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# Class Note for MATH 790 at KU

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This 5 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Kansas taught by a professor in Fall. Since its upload, it has received 23 views.

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Date Created: 02/06/15
Annihilators Linear Algebra Notes Satya Mandal September 217 2005 Let F be a led and V be vector space over F With dimV n lt 00 As usual V Will denote the dual space of V and VM V 1 De nition 01 For a subset S Q V de ne the annihilator annS f E V fu 0 for allu E S We also use the notation SO annS We may also temporarily use the notation annV S annS to underscore the fact that annV S is a subspeee of V 2 First note that for S Q V7 the annihilator annS Q V is a subspace of the dual space 3 Also note for S Q V7 if W SpanS then annS annW 4 Now let S Q V Then according to above de nition7 annihi lator of S is annS annV S is a subspace of V In this case7 there is another natural annihilator of S as a subspace of V as follows annVSuEVfv0f0r all fES It is possible to mix up two annihilator of S The rst one is a subspace of the double dual VW and the second one is the subspace of V We Will justify that these two annihilators are same Via the natural identi cation of V and V Let L V VW be the natural isomorphism Let S Q V Then La7mV 8 annV 8 Proof Let 1 E annV S Then fv 0 for all f E 8 Hence Lvf fv 0 for all f E S So7 Lv E annV 8 Therefore La7mV 8 Q annV 8 Now let G E annV 8 Since L is an isornorphisrn7 we have G Lv for some 1 E V We need to show that v E annV 8 But for f E S we have 0 Cf Lvf fv Therefore 1 E annV S So7 annV S Q LannV Hence the proof is complete Now7 for asubspace W Q V7 the annihilator WO annV W has two annihilators They are annV W0 and annV W0 It follows from above that LannV W0 annV W0 Since7 L is a linear isornorphisrn7 we also have dirna7mV W0 dirnamzV W0 8 Theorem 01 For a subspace W Q V we have W annV a7mV7 also written as W W00 Proof Write U annV annV It is easy to see that W Q U Therefore7 it is enough to show that dimW dimU We have dimW dimannV dimV Also7 by the same theorem7 dimannV Wdima7mV a7mV7 dimV dimV It follows that dimW dimannV a7mV7 dimannV annV dimU So the proof is complete Lemma 01 Suppose V is vector space of nite dimension dirnV it over F Let fg E V be two linear funetionals let Nf be the null space off and N9 be the null space ofg Then Nf Q N9 if and only ifg of for some 0 E F Proof Obvious i If g 0 then 9 of with c 0 So we assume that g a 0 So dirnNg 7271 Since Nf Q N9 we have f a 0 and dirnNf 7271 Therefore Nf N9 N3ay Now pick 6 N Since dirnV n it follows that V N Fe Also fe a 0 and 96 a 0 Write c gefe Claim that 90 First note 96 cfe Now for n E V we have n y A6 for some y E N and A E F Therefore 9 921 we W6 Ac e Cfy Ae cffv So the proof is complete Following is Theorem 207 page 110 Theorem 02 Suppose V is vector space af nite dimension dim V n7 over F Let g7f17 fT E V be linear functionals Let N be the null space ofg and Ni be the null space of f2 Then N1 0 N2 0 0 NT Q N if and only ifg ELI cifi for some Ci 6 F Proof Obvious i We use induction on r to prove this part Case 7 17 is the above Lemma 01 Now7 we assume the validity of the thoerem for r 7 1 functionals and prove it for 7 Write V NT Let 9 Givnfl f1V77 Ll finkm be the restrictions of the respective functionals to V lnduction ap plies for these functionals and it follows that 9 Ti Cifi 21 for some Ci 6 F This means7 for all m E V NT7 we have gm This means7 for all m E V NT7 we have gm Write 171 h 9 i Cifi Then NT V Q Null 7 Spaceh By the case r 1 or by Lemma 01 It follows that h chT for some CT 6 F Hence T71 7 9 Zcifi h Zcifi i1 i1 and the proof is complete

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