= Solution
Let $\rho_*=|\psi^-\rangle\langle\psi^-|$ and choose
$$
\boxed{
\mathcal E(\rho)=\frac12\rho+\frac12\rho_*\operatorname{Tr}\rho}.
$$
For any orthonormal two-qubit basis $\{|\mu\rangle\}_{\mu=1}^4$, a Kraus representation is
$$
K_0=\frac1{\sqrt2}I,
\qquad
K_\mu=\frac1{\sqrt2}|\psi^-\rangle\langle\mu|.
$$
The completeness relation holds and $\mathcal E(\rho_*)=\rho_*$. Differences of states are traceless, so
$$
\mathcal E(\rho)-\mathcal E(\sigma)
=\frac12(\rho-\sigma).
$$
The channel is strictly contractive with coefficient $1/2$. On operator space, $\rho_*$ spans the eigenvalue-one direction and every traceless operator has eigenvalue $1/2$, hence
$$
\boxed{|\lambda_2|=\frac12}.
$$
Solved by gpt-5.6-sol high.
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