Question

Problem 1: Let W = {p(t) € Pz : p'le) = 0}. We know from Problem 1, Section 4.3 and Problem 1, Section 4.6 that W is a subspace of P3. Let T:W+Pbe given by T(p(t)) = p' (t). It is easy to check that T is a linear transformation. (a) Find a basis for and the dimension of Range T. (b) Find Ker T, a basis for Ker T and dim KerT. (c) Is T one-to-one? Explain. (d) Is T onto? Explain. (e) Is T invertible? Explain.

Answer 1

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Answer: be a linear Transfor W = pptt)eP: ple-ot a subspace of P3. T: W P2 given by T(Plt)) = pllt) -mation. (a) To find for Range Let f(t) = at²+ b ť+ci+d en Then Tlflt)) = f'lt) = ( a++ bt? + ct +d) a basis T. d dt t'lt) = zat + 2 bttc zat + abttc = (3; 0,0 ) at' + (0,2,0) bt +10,0,11C Thus Range of 7 spanned by $13,0,0), (0,2,0), (0,0,1)} Since all three vectors are linearly independent. forms a basis for Range T. dimension of Range = 3. kert følt) e v5":"5+' (bit)) = 0 plt) € keit plt) = ax²+bx² + cx ad treet T(Plt)) bax + be? + ex+d) Thes (b) ut - 3 ax² + 2bx tiros 0,22 +0.X+0 -) Ano bio C=0 So any polynomialew of the form (0.28+ 0.2°+ 0.4+ d) be to pert, belong Rert spanned by {d(0, 0, 0, 1)} also LI. CS Scanned with CamScanner

basis for Rest= $ 10,0,0, 1} dimension of kert 1. (C) T is not one one. bes constant polynomialsdewi any T(2) d 2) = 0 It it's not one-to-one map. So, (d) since dun (P2) 23 = dim & Range T) (PLA) ic T is onto. a polynomial af ft) E P2 take we can degree 3ew st any Tlblt))= Alt). DA one one Seince T is not inverteble. it cann't be CS Scanned with CamScanner