Prove $T$ is a linear transformation and find $\ker T$ and $\mathrm{im} T$ Let $P_n$ be the vector space of all real polynomials of degree less than or equal to n.
Let $T : P_3 \rightarrow P_3 $ be defined by $T(p(x)) = xp'(x) - 3p'(x)$, where $p'(x)$ represents the formal derivative of $p(x)$.
I am not sure if I am doing correctly please correct me if I did anything wrong.
Let $T(p_1(x)) = xp_1'(x)-3p_1'(x)$ and $T(p_2(x)) = xp_2'(x)-3p_2'(x)$
since $ xp_1'(x)-3p_1'(x) + xp_2'(x)-3p_2'(x) = (xp_1'(x) + xp_2'(x)) + (-3p_1'(x) -3p_2'(x)) = x(p_1'(x) +p_2'(x)) -3(p_1'(x) + p_2'(x))$
so that $T(p_1(x) +p_2(x)) = T(p_1(x)) + T(p_2(x)).$
and $T(λp(x)) =  λxp'(x) - λ3p'(x) =λ(xp'(x) - 3p'(x))$
Now I can say T is a linear transformaion.
Any hint on how to find kerT and imT?

Actually, $T$ is composited by two linear transformation.
$T_1:P_2\rightarrow P_3$ which satisfies
$$T_1(p(x))=(x-3)p(x)$$
$T_2:P_3\rightarrow P_2$ which satisfies 
$$T_2(p(x))=p'(x)$$
We have $T=T_1\circ T_2$.

Because $T_1$ is injection and $T_2$ is surjection, then we get 
$$\text{Ker}(T)=\{p(x)\in P_3:p'(x)=0\}=\mathbb R$$ 
and 
$$\text{Im}(T)=\{(x-3)p(x)\in P_3:p(x)\in P_2\}=\{p(x)\in P_3:x-3|p(x)\}$$
