 Nilpotent commutator preserves generalized eigenspaces

For endomorphisms of a finite-dimensional complex vector space, suppose repeated commutation with $T$ annihilates $S$. Write the blocks of $S$ using the generalized eigenspaces of $T$. On $\mathrm{Hom}(V_{\lambda },V_{\mu })$, the operator $\mathrm{ad}T$ is $(\mu -\lambda )I$ plus a nilpotent endomorphism. If $\lambda \ne \mu$, it is invertible, so $(\mathrm{ad}T{)}^{N}S=0$ forces that off-diagonal block of $S$ to vanish. Thus $S$ preserves each generalized eigenspace. This allows a nilpotent Lie algebra representation to be split by generalized eigenvalues even when its acting operators do not commute.