Simultaneous upper-triangularization for two nilpotent commuting matrix Given two nilpotent matrix B1 and B2 over complex numbers which commute i.e. [B1,B2]=0, we know that they can be conjugated to upper-triangular ones (even strictly-triangular since they're nilpotent).
But, can we conjugate them to upper-triangular ones so that one of them e.g. B1 gets into its Jordan normal form?
Thanks for any help!
 A: Yes, we can conjugate $B_1$ and $B_2$ simultaneously by a matrix $S$, whose  columns are a basis of generalized eigenvectors of $B_1$, such that $S^{-1}B_1S$ is in Jordan normal form, and both $S^{-1}B_1S$, $S^{-1}B_2S$ are upper-triangular. This is possible, because commuting operators leave invariant generalized eigenspaces. On the other hand, not both of $B_1$ and $B_2$ can be simultaneously "Jordanized" in general.
For non-commuting strictly upper-triangular matrices $B_1$ and $B_2$, we may still apply 
$S^{-1}B_1S$ with those invertible $S$ which keep $B_2$ upper-triangular.
Then we may not arrive at a Jordan form for $B_1$, but to an "almost-Jordan-form", see section $2$ of the article Modules for certain Lie algebras of maximal class.
A: The more general fact is that, if $N \subset \mathrm{Mat}_{n \times n}(k)$ is a subrng (ring without identity) whose every element is nilpotent, then we can choose a basis to make every element of $N$ simultaneously strictly upper triangular. This is a useful lemma which, as far as I know, doesn't have a standard name.
Note that, if $N_1$ and $N_2$ are commuting nilpotent matrices then any polynomial without constant term in $N_1$ and $N_2$ is clearly nilpotent, so this applies to the rng generated by such an $N_1$ and $N_2$.
Proof: Let $R$ be the $k$-algebra (with identity) generated by $N$, so $R = k \mathrm{Id} \oplus N$ and $N$ is the radical of $R$. Let $V$ be the vector space $k^n$, which we think of as an $R$-module. By the non-commutative Nakayama Lemma, also known as the Jacobson–Azumaya theorem, $NV \subsetneq V$. So $V \supsetneq NV \supsetneq N^2 V \supsetneq \cdots$ until we reach a power for which $N^r V=0$. Then choosing a basis of $V$ to respect the filtration by $N^r V$, we find that $N$ acts strictly upper-triangularly.
