Let $A,B$ be two $3\times 3$ matrices with complex entries, such that $A^2=AB+BA$. Prove that $\det(AB-BA)=0$

Nice problem, and I want to find a solution.

$AB-BA=A^2-2BA=(A-2B)A$ so if $|A|=0$ we have done, if $|A| \not=0$ I can't prove.


$$\begin{align} \det(AB-BA)&= \det(A^2-2BA)\\ &= \det(A-2B)\det(A) \\ &= \det(A)\det(A-2B) \\ &= \det(A^2-2AB) \\ &= \det(BA-AB) \\ &= (-1)^3 \det(AB-BA) \\ &= -\det(AB-BA) \end{align}$$ So $$\det(AB-BA)=0$$

  • $\begingroup$ I get the trick. Use the fact that matrices "commute under determinants". +1 $\endgroup$ – abnry Apr 9 '15 at 16:34
  • $\begingroup$ nice. the first equality holds because $A^2 - 2AB = (AB+BA) - 2AB = AB - BA.$ i don't if it helps to add this step to yours. i had to think about it for a while. $\endgroup$ – abel Apr 9 '15 at 16:45
  • $\begingroup$ @abel I just added that step. $\endgroup$ – Kitegi Apr 9 '15 at 17:10

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