 Polynomial projections for a coprime decomposition

Suppose $p(T)q(T)=0$ for a linear operator $T$, and Bézout's identity for polynomials gives $ap+bq=1$. The kernel decomposition for coprime polynomials is $V=\ker p(T)\oplus \ker q(T)$. The displayed operators are the corresponding projection operators: $P$ is the identity on $\ker p(T)$ and zero on $\ker q(T)$, while $Q$ has the opposite action. Thus $P+Q=I$, $P^2=P$, $Q^2=Q$ and $PQ=QP=0$. These actions also prove independence from the choice of Bézout coefficients. Every summand is an invariant subspace of $T$, since the projections are polynomials in $T$ and commute with it.