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($r(T)$).
$r(x)$ 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.