= Solution
A <linear map> $\Phi$ is a <completely positive map> if $\operatorname{id}_R\otimes\Phi$ sends every <positive operator> to a <positive operator> for every finite-dimensional auxiliary system $R$. Positivity of $\Phi$ alone tests only inputs without an auxiliary system and is weaker.
For the <Kraus representation>, let $\Omega\ge0$ be any operator on the auxiliary system and the input system. Then
$$
(\operatorname{id}_R\otimes\Phi)(\Omega)=\sum_k(I_R\otimes A_k)\Omega(I_R\otimes A_k^\dagger).
$$
For any vector $v$, its expectation is $\sum_k\langle(I_R\otimes A_k^\dagger)v,\Omega(I_R\otimes A_k^\dagger)v\rangle\ge0$. Every amplification is therefore positive, proving \b[the Kraus-form map is completely positive]. No normalization condition on the $A_k$ is needed for this positivity proof. To make it a deterministic <quantum channel>, one additionally requires $\sum_kA_k^\dagger A_k=I$; a physical outcome branch instead obeys $\sum_kA_k^\dagger A_k\le I$.
Transposition is positive on one system: for any positive $\rho$ and any $v$, $v^\dagger\rho^Tv$ is the complex conjugate of $\bar v^\dagger\rho\bar v$, hence is real and nonnegative. It fails complete positivity already for a <qubit>. Apply partial transposition to the second subsystem of the <Bell state> $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$. The resulting operator is
$$
(\operatorname{id}\otimes T)(|\Phi^+\rangle\langle\Phi^+|)=\frac12\bigl(|00\rangle\langle00|+|01\rangle\langle10|+|10\rangle\langle01|+|11\rangle\langle11|\bigr).
$$
On the antisymmetric vector $(|01\rangle-|10\rangle)/\sqrt2$, its eigenvalue is $-1/2$. Thus the <partial transpose> is not positive on this entangled input, and
$$
\boxed{T\text{ is positive but not completely positive in dimension at least two}.}
$$
The same witness embeds into any larger input dimension; in the exceptional one-dimensional case transposition is simply the identity.
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