An algebra of nilpotent linear transformations is triangularizable How to prove "An algebra of nilpotent linear transformations is triangularizable" using linear algebra only? 
 A: The following argument is a simplified version of an earlier argument posted here:
Let $A$ be the algebra in question, and let $V$ be the finite-dimensional vector space on which $A$ acts.  Suppose that $V \neq 0$ (as we may, since otherwise $A = 0$ and there is nothing to prove).  I claim that $A V$ is a proper subspace of $V$.  Indeed, suppose not, i.e. suppose that $V = AV$, and choose a minimal generating set $v_1,\ldots, v_n$ for $V$ over $A$.
Then in particular  we may find $a_{i} \in A$ such that $v_1 = \sum_i a_{i} v_i$, and hence so that
$$(1-a_{1})v_1 = \sum_{i> 1} a_{i}v_i.$$
  Since $a_{1}$ is nilpotent,
say $a_{1}^N = 0$, the operator $1 - a_{1}$ is invertible, with inverse
$1 + a_{1} + \cdots + a_{1}^{N-1}$.  Acting this on both sides of the displayed
identity, we find that 
$v_1 = \sum_{i>1} b_{i} v_i$
for appropriately chosen $b_{i} \in A$, and so in fact our original generating set was not minimal ($v_1$ was superfluous), a contradiction.  
Thus $AV$ is a proper subspace of $V$.  Similary $A(AV)$ is a proper subspace of $AV$, and continuing, we find a strictly decreasing sequence of $A$-invariant
subspaces $$V \supset AV \supset A^2V \supset \ldots \supset A^NV \supset \cdots$$
which must eventually reach $0$ (just for dimension reasons).
Choosing a basis for $V$ compatible with this descending sequence, 
we obtain a
triangularization of the action of $A$ on all of $V$. QED
This argument uses nothing more than the notion of dimension and basic algebra, and so in principal is just linear algebra, but of course it is more sophisticated than 
the term "linear algebra" might suggest.  You could probably make it less sophisticated on a line-by-line basis at the expense of adding (many) more lines.  My feeling is that the result you are asking about is sufficiently non-trivial as a statement of algebra that any proof is going to be either somewhat sophisticated (using a view-point of abstract algebra, even if it only uses facts from commutative algebra) or else quite painful to write down.  But maybe I'm wrong!
A: For the record, the non-«linear algebra» proof of this is: 
Consider the local artinian algebra $B=k1_V\oplus A\subseteq\mathrm{End}(V)$. Its radical is $A$ with $1$-dimensional semisimple quotient $B/A\cong k$, so the unique simple module is of dimension $1$ over the base field. The vector space $V$ is tautologically a $B$-module which has finite length, so it has a composition series and each subquotient, being simple, is of dimension $1$. Any ordered basis of $V$ such that each of its prefixes is a basis of one of the layers of this composition series triangularizes $A$.
