 Solution

A steady state is a time-independent density operator satisfying $\mathcal{L}({\rho }_{\ast })=0$, equivalently a constant physical Bloch vector $s_{\ast }$ satisfying $\boxed{A{s}_{\ast }=-c}$. Let $K=N^2-1$. By the rank criterion for a linear system, this equation has a solution exactly whenWhen it is consistent, every algebraic solution has the form $s_0+v$ with $v\in \ker A$, so the solution affine space has dimension $K-\mathrm{rank}A$ by the rank-nullity theorem. Physical steady states are its intersection with the positive density operator body. The algebraic equilibrium is unique exactly when $\mathrm{rank}A=K$, in which case $\boxed{s_{\ast }=-A^{-1}c}$.

An arbitrary affine differential equation need not have an equilibrium: $A=0$, $c\ne 0$ is inconsistent. For the finite-dimensional, time-independent Lindblad equation under discussion, however, a physical steady state always exists. Indeed, the averages ${\bar{\rho }}_T=T^{-1}{\int }_0^T\rho (t)dt$ remain positive with trace one. Compactness supplies a convergent subsequence as $T\to \infty$, andThe limit is a physical steady state, proving that Finite-dimensional Lindbladians have stationary states and that the rank consistency condition is automatically satisfied for this physical generator.

In this setting, full rank also characterizes uniqueness among physical steady states. To justify the converse when $A$ is singular, a nonzero $v\in \ker A$ gives a fixed traceless Hermitian operator $X={\sum }_k{v}_k{\sigma }_k$. Write its positive-negative decomposition $X=X_{+}-X_{-}$. Time-average the evolutions of the two positive operators along a common convergent subsequence; the limits $Y_{\pm }$ are positive stationary operators and satisfy $X=Y_{+}-Y_{-}$. Their traces are equal because $\mathrm{Tr}X=0$. If there were only one stationary density operator, both $Y_{\pm }$ would be this matrix times the same trace, forcing $X=0$, a contradiction. Hence $\boxed{\text{unique physical steady state}⟺\mathrm{rank}A=N^2-1}$.