 Solution

Use the monomial basis $e_k=x^{n-k}{y}^k$, $0\le k\le n$. Direct differentiation giveswith the endpoint terms zero. Repeated application therefore gives $P^{n+1}=Q^{n+1}=0$, while $P^n{e}_n=n!e_0\ne 0$ and $Q^n{e}_0=n!e_n\ne 0$. Each nilpotency index is exactly $n+1$, soNextand hence $(PQ-QP)e_k=(n-2k)e_k$. These $n+1$ real eigenvalues are distinct. A polynomial annihilates this operator exactly when it vanishes at all of them, givingThe commutator is diagonal in the displayed monomial basis. Since $n\ge 1$, $P,Q$ are nonzero nilpotent linear map; a diagonalizable nilpotent map would have a diagonal matrix with all eigenvalues zero and thus be the zero map. Therefore only $PQ-QP$ is diagonalizable. The calculation realizes raising and lowering on binary homogeneous polynomials.