= Solution
Let $\Phi:M_d(\mathbb C)\to M_D(\mathbb C)$ be a <linear map>. It is <positive linear map>[positive] when $X\geq0$ implies $\Phi(X)\geq0$, and <completely positive map>[completely positive] when
$$
\Phi\otimes\operatorname{id}_r
$$
is positive for every ancillary dimension $r$. In finite dimensions it is enough to check $r=d$.
A finite family of <Kraus operator>[Kraus operators] $A^\alpha:\mathbb C^d\to\mathbb C^D$ defines the <Kraus representation>
$$
\Phi(X)=\sum_{\alpha=1}^R A^\alpha X(A^\alpha)^\dagger.
$$
This map is completely positive because, for every <positive semidefinite matrix> $Y$ on the enlarged space,
$$
(\Phi\otimes\operatorname{id}_r)(Y)
=\sum_\alpha(A^\alpha\otimes I_r)Y((A^\alpha)^\dagger\otimes I_r)\geq0.
$$
Conversely, use the unnormalized <maximally entangled state>[maximally entangled vector] $|\Omega\rangle=\sum_{j=1}^d|j\rangle|j\rangle$. Complete positivity makes the <Choi matrix>
$$
C_\Phi=(\Phi\otimes\operatorname{id}_d)(|\Omega\rangle\langle\Omega|)
$$
positive semidefinite. By the <spectral theorem for normal operators>, $C_\Phi=\sum_{\alpha=1}^R|v_\alpha\rangle\langle v_\alpha|$, where $R=\operatorname{rank}C_\Phi\leq dD$. Reshape each $v_\alpha\in\mathbb C^D\otimes\mathbb C^d$ into a matrix $A^\alpha$ by $|v_\alpha\rangle=\sum_{\mu j}A^\alpha_{\mu j}|\mu\rangle|j\rangle$. The <Choi matrix> inversion formula
$$
\Phi(X)=\operatorname{Tr}_{\rm in}\!\left[C_\Phi(I_D\otimes X^T)\right]
$$
then gives $\Phi(X)=\sum_\alpha A^\alpha X(A^\alpha)^\dagger$. Thus finite <Kraus representation>[Kraus representations] characterize finite-dimensional completely positive maps.
The additional normalization
$$
\sum_\alpha(A^\alpha)^\dagger A^\alpha=I_d
$$
makes $\Phi$ trace preserving and hence a <quantum channel>; $\sum_\alpha A^\alpha(A^\alpha)^\dagger=I_D$ instead makes it unital.
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