1 Motivation We motivate the chapter on eigenvalues by discussing the equ ation ax 2 hxy by c where not all of a h b are zero The expression ax 2 hxy by is called quadratic form in and and we have the identity ax 2 hxy by x y a h h b AX where an ID: 24033
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Chapter6EIGENVALUESANDEIGENVECTORS6.1MotivationWemotivatethechapteroneigenvaluesbydiscussingtheequationax2+2hxy+by2=c;wherenotallofa;h;barezero.Theexpressionax2+2hxy+by2iscalledaquadraticforminxandyandwehavetheidentityax2+2hxy+by2=xyahhb¸·xy=XtAX;whereX=xyandA=ahhb.Aiscalledthematrixofthequadraticform.Wenowrotatethex;yaxesanticlockwisethroughradianstonewx1;y1axes.Theequationsdescribingtherotationofaxesarederivedasfollows:LetPhavecoordinates(x;y)relativetothex;yaxesandcoordinates(x1;y1)relativetothex1;y1axes.ThenreferringtoFigure6.1:115 116CHAPTER6.EIGENVALUESANDEIGENVECTORS - ¾ 6 ¯¯@ xyx1y1PQROFigure6.1:Rotatingtheaxes.x=OQ=OPcos(+)=OP(coscossinsin)=(OPcos)cos(OPsin)sin=ORcosPRsin=x1cosy1sinµ:Similarlyy=x1sin+y1cos.Wecancombinethesetransformationequationsintothesinglematrixequation:xy=cossinsincos¸·x1y1;orX=PY,whereX=xy;Y=x1y1andP=cossinsincos.WenotethatthecolumnsofPgivethedirectionsofthepositivex1andy1axes.AlsoPisanorthogonalmatrix{wehavePPt=I2andsoP1=Pt.ThematrixPhasthespecialpropertythatdetP=1.AmatrixofthetypeP=cossinsincosiscalledarotationmatrix.Weshallshowsoonthatany22realorthogonalmatrixwithdeterminant 6.1.MOTIVATION117equalto1isarotationmatrix.Wecanalsosolveforthenewcoordinatesintermsoftheoldones:x1y1=Y=PtX=cossinsincos¸·xy;sox1=xcos+ysinandy1=xsin+ycos.ThenXtAX=(PY)tA(PY)=Yt(PtAP)Y:Nowsuppose,aswelatershow,thatitispossibletochooseananglesothatPtAPisadiagonalmatrix,saydiag(1;¸2).ThenXtAX=x1y11002¸·x1y1=1x21+2y21(6.1)andrelativetothenewaxes,theequationax2+2hxy+by2=cbecomes1x21+2y21=c,whichisquiteeasytosketch.Thiscurveissymmetricalaboutthex1andy1axes,withP1andP2,therespectivecolumnsofP,givingthedirectionsoftheaxesofsymmetry.Alsoitcanbeveri¯edthatP1andP2satisfytheequationsAP1=1P1andAP2=2P2:Theseequationsforcearestrictionon1and2.ForifP1=u1v1,the¯rstequationbecomesahhb¸·u1v1=1u1v1ora1hhb1¸·u1v1=00:Hencewearedealingwithahomogeneoussystemoftwolinearequationsintwounknowns,havinganon{trivialsolution(u1;v1).Hencea1hhb1=0:Similarly,2satis¯esthesameequation.Inexpandedform,1and2satisfy2(a+b)+abh2=0:Thisequationhasrealroots=a+bp (a+b)24(abh2) 2=a+bp (ab)2+4h2 2(6.2)(Therootsaredistinctifa=borh=0.Thecasea=bandh=0needsnoinvestigation,asitgivesanequationofacircle.)Theequation2(a+b)+abh2=0iscalledtheeigenvalueequationofthematrixA. 