1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.