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From this problem we know that if $A$ and $B$ are matrices such that $A, B$, and $AB$ are normal, then $BA$ is also normal.

I would like to know if this property holds in a more general way:

if $A$ and $B$ are bounded operators in a Hilbert space, such that $A, B$, and $AB$ are normal, then $BA$ is also normal

J.Doe
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  • In order to show that $(AB)^AB=AB(AB)^$ it is enough to prove that $x=0$, where $x=(AB)^AB-AB(AB)^$. Given the identity $\Vert x^x\Vert =\Vert x\Vert ^2$, it is in turn enough to prove that $x^x=0$, or even that $xx^*x=0$. When you expand this out the resulting expression might be quite big but you will have plenty of opportunities to employ your hypothesis! – Ruy Sep 12 '20 at 14:28
  • I apologize for the comment above since it does not seem to work. I tried to follow this method and got nowhere. I even tried proving that $xx^xx^=0$, in the hope that the hypothesis could be used more often, to no avail. – Ruy Sep 15 '20 at 16:27

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