= Solution
A <steady state> is attractive here if every physical initial <density operator> approaches it. Subtracting its <Affine Bloch equation> from the trajectory equation gives $\delta\dot s=A\delta s$, so $\delta s(t)=e^{At}\delta s(0)$. The necessary and sufficient condition is
$$
\boxed{\operatorname{Re}\lambda<0\quad\text{for every eigenvalue }\lambda\text{ of }A}.
$$
Such an $A$ is a <Hurwitz stable matrix>. To prove sufficiency without assuming diagonalizability, put it in <Jordan normal form>. Each <Jordan block> contributes $e^{\lambda t}\sum_{j=0}^{q-1}t^jN^j/j!$, and the negative real part makes the exponential dominate all polynomial factors. For necessity, an eigenmode with positive real part grows; an eigenmode with zero real part remains constant or oscillates, so it does not tend to zero. A nontrivial Jordan block on the imaginary axis also prevents decay.
The physical <Bloch vector> body has nonempty interior, so differences of physical initial vectors from $s_*$ span the full real coordinate space. Thus convergence for every physical initial state also forces $e^{At}\to0$ on that space. This rules out hiding a nondecaying eigenmode outside the physical set. The strict spectral condition entails uniqueness and global <asymptotic stability> of the <steady state>.
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