Spectral-parts formula for trace distance ($D(\rho ,\sigma )=\mathrm{Tr}(\rho -\sigma {)}_{+}$)

For the Hermitian difference $X=\rho -\sigma$, define its positive part of a Hermitian operator $Q=X_{+}$ and negative part of a Hermitian operator $R=X_{-}$. Their supports are orthogonal and $|X|=Q+R$. Equal traces give $\mathrm{Tr}Q=\mathrm{Tr}R$, so the trace distance is both $\frac{1}{2}(\mathrm{Tr}Q+\mathrm{Tr}R)$ and $\mathrm{Tr}Q$. This explains why the positive spectral projector attains the variational characterization of trace distance.