= Solution
A <steady state> is a time-independent <density operator> satisfying $\mathcal L(\rho_*)=0$, equivalently a constant physical <Bloch vector> $s_*$ satisfying $\boxed{As_*=-c}$. Let $K=N^2-1$. By the rank criterion for a linear system, this equation has a solution exactly when
$$
\boxed{\operatorname{rank}A=\operatorname{rank}[A\mid c]}.
$$
When it is consistent, every algebraic solution has the form $s_0+v$ with $v\in\ker A$, so the solution affine space has dimension $K-\operatorname{rank}A$ by the <rank-nullity theorem>. Physical <steady states> are its intersection with the positive <density operator> body. The algebraic equilibrium is unique exactly when $\operatorname{rank}A=K$, in which case $\boxed{s_*=-A^{-1}c}$.
An arbitrary affine differential equation need not have an equilibrium: $A=0$, $c\ne0$ is inconsistent. For the finite-dimensional, time-independent <Lindblad equation> under discussion, however, \b[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 $T\to\infty$, and
$$
\mathcal L(\bar\rho_T)=\frac{\rho(T)-\rho(0)}{T}\longrightarrow0.
$$
The 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> $X=\sum_kv_k\sigma_k$. 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 $\operatorname{Tr}X=0$. 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 $\boxed{\text{unique physical steady state}\iff\operatorname{rank}A=N^2-1}$.
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