three postve roots. (v) Itermedte Vlue Property : If f ( ) s rel vlued cotuous fucto the closed tervl b the fucto tes ech vlue betwee f ( ) d f () b. I prtculr f f ( ) d f () b hve opposte sgs, the the grph of fucto y = f( ) crosses the -s t lest oce..e. f ( ) = hs t lest oe root betwee d b..e. f ξ =, < ξ < b There re geerlly two types of methods used to fd roots of f ( ) =. () Drect Methods : These methods gve the ect vlue of the roots fte umber of steps. Further the methods gve ll the roots t the sme tme. These methods requre o owledge of the tl ppromto of root of the equto f ( ) =. e.g. solutos of polyoml equto re ow for polyomls of degree upto cubc..e. for + =, = for + + =, 4 ± = () Iterrtve Methods These methods re bsed o the de of successve ppromtos.e. strtg wth oe or more tl ppromtos to the root, we obt sequece of ppromte solutos whch coverges to root of gve equto. I the et secto we descrbe some umercl methods for the soluto of equto f ( ) =. 5

6 . Bsecto Method Ths method s due to Bolzo. Step : Choose d such tht f f < suppose <. (By termedte vlue prcple root les betwee d ) Defe I (, ) =. Step : The desred root s ppromtely defed by If f ( ) = the s the desred root of f ( ) =. f f. If f ( ), clculte If f ( ) f ( ) < the defe I (, ) =. Otherwse defe I = (, ) d f f <. + =. Step 3 : Defe I = I d go to step. Thus t ech terto we ether fd the desred root to the requred ccurcy or rrow the legth of tervl to hlf of the legth of tervl t prevous step. Ths process s cotued to determe smller d smller tervl wth whch the desred root les. If he permssble error s ε, the the ppromte o. of tertos () requred my be determed from the relto ε Note : The o. of tertos requred to cheve requred ccurcy depeds upo the tl tervl I. If the legth of I s suffcetly smll we wll rech t the soluto less o. of tertos. EXAMPLES : Fd rel root of the equto, f ( ) = = Aswer : Step : Sce f () = < d f () = 5>, the root les betwee d..e. I (, ) (,) = =. Step : +.5 = =, f = f.5 = 6

7 .3 Iterto Methods Bsed o Frst Degree Equto Although the bsecto method s esy to compute t s ot very effcet. For most fuctos we c mprove the speed t whch the root s pproch through dfferet schemes. Almost every fuctos c be ppromted by strght le over smll tervl. We beg from vlue tht s er to root. Ths tl vlue c be obted by loog t the grph of fucto or from few tertos of bsecto method. Iterto methods re obted by ppromtg f ( ) by polyoml of degree oe the eghbourhood of root. Thus = + = =, f ( )... (.3..) The prmeters d re to be determed by prescrbg two ppromte codtos o f ( ) d / or ts dervtves..3. Sect Method Suppose d re two ppromtos to the root, the we determe d by usg ler ppromto. ( ) = + f f = + O solvg bove two equtos smulteously for d we get = f f d = f f from equto (.3.) we get, the et ppromte root + whch my be wrtte s ( ) f f = f f f + = f f where f = f ( ) d f f ( ), =,, 3, (.3.) =. Ths s clled the Sect or the Chord Method. 7

8 Geometrclly, ths method we choose two pots o the curve d plot the le pssg through these two pots. The pot of tersecto of the strght le wth the -s s the et ppromto to the root (Fg..). 3 Fg...3. Regul Fls Method Ths s the oldest method for fdg the rel root of equto f ( ) = d closely resembles bsecto method. Ths method s lso clled s method of flse posto. I ths method we choose two pots d such tht f ( ) d f ( ) re of opposte sgs. Sce the grph of y = f( ) crosses the -s betwee these two pots, root must le betwee these pots. Now the equto of the Chord jog the two pots, f ( ),, f f y f ( ) = 8 f s The pot of tersecto of the chord wth the -s s gve by puttg y =. Thus we get, ( ) f f f = O solvg bove equto for we obt f ( ) = f f

9 Hece the secod ppromto to the root of f ( ) = s gve by f ( ) = f f... (.3..) If ow f ( ) d f re of opposte sgs the the root les betwee d d we replce by (.3..) d obt the et ppromto. Otherwse we replce by d geerte et ppromto. The procedure s repeted tll the root s obted to the desred ccurcy. The repeted pplcto of ths procedure geertes sequece. (, f ( )) 3 (, f ( )) (, f ( )) Frst terto Suppose the ppromte soluto fter ( ) tertos s deoted by. The the sequece { } pproches to the root ξ s.e. f ( ξ ) =..3.3 Newto Rhso Method Ths method s geerlly used to mprove the result obted by oe of the prevous methods. Frstly we derve ths method by usg ler ppromto. Suppose s pot the eghbourhood of the root of f ( ) =. If we ppromte f ( ) by polyoml of degree oe we get f ( ) = +. ( ) f = + f ' =... (.3.3.) d ( ) 9

10 where prme deotes dfferetto wth respect to. O solvg for s d we get = d = f ' From equto (.3.) we get, = ' f '( ) f f f f ' = f f ( ) '( ) Thus we get the et ppromte root s f ( ) '( ) + = f, =,,, 3, (3.3.) Ths method s clled the Newto Rphso method. The method (.3.3.) my lso be obted drectly from Sect method (.3..) by tg the lmt. I the lmtg process.e. whe the chord pssg through the pots, tget t pot, f d, f ( ) coverges to the f. Thus ths cse the problem of fdg root of equto f ( ) = s equvlet to fdg the pot of teersecto of the tget to the curve y = f( ) t the pot (, f ( )) wth the -s. The Newto Rphso method requres two vlues f ( ) d ' The method s pplcble oly whe f '.e. root s smple root. f. The method c lso be derved by usg Tylor seres represetto. Let be ppromte root of f ( ) = d let h = + be the correct root so tht f = f + h by Tylor seres bout, we obt h f ( ) = f ( + h) = f ( ) + hf '( ) + f "( ) +... = Neglectg the secod d hgher order dervtves we hve.e. f + hf ' = f h = f ( ) '( ) f =. Epdg

11 Therefore = + h= f f ( ) '( ) s the better ppromto th. Successve ppromtos re gve by, 3,... where f ( ) '( ) + = f, =,,, 3,... Ths s sme s the formul (.3.3.)..4 Rte of Covergece All the methods descrbed secto.3 re tertve methods d repettve pplcto of these methods geerte sequece of ppromte solutos. Covergece of ths sequece s mportt subject tht we wll dscuss ow..4. Orders of Covergece Some specl termology s used todescrbe the rpdty wth whch sequece covergeces. Let { } be sequece of rel umbers tedg to lmt *. We sy th the rte of covergece s t lest ler f there s costt c < d teger N such tht + * C * ( N)... (.4..) The covergece s tlest qudrtr C f there re costt C (ot ecessry less th ) d teger N such tht costt. ε + * C * ( N)... (.4..) I geerl f there re postve costts C, lrgest α d teger N such tht * C * α + ( N)... (.4..3) We sy tht the rte of covergece s tlest α. The costt C s clled the symptotc error If s ppromte root of f ( ) = d ξ s soluto of equto f ( ) = the = ξ s the error the soluto. If the sequece ε s, we sy tht tertve methods dscussed secto.3 re coverget. We ssume tht ξ s smple root of f ( ) = so tht f '( ξ ).

