362Duals,naturality,bilinearformsLetVbeanite-dimensional[1]vectorspac - PDF document

362Duals,naturality,bilinearformsLetVbeanite-dimensional[1]vectorspace,withabasise1;:::;enforV.Adualbasis1;:::;nforV(andfeig)isabasisforVwiththepropertythatj(ei)=1(fori=j)0(fori6=j)Fromthedenitionaloneitisnotatallclearthatadualbasisexists,butthefollowingpropositionprovesthatitdoes.[1.0.2]Proposition:ThedualspaceVtoann-dimensionalvectorspaceV(withnapositiveinteger)isalson-dimensional.Givenabasise1;:::;enforV,thereexistsauniquecorrespondingdualbasis1;:::;nforV,namelyabasisforVwiththepropertythatj(ei)=1(fori=j)0(fori6=j)Proof:Provingtheexistenceofadualbasiscorrespondingtothegivenbasiswillcertainlyprovethedimensionassertion.UsingtheuniquenessofexpressionofavectorinVasalinearcombinationofthebasisvectors,wecanunambiguouslydenealinearfunctionaljbyj Xiciei!=cjThesefunctionalscertainlyhavethedesiredrelationtothebasisvectorsei.WemustprovethatthejareabasisforV.IfXjbjj=0thenapplythisfunctionaltoeitoobtainbi=0@Xjbjj1A(ei)=0(ei)=0Thisholdsforeveryindexi,soallcoecientsare0,provingthelinearindependenceofthej.Toprovethespanningproperty,letbeanarbitrarylinearfunctionalonV.Weclaimthat=Xj(ej)
jIndeed,evaluatingtheleft-handsideonPiaieigivesPiai(ei),andevaluatingtheright-handsideonPiaieigivesXjXiai(ej)j(ei)=Xiai(ei)sincej(ei)=0fori6=j.Thisprovesthatanylinearfunctionalisalinearcombinationofthej.===LetWbeasubspaceofavectorspaceVoverk.TheorthogonalcomplementW?ofWinVisW?=f2V:(w)=0;forallw2Wg [1]Someofthedenitionsanddiscussionheremakesenseforinnite-dimensionalvectorspacesV,butmanyoftheconclusionsareeitherfalseorrequiresubstantialmodicationtobecorrect.Forexample,bycontrasttothepropositionhere,forinnite-dimensionalVthe(innite)dimensionofVisstrictlylargerthanthe(innite)dimensionofV.Thus,forexample,thenaturalinclusionofVintoitsseconddualVwouldfailtobeanisomorphism. 364Duals,naturality,bilinearformsitfollowsthatdimIm'=dimV.WeshowedabovethatthedimensionofVisthesameasthatofV,sinceVisnite-dimensional.Likewise,thedimensionofV=(V)isthesameasthatofV,hencethesameasthatofV.Sincethedimensionoftheimageof'inVisequaltothedimensionofV,whichisthesameasthedimensionofV,theimagemustbeallofV.Thus,':V�!Visanisomorphism.===[1.0.5]Corollary:LetWbeasubspaceofanite-dimensionalvectorspaceVoverk.Let':V�!Vbetheisomorphismofthepreviouscorollary.Then(W?)?='(W)Proof:First,showthat'(W)(W?)?Indeed,for2W?,'(w)()=(w)=0Ontheotherhand,dimW+dimW?=dimVandlikewisedimW?+dim(W?)?=dimV=dimVThus,'(W)(W?)?anddim(W?)?=dim'(W)since'isanisomorphism.Therefore,'(W)=(W?)?.===Asanillustrationoftheecacyofthepresentviewpoint,wecanproveausefulresultaboutmatrices.