If \(V\) is the space of all polynomials of degree less than or equal to \(n\) over a field \(F\), prove that the differentiation operator on \(V\) is nilpotent.
Solution
Let \(D:V\to V\) be differentiation. For a monomial \(x^k\) with \(0\le k\le n\), \[ D^{n+1}(x^k)=0, \] because after at most \(k+1\le n+1\) differentiations the monomial becomes zero. Since \[ 1,x,\ldots,x^n \] form a basis of \(V\), the operator \(D^{n+1}\) vanishes on a basis and therefore \[ D^{n+1}=0. \] Hence \(D\) is nilpotent [@HK71].
More precisely, if the scalar field has characteristic zero, then \(D^n(x^n)=n!\ne0\), so the nilpotence index is exactly \(n+1\); the exercise only requires nilpotence, which holds over every field.