# Finding minimal polynomial with given operator

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 $$S^{n+1} = 0$$.