MHB Problem of the Week #90 - February 17th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: If $A$ and $B$ are self-adjoint linear operators, show that $AB$ is self-adjoint if and only if $AB = BA$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Ackbach and I like Serena. You can find Ackbach's solution below.

[sp]For this problem, I will assume we are working in a Hilbert Space $H$.

$(\Leftarrow)$ Assume that $AB=BA$, and that $A$ and $B$ are self-adjoint. We show that $AB$ is self-adjoint. Since $A$ and $B$ are self-adjoint, it follows that $\langle Ax|y \rangle= \langle x|Ay \rangle$ for all $x,y\in H$. Similarly, $\langle Bx|y \rangle= \langle x|By \rangle$ for all $x,y \in H$. Let $x,y\in H$ be arbitrary. Now
$$ \langle ABx|y \rangle= \langle Bx|Ay \rangle= \langle x|BAy \rangle= \langle x|ABy \rangle,$$
by the self-adjointness of $A$ and $B$, and by commutativity of the operators. Hence, $AB$ is self-adjoint.

$(\Rightarrow)$ Assume that $A$ and $B$ are self-adjoint, as well as $AB$. We show that $AB=BA$. Let $x,y \in H$ be arbitrary. Because $A$ and $B$ are self-adjoint, it must be that $ \langle ABx|y \rangle= \langle x|BAy \rangle$. Because $AB$ is self-adjoint, it must also follow that $ \langle ABx|y \rangle= \langle x|ABy \rangle$. Thus, $\langle x|ABy \rangle= \langle x|BAy \rangle$, and hence $ \langle x|(AB-BA)y \rangle=0$. Since this is true for arbitrary $x,y$, it will also be true if $x=(AB-BA)y$, in which case we get that $\langle (AB-BA)y|(AB-BA)y \rangle= \| (AB-BA)y \|^{2}=0$. It follows that
$$ \|AB-BA \|= \sup_{ \|y \|=1} \|(AB-BA)y \|=0,$$
ergo $AB-BA=0$, or $AB=BA$.[/sp]