 Solution

Subtract a steady state to obtain $e(t)=s(t)-s_{\ast }$ with $\dot{e}=Ae$, hence $e(t)=e^{At}e(0)$. The steady state attracts every initial condition exactly whenIn other words, $A$ must be a Hurwitz stable matrix. To see sufficiency without assuming diagonalizability, each block in its Jordan normal form contributes terms of the form $t^k{e}^{\mu t}$ with finite $k$. Negative real part makes every such term tend to zero, so the whole matrix exponential tends to zero.

For necessity, an eigenmode with positive real part grows, and a mode with zero real part persists or oscillates instead of decaying. For a complex eigenvalue of a real matrix, take the real and imaginary parts of its eigenvector to obtain real nondecaying solutions. Thus any eigenvalue in the closed right half-plane prevents convergence of all deviations. Physical density operators span the trace-one affine space, so convergence from all physical initial states also forces decay on every traceless direction.

This gives global asymptotic stability for the affine linear system and uniqueness of its steady state. A merely negative-semidefinite symmetric part proves only nonincrease of a particular norm, not this necessary-and-sufficient condition.