[1.0.6]Corollary:LetMbeanm-by-nmatrixwithentriesinaeldk.LetRbethesubspaceofknspannedbytherowsofM.LetCbethesubspaceofkmspannedbythecolumnsofM.LetcolumnrankofM=dimCrowrankofM=dimRThencolumnrankofM=rowrankofMProof:ThematrixMgivesalineartransformationT:kn�!kmbyT(v)=Mvwherevisacolumnvectoroflengthn.ItiseasytoseethatthecolumnspaceofMistheimageofT.Itisalittlesubtlerthattherowspaceis(kerT)?.Fromabove,dimkerT+dimImT=dimVandalsodimkerT+dim(kerT)?=dimVThus,columnrankM=dimImT=dim(kerT)?=rowrankMasclaimed.=== 2.Firstexampleofnaturality 366Duals,naturality,bilinearformsthediagramVV�!Vf##fWW�!Wcommutes.ThecommutativityofthediagraminvolvingaparticularMandNiscalledfunctorialityinMandinN.Thatthediagramcommutesisveriedverysimply,asfollows.Letv2V,2W.Then((fV)(v))()=(f(Vv))()(denitionofcomposition)=(Vv)(f)(denitionoff)=(f)(v)(denitionofV)=(fv)(denitionoff)=(W(fv))()(denitionofW)=((Wf)(v))()(denitionofcomposition)SinceequalityofelementsofWisimpliedbyequalityofvaluesonelementsofW,thisprovesthatthediagramcommutes.Further,forVnite-dimensional,wehavedimkV=dimkV=dimkVwhichimpliesthateachVmustbeanisomorphism.Thus,theaggregateoftheisomorphismsV:V�!Viscalledanaturalequivalence.[10] 3.BilinearformsAbstractingthenotionofinnerproductorscalarproductordotproductonavectorspaceVoverkisthatofbilinearformorbilinearpairing.Forpurposeofthissection,abilinearformonVisak-valuedfunctionoftwoV-variables,writtenv
worhv;wi,withthefollowingpropertiesforu;v;w2Vand2k(Linearityinbotharguments)hu+v;wi=hu;wi+hv;wiandhu;v+v0i=hu;vi+hu;v0i(Non-degeneracy)Forallv6=0inVthereisw2Vsuchthathv;wi6=0.Likewise,forallw6=0inVthereisv2Vsuchthathv;wi6=0.Thetwolinearityconditionstogetherarebilinearity.Insomesituations,wemayalsohave(Symmetry)hu;vi=hv;uiHowever,thesymmetryconditionisnotnecessarilycriticalinmanyapplications.[3.0.1]Remark:WhenthescalarsarethecomplexnumbersC,sometimesavariantofthesymmetryconditionisuseful,namelyahermitianconditionthathu;vi= hv;uiwherethebardenotescomplexconjugation.[3.0.2]Remark:Whenthescalarsarerealorcomplex,sometimes,butnotalways,thenon-degeneracyandsymmetryareusefullyreplacedbyapositive-denitenesscondition,namelythathv;vi0andis0onlyforv=0.WhenavectorspaceVhasanon-degeneratebilinearformh;i,therearetwonaturallinearmapsv�!vandv�!vfromVtoitsdualV,givenbyv(w)=hv;wi [10]Moreprecisely,onthecategoryofnite-dimensionalk-vectorspaces,isanaturalequivalenceoftheidentityfunctorwiththesecond-dualfunctor. 368Duals,naturality,bilinearformsThus,dimW+dimW?=dimVasclaimed.Next,weclaimthatWW??.Indeed,forw2Witiscertainlytruethatforv2W?hv;wi=hv;wi=0Thatis,weseeeasilythatWW??.Ontheotherhand,fromdimW+dimW?=dimVanddimW?+dimW??=dimVweseethatdimW??=dimW.SinceWisasubspaceofW??withthesamedimension,thetwomustbeequal(fromourearlierdiscussion).===[3.0.7]Remark:Whenanon-degeneratebilinearformonVisnotsymmetric,therearetwodierentversionsofW?,dependinguponwhichargumentinh;iisused:W?;rt=fv2V:hv;wi=0;forallw2WgW?;lft=fv2V:hw;vi=0;forallw2WgAndthentherearetwocorrectstatementsaboutW??,namely�W?;rt
