Ideals in $B(H)$ are self-adjoint

Question

It is known that every (closed two-sided) ideal in a $C^{*}$-algebra is self-adjoint. The proofs that I've seen involve functional calculus and approximate units. I am wondering whether there is a more direct approach in the particular case of $B(H)$ (for a Hilbert space $H$).

Answer 1

Let $T\in J$ where $J$ is a closed two-sided ideal in $B(H)$. Consider its polar decomposistion $T=U|T|$, where $|T|=(T^*T)^{1/2}$. Clearly $T^*T\in J$, so its square root $|T|$ is in $J$ too (granted, we use some small bit of functional calculus there). Therefore, $T^*=|T|U^*\in J$ as well.

Edit: As commented by Martin Argerami, this proof works for norm closed ideals in any von Neumann algebra.