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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. We define the minimum polynomial $m(x)$ of $T$. Suppose that $q(x)$ is any polynomial in $K[x]$ and that $r(x)$ is the highest common factor of $q(x)$ and $m(x)$. Show that Ker($q(T)$) = Ker($r(T)$).

Progress

$r(x)$ hcf of $q(x)$ amd $m(x)$ $\Rightarrow$ $q=f_1r$ and $m=f_2r$ for some $f_1, f_2 \in K[x]$.

Take $v \in$ Ker ($r(T)$), then $q(T)v=(f_1r)(T)v=f_1(T)(r(T)v)=f_1(T)(0)=0$.

That is, Ker$r(T) \subseteq$ Ker $q(T)$. Not sure how to show the opposite relation; any help would be greatly appreciated. Regards.

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3 Answers 3

up vote 3 down vote accepted

Use Bézout's identity: $r(x) = a(x) q(x) + b(x) m(x)$, for some $a,b \in K[x]$.

Now take $v\in \ker q(T)$. Then $$r(T)v = a(T)(q(T)v)+b(T)(m(T)v) = a(T)(0)+b(T)(0) = 0$$ and $v\in \ker r(T)$.

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Can we say then that, for $v \in$ Ker$r(T)$, $r(T)v=0=aq(T)v+bm(T)v=0$, and so, as $r(x)|q(x)$ and $r(x)|m(x)$, that this forces $q(T)v=0$ and concludes the argument? –  Mathmo Dec 22 '11 at 13:05
    
@TJO, the relation is for the reverse inclusion: if $v\in \ker q(T)$ then $v\in \ker r(T)$. –  lhf Dec 22 '11 at 13:53

Note that $r = f_1 q + f_2 m$ (why? Think Euclid).

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$f_1$ and $f_2$ are not the same as in the question. –  lhf Dec 22 '11 at 12:56
    
Cheers! Hadn't thought of dividing through by $r(x)$ and applying Bezout's result. –  Mathmo Dec 22 '11 at 13:07

It is a matter of taste, but one can do without explicit mention of polynomial division (Euclid, Bezout). Since $q(t)$ is a multiple of $r(t)$ one certainly has $\ker(q(T)) \supseteq \ker(r(T))$ (as you remarked). For the opposite inclusion write $q(t)=p(t)r(t)$ and suppose $v\in\ker(q(T))$ so $p(T)(r(T)(v))=\vec0$; we wish to show that already $w=r(T)(v)=\vec0$. By definition of $r(t)$, there is no non-unit common factor of $p(t)$ and $(m/r)(t)$ but $w$ is annihilated by both $p(T)$ and by $(m/r)(T)$, so the minimal polynomial in $T$ that annihilates $w$ can only be $1$, which means that $w=\vec0$, as we wished to show.

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