Question

Let V and W be finite dimensional vector spaces over R and T:V + W be linear. Let V be a subspace of V and Wo = T(V). (Select ALL that are TRUE) If T is surjective then Vo = {v EV : there is w E Wo such that T(v) = w} If T is injective then dim(VO) = dim(W). dim(ker(T) n Vo) = dim(VO) - dim(Wo).

Answer 1

V and W are finite dimensional vetor spaus. No isa a subspan of v and Wo TV.). © If Tisa senjection the french w EW, I VEVSA. {ver there is we lo such that Tul=w TCU)=0} This statement is fabe TO)= W. But Voz ut v = R², WAR. V R207 , Wo=W=R. TIN - W. TM) x then Tis onto linear map. To ) = Wo T(V) = No But vo & {NEWl there is WF Wo such that T(W) = wfv s. stake mut on FALSE and

VO If T is injutive than T ) = TM. ) be a bons of Tu To Tokis linearly independent because if assume dimosk and timran E iTui -O Treni) T(& civi) 2 = 0 Tirone-one &i=1,2,K WE WO= T(V.) them I VE Vost Tv = w indon. Thainigaw 11 So {TU ,. .,TUK} generates Wo and basinf linerily independent So Here dim Woo kadimvo. statement in TRUE 3 ht s=ker TSV. and dim sal, dimvork. wt ve kNT Ovo thn Ne ke T - TV = 0 let T:Vo W. such that T (W) 2 Tv xutro The Ti is a sujection on TV.) = Wo st.) - W.. NE Kw Ti aus te vo Ker T, 2 Kertvo Tula Tul20

nullity of Ti Now from rank rank nullity theorem rank Ti is dim va dim Tv.) a dim kenti & dimvo. dím Wo + dim (kerł Ovo) - dimro - dimror dím Wo 7 dim (ker Tovo) staff met 3 is TRUE