Given the operator $T:\mathbb C_{\le n}[x]→\mathbb C_{\le n}[x]$ such that $T(p) = p' + p$ find the minimal polynomial.
What I tried:
I found the representing matrix $$A = \begin{pmatrix} 1 & 1 & 0 & \cdots & 0 \\ 0 & 1 & 2 & \cdots & 0 \\ \vdots & 0 & \ddots & \ddots & 0\\ 0 & \cdots & 0 & 1 & n\\ 0 & 0 & 0 & \cdots & 1\\ \end{pmatrix}$$ and then I found the characteristic polynomial: $f_T(x) = (x-1)^{n+1}$.
Now I know that the minimal polynomial is $m_T \in \{(x-1),(x-1)^2,\space...\space,\space(x-1)^{n+1}\}$
My guess is that $m_T = (x-1)^{n+1}$ but I don't know how to find which one it is.
Let's denote by $S$ the derivative operator $S(p) = p'$. You noted that the minimal polynomial of $T$ has the form $(x-1)^k$ for $1 \leq k \leq n + 1$ so let $m(x) = (x-1)^k$. For $m$ to be the minimal polynomial of $T$, we must have
$$ m(T) = (T - I)^k = S^k = 0. $$
However, if $k \leq n$ then $$ S^k(x^k) = k! \neq 0$$
so we must have $k = n + 1$. And indeed, $S^{n+1}$ acts on polynomials by taking the derivative $n + 1$ times and since all the polynomials $S$ acts on are of degree $ \leq n$ we get $S^{n+1} = 0$.