12 .4. Rte of covergece of Sect Method Suppose ξ s smple root of f ( ) =.e. f ( ξ ) =. ε = ξ s error. O substtutg ξ ε ε + = ε f ( ε ε ) f ( ξ + ε) ( ξ + ε ) f ( ξ + ε ) = + (3..) we get, Epdg f ( ξ+ ε ) d f ( ξ ε ) f ( ξ ) = we get, ε + = ε = ε + Tylor seres bout the root ξ d observg tht ε ' + " +...! ' + " +... ( ε ε ) ε f ( ξ) f ( ξ) ( ε ε ) f ( ξ) ( ε ε ) f ( ξ) ε ' + " +...! ' + + " +... ( ε ε ) ε f ( ξ) f ( ξ) ( ε ε ) f ( ξ) ( ε ε ) f ( ξ) ( ξ) ( ξ) ( ξ) ( ξ) ε f " f " = ε ε ( ε ε )...! f ' f ' ( ξ) ( ξ) ( ξ) ( ξ) ε f " f " = ε ε ( ε ε )...! f ' + + f ' 3 = ( ξ) ( ξ) ( ξ) ( ξ) f " ε f " = ε ε ε ( ε + ε ) f '! f ' ( ξ ) ( ξ ) f "( ξ ) ε ( ε + ε ) +... f '( ξ ) 3 f " = εε + ε ε + ε f '

13 Thus we wrte ε+ C εε = where f "( ξ ) C f '( ξ ) = d we gore hgher powers of ε. The equto ε+ = C εε... (.4..) s clled error equto. To determe the order of coverece dscussed secto.4., we hve to determe the α umber α such tht ε+ = A ε where A d α re to be determed. If we replce by we get, ε = A α ε.e. ε = A α ε α Substtutg the vlues of ε d ε + equto (.4..) we get, α = α α Aε C ε A ε ε α CA α + = ε α Comprg the powers of ε o both sdes of bove equto we get, α = + α whch s qudrtc equto α d we get α = ± 5. The hghest vlue of α = ( + 5 ) d we fd tht the rte of covergece of sect method s α =.68 d A C α + α =..4.3 Rte of Covergece of Regul Fls Method If the fucto f ( ) the equto f = s cove the tervl (, ) tht cots the root, the oe of the pots or s lwys fed d the other pot vres wth. If the pot s fed, the the fucto f ( ) s ppromted by the strght le pssg through the pots (, f ( )) d, f, =,, 3,... 3

14 Suppose ξ s smple root of f ( ) =.e. ppromte soluto. Sce the pot, f ξ = d ε = ξ s error f s fed we c wrte = + ( )( ) f f f ξ + ε = ξ + ε + f ( ξ + ε) ξ + ε ( ξ + ε) f ( ξ + ε ) f ( ξ + ε ) ε ε f ' ξ " ξ... ε ε + f + ( )! ε+ = ε = ε ε ' ε ε f ( ξ) + f "( ξ) +... ε f '( ξ) + f "( ξ) +...!! f ( ε ε ) f ( ξ) ( ξ ) ( ξ ) "( ξ ) ( ξ ) ε f " ε ε ' ξ ε ! f ' ε + ε f ' ! f ' ( ξ) ( ξ) ( ξ) ( ξ) ε f " ε + ε f " = ε ε ! f ' +! f ' ( + ) f ( ξ) ε ( ξ) ε ε ε f " ξ f " ξ = +...! '! f ' ( ξ ) 3 f " ξ = εε + εε, ε f ' Sce ε = ξ s the error the frst ppromto d s depedet of, we c wrte ( ξ ) f " ξ C = ε f ' d we get error equto ε + = Cε Here C s symptotc error costt d by equto (.4..), we observe tht the Regulr Fls method hs t lest ler rte of covergece. 4

15 .4.4 Rte of Covergece of Newto Rphso Method Suppose ξ s smple root of f ( ) =.e. f ( ξ ) = but s error the ppromte soluto epdg f ( ξ+ ε ) d f '( ξ ε ) f ( ξ ) = d f '( ξ ), we obt ξ + ε = ξ + ε. O substtutg ξ ε f ' ξ. Suppose ε = ξ = + equto (.3.3.) d + Tylor seres bout the pot ξ d usg the fct tht ε ε f '( ξ) + f "( ξ) +...! ε f '( ξ) + ε f "( ξ) + f '"( ξ) +...! + ( ξ) ( ξ) ( ξ) ( ξ) ε f " f " ε+ = ε ε ε...! f ' + + f ' ( ξ) ( ξ) ( ξ) ( ξ) ε f " f " = ε ε ε...! f ' + f ' 3 ε ε... f " ξ f " ξ f " ξ = ε ε ε +! f ' ξ f ' ξ! f ' ξ ( ξ ) ( ξ ) ε f " = +! f ' 3 ( ε ) 3 O eglectg terms cotg ε d hgher powers of ε we get error equto ε + = Cε where f " C = f ' ( ξ ) ( ξ ) covergece. Thus by equto (.4..) we obsrve tht Newto Rphso method hs secod order Note : If the root ξ of f ( ) = s root of multplcty two or more the the rte of covergece for Newto Rphso method s oe. 5

16 .5 Iterto Methods To descrbe ths method for fdg the roots of equto f ( ) =... (.5.) We rewrte ths equto the form = φ( )... (.5.) Let be ppromte vlue of the desred root ξ. Substtutg t for o the rght hd sde of equto (.5.) we obt the frst ppromto =φ ( ) The successve ppromtos re the gve by = φ ( ) = φ ( ) ( ) =φ Thus we get sequece of ppromte solutos { }. The covergece of ths sequece depeds o the sutble choce of fucto φ ( ) d tl ppromto. The fucto φ ( ) s clled terto fucto. If the fucto φ s cotuous d the sequece { } coverges to * the sequece{ }. + =φ ( ) * = lm = Lt φ = φ Lt = φ * + Thus * s root of equto (.5.) f the terto fucto φ s cotuous fucto. The followg theorem gves ecessry d suffcet codto for the covergece of Theorem : If φ ( ) s cotuous fucto some tervl [, b] tht cots the root d φ '( ) C< ths tervl, the for every choce of [ b] from φ ( ) + =, =,,, 3,..., the sequece { }, determed 6

