 Solution

The unread measurement is the pinching map $\mathcal{P}(\rho )={\rho }^{\prime }={\sum }_i{P}_i\rho P_i$, a nonselective projective measurement. Its output is block diagonal and commutes with each $P_i$, so $\log {\rho }^{\prime }$ also commutes with the projectors on its support.

First the quantum relative entropy $D(\rho \Vert {\rho }^{\prime })$ is finite. If $v$ lies in the kernel of ${\rho }^{\prime }$, positivity givesEvery summand is zero; summing ${\rho }^{1/2}{P}_{i}v=0$ gives ${\rho }^{1/2}v=0$. Thus $\ker {\rho }^{\prime }\subseteq \ker \rho$, equivalently $\mathrm{supp}\rho \subseteq \mathrm{supp}{\rho }^{\prime }$.

The block-diagonal logarithm obeys ${\sum }_i{P}_i(\log {\rho }^{\prime })P_i=\log {\rho }^{\prime }$ on this support. Cyclicity of the trace therefore givesConsequently the relative entropy of a pinched state is exactly the entropy increase:By the proved nonnegativity of quantum relative entropy and its equality condition,Equivalently, equality holds exactly when all off-diagonal blocks $P_i\rho P_j$ for $i\ne j$ already vanish, or when $\rho$ commutes with every measurement projector. This argument proves the equality condition even when the projectors are degenerate and the states singular.