Examples from Matrix representations

Differentiation on polynomial spaces

For $n\ge 1$, differentiation is the real-linear map

Take $n=4$ and let

and use the ordered standard polynomial bases

Evaluate $T$ on each domain basis vector and find the coordinates of the result in $B_W$:

These coordinate vectors, in order, are the columns of the matrix representation:

For

This agrees with

The zero first column records $T(1)=0$, so differentiation is not injective. The four pivot columns give rank $4=\dim(P_3(\mathbb R))$, so it is surjective. In words, differentiation loses the constant direction but still reaches every polynomial of degree at most $3$.

Integration gives an injective map

Define

The integral is linear in $p$, so $I$ is a linear map. If $I(p)$ is the zero polynomial, differentiating the identity $I(p)(x)=0$ gives $p(x)=0$ for every $x$, so $\ker(I)=\{0\}$. Thus $I$ is injective. Its rank is $\dim(P_n(\mathbb R))=n+1$, which is smaller than $\dim(P_{n+1}(\mathbb R))=n+2$, so it is not surjective. Concretely, every polynomial $I(p)$ vanishes at $x=0$, so the constant polynomial $1$ is not in the image.

A scalar map and its inverse

Let $\alpha\in\mathbb R\setminus\{0\}$ and define

Relative to the standard polynomial basis $(1,x,\ldots,x^n)$,

The candidate inverse is

whose matrix is

For every $p\in P_n(\mathbb R)$,

Thus $T$ is bijective with inverse $S$ and its matrix $[S]=A^{-1}$ matches the theorem.