= Nilpotent commutator preserves generalized eigenspaces
{title2=$(\operatorname{ad}T)^N S=0\ \Longrightarrow\ S(V_\lambda)\subseteq V_\lambda$}
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 $\operatorname{Hom}(V_\lambda,V_\mu)$, the operator $\operatorname{ad}T$ is $(\mu-\lambda)I$ plus a <nilpotent endomorphism>. If $\lambda\ne\mu$, it is invertible, so $(\operatorname{ad}T)^N S=0$ 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.
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