Kernel and image of coprime factors of the minimal polynomial

Question

Let given an operator $T$ with minimum polynomial $m(x)=p(x)q(x)$ such that $\gcd(p(x),q(x))=1$, then $\ker(p(T))= \text{Im}(q(T))$

So far I got $\exists r(x),s(x)$ such that $r(x)p(x)+s(x)q(x)=1$

Now we can obtain the following relations:

$r(T)p(T)+s(T)q(T)=I$

and

$m(T)=p(T)q(T)=0$ by definition of the minimum polynomial.

I do not know if I'm on the right track because I cannot seem to find any proofs of this online. Any help would be appreciated

Answer 1

If $p(T)x=0$ then

$$q(T)s(T)x=Ix-r(T)p(T)x=x$$ so $x\in\text{Im}(q(T))$.

If $x\in q(T)$ then $x=q(T)y$ and $p(T)x=p(T)q(T)y=0$, so $x\in\ker(p(T))$.

Answer 2

Hint:

• You have to show first that $\ker p(T)\subseteq \operatorname{Im}q(T)$.

If $\;p(T)(x)=0$, then $x=I(x)=\bigl(s(T)\circ q(T)\bigr)(x)=\bigl(q(T)\circ s(T)\bigr)(x)$, since $q(T)$ and $s(T)$ commute. This $x$ is the image by $q(T)$ of $y=s(T)(x)$.

• Next, let's show $ \;\operatorname{Im}q(T) \subseteq\ker p(T)$.

Indeed, if $x=q(T)(y)$ for some $y$, then $p(T)(x)=\bigl(p(T)\circ q(T)\bigr)(y)=m(T)(y)=0$.