Let V = Rn[x], the vector space of polynomials of degree <= n. Define D : V → V to be differentiation with respect to X. Find the rank and the nullity of D.
how to find the kernel of the differential operator? I think the dimension of the ker is supposed to be n. and the image, is it also n?
Encoding the polynomials as coordinate vectors $$ p_n(x) = \sum_{k=0}^n c_{k+1} x^k \\ c = (c_k) \in \mathbb{R}^{n+1} \\ e(p) = c $$ e.g. $$ p(x) = 2 x^3 - x + 5 = \sum_{k=0}^3 c_{k+1} x^k \in V \\ c = (5, -1, 0, 2)^T \in \mathbb{R}^4 $$ or another example $$ p(x) = 1 \\ c = (1,0,0,0)^T = e_1 $$ we have the differentiation operator $d/dx$ acting on the coordinates as $$ D : \mathbb{R}^{n+1} \to \mathbb{R}^n $$ with matrix representation $$ D = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & \dotsb & 0 \\ 0 & 0 & 2 & 0 & 0 & & 0 \\ 0 & 0 & 0 & 3 & 0 & & 0 \\ \vdots & & & & & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0 & \dotsb & n \\ \end{pmatrix} \quad (*) $$ e.g. $$ n = 3 \\ D = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix} \\ D \, (2x^3-x+5) = D \, (5, -1, 0, 2)^T = (-1, 0, 6)^T = 6 x^2 - 1 $$
Then we can infer from $(*)$ $$ \ker D = \{ c e_1 \mid c \in \mathbb{R} \} $$ which consists of all constant polynomials, a one dimensional subspace of $V$, and $$ \DeclareMathOperator{img}{img} \img D = \mathbb{R}^n $$ which are the polynomials of maximal degree $n$.