Minimal polynomial of the shift operator in $\mathbb{k}_{10}[x]$

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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 $m\leq n<p$.