Question

Q4 For the homomorphism from P2, the vector space of polynomials of degree two or less to P3, the vector space of polynomials of degree three or less given by : P→ P(t + 1)dt. a) Find : 0(1), 4(x), (x2) b) Find the range space and the kernel of o c)Prove that the range of O is {P € P3 / P(0) = 0} d) Prove that is a isomorphism from P2 to the range space. Let's St+1)dt = [x+2] x = (x+1)= (0-1) 14

Answer 1

homomomorphism on vector spaces is just a linear map.and a one one onto i.e. bijective linear map is called an isomorphism
& P. P. st & PEP. *( P(x)) = (p(trijdt = x- o = x - = lidt = $(1) () *) * Pictorat (6) Range space { R (*) het 9(x) E R(*) J P(x) e P2 sot • 2(a) a locttilat I scalers a, a, a, st. Now as P(x) EP P(x) = a, tqxta x²

96) itą (t+1) +9 20 ha 13.00-adi satt ens 3.673"} = ,2 +9:(*412*7 92 (*+1 - 262) = (a +92 + a) x + ( a +93) 2 + 2 x a, tęta = by az = bz let and antagabe [ 26m) = box + bez de Trage so for any polynomial i el 7 scalens 4, 2, 23, 24 (x) = a, ta x + 2 x² + 2x² for x=0, we have d(0)=0 Hence the range space is 1 TR(6)= { 2 (m) E P3 | 2(0)=0} sprost of tes} as well Kernel : let & P(x) € Ker(6) &(Plus) = 0

LP(t+1)dt =o pleidt=0 - let P(x) = a, tak tau? them from (2, +42+ + 9st) 1 = 0 from both > (9149243)*+( 403)2 + aneste By equating coefficients of like powers of a the sides, we get a, tątag=0 By back-substitution, we get azo, a=0 =0. Hence P(x) = 0 ker (4) = {0} I since ker(&) = {0} & is one one dainear map (given homomorphism). we know that any one-one map from its domain to its range is always onto. Hence & is an isomorphism from P2 to the range space.