?;lft=W�W?;lft
?;rt=WTheseareproveninthesamewayasthelastcorollary,butwithmoreattentiontothelackofsymmetryinthebilinearform.Infact,tomorescrupulouslyconsiderpossibleasymmetryoftheform,weproceedasfollows.Formanypurposeswecanconsiderbilinearmaps[11](thatis,k-valuedmapslinearineachargument)h;i:VW�!kwhereVandWarevectorspacesovertheeldk.[12]Themostcommoninstanceofsuchapairingisthatofavectorspaceanditsdualh;i:VV�!kbyhv;i=(v)Thisnotationandviewpointhelpstoemphasizethenear-symmetry[13]oftherelationshipbetweenVandV. [11]Alsocalledbilinearforms,orbilinearpairings,orsimplypairings.[12]Notethatnowthesituationisunsymmetrical,insofarastherstandsecondargumentstoh;iarefromdierentspaces,sothatthereisnoobvioussensetoanypropertyofsymmetry.[13]TheseconddualVisnaturallyisomorphictoVifandonlyifdimV1. 370Duals,naturality,bilinearformsThedimensionofthespanWofisstrictlylessthandimV,whichwe'veprovenisdimV=dimV.WemayalsoidentifyVVviathenaturalisomorphism.Withthatidentication,wemaysaythatthesetofsolutionsisW?,anddim(W?)+dimW=dimV=dimVThus,dimW?�0,sotherearenon-zerosolutions.===[25.2]Letkbeaeld,andVanite-dimensionalk-vectorspace.LetbealinearlyindependentsubsetofthedualspaceV.Let�!abeasetmap�!k.Showthataninhomogeneoussystemofequations(v)=a(forall2)hasasolutionv2V(meetingalltheseconditions).Letm=jj,=f1;:::;mg.Onewaytousethelinearindependenceofthefunctionalsinistoextendtoabasis1;:::;nforV,andlete1;:::;en2VbethecorrespondingdualbasisforV.Thenletv1;:::;vnbetheimagesoftheeiinVunderthenaturalisomorphismVV.(Thisachievestheeectofmakingtheibeadualbasistothevi.Wehadonlyliterallyproventhatonecangofromabasisofavectorspacetoadualbasisofitsdual,andnotthereverse.)Thenv=X1imai
viisasolutiontotheindicatedsetofequations,sincej(v)=X1imai
j(vi)=ajforallindicesjm.===[25.3]LetTbeak-linearendomorphismofanite-dimensionalk-vectorspaceV.ForaneigenvalueofT,letVbethegeneralized-eigenspaceV=fv2V:(T�)nv=0forsome1n2ZgShowthattheprojectorPofVtoV(commutingwithT)liesinsidek[T].FirstwedothisassumingthattheminimalpolynomialofTfactorsintolinearfactorsink[x].Letf(x)betheminimalpolynomialofT,andletf(x)=f(x)=(x�)ewhere(x�)eistheprecisepowerof(x�)dividingf(x).Thenthecollectionofallf(x)'shasgcd1,sotherearea(x)2k[x]suchthat1=Xa(x)f(x)WeclaimthatE=a(T)f(T)isaprojectortothegeneralized-eigenspaceV.Indeed,forv2V,v=1V
v=Xa(T)f(T)
v=Xa(T)f(T)
v=a(T)f(T)
vsince(x�)edividesf(x)for6=,and(T�)ev=0.Thatis,itactsastheidentityonV.And(T�)eE=a(T)f(T)=02Endk(V)sotheimageofEisinsideV.SinceEistheidentityonV,itmustbethattheimageofEisexactlyV.For6=,sincef(x)jf(x)f(x),EE=0,sotheseidempotentsaremutuallyorthogonal.Then(a(T)f(T))2=(a(T)f(T))
(1�X6=a(T)f(T))=a(T)f(T)�0 372Duals,naturality,bilinearformsIfEwereanotherprojectortoVcommutingwithT,thenEstabilizesVPforallirreduciblesPdividingtheminimalpolynomialfofT,andEis0onVQforQ6=(x�),andEis1onV.Thatis,E=1
