Solution.FirstsupposethatthereexistsT2L(V;W)suchthatnullT=U.ThendimU=d - PDF document

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Solution.FirstsupposethatthereexistsT2L(V;W)suchthatnullT=U.ThendimU=d - PPT Presentation

4Problem14onpage60ofAxlerHintyoumightneedproblem3onpage59ofAxlerSolutionFirstsupposethatTisinjectiveDe12neS0rangeT VbyS0TvvBecauseTisinjectiveeachelementofrangeTcanberepresentedinth ID: 828253

dimv solution dimnullt kxk solution dimv kxk dimnullt nxk dimw dimnullt0 dimnulls

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Solution.FirstsupposethatthereexistsT2L(
Solution.FirstsupposethatthereexistsT2L(V;W)suchthatnullT=U.ThendimU=dimnullT=dimV�dimrangeTdimV�dimW,wherethesecondequalitycomesfrom3.4.Toprovetheotherdirection,nowsupposethatdimUdimV�dimW.Let(u1;:::;um)beabasisforU.Extendtoabasis(u1;:::;um;v1:::;vn)ofV.Let(w1;:::;wp)beabasisforW.Fora1;:::;am;b1;:::;bn2FdeneT(a1u1++amum+b1v1++bnvn)asb1w1++bnwm.BecausedimWdimV�dimU,wehavepnandsownontherightsideoftheequ

ationabovemakessense.ClearlyT2L(V;W)andn
ationabovemakessense.ClearlyT2L(V;W)andnullT=U.(4)Problem14onpage60ofAxler.(Hint:youmightneedproblem3onpage59ofAxler.)Solution.FirstsupposethatTisinjective.DeneS0:rangeT VbyS0(Tv)=v.BecauseTisinjective,eachelementofrangeTcanberepre-sentedintheformTvinonlyoneway,soS0iswelldened.Ascanbeeasilychecked,S0isalinearmaponrangeT.FurthermoreS0canbeextendedtoalinearmapS2L(W;V).Ifv2V,then(ST)v=S(Tv)=S0(Tv)=vthusSTistheidentitymaponV.Intheotherdirection,issuchanSexiststhenfor

u;v2W,ifTu=TvthenSTu=STvimpliesu=v.(5)P
u;v2W,ifTu=TvthenSTu=STvimpliesu=v.(5)Problem16onpage60ofAxler.Solution.DenealinearmapT0:nullST VbyT0u=Tu.Ifu2nullSTthenS(Tu)=0whichmeansthatTu2nullS.Inotherwords,rangeT0nullS.NowdimnullST=dimnullT0+dimrangeT0dimnullT0+dimnullSdimnullT+dimnullS;wheretherstequalityfollowsfrom3.4,therstinequalityfromrangeT0nullSandthesecondfromnullT0nullT.(6)Problem26onpage61ofAxler.Solution.DeneT2L(Fn)byT(x1;:::;xn)=(nXk=1a1;kxk;:::;nXk=1an;kxk)