118CHAPTER6.EIGENVALUESANDEIGENVECTORS6.2De¯nitionsandexamplesDEFINITION6.2.1(Eigenvalue,eigenvector)LetAbeacomplexsquarematrix.ThenifisacomplexnumberandXanon{zerocom-plexcolumnvectorsatisfyingAX=¸X,wecallXaneigenvectorofA,whileiscalledaneigenvalueofA.WealsosaythatXisaneigenvectorcorrespondingtotheeigenvalue.SointheaboveexampleP1andP2areeigenvectorscorrespondingto1and2,respectively.WeshallgiveanalgorithmwhichstartsfromtheeigenvaluesofA=ahhbandconstructsarotationmatrixPsuchthatPtAPisdiagonal.Asnotedabove,ifisaneigenvalueofannnmatrixA,withcorrespondingeigenvectorX,then(A¸In)X=0,withX=0,sodet(A¸In)=0andthereareatmostndistincteigenvaluesofA.Converselyifdet(A¸In)=0,then(A¸In)X=0hasanon{trivialsolutionXandsoisaneigenvalueofAwithXacorrespondingeigenvector.DEFINITION6.2.2(Characteristicpolynomial,equation)Thepolynomialdet(A¸In)iscalledthecharacteristicpolynomialofAandisoftendenotedbychA().Theequationdet(A¸In)=0iscalledthecharacteristicequationofA.HencetheeigenvaluesofAaretherootsofthecharacteristicpolynomialofA.Fora22matrixA=abcd,itiseasilyveri¯edthatthecharacter-isticpolynomialis2(traceA)+detA,wheretraceA=a+disthesumofthediagonalelementsofA.EXAMPLE6.2.1FindtheeigenvaluesofA=2112and¯ndalleigen-vectors.Solution.ThecharacteristicequationofAis24+3=0,or(1)(3)=0:Hence=1or3.Theeigenvectorequation(A¸In)X=0reducesto2112¸·xy=00; 6.2.DEFINITIONSANDEXAMPLES119or(2)x+y=0x+(2)y=0:Taking=1givesx+y=0x+y=0;whichhassolutionx=y;yarbitrary.Consequentlytheeigenvectorscorrespondingto=1arethevectorsyy,withy=0.Taking=3givesx+y=0xy=0;whichhassolutionx=y;yarbitrary.Consequentlytheeigenvectorscorre-spondingto=3arethevectorsyy,withy=0.Ournextresulthaswideapplicability:THEOREM6.2.1LetAbea22matrixhavingdistincteigenvalues1and2andcorrespondingeigenvectorsX1andX2.LetPbethematrixwhosecolumnsareX1andX2,respectively.ThenPisnon{singularandP1AP=1002:Proof.SupposeAX1=1X1andAX2=2X2.WeshowthatthesystemofhomogeneousequationsxX1+yX2=0hasonlythetrivialsolution.Thenbytheorem2.5.10thematrixP=[X1jX2]isnon{singular.SoassumexX1+yX2=0:(6.3)ThenA(xX1+yX2)=A0=0,sox(AX1)+y(AX2)=0.Hencex¸1X1+y¸2X2=0:(6.4) 120CHAPTER6.EIGENVALUESANDEIGENVECTORSMultiplyingequation6.3by1andsubtractingfromequation6.4gives(21)yX2=0:Hencey=0,as(21)=0andX2=0.Thenfromequation6.3,xX1=0andhencex=0.ThentheequationsAX1=1X1andAX2=2X2giveAP=A[X1jX2]=[AX1jAX2]=[1X1j2X2]=[X1jX2]1002=P1002;soP1AP=1002:EXAMPLE6.2.2LetA=2112bethematrixofexample6.2.1.ThenX1=11andX2=11areeigenvectorscorrespondingtoeigenvalues1and3,respectively.HenceifP=1111,wehaveP1AP=1003:Therearetwoimmediateapplicationsoftheorem6.2.1.The¯rstistothecalculationofAn:IfP1AP=diag(1;¸2),thenA=Pdiag(1;¸2)P1andAn=P1002P1n=P1002nP1=Pn100n2P1:Thesecondapplicationistosolvingasystemoflineardi®erentialequationsdx dt=ax+bydy dt=cx+dy;whereA=abcdisamatrixofrealorcomplexnumbersandxandyarefunctionsoft.Thesystemcanbewritteninmatrixformas_X=AX,whereX=xyand_X=_x_y=dx dtdy dt: 6.2.DEFINITIONSANDEXAMPLES121WemakethesubstitutionX=PY,whereY=x1y1.Thenx1andy1arealsofunctionsoftand_X=P_Y=AX=A(PY);so_Y=(P1AP)Y=1002Y:Hence_x1=1x1and_y1=2y1.Thesedi®erentialequationsarewell{knowntohavethesolutionsx1=x1(0)e1tandy1=y1(0)e2t,wherex1(0)isthevalueofx1whent=0.