How to prove that $T: p(x) \longrightarrow p(x) + xp'(x)$ is surjective? Let $\mathbb{R}_n$ denote the space of polynomials in $\mathbb{R}$ of degree at most $n$.
In an exercise for school, I have to prove that $T:\mathbb{R}_n \longrightarrow \mathbb{R}_n$ is surjective with $T: p(x) \longrightarrow p(x) + xp'(x)$. However, I don't know how to prove this...
Thanks in advance for your help!
 A: Choose a basis, $\{1,x,x^2,\dots,x^n\}$. Then let some arbitrary $p(x) = a_0+a_1x +a_2x^2+ \dots+a_nx^n$.
It's easy to see that $T(p(x)) = a_0 + 2a_1x + 3a_2x^2 + \dots + (n+1)a_nx^n$. In other words the image of $(a_0,a_1,\dots,a_n)$ under $T$ in $\mathbb{R}^n$ is $(a_0,2a_1,\dots,(n+1)a_n)$. Clearly one can choose $(a_0,a_1,\dots,a_n)$ such that $(a_0,2a_1,\dots,(n+1)a_n)$ maps to any point in $\mathbb{R}^n$
Thus T is surjective.
A: Take $q(x):=\sum_{k=0}^{n} a_k x^k\in\mathbb{R}_n$.
You want to find a polynomial $p(x):=\sum_{k=0}^{n}b_k x^k\in\mathbb{R}_n$ (or more precise $b_k\in\mathbb{R}$) such that $q(x)=p(x)+xp'(x)$.
Since $p'(x)=\sum_{k=0}^{n-1} (k+1)b_{k+1}x^k$ you get $xp'(x)=\sum_{k=0}^{n-1} (k+1)b_{k+1}x^{k+1}=\sum_{k=1}^{n}kb_kx^k=\sum_{k=0}^{n}kb_kx^k$.
So finally you have $\sum_{k=0}^{n} a_k x^k\overset{!}{=}\sum_{k=0}^n (b_k+kb_k)x^k$.
If you compare the coefficients you get $a_k=b_k+kb_k=(1+k)b_k$ so take $b_k=\frac{a_k}{1+k}$.
Does this help?
A: Alternatively, choose a basis $\mathbf{v}=(1,x,x^2,\dots,x^n)$. Suppose that $p\in\ker(T)$. Then $T(p)=p(x)+xp'(x)=0(x)$ so $p(x)=-xp'(x)$ (I denote $0(x)$ to be the zero polynomial).
Write $p(x)=\sum_{k=0}^n a_kx^k$. Then $p'(x)=\sum_{k=0}^{n-1}(k+1)a_{k+1}x^k$ thus $$\sum_{k=0}^{n}a_kx^k=-\sum_{k=0}^{n-1}(k+1)a_{k+1}x^{k+1}=-\sum_{k=1}^{n}ka_{k}x^k$$ Note that we must have $a_0=0$. Comparing coefficients, we have that $a_k=-ka_k$ for each $k\in\{1,2,\dots,n\}$. Then $a_k(k+1)=0$. Since $k\neq -1$, then $a_k=0$ for all $k$, thus $p(x)=0(x)$. Then we see that $\ker(T)=\{0(x)\}$ so $T$ is injective.
A: If $p\in\ker(T)$, then
$$p(x)=-xp'(x).$$
Show by induction that
$$(k+1)p^{(k)}(x)=-x p^{(k+1)}(x)$$
for all $k=0.1.2,....$
Hence $p^{(k)}(0)=0$ for all $k=0.1.2,....$. If $p(x)=\sum_{k=0}^{n} a_k x^k,$ it follows that
$$a_0=a_1=...=a_n=0.$$
Hence $\ker(T)=\{0\}.$ This shows that $T$ is injective and hence surjective.