17 Coverges to the root ξ of = φ ( ). Proof : Sce ξ s root of equto = φ ( ), ξ = φ ξ... (.5.3) φ ( ) + =... (.5.4) From equto (.5.3) d (.5.4) we get, + ( ) ξ = φ ξ φ, =,,, 3,... Usg the me vlue theorem, we get where L( ) ξ φ θ ξ + = '( )( ) θ L( ξ, ) ξ represets le segmet jog ξ d., Smlrly, we obt = '( ) θ L( ξ, ) ξ φ θ ξ = '( ) θ L( ξ, ) ξ φ θ ξ = '( ) θ L( ξ, ) ξ φ θ ξ + Sce ech θ [ b, ], φ' ( θ ) C < d ξ C ( ξ ) + Thus, ε + + C ε Sce C <, the rght hd sde of bove equlty goes to zero s becomes lrge d t follows tht the sequece of ppromtos { }. Coverges to the root ξ f C <. Note : The root obted by ths method s uque. Suppose ξ d ξ re two dstct roots of equto (.5.)..e. ξ = φ( ξ ) d ξ φ( ξ ) The we get ξ ξ = φξ φ ξ = φ' θ ξ ξ ( ξ ξ )[ φ ( θ) ] ' = =. 7

18 But, φ' ( θ) C < φ' ( θ) ξ ξ = d therefore ξ = ξ, hece the root s uque. I geerl the speed of covergece depeds o the vlue of C; the smller the vlue of C, the fster would be the covergece. Therefore, the speed of covergece depedet upo the choce of φ ( ). There re my wys of rewrtg f ( ) = the form = φ ( ). For emple the equto 3 f ( ) = + = c be epressed s = ( + ) = φ ( ) (sy) 3 = = φ (sy) 3 = = φ3() (sy) 3 = + = φ ( ) (sy) 4 We hve to choose tht fucto φ ( ) for whch φ '( ) <. Sce f () = d f () =, we ow tht root les betwee d. 4 ' 3 φ = for [,] φ ' = 3 ( 3) 3 3 = s Observe tht the fuctos φ, φ 3, φ 4 re ot the epected choces of terto fucto, s φ ' ( ) for =, 3, 4 the tervl [, ]. If we choose φ = + the φ '( ) = ( + ) 3 d φ '( ) ( ) 3 = + < [,] m φ ' = =.7678< 8 d the terto method φ ( ) + = coverges to the root s ( ) φ ' <. 8

19 .6 Polyoml Equtos Polyoml fuctos re of specl mportce. They re everywhere cotuous, they re smooth, ther dervtves re lso cotuous d smooth d they re redly evluted. Descrte s rule of sgs predct the umber of postve roots. Polyomls re prtculrly well dpted to computers becuse the oly mthemtcl opertos they requre for evluto re ddto, subtrcto d multplcto. () () To determe the roots of polyoml equto t s ecessry to hve the followg formto The ect umber of rel d comple roots log wth ther multplcty. The tervl whch ech rel root les. By fudmetl theorem of lgebr we ow tht polyoml of degree hs ectly roots. Decrte s rule of sgs gves oly upper lmt of o. of + ve d ve rel roots. Ths rule does ot gve the ect umber of postve d egtve rel roots. The ect umber of rel roots of polyoml c be foud by Sturms theorem. Let f ( ) be polyoml of degree. Let f ( ) represet ts frst order dervtve. The remder of f ( ) dvded by f ( ) te wth the reverse sg s deoted by f ( ). Let f 3 ( ) deotes the remder of f ( ) dvded by f ( ) wth the reverse sg. Cotue ths process tll we rrve t costt. We thus obt sequece of fuctos f ( ), f ( ), f ( ),..., f ( ) Ths sequece s clled Sturm sequece. Sturm Theorem : The umber of rel roots of the equto f ( ) = o [, b] equls the dfferece betwee the umber of chges of sg the Sturm sequece t = d = b, provded tht f ( ), f () b. Sce polyoml of degree hs ectly roots, the umber of comple roots equls ( - umber of rel roots), where rel root of multplcty r s couted r tmes. If,, 3,..., re rel roots of f ( ) the f = 3... Comple roots occur pr. If, re comple roots the ( )( ) of degree two wth rel co-effcets d ths cse f p q = s polyoml 9

20 Thus t s obvous tht the methods of fdg roots of polyoml equto should clude the determto of ether ler fctor ( p) fctor ( p) or qudrtc fctor + p+ q. I ths secto two methods re preseted. Brge Vet method s used to determe the ler wheres Brstow method s used to determe the qudrtc fctor + p+ q..6. Brge-Vet Method Ths method s used to determe rel root of polyoml equto P ( ) = =... (.6..) If P s root of polyoml P the ( p) ppromte root of P ( ). If we dvde s fctor of polyoml P ( ). Suppose p s P by fctor ( p) the we get quotet Q polyoml of degree ( ) d remder. The remder R depeds o choce of p..e. f we chge the vlue of p, R wll get chge. Brge Vet method gves procedure to me R zero. Suppose p s ppromte root of P ( ). P ( ) = p Q ( ) + R... (.6..) 3 where Q ( ) = b + b + b b + b... (.6..3) 3 polyoml of degree ( ) d R s remder. The coeffcets polyoml Q.e. b d remder R re fuctos of p. Brge Vet method gves procedure to determe p such tht R( p ) =. P ( p) = p p Q ( p) + R( p) = R( p)... (.6..4) R( p ) = Equto (.6..4) s the equto vrble p d y tertve method dscussed secto.3 c be used to determe the root p. Applcto of Newto Rphso method dscussed secto.3.3 for equto (.6..4) gves p P ( p) '( p ) + = p P... (.6..5) For polyoml equto, the computto of P d P ' c be obted wth the help of sythetc dvso. O comprg the coeffcets of le powers of o both sdes of equto (.6..) d usg equtos (.6..) d (.6..3) we get, = p b + b + + b + R

21 Thus = b b = = b pp b = + pb = b pp b = + pb = b pb b = + pb I geerl = R pb R= + pb b pb wth b = d b = +, =,, 3,...,... (.6..6) = R. From equto (.6..4) we hve P ( p) = R= b... (.6..7) To determe P '( p ), dfferette (.6..6) wth respected p. db dp db = b + p... (.6..8) dp Equto (.6..8) c be wrtte s C = b + pc db where = C, =,, 3,..., dp Above equto c lso be represeted by C b pc = +, =,,,... (.6..9) db d C = = + pb = b dp dp d (Sce b = d s depedet of p, dfferetto of b wth respect to p s zero).

