# Kernel in a linear transformation

The question is

Let $$P_3(\mathbb{R})$$ denote the vector space containing real polynomials of the form $$f = aX^2 + bX +c$$.

Let L be the linear map

$$L: P_3(\mathbb{R}) → P_3(\mathbb{R})$$ where $$f ↦ X \cdot f' - f$$.

Find the image and kernel of L.

So in a previous assignment I have already shown that L is a linear transformation.

Regarding the image I have already found it: $$L(f) = X(aX^2 + bX + c)' - (aX^2 + bX + c) = X(2aX + b) - aX^2 + bX + c = 2aX^2 + bX - aX^2 - bX - c = aX^2 - c$$

So the image of L is the set $$\{aX^2 - c|a,c\in\mathbb{R}\}$$

I'm not sure what to do about the kernel. I know that I have to find all the $$f\in P_3(\mathbb{R})$$ such that $$L(f) = 0$$ where $$0$$ is the null/neutral element, but I'm not sure how to proceed.

You proved that $$L(aX^2+bX+c)=aX^2-c$$. Therefore,$$L(aX^2+bX+c)=0\iff a=c=0.$$In other words, $$\ker L=\{bX\,|\,b\in\mathbb R\}$$.
• @Nikolaj Feel free to mark this answer or the other answer as accepted $(\large{\color{limegreen}{\checkmark}})$. – callculus Mar 9 at 14:32
Well $$f=aX^2 + bX + c$$ and $$f'= 2aX + b$$, and we are asking for which $$f$$ does $$L(f) = Xf' - f = (2aX^2 + bX) - (aX^2 + bX + c) = aX^2 - c = 0$$ for all $$X$$. And this is all polynomials where $$a=c=0$$, so $$f=bX$$