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 $V_3$?

Hint $\{1,x,x^2,x^3,x^4\}$ be a basis for $V_3$
find the matrix in the following way.
$f(1)=1=1.0+0.x+0.x^2+0.x^3+0.x^4$
$f(x)=x-1=(-1).1+1.x+0.x^2+0.x^3+0.x^4$
$f(x^2)=x^2-2x=0.1+(-2).x+1.x^2+0.x^3+0.x^4$
$$\dots$$