22 O dfferettg (.6..7) wth respect to p we get dp( p) dr db = = = C... (.6..) dp dp dp By substtutg the vlues of P ( p ) d P '( p ) from equto (.6..7) d (.6..) equto (.6..5) we get p where + b = p, =,,, (.6..) C p s tl ppromto of fctor ( p). The method (.6..) s clled the Brge Vet Method. The clcultos of the coeffcets b d C re crred out by usg sythetc dvso. p 3... pb pb pb... 3 p b b b 3... pc pc pc... 3 C C C C 3... pb pb pb b pc pc C b b = R = P ( p) dr dp ( p C ) = dp = dp Note : If the polyoml P ( ) do ot cot the term, wrte =. Oce p s clculte wth desred ccurcy the repet the procedure for Q () to determe secod fctor of P ( ). The cotuous pplcto of Brge-Vet method produces ll rel roots..6. Brstow Method Ths method s used to etrct qudrtc fctor from polyoml P ( ), whch my gve pr of comple roots or pr of rel roots. If we dvde the polyoml P ( ) defed by equto (.6..) by the qudrtc fctor + p+ q the the quotet s polyoml of degree ( ) d remder s polyoml of degree oe. Thus P ( ) = + p+ qq ( ) + R+ S... (.6..) 3 4 where Q ( ) = b + b + b b + b... (.6..) 3

23 The coeffcet b, b, b,... b 3, b, R d S re fuctos of p d q. + p+ q s fctor of P ( ), f S( pq) R p, q =, =... (.6..3) Suppose ( p, q ) s tl ppromto for equto (.6..3) d ( p + pq, + q) s the true soluto of equto (.6..3). The R R R( p + pq, + q) = R( p, q) + p+ q = p q S S S( p + pq, + q) = S( p, q) + p+ q = p q O solvg bove equtos smulteously for p d q we get RS p = RS q SR q RS p q q p, RpS RS p q = RS RS p q q p... (.6..4) where R p, R q, S p, S q re prtl dervtves of R d S wth respect to p d q respectvely evluted t ( p, q ). Fuctos R d S re lso evluted t (, ) Thus to determe the true soluto t s ecessry to clculte p d q re ow terms of R, S, R p, R q, S p, q p q. p d q. The cremets S evluted t tl ppromto (, ) p q. These fuctos R, S d ther prtl dervtves re obted by comprg the coeffcets of equl powers of equto (.6..). From equto (.6..) d (.6..) we get : = b b = : = b + pb b = pb : = b + pb + qb b = pb b : = b + pb + qb b = pb qb R pb qb R pb qb : = = 3 : = + = S qb S qb 3

24 I geerl we wrte b = pb qb, =,, 3,...,... (.6..5) where b = d b =. From lst two equtos we get R = b d S b pb The prtl dervtves respect to p d q. From equto (.6..5) we get, = +... (.6..6) R p, R q, S p, S q c be determed by dfferettg (.6..5) wth b b b = + + p p p b p q ; b p b p = =... (.6..7) b b b = + + q q q p b q ; b q b q = =... (.6..8) b Substtuto = C, =,, 3,..., coverts equto (.6..7) to p C = b pc qc... (.6..9) 3 b If we wrte C = the equto (.6..8) becomes q C = b pc qc... (.6..) 3 4 From equto (.6..9) d (.6..) we get recurrece relto where C = b pc qc, =,, 3,..., b C = = pb = b p p C = d (Sce d b = re depedet of p d q) From equto (.6..6), (.6..7) we get R p R b = = = C p p b b S b p b C pc p p p = + + = From equto (.6..6) d (6..8) we get 4

25 R q R b = = = C q q 3 S b b S p C pc q q q q = = + = 3 O substtutg the bove vlues of R, S, equto (.6..6) we get, R p, R q, S p, S q equto (.6..4) d usg b ( C pc 3) ( b + pb )( C 3) ( C )( C pc ) ( C )( b C pc ) p = 3 3 bc b C = C C C b 3 3 ( ) d ( C )( b + pb ) ( b )( b C pc ) ( C )( C pc ) ( C )( b C pc ) q = 3 3 ( ) ( ) b C b bc = C C C b 3 The mproved vlues of p d q re ow p = p + p d q = q + q Repet the procedure by replcg the tl ppromto ( p, q ) by ( p, q ) the requred ccurcy of p d q., tll we get For computg b ' s d C ' s we use the followg represetto p... q pb pb... pb 3 qb... qb 4 pb pb qb 3 qb p b b b... b b b q pc pc... pc 3 qc... qc 4 C C C... pc qc 3 C C 5

26 P Whe p d q re obted to the desred ccurcy the polyoml ( Q ) = s + p+ q obted from the sythetc dvso procedure. The et qudrtc fctor of Q (d hece of P ( ) ) s obted by pplyg Brstow method to Q ( ). 6

27 UNIT - II SYSTEM OF LINEAR ALGEBRAIC EQUATIONS AND EIGEN VALUE PROBLEMS. Itroducto : System of ler equtos rse lrge umber of res, both drectly modelg physcl stutos d drectly the umercl soluto of other mthemtcl models. These pplctos occur vrtully ll res of the physcl, bologcl d socl sceces. Ler systems re volved optmzto theory, umercl solutos of boudry vlue problems, prtl dfferetl equtos, tegrl equtos d umerous other problems. geerlly s, The preset chpter dels wth smulteous ler lgebrc equtos whch c be represeted = b = b... (..) b = where j (, j =,,..., ) re the ow coeffcets, b ( =,,..., ) re the ow vlues d ( =,,..., ) re the uows to be determed. wrtte s I the mtr otto, the bove system of smulteous ler lgebrc equtos c be A = b... (..) where A s squre mtr of order, s colum vector wth elemets, =,,..., d b s colum vector wth elemets b, =,,...,.. Iterto Methods My ler systems re too lrge to be solved by drct methods bsed o Guss elmto or mtr verso. For these systems, terto methods re ofte the oly possble method of soluto, s well s beg fster th elmto my cses. I ths secto we dscuss two tertve methods. vz. Jcob terto method d Guss Sedel terto method. 7

28 A geerl ler tertve method for the soluto of the system of equtos A defed the form ( + ) ( ) = b my be X = HX + c, =,,, (..) where X + d X re the ppromtos for X t the ( + )th d th tertos respectvely. H s clled the terto mtr d c s colum vector. I the lmtg cse X coverges to the ect soluto X = A b... (..) Whe the system of equtos c be ordered so tht ech dgol etry of the coeffcet mtr s lrger mgtude th the sum of the mgtudes of the other coeffcets tht row - such system s clled dgolly domt. For such system the terto wll coverge for y strtg vlues. Formlly we sy tht mtr A s dgolly domt f d oly f for ech =,, 3,..., > j= j j =,, 3,..., For tertve methods we rerrge the system of equtos so tht the dgol etres of the coeffcet mtr A become dgolly domt. If ot, we rerrge the system of equtos such wy tht the dgol etres of mtr A re o-zero d possbly lrge mgtude. Such rerrgemet s clled pvotg.. Prtl Pvotg I the frst stte, the frst colum s serched for the lrgest elemet mgtude d brought s the frst dgol elemet by terchgg the frst equto wth the equto hvg the lrgest elemet mgtude. I the secod stge, the secod colum s serched for the lrgest elemet mgtude mog the ( ) elemets levg the frst elemet, d ths elemet s brought s the secod dgol etry ( ) by terchge of the secod equto wth the equto hvg the lrgest elemet mgtude. Ths procedure s cotued utl we rrve t the lst equto. Emple : Cosder the system of equtos + y+ z+ u = 7 y u = 3 y z u = 3 u = 8

