= 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 satisfy
$$
N|_U=(T-\lambda I)|_U,\quad (N|_U)^\alpha=0,\qquad
N|_W=(T-\mu I)|_W,\quad (N|_W)^\beta=0.
$$
Put $k=\max(\alpha,\beta)$. Both restrictions of $N^k$ vanish, so the <direct sum> gives
$$
\boxed{(T-D)^k=0.}
$$
Hence \b[$T-D$ 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.
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