I want to find a counterexample for the following "Theorem": Let $V \neq 0$ be a finite dimensional $K$ - vector space and $L \subset \mathfrak{gl}(V)$ a linear subspace. If all $x \in L$ are nilpotent as maps $V \rightarrow V$ then there is a $v \in V, v \neq 0$ such that $\forall x \in L: x(v) = 0$.
When $L$ is a Lie subalgebra of $\mathfrak{gl}(V)$ this statement is supposedly true, according to my lecture notes, but not true if it's just a linear subspace of $\mathfrak{gl}(V)$. I couldn't come up with a counterexample though, so that's why I am asking for help.
Certainly counterexamples can't be abelian, as then they'd automatically be Lie subalgebras. So they also, in particular, can not be one dimensional subspaces.
I tried to construct something with two nilpotent matrices where their anti-commutator vanishes so that any power of their linear combinations consists of only powers of themselves again, but so far to no avail. There was always a vector in V mapped to 0 by all elements of $L$
Thank you in advance :)
Some searching for "Linear spaces of nilpotent matrices" brings up Causa, Antonio, Riccardo Re, and Titus Teodorescu, "Some remarks on linear spaces of nilpotent matrices." Le Matematiche 53.3 (1998): 23-32, which contains the following space giving the desired counterexample:
$$V=\{ M(s, t)= \begin{pmatrix} 0 & s & 0\\ -t & 0 & s \\ 0 & t & 0 \end{pmatrix}| s, t \in \mathbb{R} \}.$$
Let's check. Clearly, $V$ is a vector space. Now, we compute
$$M(s,t)^2= \begin{pmatrix} -st & 0 & s^2\\ 0 & 0 & 0 \\ -t^2 & 0 & st \end{pmatrix}, M(s,t)^3= \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$
So all $M(s,t)$ are indeed nilpotent. Finally, the null space of $M(1,0))$ is spanned by $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ and null space of $M(0,1)$ by $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$.
Note:
This counterexample is the smallest possible, as for 2 by 2 matrices the maximal dimension of vector space of nilpotent matrices is 1, as explained here.