I have the following transformation $L$, that I have shown is linear by showing its closure properties under addition and multiplication by a scalar:
$$
L:P_3(\mathbb{R})\rightarrow P_3(\mathbb{R})\\
f\mapsto X\cdot f'-f
$$
Where $f=aX^2+bX+c$, so I get that:
$$L(aX^2+bX+c)=aX^2-c$$
I now need to determine the kernel and the image of $L$. For now I'm thinking, that I need to solve $L(aX^2+bX+c)=aX^2-c=0$, but does this mean, that $a=0$ and $c=0$ therefore $a=c$ and the kernel is then $aX^2+a$ because it does not depend on $b$ at all, or am I going in a wrong direction here? Furthermore, I am pretty lost in finding the image og $L$.
I'm new to linear algebra, and English is not my first language, so please if I use some terms wrong, feel free to correct me.
Best Answer
for $a=c=0$ the polynomial $aX^2+bX+c$ becomes $f=bX$, so the kernel is the subspace of monomials of first degree in $X$.
And you have just found the image, it is the space of polynomials of the form $f=aX^2+c$ for $a,c \in \mathbb{R}$
Note that the kernel has dimension $1$ and the image has dimension $2$ as the rank-nullity theorem wants.