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Problem

Let $V$ be a finite-dimensional vector space over a field $K$ and let $T$ be a linear transformation of $V$ to itself. Define the minimum polynomial of $T$ to be $m(x)$.

Show that:

$m(x)$ has no repeated factors $\Rightarrow$ for any polynomial $p(x) \in K[x]$, Ker($p(T)^n$) = Ker($p(T)$) for all $n > 1$.

Progress

Can we say that if $m$ has no repeated roots, then we can express $m$ as $m(x)=\prod_{i=1}^n (x- \lambda_i)$ for distinct $\lambda_i \in K[x]$?

Even if we can, not sure how this would help. Any assistance would be appreciated. Regards.

EDIT 1

$m(x)$ has no repeated factors $\Rightarrow$ $gcd(m,p)=gcd(m,p^n)$. As such Ker($p^n(T))$=Ker$(p(T))$=Ker$(gcd(m,p)(T))$ from an earlier result, here.

Further Problem

Can we show the reverse? That $m(x)$ has no repeated factors $\Leftarrow$ for any polynomial $p(x) \in K[x]$, Ker($p(T)^n$) = Ker($p(T)$) for all $n > 1$.

We can't now assume that $m(x)$ has no repeated roots so the above argument won't hold. Any thought are appreciated. Regards.

2 Answers 2

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Use your previous question. If $m$ has no repeated factors then $\gcd(m,p^n)=\gcd(m,p)$ and so $\ker p(T)^n = \ker r(T) = \ker p(T)$, where $r=\gcd(m,p)$.

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    Thanks anyway; I'll see what I can come up with...2011-12-23
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$m(x)$ is a product of distinct irreducible factors, but these need not be linear unless our field $K$ is closed.

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    Vert true, if we write that $m(x)=\prod_{i=1}^n p_i(x)$ where for each $i$, p is ireeduicble in $K[x]$, does that bring us closer?2011-12-23