Prove that $f(p(X))=p(X)-p'(X)$, where p is a polynomial, is bijective Let $V_d$ be the vector space of all polynomials with real coefficients of degree less than or equal to $d$. The linear map $f: V_3 \rightarrow V_3$ is given by: $f(p(X)) := p(X)-p'(X)$. Show that $f$ is bijective.
To show that $f$ is injective I could find that $\dim(\operatorname*{null}(f))=0$. So I guess I would need to show that a polynomial of degree less than or equal to $3$ is equal to its derivative only at $x=0$. But I don't know how to do that.
If I would simply set up this equation: $$\sum_{j=0}^3(\alpha_j - (j+1)\alpha_{j+1})X^j = 0$$ how could I show that the trivial solution is the only solution?
And to show that $f$ is surjective, would it be sufficient to say that $p(X)-p'(X)$ yields an equation that contains a linear combination of the basis of $V_3$ and with arbitrary coefficients that linear combination spans $V_3$?

Your linear map $f:V_3\to V_3$ is equal to $f=1-L$ with $L:p\in V_3\mapsto p'\in V_3$.
Notice that $L$ is nilpotent, that is, that some power of $L$ is zero. 
It follows by a calculation then that $g=1+L+L^2+\cdots+L^k$, for any $k$ sufficiently large so that $L^{k+1}=0$, is an inverse map to $g$: indeed, you can simply compute $fg$ and $gf$.
