 Solution

Let $\mathcal{G}=-i[H,\cdot ]+{\sum }_d\mathcal{D}[V_d]$ denote the Lindbladian. Substituting $\rho ={\sum }_n{r}_n{\sigma }_n$ and projecting with the Hilbert-Schmidt inner product givesFor the Quantum Hamiltonian term, cyclicity of the trace yieldsFor each Lindblad dissipator, the same cyclic rearrangement givesThese are the required coefficient matrices, so $\dot{r}=(L+{\sum }_d{D}^{(d)})r$. They are real because both generators map Hermitian matrices to Hermitian matrices, whose coordinates are real by (a).

The trace of a commutator vanishes. AlsoThus the last row of $L+{\sum }_d{D}^{(d)}$ is zero and $\boxed{{\dot{r}}_{N^2}=0}$. It is trace preservation, not preservation of the purity of a density operator, that produces this constant coordinate.

Write $G=L+{\sum }_d{D}^{(d)}$ and $n=N^2-1$. Since $r_{N^2}=1/\sqrt{N}$, the remaining coordinates obey the Affine Bloch equationBecause $[H,I]=0$ and $\mathcal{D}[V](I)=[V,V^{†}]$, its offset can also be writtenIt vanishes for unital dynamics, but need not vanish for a general dissipative evolution.