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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!

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  • $\begingroup$ Yes - sorry not to mention that clearly; I will edit it now. Thanks! $\endgroup$ – Filip92 Mar 21 '18 at 12:44
  • $\begingroup$ Not only eigenspaces, but generalized eigenspaces, are stabilized by commuting operators. $\endgroup$ – paul garrett Mar 21 '18 at 12:53
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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.

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  • $\begingroup$ Thanks for the answer! However, I think I have got a counterexample for this - sorry if I am wrong: Let $B_1=\begin{pmatrix} 0 & 0 & b \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $ and $B_2=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}.$ Then using matrix $$S=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1/b & 0 \end{pmatrix} $$ that consists of generalised eigenvectors, we see that $S^{-1}B_1S$ is in its Jordan normal form and $$S^{-1}B_2 S=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 1/b & 0 \end{pmatrix}.$$ $\endgroup$ – Filip92 Mar 22 '18 at 18:34
  • $\begingroup$ But $B_2$ is already in Jordan form, so it is not a counterexample. These two matrices cannot be Jordanized simultaneously. This is what I meant in the answer. $\endgroup$ – Dietrich Burde Mar 22 '18 at 19:02
  • $\begingroup$ True. However, instead of this $B_2$ one could use the general one (for which the question was) $$B_2=\begin{pmatrix} 0 & d & e \\ 0 & 0 & f \\ 0 & 0 & 0 \end{pmatrix}. $$ Then the conjugated matrix will be $$S^{-1}B_2 S=\begin{pmatrix} 0 & e/b & d \\ 0 & 0 & 0 \\ 0 & f/b & 0 \end{pmatrix},$$ which is not upper-diagonal in general. $\endgroup$ – Filip92 Mar 22 '18 at 19:27
  • $\begingroup$ Again, we only want one matrix in Jordan form. So interchange the roles of $B_1$ and $B_2$. For example, $S=\begin{pmatrix} df & e & 0 \cr 0 & f & 0 \cr 0 & 0 & 1\end{pmatrix}$ will put $B_2$ into Jordan form and keep both strictly upper-triangular. $\endgroup$ – Dietrich Burde Mar 22 '18 at 19:53
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    $\begingroup$ No, $B_2$ is already in Jordan form, and both are strictly upper-triangular. $\endgroup$ – Dietrich Burde Mar 23 '18 at 16:41
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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.

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    $\begingroup$ David, your result stands between the Engel's theorem and this one (due to Radjavi): Let $\mathcal{L}$ be any subset of $M_n$ that is closed under commutation $[.,.]$. Then $\mathcal{L}$ is triangularizable IFF each commutator $[.,.]$ of elements of $\mathcal{L}$ is nilpotent. cf. corollary 1.7.8 in springer.com/gp/book/9780387984674 $\endgroup$ – loup blanc Apr 8 '18 at 14:32

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