29 As. : equto vlue. 7 y = 3 z 3 u m. {,,3,} = 3 ppers thrd equto. Therefore terchge frst d thrd 3 y z u = 3 y u = + y+ z+ u = 7 u = 3 3 y = z 7 u Cosder secod colum ecludg ( ) etry. m{,,} = we c eep secod equto s t s, sce the lso gves the mmum Cosder thrd colum ecludg frst two etres d clculte m{,} =. There s o eed to terchge thrd d fourth equto d prtl pvotg s complete. The rerrgemet of system s geerlly crres out o the ugmeted mtr [ A, b]. Complete Pvotg : I ths procedure we serch the mtr A for the lrgest elemet mgtude d brg t s the frst pvot. Ths requres ot oly terchge of equtos but lso terchge of posto of the vrbles. 9

30 .. Jcob Iterto Method Ths method s terto method d s used to determe the soluto of system of ler equtos. I the system A = b, we ssume tht the quttes re o-zero d suffcetly lrge. Ths c be doe by prtl or complete pvotg. The system of equtos (..) my be wrtte s (... ) = b 3 3 = b... (...) (... ( ) ) = b From equto (...) we hve terto method b (... ) ( + ) ( ) ( ) ( ) = = ( ) + b... (...) ( + ) ( ) ( ) ( ) 3 3 (... ( ) ) b ( + ) ( ) ( ) ( ) = () () () Itlly we c ssume tht = b, = b = b.... Sce we replce the complete vector ( ) the rght sde of (...) t the ed of ech terto, ths method s clled the method of smulteous dsplcemet. I the mtr form equto (...) c be wrtte s D = ( L+ U) + b where L d U re respectvely lower d upper trgulr mtres wth zero dgol etres, D s dgol mtr such tht A = L + D + U. The mtr form of equto (...) s used to wrte terto method the form, () = b ( + ) ( ) = D L+ U + D b, =,,, 3, (...3) s the tl ppromto. Altertvely equto (...3) c be wrtte s ( + ) ( ) ( ) ( ) = D L + U + D b 3

31 ( ) ( ) = D D+ L + U + D b [ ] ( ) = D A b [ ] ( ) = + D b A Defe ( ) ( + ) ( ) V = D r = s the error the ppromto d ( ) ( ) r = b A s the resdul vector. ( ) We solve ( ) ( ) DV = r for V d fd ( + ) ( ) ( ) = + V... (...4) These equtos descrbe the Jcob terto method error formt. Thus to solve the system of equtos by Jcob terto method error form we hve the followg procedure ( ) ( ) r = b A ( ) ( ) V = D r... (...5) ( + ) ( ) ( ) = + V.. Guss Sedel Iterto Method We ow tht every mtr A c be uquely represeted s the sum of lower d upper trgulr mtr wth zero dgol etres d dgol mtr. The system of equtos A = b c be represeted by ( L+ D+ U) = b D = ( L+ U) + b D = L U + b From bove equto we hve terto method ( + ) ( + ) ( ) D = L U + b, =,,, 3, (...) () Itlly, we ssume tht = b. Method (...) s clled Guss Sedel terto method. I the eplct form equto (...) c be wrtte s ( + ) ( ) L+ D = U + b ( + ) ( ) = L+ D U + ( L+ D) b 3

32 ( ) ( ) ( ) = L+ D U + ( L+ D) b [ ] ( ) ( ) = L+ D L+ D + U + L+ D b [ ] ( ) = L+ D A b [ ] ( + ) ( ) ( ) = + L+ D b A Thus to solve the system of equtos by Guss Sedel terto method error form we hve the followg procedure. ( ) ( ) r = b A ( ) Solve ( ) ( ) L+ DV = r for V by forwrd substtuto ( + ) ( ) ( ) = + V, =,,...,... (...) System (...) descrbe the Guss Sedel method error form...3 Covergece Alyss of Itertve Methods Itertve methods re methods of successve ppromtos. Covergece of tertve methods s studed through error lyss. To dscuss the covergece of tertve method. ( + ) ( ) = H + c, =,,,3, (..3.) where ( ) + d ( ) re the ppromtos for t the ( + ) th d th tertos respectvely, we study the behvour of the dfferece betwee ect soluto d ppromto ( ). The ect soluto of tertve method wll stsfy = H+ c... (..3.) Sbtrctg (..3.) from (..3.) d substtutg ( ) ( ) ε = + ε ( + ) ( ) we get = Hε, =,,, (..3.3) Repettve pplcto of (..3.3) for ( ) ε, =,, 3,... gves ε ( + ) () = H ε, =,,, 3,... For Jcob tertve method H = D ( L+ U) d c = D b wheres for Guss Sedel tertve methods H = ( L+ D) U d c = ( L+ D) b. For both the methods terto mtr H rems costt for ech terto. 3

33 If the error sequece { ( ) } ε + coverges to zero s, we sy tht the tertv method coverges. To study the covergece of error sequece we use the followg theorems. Theorem. : Let A be squr mtr. The m lm A = f A < or ff 9 ( A ) <. m Before provg ths theorem we epl the ottos d deftos used the sttemet of the theorem. Defto : Mtr Norm The mtr orm A s o-egtve umber whch stsfes the propertes () A f A d O = where O s zero mtr. () ca = c A for rbtrry comple umber c. () A+ B A + B (v) AB A B The most commoly used orms re () Euclde orm or frobeus orm F( A) = j where A= j, j= j= () Mmum orm () A = A = m (mmum bsolute row sum) A = = = m (mmum bsolute colum sum) Hlbert orm or Spectrl orm The lrgest ege vlue modulus of mtr A s clled the spectrl rdus of the mtr A d s deoted by 9 ( A). The spectrl rdus s defed oly for squre mtrces. A = λ where ( A* A) A s the comple cojugte of A. λ = 9 d 33 T A* = A,

34 Proof of Theorem. : If A < the by defto of orm of mtr. m A m A ( AB A B ) d sce orm s cotuous fucto, m lm A lm A lm A m m m m m = = A < For smplcty, we ssume tht ll the ege vlues of the mtr A re dstct. The there est smlrty trsformto S such tht A= S DS where D s dgol mtr hvg ege vlues of A o the dgol. Therefore where m m ( )( ) (... ) A = S DS S DS S DS D m = S D SS D... SS DS m = S D S m λ m λ = m λ ( m ) m tmes lm A = S lm D S = ff ll the ege vlues stsfy λ <,.e. 9 ( A ) <. m m Theorem. : The fte seres 3 + A+ A + A +... coverges ff lm A =. The seres coverges to ( I A). Proof : From the defto of coverget seres we hve m lm A =. m m Suppose lm A = the by theorem. we hve m 9( A ) < m m 34

