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Let $\Phi :M_d(\mathbb{C})\to M_D(\mathbb{C})$ be a linear map. It is positive when $X\ge 0$ implies $\Phi (X)\ge 0$, and completely positive whenis positive for every ancillary dimension $r$. In finite dimensions it is enough to check $r=d$.

A finite family of Kraus operators $A^{\alpha }:{\mathbb{C}}^d\to {\mathbb{C}}^D$ defines the Kraus representationThis map is completely positive because, for every positive semidefinite matrix $Y$ on the enlarged space,

Conversely, use the unnormalized maximally entangled vector $|\Omega ⟩={\sum }_{j=1}^d|j⟩|j⟩$. Complete positivity makes the Choi matrixpositive semidefinite. By the spectral theorem for normal operators, $C_{\Phi }={\sum }_{\alpha =1}^R|v_{\alpha }⟩⟨v_{\alpha }|$, where $R=\mathrm{rank}{C}_{\Phi }\le dD$. Reshape each $v_{\alpha }\in {\mathbb{C}}^D\otimes {\mathbb{C}}^d$ into a matrix $A^{\alpha }$ by $|v_{\alpha }⟩={\sum }_{\mu j}{A}_{\mu j}^{\alpha }|\mu ⟩|j⟩$. The Choi matrix inversion formulathen gives $\Phi (X)={\sum }_{\alpha }{A}^{\alpha }X(A^{\alpha }{)}^{†}$. Thus finite Kraus representations characterize finite-dimensional completely positive maps.

The additional normalizationmakes $\Phi$ trace preserving and hence a quantum channel; ${\sum }_{\alpha }{A}^{\alpha }(A^{\alpha }{)}^{†}=I_D$ instead makes it unital.