If $L \subseteq \mathfrak{gl}(V)$ is an abelian, nilpotent Lie algebra is it true that $\{v \in V\setminus \{0 \} : x(v)=0 \forall x \in L \} \neq 0$?

Question

I have the following question but I can’t see to find an answer floating about anywhere.

In this setting $V$ is a finite dimensional vector space and is nonzero.

If $L \subset \mathfrak{gl}(V)$ is abelian and consists only of nilpotent maps, is there some nonzero $v \in V $ such that $x(v)=0$ for all $x \in L$?

EDIT

In fact I have just looked and seen that more generally this result is true even without $L$ being abelian but the proof is a bit longer than I would like and I think having this abelian condition should make a short proof. Is this true?

Answer 1

From Engel's theorem it follows without the assumption of abelian.

If you don't want to use the theorem (as you said in your edit), you can use the following argument.

Let $\{x_1,\dots,x_n\}$ a basis of $L$. Define $V_1=\{v\in V:x_1 v=0\}$, this is a non trivial subspace of $V$, now $x_2(V_1)\subseteq V_1$ because if $v\in V_1$ then $x_1(x_2)v=x_2(x_1v)=0$. $x_2$ is still a nilpotent map on $V_1$, and so have a null vector.

The subspace $V_2=\{v\in V_1: x_2 v=0\}$ is invariant under $x_3$ (from the same reasoning using that $x_3$ commute with both $x_1$ and $x_2$) and not trivial. Continuing this argument to get that $$V_n=\{v\in V:\forall i\; x_iv=0\}\neq \{0\}$$ as needed (because the $x_i$ form a basis).

The main point here is that if $A, B$ are commuting linear maps then the subspace of eigenvectors (with a given eigenvalue) of A is B-invariant.