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A steady state is a time-independent physical state: its derivative vanishes. With $n=N^2-1$, the Affine Bloch equation therefore requiresAs a real linear equation, it has a solution exactly when $\mathrm{rank}A=\mathrm{rank}[A\mid -c]$. If $s_p$ is one solution, all algebraic solutions are $s_p+\ker A$, an affine set of dimension $n-\mathrm{rank}A$. Physical steady states are its intersection with the set of Bloch vectors corresponding to positive trace-one density operators. An arbitrary affine equation need not have a steady state: $A=0$, $c\ne 0$ is a counterexample.

The equation here comes from a finite-dimensional time-independent Lindblad equation, however, so it always has at least one physical steady state. To prove this, start from any density operator and average its trajectory:These averages remain positive with trace one. That set is a compact set, so some sequence has a limit ${\rho }_{\ast }$. If $\mathcal{L}$ is the time-independent generator, thenbecause density operators are bounded. Continuity gives $\mathcal{L}({\rho }_{\ast })=0$. This proves directly that Finite-dimensional Lindbladians have stationary states.

If $A$ has full rank, its algebraic solution $s_{\ast }=-A^{-1}c$ is unique and existence makes that solution physical. For these Lindblad dynamics the converse also holds, even for a boundary steady state. If $A$ is singular, choose a nonzero traceless Hermitian stationary direction $X$. The two parts of its positive-negative decomposition have the same trace $\tau >0$, so $X=\tau ({\rho }_{+}-{\rho }_{-})$ for two density operators. Time-average their trajectories along a common convergent subsequence. Their limits are stationary by the preceding argument, and their difference remains $X/\tau$, since $X$ itself is stationary. The limits are therefore distinct physical stationary states.

Thus the finite-dimensional physical Bloch equation has a unique steady state exactly when $\mathrm{rank}A=n$. These steady states of an affine Bloch equation must be distinguished from convergence to that steady state, which requires the additional spectral condition in (b). The possibly nonorthonormal coordinate basis does not affect existence or uniqueness, since it still gives an invertible parametrization of the traceless Hermitian space.