1 jordan.mcd Eigenvalue-eigenvector approach to solving first order ODEs. -- Jordan normal (canonical) form. Instructor: Nam Sun Wang

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1 jordnmcd Eigenvlue-eigenvecor pproch o solving firs order ODEs -- ordn norml (cnonicl) form Insrucor: Nm Sun Wng Consider he following se of coupled firs order ODEs d d x x 5 x x d d x d d x x x 5 x x Combine he given equions ino "vecor-mrix" form d d x d d x x 5 x x x x 5 x d x d x The equion cn be bbrevied s 5 5 x x x ORIGIN d d x A x where x x x x A Find eigenvlue-eigenvecor for he liner rnsform A (nd his is where mny sudens sop) eigenvls( A ) V eigenvecs( A ) Check: Mke sure he eigenvlues nd eigenvecors sisfy he eigenvlue-eigenvecor equion (bu misleding) A V < compre V yes! A V < 6 compre V 6 yes! A V < compre V yes!

2 jordnmcd Cuion: The bove hree "yes" show h ll hree vecors sisfy he eigenvlue-eigenvecor equion A v v, bu one (or some) of hese is no linerly independen eigenvecor(s)! How cn we ell? When eigenvlues re repeed, lmos ll mhemicl pckges (including Mhcd here) fil o find similriy rnsform in wy we generlly expec Alwys crefully inspec he resuls reurned by ny mhemicl pckges o mke sure h hey mke sense Specificlly he ls column of V is linerly dependen on he firs wo columns The firs wo columns re indeed wo linerly independen eigenvecors The following re some ellle signs h somehing is miss in V If V were indeed composed of eigenvecors s is columns, we expec he following: rnk( V) rnk( Re( V) ) This should reurn An error hins somehing is fishy?? mus be rel V This mrix should be nonsingulr For ll prcicl purposes, i is singulr here This is furher confirmed by he condiion number cond( V) lrge condiion number singulr mrix V T Noe h V is no necessrily composed of orhonorml columns Orhonorml columns led o n ideniy mrix: V T V I V A V dig( ) V We expec his o be digonl mrix wih he eigenvlues s he digonl elemens: V - A VΛdig() The digonl elemens here re no ll eigenvlues of Since V is singulr, V - does no cully exis The resul V - A V reurned here by Mhcd, lhough hving "nice" numbers, is simply numericl rifc, nohing more hn n illusion I is grbge-in & grbge-ou We expec his o be he originl mrix A: V dig() V - A Agin, he resul V - A V reurned here by Mhcd, lhough hving "nice" numbers, is simply numericl rifc, nohing more hn n illusion I is grbge-in & grbge-ou Since we re shor of n eigenvecor, we my wish o find vecor h is orhogonl o he firs wo eigenvecors (Irrespecive of he number of eigenvecors, we cn lwys find se of linerly independen vecors h re equl in number o he dimension of he LVS In fc, he number of linerly independen vecors sricly comes from he dimension of he LVS -- h is how he dimension of LVS is defined; he number of linerly independen vecors hs nohing o do wih liner rnsformion) This ddiionl vecor fills he ls column of V in he following similriy rnsform A V V B Find vecor h is orhogonl o he firs wo V V < > V < > V 7 Doing so yields nonsingulr, orhonorml similriy rnsform mrix V 9 V 9 7 T compre