[Ifdx dt=kx,whereisaconstant,thend dtektx=kektx+ektdx dt=kektx+ektkx=0:Henceektxisconstant,soektx=ek0x(0)=x(0).Hencex=x(0)ekt.]Howeverx1(0)y1(0)=P1x(0)y(0),sothisdeterminesx1(0)andy1(0)intermsofx(0)andy(0).Henceultimatelyxandyaredeterminedasexplicitfunctionsoft,usingtheequationX=PY.EXAMPLE6.2.3LetA=2345.UsetheeigenvaluemethodtoderiveanexplicitformulaforAnandalsosolvethesystemofdi®erentialequationsdx dt=2x3ydy dt=4x5y;givenx=7andy=13whent=0.Solution.ThecharacteristicpolynomialofAis2+3+2whichhasdistinctroots1=1and2=2.We¯ndcorrespondingeigenvectorsX1=11andX2=34.HenceifP=1314,wehaveP1AP=diag(1;2).HenceAn=Pdiag(1;2)P1n=Pdiag((1)n;(2)n)P1=1314¸·(1)n00(2)n¸·4311 122CHAPTER6.EIGENVALUESANDEIGENVECTORS=(1)n1314¸·1002n¸·4311=(1)n132n142n¸·4311=(1)n432n3+32n442n3+42n:Tosolvethedi®erentialequationsystem,makethesubstitutionX=PY.Thenx=x1+3y1;y=x1+4y1.Thesystemthenbecomes_x1=x1_y1=2y1;sox1=x1(0)et;y1=y1(0)e2t.Nowx1(0)y1(0)=P1x(0)y(0)=4311¸·713=116;sox1=11etandy1=6e2t.Hencex=11et+3(6e2t)=11et+18e2t;y=11et+4(6e2t)=11et+24e2t:Foramorecomplicatedexamplewesolveasystemofinhomogeneousrecurrencerelations.EXAMPLE6.2.4Solvethesystemofrecurrencerelationsxn+1=2xnyn1yn+1=xn+2yn+2;giventhatx0=0andy0=1.Solution.ThesystemcanbewritteninmatrixformasXn+1=AXn+B;whereA=2112andB=12:ItisthenaneasyinductiontoprovethatXn=AnX0+(An1+¢¢¢+A+I2)B:(6.5) 6.2.DEFINITIONSANDEXAMPLES123AlsoitiseasytoverifybytheeigenvaluemethodthatAn=1 21+3n13n13n1+3n=1 2U+3n 2V;whereU=1111andV=1111.HenceAn1+¢¢¢+A+I2=n 2U+(3n1+¢¢¢+3+1) 2V=n 2U+(3n1) 4V:Thenequation6.5givesXn=1 2U+3n 2V¶·01+n 2U+(3n1) 4V¶·12;whichsimpli¯estoxnyn=(2n+13n)=4(2n5+3n)=4:Hencexn=(2n+13n)=4andyn=(2n5+3n)=4.REMARK6.2.1If(AI2)1existed(thatis,ifdet(AI2)=0,orequivalently,if1isnotaneigenvalueofA),thenwecouldhaveusedtheformulaAn1+¢¢¢+A+I2=(AnI2)(AI2)1:(6.6)HowevertheeigenvaluesofAare1and3intheaboveproblem,soformula6.6cannotbeusedthere.Ourdiscussionofeigenvaluesandeigenvectorshasbeenlimitedto22matrices.Thediscussionismorecomplicatedformatricesofsizegreaterthantwoandisbestlefttoasecondcourseinlinearalgebra.Neverthelessthefollowingresultisausefulgeneralizationoftheorem6.2.1.Thereaderisreferredto[28,page350]foraproof.THEOREM6.2.2LetAbeannnmatrixhavingdistincteigenvalues1;:::;¸nandcorrespondingeigenvectorsX1;:::;Xn.LetPbethematrixwhosecolumnsarerespectivelyX1;:::;Xn.ThenPisnon{singularandP1AP=2666410¢¢¢002¢¢¢0............00¢¢¢n37775: 124CHAPTER6.EIGENVALUESANDEIGENVECTORSAnotherusefulresultwhichcoversthecasewheretherearemultipleeigen-valuesisthefollowing(Thereaderisreferredto[28,pages351{352]foraproof):THEOREM6.2.3SupposethecharacteristicpolynomialofAhasthefac-torizationdet(¸InA)=(c1)n1¢¢¢(ct)nt;wherec1;:::;ctarethedistincteigenvaluesofA.Supposethatfori=1;:::;t,wehavenullity(ciInA)=ni.Foreachsuchi,chooseabasisXi1;:::;XinfortheeigenspaceN(ciInA).ThenthematrixP=[X11j¢¢¢jX1n1j¢¢¢jXt1j¢¢¢jXtnt]isnon{singularandP1APisthefollowingdiagonalmatrixP1AP=26664c1In10¢¢¢00c2In2¢¢¢0............00¢¢¢ctInt37775:(Thenotationmeansthatonthediagonaltherearen1elementsc1,followedbyn2elementsc2,...,ntelementsct.)6.3PROBLEMS1.LetA=4310.FindaninvertiblematrixPsuchthatP1AP=diag(1;3)andhenceprovethatAn=3n1 2A+33n 2I2:2.IfA=0:60:80:40:2,provethatAntendstoalimitingmatrix2=32=31=31=3asn!1. 