Given the linear mapping:
$f: \mathbb{R}_3[X] \rightarrow \mathbb{R}_5[X]$
defined by $f(P(X)) = P''(X) - P'(0) + P'(X)X^2 + (P(1) - P(0))X^5$
Find the image of this function. How should I handle this? I set
$f(P(X)) = P''(X) - P'(0) + P'(X)X^2 + (P(1) - P(0))X^5 = aX^5 + bX^4 + cX^3 + dX^2 + eX + f $ and tried to find restrictions on P(X) but I got nowhere. Should I translate this problem to coordinates, where this is easier to solve, and then translate back? Thanks in advance.
I haven't worked it out, but I can offer two hints, i.e. two possible ways to approach this problem.
(1) Use a basis of the domain vector space, see what the basis elements map to, and then the image will be the span of the images of the basis elements. For $\mathbb{R}_3[X]$ (although I'm more used to something like $P_3[X]$ as the notation for this space), use the standard basis $\{1,X,X^2,X^3\}$, find $f(\cdot)$ for each one of them, and then the answer is their span.
(2) Set up a generic element of the domain rather than the codomain space. A generic element of $\mathbb{R}_3[X]$ is a polynomial $P(X)=a+bX+cX^2+dX^3$. Find $f(P(X))$ and see how it looks.