If $T(p(t)) = p(t+1)$ then find its minimal polynomial where $T$ is a linear operator from $\Bbb{P_n} \rightarrow \Bbb{P_n}$ I tried substituting $p(t)$ with $p(t-1)$ and then taking the transformation to get some kind of annhilating polynomial but that just gave me trivial solutions. Also after spending an hour I think it can be done without knowing the annihilating polynomial itself.
Is it even possible to find a polynomial of $T$ in this case?
 A: If $p \ne 0$ then from $p(t+1) = \lambda p(t)$ by comparing leading coefficients we get that necessarily $\lambda = 1$. Therefore the only eigenvalue of $T$ is $1$ so the minimal polynomial must be of the form $(T-I)^k$. Indeed by Cayley-Hamilton certainly $(T-I)^{n+1}=0$ since $\dim \mathbb{P}_n=n+1$. Furthermore, for $1 \le k \le n$ and the polynomial $p(t) = t^k$ we have
$$((T-I)^kp)(t) = \sum_{j=0}^k {k \choose j} (T^jp)(t) = \sum_{j=0}^k {k \choose j} 
 p(t+j) = \sum_{j=0}^k {k \choose j} (t+j)^k$$
and for $t = 0$ it is
$$((T-I)^kp)(t) = \sum_{j=0}^k {k \choose j} j^k > 0$$
so clearly $(T-I)^k \ne 0$. Therefore the minimal polynomial has to be $(T-I)^{n+1}$.
A: Consider the basis of polynomials $\{1,t,t^2,\ldots,t^n\}$. Then, from$T(t^k)=(t+1)^k=1+kt+\binom{k}{2}t^2+\cdots+1$, the matrix of $T$ with respect to this basis is $$\begin{pmatrix}1&1&1&1&\cdots&&\cr 0&1&2&3&\cdots&&\cr
0&0&1&3&\cdots&&\cr0&0&0&1&&\cr \vdots&&&\ddots&\ddots&\cr0&\cdots&&&0&1\end{pmatrix}$$
Since the matrix is triangular, all the eigenvalues are $1$ and its minimal polynomial is of the type $(I-A)^k$. In fact, since none of the terms of the supra-diagonal is zero, the minimum value of $k$ that makes it zero is $n$, $(I-A)^n=0$.
