Let $f$ and $g$ be two endomorphisms of a complex vector space $E$ such that $f \circ g = g \circ f $. Show that there exists a basis of $E$ in which $f$ and $g$ are upper-triangular matrices.
The proof is by recurrence.
The result is true for $ n = 1 $.
We suppose that there exists a basis $ B = (e_1,...,e_n) $ in which $f$ and $g$ are upper triangular and we prove this for the case of an $n+1$ dimensional space.
$f$ and $g$ share a common eigenvector $v_\lambda$.
We will have another basis, $ B'= \{v_{\lambda} \} \cup {B} $, in which $f$ and $g$ are:
$ A(f,B') = \begin{pmatrix} v_\lambda & a \\ 0 & A' \end{pmatrix} $
and
$ C(g,B') = \begin{pmatrix} v_\lambda & b \\ 0 & C' \end{pmatrix} $
We have $f \circ g = g \circ f \iff A'C' = C'A' $.
From the recurrence hypothesis, $A'$ and $C'$ are upper triangular in the basis $B$, thus $A$ and $C$ are upper triangular in the basis $B'$.
I'm not sure if they are upper triangular in the same basis.