Show that $T(p)=p+p'$ is invertible 
We have a linear transformation $T:P\rightarrow P$ where $T(p)=p+p'$ and $P$ is the vector space of all real polynomials. I want to show that $T$ invertible.

I was able to prove that its injective, but I have difficulty to show that is surjective or is possible even to find $S$ such that $T\circ S=I$.
 A: Note that your operator can be written as
$$T(a_n x^n + \dots + a_0) = a_n x^n + (n a_n + a_{n - 1}) x^{n - 1} + \dots + (2a_2 + a_1) x + (a_1 + a_0)$$
So given a polynomial $p = b_n x^n + \dots + b_0$, you're looking for a simultaneous solution to
\begin{align*}
a_1 + a_0 &= b_0 \\
2a_2 + a_1 &= b_1 \\
&\vdots \\
n a_n + a_{n - 1} &= b_{n - 1} \\
a_n &= b_n
\end{align*}
This can be easily solved by back-substitution, starting with the (known) value for $a_n$.
A: Alternatively, for any polynomial $q(t)$, set $p(t)$ to be $\exp(-t)\,\int_{-\infty}^{t}\,q(s)\,\exp(s)\,\mathrm{d}s$.  Show that $p$ is indeed a polynomial and $T(p)=q$.
We can also prove injectivity in this manner. Prove that $p(t)$ is in the kernel of $T$ iff $p(t)\,\exp(t)$ is constant, or equivalently $p\equiv0$.
A: Three possible solutions.

1) Show that, for all $n$, $T$ is invertible when restricted to a transformation on the vector space of all polynomials of degree $\leq n$.  Since $P$ is the union of these spaces, it follows that $T$ is injective and surjective on $P$.
Since a transformation of a finite-dimensional space is invertible exactly when it has trivial null space, it is enough to solve the equation $p + p' = 0$.  But we can easily see that this cannot be solved by a nonzero polynomial (for example, by comparing the degree of $p$ and $p'$).

2) We have $T = I + D$, where $I$ is the identity and $D$ is the derivative.  Formally, we have $1/(I+D) = I - D + D^2 - D^3 + \cdots$, so we can reasonably guess that the inverse of $T$ is $p\mapsto \sum_n (-1)^n \frac{d^n}{dx^n} p(x)$—note that there are only finitely many terms in this sum, since $p$ is a polynomial—and verify that this is indeed a two-sided inverse for $T$ by manipulation of power series (again, these are finite power series, so no worries about convergence when doing this).

3) Note that $(e^x p)' = e^x (p+p') = e^x T(p)$.  Since the derivative is an invertible linear map on the space $\{e^x p \mid p\in P\}$ (which we can check by verifying that exactly one antiderivative is of the correct form) , the conclusion follows.
