Theorem: Exact complete-complement criterion

For a finite complete isometry $V:L\to A\otimes B$, the following are equivalent:

If $s=\mathrm{rank}{\tau }_B$, then $\dim A\ge ds$. There are an isometry $T:L\otimes {\mathbb{C}}^s\to A$ and a state $\sigma$ such that

An exact decoder on the entire accessible space is

for any fixed state ${\omega }_L$.

Proof. Suppose a decoder is exact and dilate it by $W:A\to LG$. Then $(W\otimes I_B)V$ dilates the identity channel. Its Choi operator has rank one, so every Kraus operator is a scalar multiple of identity. The composite therefore has the form $|\psi ⟩\mapsto |\psi ⟩|\chi {⟩}_{GB}$, with fixed $\chi$. Its residual is constant.

Conversely, diagonalize ${\tau }_B={{\sum }_{\alpha =1}}^s{\lambda }_{\alpha }|\alpha ⟩⟨\alpha |$, where ${\lambda }_{\alpha }>0$, and write

Constancy on every matrix unit gives

These $ds$ vectors define $T$. Equations (eq:normalform) and (eq:exactdecoder) follow directly. The decoder is CP, its traces sum to $\mathrm{Tr}Y$, and it returns every logical matrix unit.