I have some questions about the proof of this statement in the book "Elements of Noncommutative Geometry" by Garcia-Bondía.
A ideal $\mathfrak{a}\subset K(\mathcal{H})$ is called symmetrically normed, if $\mathfrak{a}$ is complete with respect to a norm $||\cdot||_\mathfrak{n}$ on $\mathfrak{a}$, which satisfies that $ ||RTS||_\mathfrak{a}\leq ||R||\, ||T||_\mathfrak{a} \,||S||$ for all $ T\in\mathfrak{a}; R,S\in\mathcal{L}(\mathcal{H})$.
Proof of the statement: We need to show that every selfadjoint element can be written as linear combination of positive elements (since we can write every elements as linear combi of selfadjoint elements). For $T\in\mathfrak{a}$ follows that $|T|\in\mathfrak{a}$. Then we define $F$ by $F=1$ on $\ker T$ and $F=U$ on $(\ker T)^\perp$. This $F$ commutes with $|T|$. The projectors $P_\pm=\frac{1}{2}(1\pm F)$ commute with $|T|$. $P\pm$ are spectral projections, so $P_\pm|T|=P_\pm|T|P_\pm$ are positive and lie in $\mathfrak{a}$. It holds that $T=U|T|=(P_+-P_-)|T|=P_+|T|P_+-P_-|T|P_-$.
To show that $|T|\in\mathfrak{a}$, I tried to show that $||T||_\mathfrak{a}=||\,|T|\,||_\mathfrak{a}$. For that I tried to adjust the unitarily invariant of symmetric norms, i.e. $||UTV||_\mathfrak{a}=||T||_\mathfrak{a}$, for partial isometries (some kind of $||W\,|T|\,||_\mathfrak{a}=||\,|T|\,||_\mathfrak{a}$) but this does not work. I don't see why $F$ commutes with $|T|$ and why $P_\pm$ are orthogonal projections. If that is clear, then $P_\pm$ are positive and commute with $|T|$, hence $P_\pm|T|$ is positive and the rest makes sense for me.
Thanks for your help.
It looks like you are writing $T=U|T|$ for the polar decomposition. It is also not clearly stated in the proof, but you are assuming that $T$ is selfadjoint.
$|T|\in\mathfrak a$ follows directly from $|T|=U^*T $.
From the polar decomposition you know that the closure of the range of $|T|$ is $\operatorname{ran}T ^*=(\ker T)^\perp$. So $F|T|=U|T|=T $. It also follows from $T$ selfadjoint that $U$ is selfadjoint, and that $F$ is selfadjoint. Thus $$|T|F=(F|T|)^*=T ^*=T=F|T|.$$
Now you get that $P_\pm$ are selfadjoint, hence orthogonal projections.