Question

8. (20 points) Consider the vector space V = P of polynomials in one variable. Consider the operator L: VV given by d L(p(x)) = (22p(x)) - xp(x). (a) (10 points) Prove that this operator is a linear operator. (b) (10 points) Given that L(.) = 2x2, find all other polynomial solutions p(x) to the equation L(p(x)) = 2x², if there are any.

Answer 1

6 an 8. L: NV is given by LCP) = d. (xp) – ap 63 P, (R) and P8 EV. then L(Pig) + bee) d (u (P&O) + *.) dd (a p(a) _ xp,20}+2 Let du -« (py@es +pco) a 6 P 22 ) _x Pa idre da he cou? - LC B, C) +L(B2003 Now, Let CEIR. then L (epers) at a (x² ep (*) _ x (epa) {alla dr Pion) p, ges} =cL(8,9) x C ! 1 )

So, LCÞ,80 +BⓇ) and LCP, @)+L(P2@es) for all p,co), ß ® EVEP) L(CHC)) = CLCF,®) for all CER, P,M) Ev Thus we can conclude that Lisa Linear L is a linear operator, operator so , b Given that Lx) = 2x² Let þ (W) = axtb be the polynomial such that L ($9).) 2x² Now, L(antb) an (ax3 + bx²) .. anne & (47(antb)) - n (anth d dx Ban²+ 2br -ax²_be = 2ax²tbu

50, L (axtb) - 2x2 gan²+bx 222 ラ 2a = 2 and b.co bao so, antb = 2. thus 5a =1, þ (B) = po = is the only solution for L(p«)) = x2 Notes We see that LC1 = x, L(x) = 2x%, L (x2) = 323, L (X") = (n+1) at Since. qx, 2x², 3x3, ...} is linearly independent set, so Lis one to one Then L(P(4)) = x² has unique solution pa)= x