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Let $A,B \in M_3(\mathbb{C})$ such that $(AB)^2 = A^2B^2$ and $(BA)^2 = B^2A^2$. Prove that $\det(AB-BA) = 0$.

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    $\begingroup$ What have you tried? $\endgroup$ – CiaPan Apr 4 at 9:43
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    $\begingroup$ I tried to prove that rank(AB - BA) is less than 3. $\endgroup$ – Ddang Apr 4 at 9:46
  • $\begingroup$ And this could work perhaps like here. $\endgroup$ – Dietrich Burde Apr 4 at 9:47
  • $\begingroup$ I don’t think so. If C = AB - BA, we get from the conditions that ACB = 0 = BCA. I don’t see how to go on from here. Maybe the rank doesn’t matter at all... $\endgroup$ – Ddang Apr 4 at 9:53
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Hints. Let $C=AB-BA$. The given conditions imply that $ACB=0$ and $BCA=0$. We don't need both of them. One --- say, $ACB=0$ --- is enough:

  • If at least one of $A$ or $B$ has rank $\le1$, argue that $\operatorname{rank}(C)=\operatorname{rank}(AB-BA)\le2$.
  • If $\operatorname{rank}(A),\operatorname{rank}(B)\ge2$, then $\operatorname{rank}(CB)\ge2$ when $C$ is non-singular. Now consider Sylvester's rank inequality $ \operatorname{rank}(A)+\operatorname{rank}(CB)-3\le\operatorname{rank}(ACB)$.
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