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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 :)

tohann123
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  • Food for thought in $\dim V=2$: take $L=\textrm{span}{A,B}$ where $A(\lambda e_1+\mu e_2)=\mu e_1$ and $B(\lambda e_1+\mu e_2)=\lambda e_2$. The kernels of $A,B$ share only the zero vector but $AB-BA\notin L$. I don't know what to do if $L$ is already given though. Do you want to disprove the claim for all $L$, or a counterexample for each $V$ suffices? – Al.G. Feb 18 '24 at 10:28
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    @Al.G. That's a good idea! So to clarify what I'm asking for, I want to find a linear subspace of $\mathfrak{gl}(V)$ such that all elements of the subspace are nilpotent as maps on V but there is no non zero $v$ such that $\forall x \in L: x(v) = 0$. So I think your example is a great idea already but the only issue I see is that A+B is in $L$ and isn't nilpotent except if either $\mu$ or $\lambda$ are zero. But then I find such a $v$ again. – tohann123 Feb 18 '24 at 10:43
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    For reference: The statement you call "supposedly true" is (one version of) Engel's Theorem. The proof of that uses that our linear subspace is closed under the Lie bracket several times, so the suspicion is, indeed, that it's no longer true in the generalized version. But on the other hand, I have failed to find a counterexample (in fact, I do not even have an example of a subspace as demanded that is not contained in a nilpotent subalgebra of $gl(V)$), but from the number of upvotes I guess I'm not alone. Interesting question! – Torsten Schoeneberg Feb 23 '24 at 20:57

1 Answers1

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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.

Max
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