 Solution

Write $\rho ={\sum }_n{r}_n{\sigma }_n$ in the Generalized Bloch representation. Linearity of the Lindbladian gives ${\dot{r}}_m={\sum }_n\mathrm{Tr}({\sigma }_m\mathcal{L}({\sigma }_n))r_n$. For the Hamiltonian term, cyclicity yieldsFor one Lindblad operator $V_d$, the jump term becomes $\mathrm{Tr}({\sigma }_m{V}_d{\sigma }_n{V}_d^{†})=\mathrm{Tr}(V_d^{†}{\sigma }_m{V}_d{\sigma }_n)$, and the two remaining terms giveThusEvery coefficient is real because $\mathcal{L}({\sigma }_n)$ is Hermitian. In addition, $L_{mn}=-L_{nm}$, although a dissipative $D^{(d)}$ need not be symmetric.

The Hamiltonian commutator has zero trace, and each dissipator obeys $\mathrm{Tr}(V\rho V^{†})=\mathrm{Tr}(V^{†}V\rho )$, so $\mathrm{Tr}\mathcal{L}(\rho )=0$. Consequently $\boxed{{\dot{r}}_{N^2}=0,r_{N^2}=1/\sqrt{N}}$ for normalized density operators. Orthogonality to $I/\sqrt{N}$ makes every other ${\sigma }_k$ traceless. Define $s=(r_1,\dots ,r_{N^2-1}{)}^T$. With $M=L+{\sum }_d{D}^{(d)}$, trace preservation gives the block formHere the indices of $A$ range from $1$ to $N^2-1$. Since $\mathcal{L}(I)={\sum }_d(V_d{V}_d^{†}-V_d^{†}{V}_d)$, the offset can also be written $c_m=N^{-1}\mathrm{Tr}({\sigma }_m{\sum }_d[V_d,V_d^{†}])$. This is the Affine Bloch equation; its offset is zero for unital dynamics, which preserve the identity.