Write . The generalized eigenspace decomposition of the linear map isfor any sufficiently large . Because is a derivation, the generalized-eigenspace bracket lemma givesConsequently is a Lie subalgebra.
The set is the normalizer of a Lie subalgebra . Certainly . Conversely, if , then givesOn the direct sum of the nonzero generalized eigenspaces, is invertible. Hence the nonzero-eigenvalue component of vanishes, and
Now let be a Lie subalgebra containing . Since , the subspace is -invariant. The generalized zero eigenspace of the induced map on is the image of , hence is zero. If , then , so lies in that zero eigenspace. Thus and
A Nilpotent Lie algebra is one whose lower central serieseventually reaches zero. Suppose is nilpotent and . Choose the least for which . Then , and anysatisfies . Therefore , proving the normalizer condition for a nilpotent Lie algebra
It remains to prove the converse needed here. The Engel lemma states that if a finite-dimensional Lie algebra of linear maps consists of nilpotent maps, then the maps have a common nonzero vector in their kernels. To prove it, induct on the dimension of the algebra. For a maximal proper subalgebra , induction applied to the action of on produces with . Thus is an ideal of codimension one. Induction also gives a nonzero common kernelThe ideal property makes invariant under ; a nilpotent representative of a basis of has a nonzero kernel on , yielding a vector killed by all of .
Apply the lemma to the Adjoint representation. It produces a nonzero element of the Center of a Lie algebra. Induction on , followed by passage to the quotient by this center, proves Engel theorem: if every is nilpotent, then is nilpotent. The hypothesis says exactly that every is nilpotent, so
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