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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.

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2 Answers 2

up vote 2 down vote accepted

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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Many thanks! To prove the opposite relation ($\Leftarrow$), can we apply the Primary Decomposition Theorem somehow? –  Mathmo Dec 23 '11 at 13:58
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To prove the $\Leftarrow$ direction for this question though, we can't assume that $m(x)$ has no repeated roots, and therefore can't say that $gcd(m,p^n)=gcd(m,p)$ surely. –  Mathmo Dec 23 '11 at 14:10
    
Actually, I really don't understand how the kernel relationship implies the relationship between the gcd's. We have only shown that in one direction, and we relied on no repetition of roots in the minimum polynomial. If you could explain I'd be very grateful. –  Mathmo Dec 23 '11 at 15:59
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@TJO, I now cannot see a way to do the reverse implication. Sorry for the noise. –  lhf Dec 23 '11 at 16:04
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Thanks anyway; I'll see what I can come up with... –  Mathmo Dec 23 '11 at 16:10
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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? –  Mathmo Dec 23 '11 at 13:48
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