Let $A,B \in \mathbb{M}_n (\mathbb{C})$. If $A^2B + BA^2 = 2ABA$ then exist $k \in \mathbb{N}$ where $(AB-BA)^k = 0$.
I tried solve with minimal polynomial, but I did not have much effect.
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3The hypothesis is that $[A,[A,B]]=0$, and the conclusion that $[A,B]$ is nilpotent. This is a clearer way to state this, IMO. This result is known as Jacobson's lemma; googling should find proofs —for example, this one – Mariano Suárez-Álvarez Apr 10 '13 at 21:17
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1This is hardly an exercise... It always helps to explain in what context your question occured, @jonjones. – Mariano Suárez-Álvarez Apr 10 '13 at 21:24
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Wow, that seems a lot more difficult than I initially thought. – Christopher A. Wong Apr 10 '13 at 21:28
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I'm preparing for my test. I already spent two days in this question. – jon jones Apr 10 '13 at 21:45
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Well, as I said, this is a very cruel exercise in linear algebra, if that is how you encountered it. – Mariano Suárez-Álvarez Apr 10 '13 at 21:47
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2Does this answer your question? $AB-BA$ is a nilpotent matrix if it commutes with $A$ – user26857 Dec 31 '19 at 18:09
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Let's denote by $C$ the matrix $AB-BA$.
From this is enough to show that $\operatorname{Tr}(C^n)=0$ for all $n\in\mathbb N$.
From Mariano's comment we easily see that $AC=CA\Rightarrow AC^2=CAC=C^2A\Rightarrow\ldots \Rightarrow AC^m=C^mA, \ \forall m\in\mathbb N$. Therefore
$$
C^n=C^{n-1}C=C^{n-1}(AB-BA)=\\
AC^{n-1}B-C^{n-1}BA=A(C^{n-1}B)-(C^{n-1}B)A.
$$
We conclude that $\operatorname{Tr}(C^n)=0$ for all $n\in\mathbb N$.
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Nice one (+1): But I did not know that there was this same problem posted in the past. There are some other insights too. http://math.stackexchange.com/questions/299640/ab-ba-is-a-nilpotent-matrix-if-it-commutes-with-a?rq=1 – Sungjin Kim Apr 10 '13 at 22:41