Question #154595

Let V be the vector space of polynomials of degree less than of equal to n.Show that the derivative operator on V is nilpotent of index n+1.


1
Expert's answer
2021-01-12T16:03:46-0500

Let P denotes a polynomial and D denotes the derivative operator. Then let P=a0+a1X++akXkP=a_0 +a_1X+\cdots+a_kX^k for kn.k\leq n. Then DP=a1+2a2X++kakXk1.DP=a_1+2a_2X+\cdots+ka_kX^{k-1}. Then degDP=k1.deg DP=k-1. Hence applying this repetitively we get, deg DP=kiDP=k-i for iki\leq k . For i=k,i=k, the degree is 0 hence a constant polynomial. So for i>ki>k the polynomial vanishes. Hence for any polynomial of degree n,\leq n, Dn+1(P)=0.D^{n+1}(P)=0. Also for the ploynomial XnX^{n} , we see, Dn(Xn)=n!.D^{n}(X^{n})=n!. Hence DnD^{n} is not the zero operator on V. Hence it is nilpotent of index n+1.


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