3 V T V V B V A V B V T 5 5 jordnmcd These seem nice on he surfce The new mrix hs off-digonl elemens V - A V is nohing more hn n lernive descripion of he originl problem wih new bsis (ie, he new coordine sysem) Unforunely, his new descripion (B) conining off-digonl elemens is no much beer hn he originl one (A) Verdic: mking he rd V orhogonl o he firs wo does nohing useful Swpping columns of V only moves he off-digonl elemens round nd does no mke he mer ny beer To demonsre his, we swp he second column wih he hird column Swp columns emp V V < > V < > V emp V A 5 5 Swp columns bck o resume work The new mrix sill hs off-digonl elemens, nd no moun of swpping of V will mke V - A V excly digonl or ny closer o being digonl emp V V < > V < > V Wihou full se of eigenvecors, we will no be ble o digonlize V - A V The ordn norml form comes closes o being digonl A V Λ V A V V emp we cnno chieve fully Λdig() due o less-hn-full se of eigenvecors ordn norml form close o being dig() Unforunely, Mhcd does no hve nice build-in funcion h reurns he similriy rnsform mrix leding o he ordn norml form A V V We re forced o genere i ourselves Since here re wo eigenvecors, here re wo ordn blocks We ssume i kes he following form (We will ler vlide invlide his ssumpion) where is ssocied wih he firs eigenvecor is ssocied wih he second eigenvecor In oher words, we ssume he missing, hird vecor in he similriy rnsform is genered from he firs eigenvecor wo vecors A V <, > V, V, V < > V A V < > V < > V V V V A V < > A V < > V V < > V A I A I V V <> is he eigenvecor for he eigenvlue V V < > V <> is genered from eigenvecor V <> from his equion

4 A 5 5 V v v v 5 5 V jordnmcd le V 5 v v The firs eqn nd he second eqn re v v 5 v v Remove conrdicion by modifying he righ hnd side of he nd eqn nd rd eqn ccordingly V v v v conrdicory (Thus, he ssumpion h he x ordn block is ssocied wih V <> is wrong) The second eqn nd he hird eqn re idenicl 5 v v 5 5 v v v v v 5 v v Side Noe Noe h in his problem, we cn sfely conclude h we hve mde wrong ssumpion he beginning, simply swp V <> wih V <>, nd proceed, hoping h hings would work ou In removing he conrdicion, we demonsre h we cn cively deermine which eigenvecor is responsible for which ordn block In he bove clculion, we find h V is liner combinion of sclr muliple of V <> nd V <>, nd i is he eigenvecor corresponding o he ordn block linerly independen eigenvecors: V V 9 7 A liner combinion of V <> & V <> gives V α V < ' > β V V α β α V β V < 667 compre V 667 Thus, he bove numericlly demonsres h V we hve clculed is combinion of wo V β 75 V, eigenvecors V <> & V <> Thus, V is lso n eigenvecor, becuse ny sclr muliples of n eigenvecor or ny liner combinions of eigenvecors of repeed eigenvlues re lso eigenvecors of he repeed eigenvlues A V < A V < 6 A V compre V compre V 6 compre V Of course, we hve lredy previously esblished h V <> is n eigenvecor, so is V <> Yes, V is lso n eigenvecor I is he eigenvecor responsible for he ordn block

5 5 jordnmcd We cnno simply conclude h V <>, no V <>, should hve been he eigenvecor responsible for he ordn block The process of removing he conrdicion llows us o rrive he correc eigenvecor V responsible for he ordn block under considerion We hve found wo linerly independen eigenvecors he beginning of his workshee However, we mus emphsize h he eigenvecors re no unique; ny wo of he hree or liner combinion hereof is lso n eigenvecor When here re only wo ordn blocks, i is emping o conclude h if i is no he s eigenvecor h is responsible for he ordn block (s we hve shown vi conrdicion), hen i mus be he nd eigenvecor This logic is flse becuse here re mny combinions of vecors h qulify s eigenvecors, nd i is flse o consider one o be "more qulified" eigenvecor hn noher Furhermore, here exis cses where here exis more possibiliies -- sy, we hve four repeed eigenvlues nd hree eigenvecors In h cse, we know here re hree ordn blocks (becuse of hree eigenvecors):,, nd Bu which one of he hree eigenvecors is responsible for he ordn block? Acully, none of he originl hree (unless we re exremely lucky nd hppen o hi correc one), bu V, combinion of he firs wo, is he nswer The conrdicion removl process nswers he quesion wihou guess work Bck o our problem If we do no proceed o remove conrdicion, we conclude h he originl ssumpion h he missing vecor is genered from he firs eigenvecor is wrong Insed, we ssume he ordn norml form is he following (he oher lernive) where is ssocied wih he firs eigenvecor is ssocied wih he second eigenvecor This is equivlen o ssuming h he missing vecor V <> is genered from he second eigenvecor V <> A I V V A 5 V 9 le V v v 5 7 v 5 5 v v v v v v v 9 5 v v 7 The firs eqn nd he second eqn re conrdicory The second eqn nd he hird eqn re idenicl Agin, we hve se equions whose LHS re idenicl o hose when we ssumed V <> ws he eigenvecor responsible for he ordn block This is necessrily so becuse he LHS in boh cses re idenicl: (A-I) V <> Conrdicion removl gin leds o n eigenvecor vecor V h is idenicl o he one before Bsiclly, if we re exremely lucky (which is rre), we hppen o sr wih correc eigenvecor, nd we re ble o consruc ordn block If we re unlucky (which is more likely), irrespecive of he eigenvecor we hppen o pick iniilly, he conrdicion removl process described here leds o correc one V 9 7 remove conrdicion V Sme s before

