definition.) Proof We prove the following form of the theorem: if is a Lie subalgebra such that every is a nilpotent endomorphism and if V has positive dimension, then there exists a nonzero vector v in V such that for each X in . The proof is by induction on the dimension of and consists of a few steps. (Note the structure of the proof is very similar to that for Lie's theorem, which concerns a solvable algebra.) The basic case is trivial and we assume the dimension of is positive. Step 1: Find an ideal of codimension one in . This is the most difficult step. Let be a maximal (proper) subalgebra of , which exists by finite-dimensionality. We claim it is an ideal of codimension one. For each , it is easy to check that (1) induces a linear endomorphism and (2) this induced map is nilpotent (in fact, is nilpotent as is nilpotent; see Jordan–Chevalley decomposition#Lie algebras). Thus, by inductive hypothesis applied to the Lie subalgebra of generated by , there exists a nonzero vector v in such that for each . That is to say, if for some Y in but not in , then for every . But then the subspace spanned by and Y is a Lie subalgebra in which is an ideal of codimension one. Hence, by maximality, . This proves the claim. Step 2: Let . Then stabilizes W; i.e., for each . Indeed, for in and in , we have: since is an ideal and so . Thus, is in W. Step 3: Finish up the proof by finding a nonzero vector that gets killed by . Write where L is a one-dimensional vector subspace. Let Y be a nonzero vector in L and v a nonzero vector in W. Now, is a nilpotent endomorphism (by hypothesis) and so for some k. Then is a required vector as the vector lies in W by Step 2. See also Lie's theorem Heisenberg group Notes Citations Works cited Representation theory of Lie algebras Theorems in representation theory