Question

Prove the following

→ V such that (a) If T:V + W is linear and injective, then there exists a linear map S: W ST = I. (b) If S: W → V is linear and surjective, then there exists a linear map T:V ST = 1. W such that

Answer 1

Answer : Suppose To V W is a linear f injective map. we define soWv by SCT ()) = v because I is injective, each element of W can be represented as in the form TCW) in only one way. pie. T(W) = T(u) = NEU so T is well defined, linear & injective map. first we check that s is linear map from W to v. for any relf & T (oc) ,T(Y) E W , we have s(Text + Tey)) = S( 7 (2+y)) S:T is lineas = (ST)(a + (ST)(4) = x+Y = S(TC»c)) + 5 (7(41) f Sert(a) = S(T()) = 72 = r s (T(x)) s is linear map ..

we know that s can be extended to S'e L(w,v) such that for any new, we hawe scu) = s'(u). For any Vev, gince TCU) EW (ST) (V) = 5 (TCVD) is is linear Thus map. ST is identity i.e. st=I. 1 Suppose I: W - W is linear 4. surjective map. Thus , which equals range of s is finite dimensional. let V, U2, ...... Vm be a basis of V, since s is surjective , for each j there exists wj e W such that Swi) = 0; #j Then there exists a unique linear map To V W such that Tous) = W;

Hence for any VE W .. v=a, Vita₂V2 torront amum for some de, az,oreng am GIF we have , (s T}(v1 = $((a. Vita,vet. ---+amvm)) = s (, 5cv) +az T(v.) toont am TCvm) = (a.Wit arrint amwm) ( linear IT is = a, Slove) t az S(we) +- tam SCW m) = aivit a₂Vztrroot amVm SiS is , linear (StXv)' =V .. ST = E V . Hence we have ST is an identity map.