= Spectral Kraus decomposition
{title2=$J(\Phi)=\sum_k|v_k\rangle\langle v_k|$}
Use the unnormalized <Choi matrix> $J(\Phi)=(\Phi\otimes\mathrm{id})(|\widetilde\Psi\rangle\langle\widetilde\Psi|)$, which has trace $d$ for a channel. A positive such matrix has a spectral decomposition with vectors $v_k=\sqrt{\lambda_k}u_k$. In paired bases define the <Kraus operator> $A_k$ by $(A_k)_{ij}=(v_k)_{ij}$, so $v_k=(A_k\otimes I)\sum_j|j\rangle|j\rangle$. Partial contraction with a <conjugate index vector> gives $A_k|\phi\rangle$, hence $\Phi(|\phi\rangle\langle\phi|)=\sum_kA_k|\phi\rangle\langle\phi|A_k^\dagger$. Rank-one projectors span the operator space, proving the <Kraus representation> on all inputs. Trace preservation is equivalent to $\sum_kA_k^\dagger A_k=I$. This construction needs at most $\operatorname{rank}J$ nonzero Kraus operators.
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