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First prove monotonicity for a partial trace. Let $r=\dim {\mathcal{H}}_B$ and define the Heisenberg-Weyl twirling channelThe supplied identity on individual operators extends to bipartite operators by expanding them in a tensor-product basis. The joint convexity of quantum relative entropy and its unitary invariance implyThe additivity of quantum relative entropy makes the left side $D({\rho }_A\Vert {\sigma }_A)$, proving the partial-trace case.

For a general deterministic quantum operation, meaning a quantum channel, use a Stinespring dilation $\Lambda (X)={\mathrm{Tr}}_E(VX{V}^{†})$ with $V^{†}V=I$. An isometric embedding preserves quantum relative entropy, because restricting the output to the common image of $V$ preserves the eigenvalues and trace formula. Applying the partial-trace result givesThe other properties used are unitary invariance and invariance under adjoining an identical ancillary state; both follow directly from the relative-entropy trace formula. This is the Lindblad-Uhlmann monotonicity theorem. The proof applies to deterministic channels; normalized postselection is not such a linear channel.

Both states are block diagonal in $X$. On the support of each block,Substitution into quantum relative entropy givesTaking $-\mathrm{Tr}{\rho }_{XB}\log {\rho }_{XB}$ similarly yields the entropy of a classical-quantum state:

The pinching map is self-adjoint for the trace pairing, and $\log \mathcal{P}(\rho )$ is block diagonal. Thus $\mathrm{Tr}\rho \log \mathcal{P}(\rho )=\mathrm{Tr}\mathcal{P}(\rho )\log \mathcal{P}(\rho )$. The kernel of $\mathcal{P}(\rho )$ is contained in that of $\rho$, so the relative entropy is finite on its support. Expanding the quantum relative entropy gives the displayed identity. Its nonnegativity proves entropy increase, with equality exactly when the original state is already block diagonal.

For common classical weights, the matrix logarithm on each positive-weight block is $\log (p_i{\rho }_i)=(\log p_i)I+\log {\rho }_i$ on its support. The two logarithms of $p_i$ cancel in the quantum relative entropy, giving the displayed identity. Zero-weight blocks are omitted, including if their unweighted relative entropy is infinite. If a positive-weight block violates support inclusion, both sides are infinite. Applying the data-processing inequality for quantum relative entropy to the partial trace over the flag gives joint convexity of quantum relative entropy.

The relative entropy of entanglement is the smallest quantum relative entropy from a density operator to a separable quantum state, with the separable set specified for the relevant division into parties. Nonnegativity of quantum relative entropy implies $E_R\ge 0$. For a separable quantum state, choosing $\sigma =\rho$ gives $E_R=0$. For finite-dimensional bipartite states, compactness of the separable set and lower semicontinuity of the relative entropy give the converse. The support convention is $D(\rho \Vert \sigma )=+\infty$ unless $\mathrm{supp}\rho \subseteq \mathrm{supp}\sigma$.

If $p_i=⟨{\psi }_i|\rho |{\psi }_i⟩=0$, positivity gives $\sqrt{\rho }|{\psi }_i⟩=0$, so the corresponding basis vector is in $\ker \rho$. Therefore $\ker \Delta (\rho )\subseteq \ker \rho$, and $\mathrm{supp}\rho \subseteq \mathrm{supp}\Delta (\rho )$. This makes the quantum relative entropy of $\rho$ relative to its rank-one dephasing finite.

The support of a positive operator is the orthogonal complement of its kernel. In finite dimension it is the span of eigenvectors with positive eigenvalues, equivalently the operator's image. Quantum relative entropy is finite only when the first state's support lies in the reference state's support.