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Let $A$ and $B$ in $M_n(\mathbb C)$ such that the rank of $AB-BA$ is $1$. Prove that $A$ and $B$ are simultaneously triangularisable.

This generalizes the classical case $AB = BA$.

By induction on $n$, it suffices to show that $A$ and $B$ have a common eigenvector. So, it would be sufficient to find a eigenspace of $A$ which is stable by $B$ since matrices are complex.

Do you have ideas for that? Thank you.

user26857
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2 Answers2

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That follows from the book Simultaneous Triangularization by Radjavi and Rosenthal (page 8). The original proof is due to Thomas Laffey.

Let $\{y\}$ be a basis of $\mathrm{Im}(AB-BA)$. Let $\lambda\in\mathrm{Spec}(B)$. If $B=\lambda I$, then there is almost nothing to do. Otherwise $F=\ker(B-\lambda I)$, $G=\mathrm{Im}(B-\lambda I)$ are non-trivial $B$-invariant subspaces. If we show that $F$ or $G$ is $A$-invariant, then we are the kings of oil.

Assume that $F$ is not $A$-invariant. Then there is $x$ s.t. $(B-\lambda I)x=0$, $(B-\lambda I)Ax\not= 0$. We have $$A(B-\lambda I)x-(B-\lambda I)Ax=ABx-BAx=-(B-\lambda I)Ax\in\mathrm{Im}(AB-BA)\cap\mathrm{Im}(B-\lambda I)\setminus\{0\}.$$ Thus $y\in G$.

Let $z\in \mathbb{C}^n$. Then $A(B-\lambda I)z$ is in the form $(B-\lambda I)Az+\alpha y$. Therefore, $G$ is $A$-invariant and we are done. $\square$

user26857
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    "we are the kings of oil." I've never heard that expression before. By context it means "we're done" right? – N8tron May 27 '18 at 18:53
  • @N8tron . The french peoples are very pretentious. They say "we do not have oil but we have ideas." Unfortunately, as would have said De Gaulle: "Researchers who research, we find them; researchers who find, we search them." In fact, we dream of being kings of oil. –  May 27 '18 at 19:09
  • I have never heard that in France. –  May 29 '18 at 10:43
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    @Nûr . My friend, you are too young. In the 70s, it was a joke that was done with "black feet" (repatriated from Algeria) in the south of France. Since the loss of Algeria and its oil wells, the French had to pay for oil at a high price, especially as they were very attached to their car. In this vein, the riots of May 1968 in Paris stopped as soon as the oil came back in the city pumps... –  May 29 '18 at 17:45
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Lets prove that $A$ and $B$ have at least one common eigenvector. If $x\in \ker(A-\lambda I)$, we have $$ (AB-BA)x = (A-\lambda I)Bx. $$ For $x\in \ker (AB-BA)$ this imply $Bx\in \ker(A - \lambda I)$, which means that $\ker(A-\lambda I)$ is an invariant subspace for operator $B$, so there is an eigenvector $y$ of $B$ lying in it.

We win when at least one eigenvector of $A$ or $B$ lying in $\ker(AB-BA)$. Consider the case when all the eigenvectors of $A$ and $B$ are not in $\ker(AB-BA)$. From well-known identity $$ \dim\ker(AB-BA) = n - \mathrm{rank}\,(AB-BA) = n-1, $$ so eigenvectors of both $A$ and $B$ lying in the subspace of dimension $1$, which means they have common eigenvector.

Now, lets $v$ be a common eigenvector of $A$ and $B$ in some basis $\{e_1,\dots, e_n\}$. Then in basis $\{v,e_2,\dots, e_n\}$ matrices of $A$ and $B$ become $$ A = \begin{pmatrix} \lambda & X \\ 0 & A_1 \end{pmatrix}, \quad B = \begin{pmatrix} \mu & Y \\ 0 & B_1 \end{pmatrix}. $$ This yields $\mathrm{rank}(A_1B_1-B_1A_1) \leq \mathrm{rank}(AB-BA)$, so we have the same situation for $A_1$ and $B_1$. Repeating this reasoning $n$ times we obtain the upper-triangular forms of $A$ and $B$ in the same basis.

Anton Grudkin
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    Even if all the eigenvectors of $A$ and $B$ are not in $\ker(AB-BA)$ and $\dim\ker(AB-BA) = n-1$,we still can not have eigenvectors of both $A$ and $B$ lying in the subspace of dimension 1 (because the complementary space of $\ker(AB-BA)$ is not unique) .You need to find other reasons to verify $A$ and $B$ have a common eigenvector in this condition. – Elliot Aug 03 '20 at 07:42