Coherent information equals output entropy minus environment entropy

Given: Nonzero finite-dimensional complex inner-product spaces $A,B,E$, a channel $\mathcal{N}:A\to B$, a Stinespring isometry $V:A\to B\otimes E$ for $\mathcal{N}$, and a density operator $\rho$ on $A$.

Complementary channels and coherent information: for a purification $\mathrm{\mid }\psi ⟩\in A\otimes R$, $I_c(\rho ,\mathcal{N})=S({\omega }_B)-S({\omega }_{BR})$, with ${\omega }_{BR}=(\mathcal{N}\otimes {\mathrm{id}}_R)(\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid })$; the chosen complement is ${\mathcal{N}}^c(X)={\mathrm{Tr}}_B(VX{V}^{\ast })$.

Ensemble and purification and Purifications exist and are unitarily equivalent after padding: a purification is a unit vector $\mathrm{\mid }\psi ⟩\in A\otimes R$ with ${\mathrm{Tr}}_R\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }=\rho$, and such a finite purification exists.

Schmidt decomposition: a finite pure bipartite vector has a Schmidt decomposition, so its two reduced density operators have the same positive eigenvalues with multiplicities.

Every finite-dimensional channel has a finite isometric dilation: $\mathcal{N}(X)={\mathrm{Tr}}_E(VX{V}^{\ast })$ for every input operator $X$.

Finite-dimensional von Neumann entropy: $S(\tau )=-{\sum }_{{\lambda }_j>0}{\lambda }_j\ln {\lambda }_j$ for the eigenvalues of a finite-dimensional density operator; zero eigenvalues contribute zero.

Partial trace over a finite tensor factor, Partial trace over the first tensor factor, Ordered tensor registers, their inner product, and wire permutations, and Finite wire permutations are unitary basis permutations: ordered partial traces sum equal traced-factor indices; the factors of $B\otimes E\otimes R$ can be reordered by a unitary wire permutation.