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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.

Thank you for your time!

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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, $\ker L=\{bX\,|\,b\in\mathbb R\}$.

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  • $\begingroup$ Thank you! Really appreciate the help! $\endgroup$ – Nikolaj Feb 28 at 15:52
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    $\begingroup$ @Nikolaj Feel free to mark this answer or the other answer as accepted $(\large{\color{limegreen}{\checkmark}})$. $\endgroup$ – callculus Mar 9 at 14:32
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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 $f=bX$

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  • $\begingroup$ Thank you that makes sense! $\endgroup$ – Nikolaj Feb 28 at 15:52

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