35 Sce the mgtude of lrgest ege vlue of mtr A s strctly less th, I A d therefore ( I A) ests. We ow the detty ( m... ) m I + A+ A + A + + A I A = I A + 3 ( 3 m) ( m+ ) I+ A+ A + A A = I A I A m Sce lm A = (by theorem.), we hve m 3 m I + A+ A + A A = ( I A) Theorem.3 : No ege vlue of mtr A eceeds the orm of mtr A..e. A 9( A) Proof : For ege vlue λ of mtr A we hve A = λ where s o-zero ege vector correspodg to ege vlue λ. λ = λ = A A Thus we hve λ A ( ).e. λ A.e. 9( A) A (where λ s y ege vlue) Theorem.4 : The terto method of the form ( + ) ( ) = H + c for the system of equto A = b coverges to the ect soluto for y tl vector () f H <. Proof : We te tl vector () =. The the repeted pplcto of terto method gves () = c () () = H + c = Hc + c = ( H + I) c (3) () = H + c = H H + I c + c = H + H + I c ( + ) = H + H H + H + I c 35

36 ( + ) lm = lm H + H + H H + H + I c Thus lm + = ( I H) c = ( I H) c (f H < ) I cse of Jcob terto method we hve H = D ( L+ D) d c = D b [ ] ( I H) c = I + D ( L+ U) D b [ D D D ( L U) ] = + + D b [ D ( D L U) ] = + + D b = D+ L + U D D b = A b = Thus for Jcob terto method we hve lm + = ( I H) c = I cse of Guss Sedel terto method we hve H = ( L+ D) U d c = ( L+ D) b ( I H) c = I + ( L+ D) U ( L+ D) b 36 = ( L+ D) ( L+ D) + ( L+ D) U ( L+ D) b = ( L+ D) ( L+ D + U) ( L+ D) b ( ) = A ( L+ D) ( L+ D) b ( AB) = B A = A b = Thus for Guss Sedel terto method we hve lm + I H = c = A b =

37 From theorem.3 we observe tht Jcob tertve method coverges f ( ) ( 9 H 9 D L U 9 D ( L U) ) J = + = + < d Guss Sedel tertve method coverges f 9( H ) 9( ( L D) U) 9( ( L D) U) G = + = + < Theorem.5 : A ecessry d suffcet codto for covergece of tertve method ( + ) ( ) = H + c s tht the ege vlues of the terto mtr H stsfy λ H <, =,,..., where λ ( H ) re ege vlues of mtr H. Proof : Suppose λ, λ, λ 3,..., λ re the ege vlues of mtr H d,, 3,..., be the correspodg depedet ege vectors. () ε s -vector. We wrte () ε = c + c c () ε... H = ch + ch + + c H λ λ... λ = c + c + + c = = ( ) () lmε H ε lm cλ = ( ) ε ff λ s..e. ( ) ε ff λ H <. Defto. : The rte of covergece of tertve method s gve by v= log [ 9( H) ] where 9 ( H ) s the spectrl rdus of mtr H. Theorem.6 : If A s strctly dgolly domt mtr, the the Jcob terto scheme coverges for y tl strtg vector. Proof : The Jcob terto scheme s gve by ( + ) ( ) = D L+ U + D b ( ) = D A D + D b ( ) ( ) = I D A + D b 37

38 The terto scheme wll be coverget f I D A <. Usg bsolute row sum orm we hve I D A = m j= j j Sce A s strctly dgol domt, j < for ll =,,, 3,..., d therefore j= j I D A < d therefor the Jcob terto scheme coverges for y tl strtg vector. Theorem.7 : If A s strctly dgolly domt mtr, the the Guss-Sedel terto scheme coverges for y tl strtg vector. Proof : The Guss Sedel terto scheme s gve by ( + ) ( ) = D+ L U + ( D+ L) b [ ] ( ) = D+ L A D+ L + D+ L b ( ) = I D+ L A + ( D+ L) b The terto scheme wll be coverget f 9( I ( D+ L) A) < Let λ be ege vlue of I ( D+ L) A. I ( D+ L) A = λ ( D+ L) ( D+ L) ( D+ L) A = λ ( D+ L) [ D+ L A ] = λ U = λ ( D+ L) ( A= L + D + U ).e..e. j j j j j=+ j= = λ = λ + λ j j j j j=+ j= 38

39 .e. λ = λ j j j j j= j=+ λ λ + j j j j j= j=+... (..3.) Sce s ege vector,. We ssume tht =. Choose de such tht = d j, j. From equto (..3.4) we get λ λ + j j j= j=+ λ j j j= j=+ λ j j=+ j= j Sce mtr A s dgolly domt >. j= j j > + j j j= j=+ or > j j j= j=+ j j=+ < j= j d therefore λ <. 39

40 But λ s y ege vlue of I ( D+ L) A. 9( I ( D+ L) A) < Note : The rte of covergece of Guss Sedel scheme s twce tht of the Jcob scheme. It my hppe tht for the system A = b, ( H ) < system of equto A dgolly domt. = b for ( H ) < 9 but ( H ) > G J 9 whch but ( H ) > 9. Smlrly t s possble to hve the G 9. For these systems mtr A s ot J.3 Mtr Fctorzto Method The system of equto A Cse () : A = D The system of equtos s = b c be drectly solved the followg cses. b = b = for =,, 3,..., ; Cse () : A = L (Lower Trgulr Mtr) The system of equtos s of the form = b + = b = b = b Solvg frst equto we get. If we substtute ths vlue of secod equto we get d so o. Sce uows re determed by forwrd substtuto, the method s clled forwrd substtuto method. Cse () : A = U (Upper Trgulr Mtr) I ths cse system of equto s of the form = b = b 3 3 4

41 = b = b ( )( ) ( ) = b From lst equto we get from secod lst equto o substtutg the vlue of we get d so o. Sce the uow re determed from bc substtuto, ths method s clled the bc substtuto method. Thus the equto A = b s drectly solvble f the mtr A c be trsformed to oe of the three cses dscussed bove..3. Trgulzto Method Ths method s lso ow s the decomposto method or fctorzto method. I ths method, the coeffcet mtr A of the system of equtos A = b s decomposed or fctorzed to the product of lower trgulr mtr L d upper trgulr mtr U. We wrte the mtr A = LU. where, L l l l = l3 l3 l 33 l l l 3 l d U u u u u u u u 3 3 = u33 u3 u The the system of equto A LU = b = b becomes We wrte bove equto s the followg two systems of equtos U = z Lz = b From Cse () we determe z d the form Cse () we solve U... (.3.) = z to clculte. 4

42 .3. () Doolttle s Method I ths method we wrte A = LU where the dgol elemets of mtr L re. We wrte 3 u u u3 u l u u u u33 3 = l3 l 3 u33 u3 l l l u 3 3 These re ( + ) ( + ) equtos + = wth rght hd sde product. We get compoetwse equtos. uows comprg left hd sde A = u, = u, 3 = u3,..., = u = l u, = l u + u,..., = lu + u By usg forwrd substtuto we clculte u, u, u 3,..., u, l, u, u 3,..., u,... Oce the mtrces L d U re ow the soluto s obted by represetg the system = b the form of equto (.3.).3. (b) Crout s Method I ths method we wrte A = LU where the dgol elemets of mtr U re ll. We wrte 3 l u u3 u 3 u3 u l l = l3 l3 l 33 u3 l l l l 3 3 Ag these re equtos uows. Equtg the compoetwse elemets of l.h.s. d r.h.s. we get 4

