I've been trying to prove the following statement:
Let $V$ be a vector space over general field $\mathbb F$ and $T$ a linear operator. Let $P, Q, R$ be polynomials over $\mathbb F$, such that $P=QR$. Prove that if $P(T)=0$ and $gcd(Q,R)=1$, then $V=ker(Q(T))\oplus ker(R(T))$.
I've proven so far that $ker(Q(T)) \cap ker(R(T)) = \{0\}$ by showing that if $v \in$ $ker(Q(T)) \cap ker(R(T))$ then the minimal polynomial of $v$ in relation to $T$ divides both $Q,R$ in contradiction to their gcd being 1. I've also shown that since $P(T)=0$ then $ker(P(T))=V$. I now want to show that for every $v \in V$ there exist $u,w \in ker(Q(T)), ker(R(T))$ such that $v = u + w$ but I'm not really sure how to do this.
For a linear operator $A$ the rank is equal to $$\dim{\rm Im}\, A=\dim V-\dim\ker A$$ If $P(T)Q(T)=0$ then ${\rm Im}\,Q(T)\subset \ker P(T).$ Therefore $$\dim V-\dim\ker Q(T)=\dim{\rm Im}\,Q(T)\le \dim\ker P(T)$$ Therefore $$\dim\ker P(T)+\dim\ker Q(T)\ge \dim V\qquad (*)$$ On the other hand $\ker P(T)\cap \ker Q(T)=\{0\},$ therefore $$n\ge \dim(\ker P(T)\oplus \ker Q(T))=\dim\ker P(T)+\dim\ker Q(T)\ge n$$ Thus $$\dim(\ker P(T)\oplus \ker Q(T))=n$$ which implies $$\ker P(T)\oplus \ker Q(T)=V$$
Remark I prefer to prove the property $\ker P(T)\cap \ker Q(T)=\{0\}$ as follows. As $\gcd (P,Q)=1$ there are polynomials $P_1$ and $Q_1$ such that $P_1P+Q_1Q=1.$ Hence $$P_1(T)P(T)x+Q_1(T)Q(T)x=x$$