y^{\dagger}]^{-1}.$$ Then multiply ${{\boldsymbol{U}}}^{\dagger}(t)$ on the right, and then divide both side by $i$, one could obtain $$\begin{split}
 {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{G}}}(t) {\mathcal{P}}_{{{\boldsymbol{U}}}(t)} 
 =&  (-i) {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} \left( {\mathbb{E}}_{\theta_t} \bigl[a({{\boldsymbol{U}}}(t)y,t) ({{\boldsymbol{U}}}(t) y)^{\dagger}\bigr] + \sum_{j=1}^{N} {\mathbb{E}}_{\theta_t} \bigl[b_j({{\boldsymbol{U}}}(t)y,t) b_j^{\dagger}({{\boldsymbol{U}}}(t)y,t) \bigr] \right) {{\boldsymbol{U}}}(t) {\mathbb{E}}_{\theta_t}[y y^{\dagger}]^{-1} {{\boldsymbol{U}}}^{\dagger}(t).\\
 \end{split}$$ Thus we have already obtained the expression of ${{\boldsymbol{G}}}(t) \equiv {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{G}}}(t) {\mathcal{P}}_{{{\boldsymbol{U}}}(t)} + {\text{h.c.}}$ (cf. [Eq. ]{}). [ As a remark, up to here, we have not yet used any information nor assumption about $A$ and $B_j$. ]{}

Then we plug the expression of ${{\boldsymbol{G}}}(t)$ into ${{\boldsymbol{C}}}$, $$\begin{split}
{{\boldsymbol{C}}}&= {\mathcal{P}}_{{{\boldsymbol{U}}}(t)} {\mathbb{E}}_{\theta_t} \bigl[ \bigl({{\boldsymbol{U}}}(t) A(y,t) - a({{\boldsymbol{U}}}(t)y,t)\bigr)y^{\dagger} {{\boldsymbol{U}}}^{\dagger}\bigr] + {\text{h.c.}}\\
& \qquad + \sum_{j=1}^{M} {{\boldsymbol{U}}}(t) {\mathbb{E}}_{\theta_t} \bigl[B_j(y,t) B_j^{\dagger}(y,t)\bigr] {{\boldsymbol{U}}}^{\dagger}(t)  \\
& \qquad - \sum_{j=1}^{N} \left( {\mathcal{P}}_{{{\boldsymbol{U}}}(t)} {\mathbb{E}}_{\theta_t} \bigl[b_j({{\boldsymbol{U}}}(t)y,t) b_j^{\dagger}({{\boldsymbol{U}}}(t)y,t)\bigr] {\mathcal{P}}_{{{\boldsymbol{U}}}(t)} 
 + {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {\mathbb{E}}_{\theta_t} \bigl[b_j({{\boldsymbol{U}}}(t)y,t) b_j^{\dagger}({{\boldsymbol{U}}}(t)y,t)\bigr] {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)}\right). \\
\end{split}$$ One could observe that ${{\boldsymbol{C}}}= {\mathcal{P}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{C}}}{\mathcal{P}}_{{{\boldsymbol{U}}}(t)} + {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{C}}}{\mathcal{Q}}_{{{\boldsymbol{U}}}(t)}$ with $$\left\{
\begin{split}
{\mathcal{P}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{C}}}{\mathcal{P}}_{{{\boldsymbol{U}}}(t)} &= {\mathcal{P}}_{{{\boldsymbol{U}}}(t)} {\mathbb{E}}_{\theta_t} \bigl[ \bigl({{\boldsymbol{U}}}(t) A(y,t) - a({{\boldsymbol{U}}}(t)y,t)\bigr)y^{\dagger} {{\boldsymbol{U}}}^{\dagger}\bigr] + {\text{h.c.}}+ \sum_{j=1}^{M} {{\boldsymbol{U}}}(t) {\mathbb{E}}_{\theta_t} \bigl[B_j(y,t) B_j^{\dagger}(y,t)\bigr] {{\boldsymbol{U}}}^{\dagger}(t) \\
&  - \sum_{j=1}^{N}{\mathcal{P}}_{{{\boldsymbol{U}}}(t)} {\mathbb{E}}_{\theta_t} \bigl[b_j({{\boldsymbol{U}}}(t)y,t) b_j^{\dagger}({{\boldsymbol{U}}}(t)y,t)\bigr] {\mathcal{P}}_{{{\boldsymbol{U}}}(t)}, \\
{\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{C}}}{\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} &= - \sum_{j=1}^{N} {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {\mathbb{E}}_{\theta_t} \bigl[b_j({{\boldsymbol{U}}}(t)y,t) b_j^{\dagger}({{\boldsymbol{U}}}(t)y,t)\bigr] {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)}.
\end{split}\right.$$ Notice that ${\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{C}}}{\mathcal{Q}}_{{{\boldsymbol{U}}}(t)}$ does not depend on any parameter we choose, that is, it is independent of $A$, 