43 = l, u = l, 3 = l u3,..., = lu = l, = u By usg forwrd substtuto we get, l =, l l, 3 = lu3+ l u3,..., = lu + lu u =, 3 u l 3 =,..., u = l l l =, l = l u,..., u = lu Oce the mtrces L d U re determed the system of equtos A = b s represeted the form of equto (.3.) d system of equtos (.3.) s solved from Cse () d from Cse () respectvely..4 Ege Vlues d Ege Vectors I secto. d.3 tertve methods for ler system of equtos re dscussed. Cosder the system of equto A = λ... (.4.) Equto (.4.) s clled ege vlue problem. The ege vlues of A re gve by the roots of the chrcterstcs equto det( A λi) =... (.4.) If A s squre mtr of order, equto (.4.) gves polyoml equto of degree. The roots of ths polyoml equto re clled ege vlues d my be determed by the methods gve Ut. Oce the roots λ of polyoml (.4.) re ow the o-zero vector such tht A = λ... (.4.3) s clled the ege vector or chrcterstc vector correspodg to λ. O multplyg equto (.4.3) by costt c we get where y Ac = λc Ay = λ y = c.e. y s lso chrcterstc vector of A correspodg to ege vlue 43 λ. Ths shows tht ege vector s determed oly to wth rbtrry multplctve costt. O premultplyg equto (.4.) (m ) tmes by A we obt m m m A A A ( A) A m λ λ λ λ λ A m = = = = = λ m...

44 m A m = λ for m =,, 3, 4, (.4.4) Equto (.4.4) shows tht Sce ( T ) det m λ s ege vlue of m A f λ s ege vlue of A. A λi = det A λi, A d A T hve the sme ege vlues. If u s ege T vector correspodg to the ege vlue λ, the Au = λu. Premultplcto by u gves T T λu u uau = d we get T T uau λ = uu If u s ege vector of mtr A the λ s the ege vlue of mtr A. Thus gve ege vlue λ we fd ege vector by solvg A ege vector s such tht T A T. For rbtrry u, the rto = λ d gve ege vector we fd the correspodg T uau T uu s clled the Rylegh quotet. Let A d B be two squre mtrces of sme order. If o-sgulr mtr S c be determed =,... (.4.5) B S AS the the mtrces A d B re sd to be smlr d the mtr S s clled smlrly mtr d the trsformto s clled smlrly trsformto. From equto (.4.5) we wrte A= SBS If λ s ege vlue of A d u s the correspodg ege vector the Au = λ u = λ S Au S u Put u = Sv the S ASv = λs Sv = λv..e. Bv = λv. But the ege vlues of A d B re sme d gve ege vectors u of mtr A, S u re the ege vectors of the mtr B. A smlrty trsformto, where S s the mtr of ege vectors reduces mtr A to ts dgol form. The ege vlues of A re the dgol elemets. If ege vectors of A re lerly depedet the S ests d the mtr A s sd to be dgolzble. 44

45 .4. Bouds o Ege Vlues Bruer. The bouds o the ege vlues of the mtr A re gve by the theorems by Gerschgor d Theorem (Gerschgor) The lrgest ege vlue modulus of squre mtr A cot eceed the lrgest sum of the modul of the elemets log y row or y colum. Proof : Let λ we ege vlues of A d be the correspodg ege vector. Suppose [,,..., ] T =. Sce λ s ege vlue of A d s correspodg ege vector, A = λ 3 3 = λ 3 Let = m r. Select the th equto d dvde t by. The th equto s r =λ The λ = Sce r = m r,, r =,,..., r d λ Thus f λ s ege vlue the λ for some. = λ m = Thus ech ege vlue d therefore the lrgest ege vlue modulus of squre mtr A cot eceed the lrgest sum of the modul of the elemets log y row. 45

46 Sce A d A T hve sme ege vlues, the theorem s lso true for colums. (Repet the procedure for A T ) Theorem (Bruer) : Let P be the sum of the modul of the elemets log the th row ecludg the dgol elemet of squre mtr A. Every ege vlue of A les sde or o the boudry of t lest oe of the crcles λ = P, =,, 3,...,. Proof : Let λ be ege vlue of A d be the correspodg egevector. Suppose [,,..., ] T = The A = λ c be wrtte s = λ =λ = λ Let = m. Select th equto from bove equtos. The th equto s r r = λ Dvde bove equto by d rerrge the terms. λ ( ) ( + ) = ( ) + ( + ) Sce r = m r,, r. r d λ ( + ) + ( ) λ P Thus Therefore ll the egevlues of A le sde or o the uo of the bove crcles. Sce A d A T hve sme egevlues, theorem holds for colum sum lso.e. λ... ( ) ( + ) , =,, 3,...,. 46

47 The bouds obted for rows d colums re depedet. Hece ll the ege vlues of A must le the tersecto of these bouds. These crcles re clled the Gerschgor crcles d the bouds re clled the Gerschgor bouds. Emple.4. : Estmte the egevlues of the mtr A = 3 usg the Gerschgor bouds d Bruer theorem. Aswer : By Gerschgor theorem correspodg to row we hve.e. λ 5 λ m{ + +, + +, } m{ 4,3,5} = 5 Smlrly by cosderg colum sum we get, λ m{ + +, + + 3, + + } m{ 3,6,3} = 6 By Bruer s theorem every egevlue of A les sde or o the boudry of tlest oe of the crcles λ A. Correspodg to rows we hve λ 3, λ, λ + 4 Correspodg to colums we hve λ, λ 5, λ + The uo correspodg to row sum gves { } { } λ 3 λ λ + 4 = λ 3 λ + 4 The uo correspodg to colum sum gves { } { } λ λ 5 λ+ = λ 5 λ+ Thus the requred rego s gve by { λ 5} { λ 6} { λ 3 λ+ 4} { λ 5, λ + } 47

48 Emple.4. : Estmte the ege vlue rego of the mtr 3 A = 5 3 By Gerschgor theorem λ m{ 7,9,9} = 9 Correspodg to colum we hve λ m{ 7,9,7} = 9 Sce mtr A s symmetrc ll egevlues re rel d λ 9 gves the tervl [ 9, 9]. Thus ll egevlues le the tervl [ 9, 9]. By Bruer theorem the egevlues le the rego { } correspodg to row sum d correspodg to coulm sum we hve { } λ 3 4 λ 5 4 λ 3 4 Thus we hve the rego { } λ 3 4 λ 5 4 λ 3 4 = { 4 λ λ 5 4} { λ 7 λ 9} = [,9] Thus ll roots le the tervl [,9]. λ 3 4 λ 5 4 λ Jcob Method for Symmetrc Mtrces For rel symmetrc mtr ll egevlues re rel d there est rel orthogol mtr S such tht S AS s dgol mtr D. The dgol etres of D re ll egevlues of mtr A. The dgolzto s cheved by pplyg seres of orthogol trsformtos. Computtol Procedure Let m { j : jj,,,,3,..., } = =. Cosder mtr formed by the tersecto of th & th row d th & th colum. The we get mtr A = ( = sce A s symmetrc mtr) 48

