 Solution

Since $p$ and $q$ are coprime, Bézout's identity for polynomials gives $a,b\in F[x]$ with $ap+bq=1$. For every $v\in V$,The first summand belongs to $U$: applying $p(T)$ gives $b(T)p(T)q(T)v=0$. The second belongs to $W$ by the same calculation. Here all polynomials in $T$ commute, and $p(T)q(T)=f(T)=0$. Thus $V=U+W$.

If $v\in U\cap W$, both terms on the right of the identity vanish, so $v=0$. ThereforeFinally, $p(T)Tv=Tp(T)v=0$ whenever $v\in U$; hence $T(U)\subseteq U$. Replacing $p$ by $q$ proves $T(W)\subseteq W$. These are invariant subspaces; this is the coprime-factor case of the primary decomposition theorem for linear maps.