Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2014 / 1 / b / iii / Solution

Let $P=b(T)q(T)$ and $Q=a(T)p(T)$. By part (ii), these are complementary projection operators with images $U=\ker (T-\lambda I{)}^{\alpha }$ and $W=\ker (T-\mu I{)}^{\beta }$. Thus $D=\lambda P+\mu Q$ acts as $\lambda I$ on $U$ and $\mu I$ on $W$.

The linear operator $T$ preserves both summands because it commutes with its own polynomials. Therefore $N=T-D$ preserves them too, and its restrictions satisfyPut $k=\max (\alpha ,\beta )$. Both restrictions of $N^k$ vanish, so the direct sum givesHence $T-D$ is a nilpotent linear map. The operator $D$ is diagonalizable, since its two summands are its eigenspaces, and it commutes with $T$. This illustrates the Jordan–Chevalley decomposition in the case of two distinct eigenvalues.