6 Normlize he second eigenvecor (which is no necessry) 5 v v V V V V V 6 jordnmcd 7 Le v v v Since v is rbirry, we cn chnge v o ny vlue wihou ffecing he ordn norml form Give i ry! v rbirry v v v v The wo vecors V <> nd Thus, V v V 65 V <> llow he ordn block 5 v v v o form Similriy rnsform yields ordn norml form V A V compre Noe h he new coordines re no muully orhogonl V T V 5 5 V T V I Furhermore, he bove sclr produc mrix shows h he hird vecor is no normlized Nor should we ry o normlize i Hd we normlized i, similriy rnsform yields ordn form h is no quie norml The following demonsres his poin V' V Normlize he rd column V' T V' V' A V' V' V V 56 The off-digonl erm is no Noe h he similriy rnsform mrix V h leds o ordn norml form is no unique, jus s eigenvecors re no unique The following re some exmples V' V' A V' V' V' V' A

7 V' V' V' A V' Wih ordn norml form, i is esy o find he exponenil exp(* V' A V' exp( exp exp( A ) V exp( V exp exp 7 jordnmcd exp exp exp exp exp exp 7 7 exp( ) exp( ) 7 exp( ) 7 exp( ) exp( ) exp( A ) V exp( V expa( exp( exp( ) exp( ) exp( exp( exp( exp( ) exp( ) exp( Since, s menioned previously, he columns of V do no need o be norml, nor should ll he columns be norml In fc, no normlizing he columns someimes mkes similriy rnsform less compliced Below, we ry noher V ken from one of he non-unique exmples bove exp( A ) V' exp( V' exp exp exp exp 6 6 exp( 9 exp( 6 exp( ) 6 exp( exp( exp exp 6 exp( Finlly, he soluion o he ODE is, x exp( A x

8 jordnmcd Exmple IC: x, 5 The ses re given by: x( ) expa( x Dynmic Ses x( x( x( 6 5 Phse Digrm 5 Phse Digrm x( 5 x( x( x( Phse Digrm x( x(

9 9 jordnmcd z( ) V x( Dynmic Ses z( z( z( 6 Phse Digrm Phse Digrm z( 5 z( 5 z( ) z( ) z(, z( ) z(, z( ) 5 Phse Digrm z( z( ) 5 z(, z( )

