Minimal polynomial of the shift operator in $\mathbb{k}_{10}[x]$ Let $\mathbb{k}$ be a field and denote by $\mathbb{k}_{10}[x]$ the vector space of all polynomials of $x$ of degree at most 10.
Define the shift operator $T:\mathbb{k}_{10}[x]\to \mathbb{k}_{10}[x]$ by equation $T(f(x))=f(x+1)$. Find the minimal polynomial of this operator.
My approach: Consider the standard basis of $\mathbb{k}_{10}[x]$ namely $\{1,x,x^2,\dots,x^{10}\}$. And if we write the matrix of this operator we will get $11\times 11$ matrix which is upper diagonal with $1$'s on the diagonal. 
Then the characteristic polynomial of this operator will be $P(t)=(t-1)^{11}$. Since minimal polynomial $\mu(t)$ divides it then it should be of the form $(t-1)^k$, $0<k\leq 11$. I am trying to show that $k=11$ but cannot prove it rigorously.
Suppose $k<11$ then $\mu(t)=(t-1)^k$. Hence it means that the operator $(f-\text{id})^k$ is trivial operator. I was applying it to $x^k$ trying to get contradiction but no results.
Would be very grateful if anyone can show the solution!
 A: Hint:
You can check the matrix is
$$T=\begin{pmatrix}
1&1&1&1&\dots&1&\dots&1\\
0&1&2&3&\dots& k&\dots& 10 \\
0&0&1&3&\dots &\binom k2 &\dots&\binom{10}2 \\
\vdots & &&\vdots &\ddots &\vdots &&\vdots \\
0& 0&0&0&\dots&1&\dots&\binom{10}k\\
\vdots & &&\vdots & &\vdots &&\vdots \\
0&0&0&0&\dots&0&\dots&
\binom{10}9\\
0&0&0&0&\dots&0&\dots&1
\end{pmatrix}=I+N,$$
and show (by an easy finite induction) that each power $(T-I)^r=N^r$ has exactly its first $r$ columns which are null columns.
A: It suffices to show that $(T-I)^{10}$ is not the zero operator. Indeed, we have
$$(T-I)^{10} = \sum_{k=0}^{10} {10 \choose k} (-1)^{10-k}T^k.$$
Plugging in the polynomial $x^{10}$ yields
$$(T-I)^{10}x^{10} = \sum_{k=0}^{10} {10 \choose k} (-1)^{10-k}(x+k)^{10}$$
which has constant term equal to
$$\sum_{k=0}^{10} {10 \choose k} (-1)^{10-k}k^{10} = 10!$$
This can be calculated by hand, or by using this identity. Therefore $(T-I)^{10}x^{10} \ne 0$ and hence $(T-I)^{10} \ne 0$ as well.
A: Work instead with the monomial basis of falling factorials $x^{\underline{n}}=x(x-1)\cdots (x -n+1)$, and let $\Delta=T-I $.
Then from the identity of formal power series $\mathbb{Q}[x][[t]]$
$$\Delta^m(1+t)^x = t^m(1+t)^x$$
one finds
$$\Delta^mx^{\underline{n}}=n^{\underline{m}}x^{\underline{n-m}}$$
so that by direct calculation, the minimal polynomial $m(\delta)$ of $\Delta$ is $\delta^p$ if the characteristic of the field of scalars is $p=2,3,5,7$, $\delta^{11}$ otherwise.
For in the characteristic $p$ cases given, $p|n^{\underline{p}}$ because $n^{\underline{p}}$ is a product of $p$ consecutive integers, but $n\equiv -1\pmod{p}\Rightarrow n^{\underline{p-1}}\equiv -1\pmod{p}$ by Wilson's Lemma, so that $\Delta^p=0$ but $\Delta^{p-1}\neq 0$.
And in the other characteristics, we have $\Delta^{11}=0$ but $\Delta^{10}\neq 0$, since $p$ does not divide $n^{\underline{m}}$ for $m\leq n<p$.
