= Solution
A <linear map> $\alpha$ is a <nilpotent endomorphism> if $\alpha^q=0$ for some positive integer $q$, and a <semisimple endomorphism> if it is <diagonalizable> over $\mathbb C$. Decompose $V$ into its <generalized eigenspaces> $V_\lambda=\ker(\alpha-\lambda I)^d$, where $d=\dim V$. Define $\alpha_s$ on $V_\lambda$ to be $\lambda I$, and set $\alpha_n=\alpha-\alpha_s$. Then $\alpha_s$ is diagonalizable, $\alpha_n$ is nilpotent, and both preserve these <vector subspaces> and commute. Thus
$$
\boxed{\alpha=\alpha_s+\alpha_n,\qquad[\alpha_s,\alpha_n]=0.}
$$
For uniqueness, suppose $\alpha=S+N$ with $S$ semisimple, $N$ nilpotent and $SN=NS$. Both commute with $\alpha$, so preserve each $V_\lambda$. On an <eigenspace> of $S$ of <eigenvalue> $\mu$ inside $V_\lambda$, the map $\alpha$ is $\mu I+N$ and has only the <eigenvalue> $\mu$. Since the same subspace lies in $V_\lambda$, $\mu=\lambda$. Hence $S=\lambda I$ on $V_\lambda$, proving $S=\alpha_s$ and $N=\alpha_n$. This is the additive <Jordan–Chevalley decomposition>.
We need a polynomial consequence of this decomposition. <Hermite interpolation> supplies a <polynomial> $p$ with $p(\alpha)=\alpha_s$, by prescribing $p(t)\equiv\lambda\pmod{(t-\lambda)^d}$ for each <eigenvalue>. For any endomorphism $T$, its semisimple part can likewise be expressed as a <polynomial> in $T$ with zero constant term: if zero is an <eigenvalue>, its interpolation condition already forces this; if not, add the independent condition $p(0)=0$.
On $\operatorname{End}(V)$, the maps $\operatorname{ad}\alpha_s$ and $\operatorname{ad}\alpha_n$ commute. The first is diagonalizable, with <eigenvalues> $\lambda-\mu$ on $\operatorname{Hom}(V_\mu,V_\lambda)$. The second is nilpotent, since
$$
(\operatorname{ad}N)^k(T)=\sum_{j=0}^k(-1)^j\binom kj N^{k-j}TN^j,
$$
which vanishes for $k\ge2q-1$ when $N^q=0$. Uniqueness therefore proves <adjoint compatibility of additive Jordan decomposition>: $\operatorname{ad}\alpha_s=(\operatorname{ad}\alpha)_s$.
Now assume $\alpha\in M$. The condition $[\alpha,W]\subseteq U\subseteq W$ implies that both $W$ and $U$ are invariant under $\operatorname{ad}\alpha$, and every <polynomial> in $\operatorname{ad}\alpha$ with zero constant term maps $W$ into $U$. Define $\beta$ to be multiplication by $\overline\lambda$ on $V_\lambda$. It commutes with $\alpha$. On $\operatorname{Hom}(V_\mu,V_\lambda)$, $\operatorname{ad}\beta$ acts by $\overline{\lambda-\mu}$. <Polynomial interpolation> on the finite set of differences gives $\operatorname{ad}\beta=q(\operatorname{ad}\alpha_s)$ with $q(0)=0$. The preceding paragraph then expresses $\operatorname{ad}\beta$ as a <polynomial> in $\operatorname{ad}\alpha$ with zero constant term. Consequently $[\beta,W]\subseteq U$, so $\beta\in M$.
The assumed <trace orthogonality nilpotence lemma> now follows directly. On $V_\lambda$, $\alpha=\lambda I+\alpha_n$ and $\beta=\overline\lambda I$, while the <matrix trace> of the nilpotent restriction of $\alpha_n$ is zero. Hence
$$
0=\operatorname{tr}(\alpha\beta)=\sum_\lambda\dim(V_\lambda)|\lambda|^2.
$$
Every summand is nonnegative, so every <eigenvalue> of $\alpha$ is zero. Its <Jordan–Chevalley decomposition> therefore has $\alpha_s=0$, and \b[$\alpha$ is nilpotent]. Notice that neither $U$ nor $W$ was required to be a Lie subalgebra.
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