10 jordnmcd Eigenvlue-eigenvecor clculions mnully from scrch This smll x sysem is perhps clerer o demonsre he mehod of consrucing ordn norml form For lrge sysem wih mny eigenvlues nd mny high dimensionl eigenvecors, his mnul mehod obviously becomes imprcicl, nd he conrdicion-removl pproch works fser, becuse he compuer hs lredy done mos of he work in finding eigenvlues nd eigenvecors -- someimes hese eigenvecors re jus no quie he righ ones needed o consruc ordn norml block, nd bi of humn inervenion is clled for Sep Eigenvlue clculion I ideniy( ) ssume I A ( ) 9 6 ( ) 6 9 ( ) ssume A I ( ) 9 6 ( ) 6 9 ( ) Sep Eigenvecor clculion 5 v 5 v v ll hree equions re equl ( A I) V v v v s eigenvecor: v rbirry 5 v 5 v v nd eigenvecor: v rbirry v v Noe h becuse v nd v re boh rbirry, wo eigenvecors re rbirry, s long s v v We cn le v in he firs eigenvecor nd v in he second eigenvecor V V A liner combinion of V <> & V <> is lso n eigenvecor Since we cn only deermine n eigenvecor up o muliple, we cn le β nd leve α o remin rbirry (unil i is fixed by requiring i o be he eigenvecor for ordn block) V α V < > β V < > α β α An equivlen wy is o le v & leve v o remin rbirry (unil i is fixed by requiring i o be he eigenvecor for ordn block) V rbirry Sep ordn block clculion -- ssign eigenvecors o ordn blocks nd find missing vecor(s) where is ssocied wih n eigenvecor h is linerly independen of V is ssocied wih he eigenvecor V

11 A V V missing V V missing V V missing jordnmcd This is equivlen o ssuming h he missing vecor V missing (which is no n eigenvecor) is genered from he eigenvecor V s column of he bove eqn: A V V lredy done in eigenvecor clculion Sep nd column of he bove eqn: ( A I) V missing V o be done here in Sep A 5 V α le V missing v v 5 v 5 5 v v v α 5 v v α The firs eqn nd he second eqn leds o v v 5 v v α The second eqn nd he hird eqn re idenicl v rbirry v rbirry v v Thus, given V, here remin mny choices for nd eigenvecor h is linerly independen of V, nd here re lso mny choices for missing vecor V missing The following is one exmple (bu no unique one) V Check: V missing V A V V V V missing V

12 jordnmcd Mulipliciy of Possible ordn Norml Forms If we find repeed eigenvlues bu independen eigenvecors (which mens ordn blocks), we hve severl possibiliies possibiliy # where exp exp( ) possibiliy # where exp exp( ) If we find 5 repeed eigenvlues bu independen eigenvecors (which mens ordn blocks), we hve severl possibiliies possibiliy # where exp exp( ) possibiliy # where exp exp( ) If we find 5 repeed eigenvlues bu independen eigenvecors (which mens ordn blocks), we hve severl possibiliies possibiliy # where exp exp( ) possibiliy # where

13 jordnmcd Numericl Sensiiviy The ordn norml form is very sensiive o round-off errors, nd he inies numericl rifcs will rever i o digonl form Thus, i is no used in numericl lgorihms The exmple below demonsres his sensiiviy Eigenvlues & eigenvecors for he originl problem Λ dig( eigenvls( A) ) V eigenvecs( A ) rnk( A) 56 Λ 9 rnk( V) rnk( Re( V) ) 7 mus be rel We dd iny error o one of he mrix elemens nd we end up wih (seemingly) jus one eigenvecor (cully wo eigenvecors bu no chnge in eigenvlues, which we disply wih mny digis o show here is indeed jus one eigenvlue) A' A i j A' i, j A' i, j Λ dig( eigenvls( A' ) ) V eigenvecs( A' ) rnk( Re( V) ) Λ rnk( V) V Λ V compre A' 5 no mch mus be rel 5 We dd iny error o noher one of he mrix elemens nd we end up wih hree differen eigenvlues nd hree differen eigenvecors A' A i j A' i, j A' i, j Λ dig( eigenvls( A' ) ) V eigenvecs( A' ) Λ V Λ V compre A' yes mch rnk( Re( V) ) rnk( V) mus be rel 5 5

