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Use Klein's inequality, or equivalently nonnegativity of quantum relative entropy. First check the support needed for the logarithm. If $q_y=0$, positivity gives $|{\rho }_{yz}{|}^2\le {\rho }_{yy}{\rho }_{zz}=0$ for every $z$; the corresponding row and column vanish. Thus support inclusion under rank-one dephasing gives $\mathrm{supp}\rho \subseteq \mathrm{supp}\sigma$, and the logarithms may be evaluated on this support.

Since $\log \sigma$ is diagonal in the dephasing basis,The relative-entropy identity for rank-one dephasing follows:Klein's inequality gives $D(\rho \Vert \sigma )\ge (\mathrm{Tr}\rho -\mathrm{Tr}\sigma )/\ln 2=0$. ThereforeEquality holds precisely when $\rho =\sigma$, meaning that the input was already diagonal in the chosen basis. This quantifies why rank-one dephasing removes coherence without reducing the entropy.

Put $p_i=⟨{\psi }_i|\rho |{\psi }_i⟩$. The rank-one dephasing is ${\rho }^{\prime }={\sum }_i{p}_i|{\psi }_i⟩⟨{\psi }_i|$. If $p_i=0$, positivity givesso $\rho |{\psi }_i⟩=0$. The kernel of ${\rho }^{\prime }$ is exactly the span of these zero-probability basis vectors and is therefore contained in the kernel of $\rho$. Taking orthogonal complements proves support inclusion under rank-one dephasing:Here the support of a positive operator is the orthogonal complement of its kernel.

On that support, ${\log }_2{\rho }^{\prime }$ is diagonal in the measurement basis, givingConsequently the relative-entropy identity for rank-one dephasing isKlein's inequality gives $D_{\mathrm{rel}}(\rho \Vert {\rho }^{\prime })\ge 0$, since both density operators have trace one. For singular $\rho$, first restrict to $\mathrm{supp}{\rho }^{\prime }$ where ${\rho }^{\prime }$ is positive definite, replace $\rho$ by $(\rho +\eta I)/(1+\eta \dim \mathrm{supp}{\rho }^{\prime })$, and let $\eta \downarrow 0$. The support inclusion ensures that the limit is finite. ThusThis proves entropy increase under nonselective projective measurement using Klein's inequality. Equality holds exactly when $\rho ={\rho }^{\prime }$, so the original density operator was already diagonal in the chosen basis.

Because $\log \Delta (\rho )$ is diagonal, $\mathrm{Tr}(\rho \log \Delta (\rho ))=\mathrm{Tr}(\Delta (\rho )\log \Delta (\rho ))$. Hence $D_{\mathrm{rel}}(\rho \Vert \Delta (\rho ))=S(\Delta (\rho ))-S(\rho )$. Support inclusion under rank-one dephasing handles zero probabilities, and Klein's inequality makes the difference nonnegative. Equality holds exactly when the state was already diagonal.