Ex�+XQ6=x�0
EQ=EPThisprovestheuniquenesseveningeneral.===[25.4]LetTbeamatrixinJordannormalformwithentriesinaeldk.LetTssbethematrixobtainedbyconvertingalltheo-diagonal1'sto0's,makingTdiagonal.ShowthatTssisink[T].ThisimplicitlydemandsthattheminimalpolynomialofTfactorsintolinearfactorsink[x].Continuingasinthepreviousexample,letE2k[T]betheprojectortothegeneralized-eigenspaceV,andkeepinmindthatwehaveshownthatVisthedirectsumofthegeneralizedeigenspaces,equivalent,thatPE=1.Bydenition,theoperatorTssisthescalaroperatoronV.ThenTss=X
E2k[T]since(fromthepreviousexample)eachEisink[T].===[25.5]LetM=AB0Dbeamatrixinablockdecomposition,whereAism-by-mandDisn-by-n.ShowthatdetM=detA
detDOnewaytoprovethisistousetheformulaforthedeterminantofanN-by-NmatrixdetC=X2SN()a(1);1:::a(N);Nwherecijisthe(i;j)thentryofC,issummedoverthesymmetricgroupSN,andisthesignhomomorphism.ApplyingthistothematrixM,detM=X2Sm+n()M(1);1:::M(m+n);m+nwhereMijisthe(i;j)thentry.SincetheentriesMijwith1jmandmim+nareall0,weshouldonlysumoverwiththepropertythat(j)mfor1jmThatis,stabilizesthesubsetf1;:::;mgoftheindexingset.Sinceisabijectionoftheindexset,necessarilysuchstabilizesfm+1;m+2;:::;m+ng,also.Conversely,eachpair(1;2)ofpermutation1oftherstmindicesand2ofthelastnindicesgivesapermutationofthewholesetofindices.LetXbethesetofthepermutations2Sm+nthatstabilizef1;:::;mg.Foreach2X,let1betherestrictionoftof1;:::;mg,andlet2betherestrictiontofm+1;:::;m+ng.And,infact,ifweplantoindextheentriesoftheblockDintheusualway,we'dbetterbeabletothinkof2asapermutationoff1;:::;ng,also.Notethat()=(1)(2).ThendetM=X2X()M(1);1:::M(m+n);m+n 374Duals,naturality,bilinearformsAnd,atthesametimesubtractingA1j=A11timestherstcolumnofAfromthejthcolumnofAfor2jmdoesnotchangethedeterminant,andthenewmatrixisA110DAlsobythepreviousexample,detA=detA110D=A11
detDThus,puttingthetwocomputationstogether,det(A B)=An11detB
det(D B)=An11detB
(detD)n(detB)m�1=(A11detD)ndetB
(detB)m�1=(detA)n(detB)masclaimed.Anotherapproachtothisistoobservethat,intheseterms,A Bis0BBBBBBBBBBBBBBBB@A110:::00A11......0A11:::A1m0:::00A1m......0A1m......Am10:::00Am1......0Am1:::Amm0:::00Amm......0Amm1CCCCCCCCCCCCCCCCA0BB@B0:::00B......0B1CCAwheretherearemcopiesofBonthediagonal.Bysuitablepermutationsofrowsandcolumns(withaninterchangeofrowsforeachinterchangeofcolumns,thusgivingnonetchangeofsign),thematrixcontainingtheAijsbecomes0BB@A0:::00A......0A1CCAwithncopiesofAonthediagonal.Thus,det(A B)=det0BB@A0:::00A......0A1CCA
det0BB@B0:::00B......0B1CCA=(detA)n
(detB)mThismightbemoreattractivethantherstargument,dependingonone'stastes.===Exercises25.[4.0.1]LetTbeahermitianoperatoronanite-dimensionalcomplexvectorspaceVwithapositive-deniteinnerproducth;i.LetPbeanorthogonalprojectortothe-eigenspaceVofT.(ThismeansthatPistheidentityonVandis0ontheorthogonalcomplementV?ofV.)ShowthatP2C[T].