Theorem: Exact full-code recovery

A deterministic CPTP decoder on $R$ recovers every code input, including external references, exactly if and only if

for one fixed state ${\tau }_B$ and every $X\in \mathcal{B}(C)$. If $r=\mathrm{rank}{\tau }_B$, exact recovery implies $\dim R\ge dr$.

Proof. Dilate an exact decoder by $W:R\to C\otimes G$. The composite $(W\otimes I_B)V$ dilates identity. On each basis input its decoded marginal is pure, so

Exact recovery of off-diagonal matrix units forces $⟨{\chi }_j,{\chi }_i⟩=1$; all vectors coincide. Tracing $CG$ gives the constant complement.

Conversely, diagonalize ${\tau }_B={{\sum }_{a=1}}^r{\lambda }_a|a⟩⟨a|$, omitting zero eigenvalues, and write

Constancy on all matrix units implies $⟨u_{jb},u_{ia}⟩={\delta }_{ij}{\delta }_{ab}$. These $dr$ vectors are orthonormal. On their span, the isometry $U{u}_{ia}=|i⟩|a⟩$, followed by tracing the second factor, recovers the input. Complete the decoder on the orthogonal complement by fixed-state preparation. The result is CPTP on all accessible inputs. Matrix-unit equality persists with every reference.