Question

1. Let U,V,W be vector spaces over field F, and let S ∈ L(U,V) andT ∈ L(V,W).

2. (a) Show that if T ◦ S is injective, then S is injective

3. (b) Give an example showing that if T ◦ S is injective then T need not be injective.

4. (c) Show that if T ◦ S is surjective, then T is surjective.

5. (d) Give an example showing that if T ◦ S is injective then S need not be surjective.

Answer 1

- Let SEL(UV) and TE L(v.W) and Tos is injective. Now 5: 0 V . T: V W and ToS: U maps. W 'are linear T let Tos is injective linear map / transformation let at kers, Then S(d) = 0 → T(SC) = T(0) → T ((Q)) = 0 (OT is lineas map) > To S() = 0 => d=0 (o: Tos is injeetive) in ker s ={0} Since her Sivis Kers SOS , Then 5:0 →V is inpeetive. o consider, s: R R² is defined by s(2.7) = (x,4,0) and Ti R? Ar is defined by T(4:2) = (1,7) clearly sis injective and T is not injective . But tosi Ro RT , Tos(4.7) = (x,y) is injective maps linaus. © Let Tos: U JW be surjective linas map. et qew, Then I ft U sot tos (y) = x (:: Tos is surjeefive) - Now To s(Y) = T(S(%)) = x Since Si U V is a linear map, Then S(Y) EN. Therefore it has a pre-image 8(%) in v. under the mapping T: V+W. Since x is arbitary, Then I is suspective linear map. Scanned with CamScanner

@ same as 6 consides si R R² is defined by $(2.9) = (4.7,0) and T: R² ~ Ř is defined by TC. 8,2)= (27) Then s is not surjective, but tosh ) Tos(x)) = T(*%,0) is injective linews map. = (,y) Scanned with CamScanner