6.3.PROBLEMS1253.Solvethesystemofdi®erentialequationsdx dt=3x2ydy dt=5x4y;givenx=13andy=22whent=0.[Answer:x=7et+6e2t;y=7et+15e2t.]4.Solvethesystemofrecurrencerelationsxn+1=3xnynyn+1=xn+3yn;giventhatx0=1andy0=2.[Answer:xn=2n1(32n);yn=2n1(3+2n).]5.LetA=abcdbearealorcomplexmatrixwithdistincteigenvalues1;¸2andcorrespondingeigenvectorsX1;X2.AlsoletP=[X1jX2].(a)Provethatthesystemofrecurrencerelationsxn+1=axn+bynyn+1=cxn+dynhasthesolutionxnyn=®¸n1X1+¯¸n2X2;whereandaredeterminedbytheequation=P1x0y0:(b)Provethatthesystemofdi®erentialequationsdx dt=ax+bydy dt=cx+dyhasthesolutionxy=®e1tX1+¯e2tX2; 126CHAPTER6.EIGENVALUESANDEIGENVECTORSwhereandaredeterminedbytheequation=P1x(0)y(0):6.LetA=a11a12a21a22bearealmatrixwithnon{realeigenvalues=a+iband =aib,withcorrespondingeigenvectorsX=U+iVand X=UiV,whereUandVarerealvectors.AlsoletPbetherealmatrixde¯nedbyP=[UjV].Finallyleta+ib=reiµ,wherer�0andisreal.(a)ProvethatAU=aUbVAV=bU+aV:(b)DeducethatP1AP=abba:(c)Provethatthesystemofrecurrencerelationsxn+1=a11xn+a12ynyn+1=a21xn+a22ynhasthesolutionxnyn=rnf(®U+¯V)cosnµ+(¯U®V)sinnµg;whereandaredeterminedbytheequation=P1x0y0:(d)Provethatthesystemofdi®erentialequationsdx dt=ax+bydy dt=cx+dy 6.3.PROBLEMS127hasthesolutionxy=eatf(®U+¯V)cosbt+(¯U®V)sinbtg;whereandaredeterminedbytheequation=P1x(0)y(0):[Hint:Letxy=Px1y1.Alsoletz=x1+iy1.Provethat_z=(aib)zanddeducethatx1+iy1=eat(+i¯)(cosbt+isinbt):Thenequaterealandimaginarypartstosolveforx1;y1andhencex;y.]7.(Thecaseofrepeatedeigenvalues.)LetA=abcdandsupposethatthecharacteristicpolynomialofA,2(a+d)+(adbc),hasarepeatedroot.AlsoassumethatA=®I2.LetB=A®I2.(i)Provethat(ad)2+4bc=0.(ii)ProvethatB2=0.(iii)ProvethatBX2=0forsomevectorX2;indeed,showthatX2canbetakentobe10or01.(iv)LetX1=BX2.ProvethatP=[X1jX2]isnon{singular,AX1=®X1andAX2=®X2+X1anddeducethatP1AP=10:8.Usethepreviousresulttosolvesystemofthedi®erentialequationsdx dt=4xydy dt=4x+8y; 128CHAPTER6.EIGENVALUESANDEIGENVECTORSgiventhatx=1=ywhent=0.[Tosolvethedi®erentialequationdx dtkx=f(t);kaconstant;multiplythroughoutbyekt,therebyconvertingtheleft{handsidetodx dt(ektx).][Answer:x=(13t)e6t;y=(1+6t)e6t.]9.LetA=241=21=201=41=41=21=41=41=235:(a)Verifythatdet(¸I3A),thecharacteristicpolynomialofA,isgivenby(1)(1 4):(b)Findanon{singularmatrixPsuchthatP1AP=diag(1;0;1 4).(c)ProvethatAn=1 32411111111135+1 34n2422411211235ifn1.10.LetA=2452225222535:(a)Verifythatdet(¸I3A),thecharacteristicpolynomialofA,isgivenby(3)2(9):(b)Findanon{singularmatrixPsuchthatP1AP=diag(3;3;9).