Let $A,B \in \mathbb{C}_{n\times n}$ and let $[A,B]$ be their commutator. What can we say when $$ [A,B]=A \quad ? $$ It feels like this is something interesting to study (specially in physics). Does anyone know of any resources on this?
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2Related – José Carlos Santos Sep 23 '21 at 10:46
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Is this the commutator $A^{-1}B^{-1}AB$, or is it the commutator $AB-BA$? – Gerry Myerson Sep 23 '21 at 10:47
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2@GerryMyerson if this has anything to do with physics, then $[A,B] = AB-BA$. This is sort of canonical among physicists. – achille hui Sep 23 '21 at 10:52
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1Does this answer your question? Example of two-dimensional non-abelian Lie algebra? – Conifold Sep 23 '21 at 10:54
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Does this answer your question? If $A$ and $AB-BA$ commute, show that $AB-BA$ is nilpotent – user8675309 Sep 23 '21 at 17:20
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If ${\cal A}$ is a complex Banach algebra the following is known (Bonsall, Duncan: Complete Normed Algebras, §18 Prop. 13): Let $a,b \in {\cal A}$ and $a(ab-ba)=(ab-ba)a$. Then $r(ab-ba) =0$ (here $r$ denotes the spectral radius). Hence in the case $[A,B]=A$ we have $A[A,B]=A^2=[A,B]A$ and get $r(AB-BA)=r(A)=0$. Thus $A$ is a nilpotent matrix.
Edit: As $A$ is nilpotent I wondered whether $A=0$ at least if $n$ is small, but no $(n=2)$: Let $$ A=\left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right), \quad B=\left(\begin{array}{cc} \alpha & \beta \\ 0 & 1+\alpha \end{array} \right). $$ Then $[A,B]=A$.
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This is the assumption in the Proposition used. At least you need something to conclude that $r(AB-BA)=0$ since that is false, in general. – Gerd Sep 23 '21 at 13:21
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$$[A,B]=A \implies \\ e^{-B} A e^B= A + A + A/2!+ A/3!+... = (1+e)A.$$
You are dealing with the affine group in one dimension, whose Lie algebra is the simplest 2d nonabelian Lie algebra.
Cosmas Zachos
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