k \le r$ nor $j, k > r$. Denote ${{\boldsymbol{D}}}\equiv \bigl[{{\boldsymbol{C}}}, {\mathbb{E}}_{\mu_{{LR},t}}[ x x^{\dagger}]\bigr]_{-}$, which is anti-Hermitian, that is, ${{\boldsymbol{D}}}^{\dagger} = - {{\boldsymbol{D}}}$. Plugging the expression of $\delta {{\boldsymbol{G}}}$ into the third condition in [Eq. ]{}, we could easily compute that $$0 = \lambda {\left\langleU_k, {{\boldsymbol{D}}}U_j\right\rangle} + \lambda^{*} {\left\langleU_j, {{\boldsymbol{D}}}U_k\right\rangle} = \lambda {\left\langleU_k, {{\boldsymbol{D}}}U_j\right\rangle} + \lambda^{*} {\left\langleU_k, {{\boldsymbol{D}}}^{\dagger} U_j\right\rangle}^{*} = \lambda {\left\langleU_k, {{\boldsymbol{D}}}U_{j}\right\rangle}  - c.c.$$ Hence $\text{Im}\left(\lambda {\left\langleU_k, {{\boldsymbol{D}}}U_j\right\rangle}\right) = 0$. Since $\lambda\in {\mathbb{C}}$ is arbitrary, ${\left\langleU_k,  {{\boldsymbol{D}}}U_j\right\rangle} = 0$ for all $j\le r$ and $k > r$. That is to say, ${\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{D}}}{\mathcal{P}}_{{{\boldsymbol{U}}}(t)} = 0$. By plugging the expression of ${{\boldsymbol{D}}}\equiv \bigl[{{\boldsymbol{C}}}, {\mathbb{E}}_{\mu_{{LR},t}}[ x x^{\dagger}]\bigr]_{-}$, we could compute that $$\begin{split} 
0 
&= {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{C}}}{{\boldsymbol{U}}}(t) {\mathbb{E}}_{\theta_t} \bigl[yy^{\dagger}\bigr] {{\boldsymbol{U}}}^{\dagger}(t)\\
&= {\mathbb{E}}_{\theta_t} \Biggl[ i {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{G}}}(t) {{\boldsymbol{U}}}(t) y y^{\dagger} 
- {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} a({{\boldsymbol{U}}}(t)y,t) y^{\dagger}\\
& \hspace{4em} -{\mathcal{Q}}_{{{\boldsymbol{U}}}(t)}\sum_{j=1}^{N} b_j({{\boldsymbol{U}}}(t)y,t) b_j^{\dagger}({{\boldsymbol{U}}}(t)y,t) {{\boldsymbol{U}}}(t) \Biggr] {\mathbb{E}}_{\theta_t}\bigl[yy^{\dagger}\bigr] {{\boldsymbol{U}}}^{\dagger}(t) \\
&= i {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{G}}}(t) {{\boldsymbol{U}}}(t) {\mathbb{E}}_{\theta_t} \bigl[y y^{\dagger}\bigr]^2 {{\boldsymbol{U}}}^{\dagger}(t) 
- {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {\mathbb{E}}_{\theta_t} \bigl[a({{\boldsymbol{U}}}(t)y,t) y^{\dagger}\bigr] {\mathbb{E}}_{\theta_t}\bigl[y y^{\dagger}\bigr]{{\boldsymbol{U}}}^{\dagger}(t) \\
&- {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} \sum_{j=1}^{N} {\mathbb{E}}_{\theta_t}\bigl[ b_j({{\boldsymbol{U}}}(t)y,t) b_j^{\dagger}({{\boldsymbol{U}}}(t)y,t) \bigr] {{\boldsymbol{U}}}(t) {\mathbb{E}}_{\theta_t}[ y y^{\dagger}] {{\boldsymbol{U}}}^{\dagger}(t).
 \end{split}$$ Multiply both sides by ${{\boldsymbol{U}}}(t){\mathbb{E}}_{\theta_t}\bigl[yy^{\dagger}\bigr]^{-2}$ on the right, the last equation yields $$i {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{G}}}(t) {{\boldsymbol{U}}}(t)={\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} \left( {\mathbb{E}}_{\theta_t} \bigl[ a({{\boldsymbol{U}}}(t)y,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] {{\boldsymbol{U}}}(t)\right) {\mathbb{E}}_{\theta_t}[y 