The Question:
Let $V$ be a finite dimensional vector space over $\Bbb C$, with linear map $T:V \rightarrow V$. Suppose that the minimal polynomial of $T$ is $$m_T(x) = p(x)q(x) \\ p(x)=(x-\lambda)^\alpha \quad q(x)=(x-\mu)^\beta \quad \lambda \neq \mu$$
and $a$, $b$ are polynomials such that $ap+bq=1$. Define the new linear transformation
$$D = \lambda b(T)q(T)+\mu a(T)p(T)$$
Show that $T-D$ is nilpotent.
My Attempt:
In the previous parts of the question, I have shown that
(i) $V = \text{Ker}\;p(T) \oplus \text{Ker} \; q(T)$
(ii) $E:=a(T)p(T)$ and $F:=b(T)q(T)$ are projections (i.e. $E^2=E$ and $F^2=F$) with
$$\text{Im} \; E=\text{Ker} \; q(T) \qquad \quad \text{Ker} \; E = \text{Ker} \; p(T) \\ \text{Im} \; F=\text{Ker} \; p(T) \qquad \quad \text{Ker} \; F = \text{Ker} \; q(T)$$
In order to show that a linear transformation is nilpotent, we have to show that its only eigenvalue is $0$, I guess?
Now I took a vector $v \in \text{Ker} \; p(T)$ and observed that
$$v \in \text{Ker} \; p(T) \implies (T-D)(v) = Tv - \lambda v$$
Similarly,
$$v \in \text{Ker} \; q(T) \implies (T-D)(v) = Tv - \mu v$$
Moreover, since the minimal polynomial of $T$ is $(x-\lambda)^\alpha (x-\mu)^\beta$, the only eigenvalues of $T$ are $\lambda$ and $\mu$.
And then I can't quite see what the next move should be. Any hints?
Writing $D = \lambda F + \mu E$ and noting that $EF = FE = 0$, you can show by induction that $$(T-D)^n = (T-\lambda)^n E + (T-\mu)^n F.$$ By (ii) this is zero when $n = \max\{\alpha,\beta\}$.