The differentiation operator on the space of polynomials of degree less than or equal to 3 is represented in the ‘natural’ ordered basis by the matrix \[ \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}. \] What is the Jordan form of this matrix? (F a subfield of the complex numbers.)
Solution
Let \(D\) denote differentiation on the polynomials of degree at most three. Since \[ D^4=0 \] but \[ D^3(x^3)=6\ne0, \] the nilpotency index is four. Therefore the Jordan form must contain a block of size four; as the whole space has dimension four, it is a single nilpotent Jordan block [@HK71]. Hence \[ \boxed{ J=J_4(0)= \begin{pmatrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0 \end{pmatrix}.} \]
Explicitly, in the ordered basis \[ \left(1,x,\frac{x^2}{2!},\frac{x^3}{3!}\right), \] differentiation sends each basis vector after the first to the preceding one. Thus its matrix in that basis is exactly the displayed Jordan block, which also directly verifies the similarity.