Answer to Question #154595 in Linear Algebra for Ashweta Padhan

Question

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.

Answer 1

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