Question

Problem 5: Let V and W be vector spaces and let B = {V1, V2, ..., Un} CV be a basis for V. Let L :V + W be a linear transformation, and let Ker L = {2 € V: L(x)=0}. (a) If Ker L = {0}, show that C = {L(v1), L(02), ..., L(vn) } CW is a linearly independent set in W. (b) If C = {L(01), L(V2),..., L(Un)} C W is a linearly independent set in W, show that Ker L = {0}. (Hint: Any z e KerL can be written uniquely as a linear combination of the basis vectors.) (c) If B = {V1, V2, ..., Vk } CV is a linearly independent set in V, must C = {L(v1), L(V2), ..., L(Uk) } CW be a linearly independent set in W?

Answer 1

tesin Ker La foy To show c = {210), L(V2), ..., L(Un}} CW is linearly independent set. consider GLING) + C2 L(V2) +---thaton] =0 => L(CV to ų t...thn On] =0[/L is linear ten In =0 [: Kerl=soy Cmt Ol : fv, U2, {h, U2,..., dh] sa basis of vand this, lineanty independent This proves that {214), L(1) engi linearly independent set. 2) Goth V2+ of => 6=2= LI 2 is

yo Then I 9, C2, ..., an mich that @ca { LIQ), L(V2), ---, Livn]}. Owis linean arly independent To show kerl-{o). bol If possible let kerl & {oy. Then Fre such that Llor=0 ago and oto. = GVytly v2 +--- ten In B is a basis of v} => LIU). L (Glycy z ta.. tch Vn) 27 GLEG)+(₂ L(42) to .. thn Llan) verere) ané me] Lis linear -7 G=G=....360=0 1:{4(3), --Lf ven]}, is [linearly independent => Us Greyta "B . ㄷ O ten In -0. Vyt ..to. In a 0. which is a contradiction as vzo. Thus kert = {0} letor False Let v=1R? WER. & L: R = R defined by L(X,Y)= a. Then {(1,0).(0.1)) is linearly independent in v. But & L(1,0), L10,12}={, og is not linearly independent in w.