14 jordnmcd Verify he exp( funcion D y b D y c y le y exp( b b b c c chrcerisic equion When he chrcerisic equion hs repeed roos (b c), we hve b y exp( y exp( D y z z b z c y c b y z A y z A c b b b c exmple eigenvls c b D y D y y ( ) ( ) b b c b b c mnul clculion ( ) A A v v v v v v v v v v v Boh equions re idenicl Mhcd funcions "eigenvls" nd "eigenvecs" ssign v v eigenvls( A ) V eigenvecs( A ) Since here is only one eigenvecor, here is one ordn block A V V missing V V missing V V missing s column of he bove eqn: A V V nd column of he bove eqn: ( A I) V missing V A v v V eigenvecor for V V V missing sme s eigenvecor clculion o be done here le V missing v v v The eqns re linerly dependen v v v rbirry v v v V V missing V V V V missing

15 Check: V A exp( exp exp( exp( 5 jordnmcd ok V V exp( exp( ok exp( A ) V exp( V exp( exp( exp( y y' Check: exp( A y ( exp( exp( y y' exp( ( exp( y' y C exp( ) C exp( y' C exp( ) C ( exp( ) exp( ) exp( y( ) y C exp( ) C exp( ) C y'( ) y' C exp( ) C ( exp( ) exp( ) ) C C y C C y y' C C y' y y exp( ) y' y exp( y' y' exp( ) y' y exp( y y' y

16 6 jordnmcd Verify he / exp( funcion D y b D y c D y d y le y exp( b c exmple d chrcerisic equion When he chrcerisic equion hs repeed roos, we hve D y 6 D y D y y ( ) he soluions re: y exp( y exp( y exp( D y z x z x 6 x z y 6 y z x A y z x A 6 compuer clculion /w Mhcd funcions "eigenvls" nd "eigenvecs" eigenvls V eigenvecs( A) Check: V A ok V V no! rnk( Re( V) ) bu n inspecion of V bove shows only one linerly independen eigenvecor! mnul clculion A v v 6 v v v v v v eqns & re linerly v v v v v v 6 v v v rbirry v v v v independen eqn is -x eqn plus x eqn only one eigenvecor: V Since here is only one eigenvecors, here is one ordn block A V V missing V missing V V missing V missing V V missing V missing V V s column of he bove eqn: A V V sme s eigenvecor clculion nd column of he bove eqn: ( A I) V missing V o be done here

17 le V eigenvecor V missing v v v 7 jordnmcd v v eqns & re linerly independen eqn is V missing v v -x eqn plus x eqn v v v v rbirry v v v v rd column of he bove eqn: ( A I) V missing V missing 6 v v v 6 v v v o be done here v v eqns & re linerly independen eqn is V missing v v -x eqn plus x eqn 9 v v v v rbirry v v v v 9 V V V V V missing V V missing Check: V A ok V V ok 9 6 exp( exp exp( exp( exp( exp( exp( ) exp( exp( exp( A ) V exp( V exp( exp( 9 exp( exp( exp( ) exp( 9 exp( exp( A exp( exp( exp( exp( exp( exp( exp( exp( exp(

18 jordnmcd y y' y'' Check: exp( A y y' y'' exp( exp( exp( exp( exp( exp( exp( exp( exp( y y' y'' y C exp( ) C exp( ) C exp( ) y' C exp( ) C ( exp( ) exp( ) ) C exp( ) exp( ) y'' C exp( ) C exp( ) exp( ) C exp( ) exp( ) exp( ) y( ) y C exp( ) C exp( ) C exp( ) C y'( ) y' C exp( ) C ( exp( ) exp( ) ) C exp( ) exp( ) C C y''( ) y'' C exp( ) C exp( ) exp( ) C exp( ) exp( ) exp( ) C y y' C C C C y y' y y' y y'' C C y'' y'' y' y y y exp( ) y' y exp( ) y'' y' y exp( ) y' y exp( ) y' y ( exp( ) exp( ) ) y'' y' y exp( ) exp( ) y'' y exp( ) y' y exp( ) exp( ) y'' y' y exp( ) exp(