Image and Kernel of Polynomial Linear Transformations

Question

Let $D: P\Rightarrow$ $P$ and $G: P\Rightarrow$ $P$ be linear transformations where, $D(p)=p'$ and $G(p)=q, $$q(x)=xp(x)$. How do I find $D$ and $G$'s image and kernel?

Answer 1

Welcome to math SE. On your next posts, please add any progress you have made on the problem, people here are a little peculiar on seeing that you have actually tried something. Anyway, I'm not:

I'd start by showing that these maps are actually linear (you can do it as an exercise). Now, for the kernel of $D$, it is $\text{Ker}(D)=\{p(x): p'(x)=0\}$, therefore the kernel consists of the constant polynomials, that is $\text{Ker}(D)=\{p(x): \text{ deg}(p(x))=0\}$. As about the image, our intuition says "it is probably the whole space of polynomials"; so let's try proving this: indeed, take $p(x)=a_0+\dots+a_nx^n$ and "integrate" it: consider the polynomial $q(x)=a_0x+\frac{a_1}{2}x^2+\dots+\frac{a_n}{n+1}x^{n+1}$. Then $D(q(x))=p(x)$ and our intuition was right.

About $G$, let's start with the kernel: if $p(x)\in\text{ker}(G)$, then $xp(x)=0$; if $p(x)$ is not identically zero, then $xp(x)$ has at most $deg(p(x))+1$ roots. Now i assume these are real/complex polynomials, hence we cant have $xp(x)=0$. Therefore $p(x)=0$ and $G$ is 1-1. About the image, it obviously consists of all polynomials of the form $a_0x+a_1x^2+\dots+a_nx^{n+1}$, that is all the polynomials that have $0$ as a root.