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 $V_3$?
 A: If $p(x)\in\ker f$ then 
$$f(p(x))=p(x)-p'(x)=0\iff p(x)=p'(x)\Rightarrow \deg p=\deg p'$$
but 
$$\deg p'=\deg p \iff p= 0$$
so we conclude that
$$\ker f=\{0\}$$
and then $f$ is injective so bijective in finite dimensional space.
A: Hint $\ p\in \ker f\ \Rightarrow\ p = p' \Rightarrow\ p = p'' \ldots \Rightarrow p = p^{\color{#c00}{(4)}}\! = 0\ $  by $\,p\,$ by $\,\deg p \le\color{#c00} 3$
In fact we can explicitly invert $\,f = 1 - D,\ D = d/dx\ $ using  $\,D^{\color{#c00}4} = 0\,$ on $\,V_3$
$$(1-D)(1+D+D^2+D^3) = 1-D^4 = 1$$
A: 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 $gf$.
A: Hint $\{1,x,x^2,x^3,x^4\}$ be a basis for $V_3$
find the matrix in the following way.
$f(1)=1=1.0+0.x+0.x^2+0.x^3+0.x^4$
$f(x)=x-1=(-1).1+1.x+0.x^2+0.x^3+0.x^4$
$f(x^2)=x^2-2x=0.1+(-2).x+1.x^2+0.x^3+0.x^4$
$$\dots$$
