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!



Simultaneous upper-triangularization for two nilpotent commuting matrix
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](https://homepage.univie.ac.at/Dietrich.Burde/papers/burde_03_ben.pdf).


