> 
> **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](https://math.stackexchange.com/questions/93436/linear-algebra-minimum-polynomial).

> 
> **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.