TITLE: An algebra of nilpotent linear transformations is triangularizable
QUESTION [8 upvotes]: How to prove "An algebra of nilpotent linear transformations is triangularizable" using linear algebra only?

REPLY [3 votes]: 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 $A$.