Question

Q7 8 Points Let V, W, and U be three finite dimensional vector spaces over R and T:V + Wand S : W → U be two linear transformations. Q7.1 4 Points Show that null(So T) < null(T) + null(S) Please select file(s) Select file(s) Save Answer Q7.2 4 Points Show that rank(S • T) > rank(T) + rank(S) – dim(W) (Hint: Use part (1) at some point)

Answer 1

vector space Fal Given V, W, u are finilē dimensional over R and T:V W 3 SiW-U To show mull (507) s null(1) trull is) By Rank No ility Theorem Rank (T) + dim (her (T))= din LW Rienke 8) + dimaker (5) )= ciim l W) Hank (sot) + dim (ker (so t) = dim (v) let {V1, V2 vz] is be the basis kelt a low

Then (SoT) (u;)=0 Thus we -V2191192.. yn} can fore are Then До ay + AŤ bi, bz. ayit A2Y2 t =) aiyit - bzlz=0 fore each i. So keyct) C her (soT). extend this to a basis {v, V₂ each fare keu (so 7) . Then Je to k We have (s. 1) yg) s(tly;)) Hence Tly;) & Kells) fore each j. Further of ai,...ak er scolars such that data biVpbati buz a (tly, t tak (TUN) =0 т( ay, + a292 ttany k) =0 Quyut ker (1) bz ER such that + anya bivi + b2 V2t. tbalz takyk-bivi But these vectors forem a basiq fore Ker(Sot), so in particulat , a, Thus {ly,), ... Ilyk)} is a line Inde pendent Subset Kerts) and so dim(her (s)) >k. dim (her (1)) + dim (ker(s) ədim ker/sot)) Null (Sot) < null(t) + null (5) Hence proved Foz From have Null (Sot) Mull (T) + Null is) By using din teren din hoe (so). A dim(o)= reank (507) + dim (ker (50) dom (ker(Sol) = null SoT) = Pulting this - au=0 4 above we wing Rank hullity theorem, { from 092,0%. dim va mank (Soto) value in ean ④

= ) I 7 dim (her (t))+dim (keris)) > dim(o)- Rankso) 6 Also from @ recenk (T) + dim (ner t)) = dimcy) dim (ker T)= dimev) dimev) - reant to 66 and from @ mank(s) + dim (hers) dim (w) => dim (ner(s)) dim (W) -recents. = Pulting een into G dim v - Meent LT) + dèm(w) – moenk(s) 3 din Lv) R 160 Rank (sot) > Ranke i) + reanu (S) 7 Rank (507) dim (W) Rank (507)> Rank (T) + Rank (s)- dimew), Hence preoved