49 S * * AS cosθ sθ Choose S* = sθ cosθ d fd * θ such tht S AS * s dgol mtr. S * AS cosθ sθ cosθ sθ * = sθ cosθ sθ cosθ sθ cos θ + sθ + s θ ( ) + cosθ = sθ ( ) + cosθ s θ + cos θ sθ * * Now we choose θ such tht the off dgol etry of mtr S AS becomes zero so tht becomes dgol mtr. Thus we choose θ such tht ( ) s + cosθ = θ θ = = t Ths equto produces four vlues of θ. We choose θ betwee π π d d we get, 4 4 θ t = f π = f =, > 4 π = f =, < 4 Wth ths choce of θ costruct orthogol mtr S s follows. Wrte cosθ, sθ, sθ, cosθ t (, ), (, ), (, ), (, ) postos of mtr S respectvely. Wrte remg dgol etres to be d rest of the offdgol etres. Thus we get mtr S s 49

50 S th jth th cosθ sθ = th j sθ cosθ Defe B = S AS I B ( ) d (, ),, etres re zero. Repet procedure for B. We get S. Defe = = B S BS S S AS S After mg r trsformtos we get, r = r r r B S S S S AS SS S = ( SSS... S ) ASSS (... S ) = S 3 r 3 AS where S = SSS 3... Sr As r, Ths procedure s clled Jcob method. r B r pproches dgol mtr wth the egevlues o the ledg dgol. The covergece to dgol mtr tes plce eve f the mmum of offdgol elemets re ot selected d we me y offdgol etry zero. Ths modfcto s clled the specl cyclc Jcob method. I ths method there s o serch for mmum offdgol etry..6 Householder s Method for Symmetrc Mtrces I Jcob method symmetrc mtr s coverted to dgol mtr through smlrty trsformto. I ths method symmetrc mtr A s reduced to the trdgol form by orthogol trsformtos. The orthogol trsformtos re of the form T P = I ww... (.6.) where w s colum vector such tht T w w=. 5

51 Observe tht P s symmetrc d orthogol. T T T T T T T P = I ww = I ww = I ww = P T T T P P = I ww I ww = I ww T ww T + 4ww T ww T = I ( T ww= ) At the frst trsformto we fd ' s such tht we get zero the posto (, 3), (, 4),..., (, ) d zero the correspodg postos the frst colum.e. (3, ), (4, ), (5, )... (, ). Thus oe trsformto P AP = A, brg ( ) zeros the frst row d frst colum. I the secod trsformto P AP, we get ( 3) zeros the secod colum d secod row mely (, 4), (, 3 3 5), (, 6)... (, ) d (4, ), (5, )... (, ) postos. The fl mtr s trdgol. The trdgolzto s completed wth ectly ( ) Householder trsformto. The mtr P r s costructed s follows. The vector w r s costructed wth the frst (r ) compoets s zeros. [,,,...,,,..., ] T r = r r+ w T Sce wr w r =, r + r+ + r = wth ths choce of w r, T r r r P = I ww. Let us llustrte ths procedure for 3 3 d 4 4 mtrcs. A 3 = [,, ] T 3 T w = w w, w w = w + w = P = I ww T 3 3 o = w, w, w w 3 = ww w ww 3 3 w3 5

52 T P P = I therefore T P s orthogol d P = P. 3 = ww 3 w 3 AP w ww ( ) + ( ) ( ) ( ) ( ) + ( ) w w w ww w = w ww ww + w w ww ww w T P AP s T T P AP = w ww 3 ww 3 w 3 ( ) + ( ) ( ) + ( ) ( ) + ( ) w ww ww w w ww ww w w ww ww w P AP s trdgol f (, 3) etry of the mtr P AP s zero. But (, 3) etry of ww + w = w w + w = e. 3 wr 3 = where r = w + 3w3 T ' (, ) etry of P AP deoted by s ( ) ' = 3 3 w ww = wr ' = ( wr) + ( wr) [ wr= ] = + + r w + w rw + w = r 4r T 3 3 5

53 ' Thus = wr + wr 3 3 ( wr) = + ' = + 3 Therefore, ' =± + = wr =± S (sy) Now we hve two equtos rw =... () rw =± S... () Multply equto () by w 3 d () by w d dd. Sce w + 3 w 3 = r, we hve r r Sw =±.e. r =± Sw Now from equto () we hve ± Sw =± S Thus w ± S =± S w = S, w 3 = w I computg w 3 from w, we choose w s lrge s possble. We demostrte the reducto the followg emple. Emple.6. : Reduce the mtr 3 4 A = 3 4 to trdgol form. Aswer : Here + 3 = 9+ 6 =± 5. w 3 = = + S 5 ( w s m. for S = 5 sce s +ve choose S +ve) 53

54 4 w =.e. w = 5 5 w = = = w T w =,, 5 5 T d P = I ww A 3 4 = = = T P AP A = = To llustrte the procedure for 4 4 mtr let A =

55 Sce the trsformtos T r P AP re orthogol, the sum of squres of the elemets y r row re vrt. We wll use the fct tht sum of squres of elemets y row of mtr A s sme s the sum of squres of elemets correspodg row of mtr T Choose w [,,, ] 3 4 T r P AP. T = d w w = + + =. 3 4 At the frst trsformto we fd, 3, 4 such tht we get zero the posto (, 3), (, 4) d (3, ), (4, ). I the mtr P frst row s ut vector d therefore the posto (, 3), (, 4) hve zero etry f the correspodg elemets AP re zero. The frst row of AP s gve by the followg product. r [ p, p, p ] = where p = Now we eed to fd, 3, 4 such tht 3 p 3 = d 4 p 4=. So tht (, 3) d (, 4) posto of PAP wll become zero. Sce the sum of the squres of the elemets y row s vrt uderthe orthogol trsformto we hve, ( ) ( ) ( ) = p p p = + p p =± + + =± S (sy) Thus we hve three equtos p =... (.6.)

56 p =... (.6.) 4 4 p =± S... (.6.3) Sce mtr A s gve mtr, S s ow qutty. Multply (.6.3) by, (.6.) by 3 d (.6.) by 4 d dd. We get, p p + + =± S 3 4 But = d therefore p = S Now f we put ths vlue of p equto (.6.3) the equto becomes qudrtc equto d c be solved for. ( ) S =± S =... (.6.4) S From equto (.6.) d (.6.) we get = = p d S = = p... (.6.5) S From equto (.6.4) we observe tht d therefore p posses two vlues. Sce 3 d 4 cots the deomtor, we choose the lrge root by. Ths s doe by tg sutble sg equto (.6.4). sg ( ) = + S 3 ( ) 3 sg =, S 4 = sg ( ) 4 S where sg ( ) s sg fucto whch tes vlue f < d vlue f >. Thus the trsformto PAP produces zero vlue (, 3), (, 4) d therefore (3, ), (4, ) postos. Oe more trsformto dscussed for 3 3 mtr produces zeros (, 4) d (4, ) postos. The resultg mtr wll be trdgol mtr. 56