Showing that if $||Tx|| = ||T^*x||$ in a $^*$-subalgebra of the bounded operators of a hilbert space is commutative.

Question

Let $B(\mathcal{H})$ be the bounded operators for a Hilbert space over the complex field. I know that if $||Tx|| = ||T^*x||$, then $T$ is normal since we can show that $\langle TT^*x, y \rangle = \langle T^*Tx, y \rangle$. Let $A$ be a $^*$-preserving subalgebra. I am tyring to show that if $||Tx|| = ||T^*x||$ for $T \in A$, then $A$ is commutative. I was trying to show that $\langle TSx, y \rangle = \langle STx, y \rangle$ using polarization identity, but can't seem to figure it out. Sorry if this is simple since it seems quite easy.

Answer 1

Stage 1: All self-adjoints in $\mathfrak A$ commute.

Let $S, T$ be self-adjoints in $\mathfrak A$. Let $W=S+iT$. Since $\|Wx\|=\|W^* x\|$ for all $x\in\mathcal H$, $W$ is normal. Now $W W^*=(S+iT)(S-iT)=S^2+i(TS-ST)+T^2$ and $W^* W=(S-iT)(S+iT)=S^2-i(TS-ST)+T^2$. Subtracting we get $0=W W^*-W^* W=2i(TS-ST)$ and therefore $S,T$ commute.

Stage 2: All members of $\mathfrak A$ commute.

For $S, T\in\mathfrak A$ write $S=S_1+iS_2, T=T_1+iT_2$ where $S_1=(S+S^*)/2,S_2=(S-S^*)/(2i),T_1=(T+T^*)/2,T_2=(T-T^*)/(2i)$. Then $S_1,S_2,T_1,T_2$ are self-adjoint and all mutually-commute.