= Solution
{c}
Use the <monomial basis> $e_k=x^{n-k}y^k$, $0\leq k\leq n$. Direct differentiation gives
$$
Pe_k=k e_{k-1},\qquad Qe_k=(n-k)e_{k+1},
$$
with the endpoint terms zero. Repeated application therefore gives $P^{n+1}=Q^{n+1}=0$, while $P^ne_n=n!e_0\neq0$ and $Q^ne_0=n!e_n\neq0$. Each <nilpotency index> is exactly $n+1$, so
$$
\boxed{m_P(t)=m_Q(t)=t^{n+1}.}
$$
Next
$$
PQe_k=(n-k)(k+1)e_k,\qquad
QPe_k=k(n-k+1)e_k,
$$
and 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, giving
$$
\boxed{m_{PQ-QP}(t)=\prod_{k=0}^n(t-(n-2k)).}
$$
The <commutator> is diagonal in the displayed <monomial basis>. Since $n\geq1$, $P,Q$ 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 \b[only $PQ-QP$ is <diagonalizable>]. The calculation realizes <raising and lowering on binary homogeneous polynomials>.
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