Unipotent subgroup of $GL_n(K)$ has a common eigenvector An element of $GL_n(K)$ is called unipotent if its charateristic polynomial is $(x-1)^n$. A subgroup $H$ of $GL_n(K)$ is called unipotent if every element of  $H$ is unipotent.
My question is: How to show the elements of $H$ has  a common eigenvector ?
In Alperin's Book: Groups and Representations, there's a proof (Propositon 13.28). But that requires the fact (Theorem 13.27) that an algbra generated by nilpotent elements is nilpotent.
Is there any elementary proof? Thanks.
 A: Not an answer, but here is an outline of the proof for 13.27 as presented in the book.  
Suppose that $A$ is a (finite-dimensional?) $K$-algebra generated by nilpotent elements.


*

*Without loss of generality (via extension of scalars), we suppose that $K$ is algebraically closed

*Proceed by induction on $\dim_K(A)$. For dimension $1$ this holds trivially.

*If $A$ has a non-zero nilpotent ideal $I$, then we note that $A/I$ is nilpotent by the inductive hypothesis and we're done.

*By contradiction, we argue that such an ideal must exist. If $A$ has no non-zero nilpotent ideals, then $A$ is an algebra with unit (since this holds for any algebra with no non-zero nilpotent ideals). Moreover, since $A$ is an algebra with unit and no non-zero nilpotent ideals, we can conclude that $A$ is semisimple.  Since $A$ is semisimple, it is isomorphic to a direct sum of matrix algebras (with unit).  

*Let $B \subset \mathcal M_n(K)$ denote one of these matrix algebras. Because $A$ is generated as a vector space by nilpotent elements, so is $B$.  Let $M_1,\dots,M_k$ be a spanning set for $B$.  We see that $\operatorname{trace}(M_j) = 0$ for all $j$. This implies that $\operatorname{trace}(M) = 0$ for all $M \in B$.  However, $\operatorname{id} \in B$ satisfies $\operatorname{tr}(\operatorname{id}) = n$, so we have a contradiction.

