Kernel and Image I have to find the kernel for this transformation:
$$T(p(x))=xp(1)-xp'(1)$$
My solution so far:
$p'(1)=0$
Hence $T(p(x))=xp(1)$
So $xp(1)=0$
Either $x=1$ or $p(1)=0$?
How do you solve the kernel from this point?
Thank you!

A polynomial $p(x)$ is in the kernel of $T$ iff it is mapped to the zero polynomial $q(x)=(0,0)^T$.
So, in order to be in the kernel, $(Tp)(x)=xp(1) - xp'(1)= x(p(1) - p'(1))$ needs to be equal to $q$ for all $x \in \mathbb R$. This is only possible if $p(1)=p'(1)$. 
For $p(x)=ax^2 + b x + c$, which values of $a,b,c\in \mathbb R^2 $ allow this identity?
