Vector space of finite dimension, and a subalgebra of $L(V)$ whose all its elements are nilpotent. 
Let $V$ be a vector space of finite dimension, and let $A$ be a subalgebra of $L(V)$ where all its elements are nilpotent. Show that $\bigcap_{f\in A} \ker(f)$ is not reduced to $\{0\}$.

$L(V)$ means linear maps in $V$.
 A: First, we prove the claim in a simple setting, which which allows us to understand the general proof.
Let $V$ and $A$ be as in the problem.

Simple case:
Suppose that $A$ is generated (as an algebra) by the nilpotent elements $a,b \in L(V)$. Suppose furthermore that all $a^{2} = 0$, $b^{2} = 0$, $(ab)^{2} = 0$, and $(ba)^{2} = 0$. Then, $A$ is the linear span of the elements
\begin{align}
a, b, ab, aba,\text{ and } bab.
\end{align}
These elements suffice, since by assumption we have $abab = baba = 0$. First of all, note that
\begin{equation}
\cap_{f \in A} \ker(f) = \ker(a) \cap \ker(b).
\end{equation}
The idea is now to construct a non-zero vector $v \in \ker(a) \cap \ker(b)$.
Now, since $a$ is nilpotent, it is clear that there is a non-zero element $x \in V$ with $ax \neq 0$. Furthermore, we have that $ax \in \ker(a)$.
Now, if $bax = 0$, then $ax \in \ker(a) \cap \ker(b)$. If $bax \neq 0$, then $bax$ is a non-zero element in $\ker(b)$.
Now, if $abax= 0$, then $bax \in \ker(a) \cap \ker(b)$. If $abax \neq 0$, then $abax$ is a non-zero element in $\ker(a)$.
Our procedure terminates, because $baba = 0$, hence $babax = 0$, hence $abax \in \ker(b)$.
We conclude that
\begin{equation}
 \cap_{f\in A} \ker(f) = \ker(a) \cap \ker(b),
\end{equation}
contains at least one non-zero element.

General case:
Let us now consider the more general case. As a vector space $A$ is the linear span of a finite number of elements $a_{1},\dots, a_{k} \in A$.
By assumption, there exist numbers $n_{1},\dots,n_{k} \in \mathbb{N}$, such that $a_{i}^{n_{i}} = 0$ for all $i \in \mathbb{N}$.
We will need the following lemma, for which I do not have a proof yet.

There exists a number $N < \infty$, such that the following holds.
  Let $\sigma:\mathbb{N} \rightarrow \{1, \dots, n\}$ be arbitrary, then the element
  \begin{equation}
 a_{\sigma(N)} a_{\sigma(N-1)} \dots a_{\sigma(2)} a_{\sigma(1)} = 0.
\end{equation}

Now we can prove the claim in a way very similar to the proof above.
Let us write $X = \cap_{f\in A} \ker(f) \subseteq V$.
Let $x \in V$ be arbitrary. Either $x \in X$, or there exists an element $a_{i}$ such that $a_{i}x \neq 0$. In the first case we are done, so suppose that $a_{i}x \neq 0$. Define $\sigma(1) = i$. Let $0 \neq y \in V$ be of the form $a_{\sigma(l)} \dots a_{\sigma(1)} x$. Then either $y \in X$, or there exists an element $a_{j}$ such that $a_{j}y \neq 0$. Define $\sigma(l+1) = j$. In this way, we inductively define $\sigma: \mathbb{N} \rightarrow \{ 1, \dots n \}$.
The lemma tells us that $a_{\sigma(N)} \dots a_{\sigma(1)}x = 0$, hence, this inductive process stops.
