Question

Exercises 1. The set P2 of polynomials of degree less than or equal to two is a vector space under polyno- mial addition and scalar multiplication by real numbers. (a) (5 points) Show that the set A = {1, 2, 22) is a basis for P2. (b) (2 points) Find the coordinate vector of an arbitrary polynomial of degree 2 or less (i.e., vector) p(x) = ax? +bx+c relative to the basis A. (c) (3 points) The derivative transformation D:P2 → P, defined by DS) = f' is a linear transformation we know this from our Calculus I course). Without writing down any matrix, how do you know that D cannot be an invertible transformation? (d) (5 points) Find the matrix [D]A associated to differentiation. Then determine its null space and the polynomials in P2 to which Nul [D] A corresponds. Does your answer make sense given what you learned in Calculus? 2. The set PE, of functions of the form palet with p(x) in P2 is also a vector space under function addition and scalar multiplication by real numbers. (a) (5 points) Show that C = {e", ze", zer} is a basis for PE, (b) (2 points) Find the coordinate vector of an arbitrary element i.e. vector) f(3) = (ax2 + bx + c)e in PE2 relative to the basis C. (c) (3 points) Find the matrix [D]c of the derivative transformation D : PE, PE, (d) (5 points) Note that [Dc is invertible! Find its inverse and use it to compute (32?e" - 6xe® +e") di.

subject: Linear Algebra

if someone could answer and explain why the answers are correct that would be much appreciated. Thanks in advance!!

Answer 1

sol 1: Poat Qot Qix + Qgx : Qo, Q1, Qge R3 a et ple) e la then p(x)= aot aix+ agx for some do ,a,,ad * p(x) is linear Combination of lixx. Thus This is true for all p(x) E Pa Consider Q+ But tuzo B + 271 = 0 (Taking derivative on sotto TT side) T 21=0 a Y=o TT B=0 & BSO, V-o and + Bx+ yardzo = 20 Acqlix,x²3 ů a basis for Pa & b Gt P(x1= ax + 500+ c = Colt 3x + a.? [p(x)] = [c, b,]" = ie Cordinate vector of pixla cha- e Define Di Pa 1. clearly Pa by D(f) = f" D is linear transformation

As Pg is vector space of all atmost 2-degree poly. and derivative of these polynomials is almost any zero degree polynomial ce. Constant polynomial. So, polynomials of degree and I does not have П ри image This implies that Du not onto - and Therefore not invertible d. Now D: Pg + lg by Dl P(x)) = þ" Now A= 91,8x²3 so, D(1)= 0 = 0.1 + 0.X + 0.? D(x) 0 = 0./+ 0.8+ 0.? D(72) = 2 - 2.1+ 0-+ 0.? So, [D]A= 10A [D[+] A DERYA) 1. = 10 001 loool Now Null D = { PB) E Pal | D(pl) = 0 3 = ſao+(18+ agar? (Ro+ajx+ ag x?)"=03 = g qut Qix + agx? 20,503 - î Got Qix I 20141 ER3 = pi.

2 соо Mulon{ (6): 6. 9) 8: (13 33 212 - NUIDA] i Goresponds to Pr. in Pa.