Why does $\omega((I-E)T)=0$ here?

Question

I am reading Kadison-Ringrose-Vol2. The authors defined the completely additive states and their supports in Definition 7.1.1. And just after the definition they remarked the following:

If $E$ is a support of a state $\omega$, then $\omega(I-E)=0.$

This is fine. Then they used Proposition 4.5.1 to conclude the following:

Since $I-E\ge 0$ and $\omega$ is a state then for any $T \in \mathscr{R}$, $$0=\omega(T(I-E))=\omega((I-E)T)$$

Here the first equality follows from Proposition 4.5.1 by using the fact that for any state $\rho$ we have $\rho(B^*A)=0$ whenever $B\in \mathscr{R}$ and $A\in \mathscr{L}_\rho:=\{T\in \mathscr{R}~:~\rho(T^*T)=0\}$.

But how did the second equality follow? Why $\omega((I-E)T)=0$? $T$ may not commutes with $I-E$.

Any help is appreciated. Thanks

Answer 1

Here are two arguments.

1. Because $\omega\geq0$, it satisfies Cauchy-Schwarz: $$ |\omega(B^*A)|≤\omega(A^*A)^{1/2}\omega(B^*B)^{1/2}. $$ Then $$ |\omega((I-E)T)|^2≤\omega((I-E)(I-E)^*)\omega(T^*T)=\omega(I-E)\omega(T^*T)=0. $$

2. The second way is to notice that $$\omega(T^*)=\overline{\omega(T)} $$ for all $T$. Here if $S=S^*$ then $S=S^+-S^-$, and so $\omega(S)=\omega(S^+)-\omega(S^-)\in\mathbb R$. Then writing $T=A+i B$ with $A,B$ selfadjoint, $$ \omega(T^*)=\omega(A-i B)=\omega(A)-i\omega(B)=\overline{\omega(A)+i\omega(B)}=\overline{\omega(A+i B)}=\overline{\omega(T)}. $$ And now $$ \omega((I-E)T)=\overline{\omega(T^*(I-E))}=0. $$