Linear mapping between vector spaces. I'm curious to see if the following mapping is in fact bijective. Let $P(\mathbb{R})$ be the space of all polynomials with real coefficients. Let $f\in P(\mathbb{R})$. Then is
$f(x)\mapsto (x-1)f'(x)+f(x)$ a bijective map?
 A: Note that $f(t \mapsto \sum_{k=0}^n p_k t^k)(x) = \sum_{k=0}^{n-1} (1+k) (p_k-p_{k+1}) x^k + (1+n) p_n x^n$ .
Hence if $f(p) = q$, we have $(1+n)p_n = q_n$ and $(1+k)(p_k-p_{k+1})  = q_k$ for $k < n$. This is solved uniquely by $p_n = {1 \over n+1} q_n$ and
$p_k = {1 \over 1+k}q_k +p_{k+1}$ for $k <n$.
Aside: To show uniqueness, suppose $r(x) = \sum_{k=0}^n r_k x^k = 0$ for all $x$. Then $r(0) = r_0 = 0$, and by differentiating, we get $r_1 = 0$, etc. Hence if $a =b$, then the corresponding coefficients must equal, that is $a_k = b_k$.
Alternative: Here is another way. Pick some $n$, the consider the restriction of $f$ to $P^n(\mathbb{R})$. Clearly the restriction satisfies $f: P^n(\mathbb{R}) \to P^n(\mathbb{R})$, both finite dimensional spaces. Now consider $f(p) = 0$. From above we see that $p_n = 0$ and so $p_k = 0$ for $k<n$ and hence $p=0$. Hence $f$ is invertible (on $P^n(\mathbb{R})$).
Since $n$ was arbitrary, we see that we must have ${\cal R} f = P(\mathbb{R})$, and if $p \in \ker f$, then we must have $p \in P^n(\mathbb{R})$ for some $n$, hence $p = 0$, hence $\ker f = \{0\}$.
A: To show it's invertible, solve the appropriate differential equation and choose the constant of integration so that you end up with a polynomial solution.
