How do I isolate $S$ in $P = \overline S - RSR^{\mathsf H}$?

Question

I have

\begin{align} R &= U^{\mathsf H}S^{-1} \\ P &= \overline{S} - RU \end{align}

I know $R$ and $P$. $P$ and $S$ are Hermitian and invertible. $U$ is symmetric. How do I find $S$ and $U$?

I got halfway:

\begin{align} R &= U^{\mathsf H}S^{-1} \\ &\Downarrow \\ U &= (RS)^{\mathsf H} \\ &= S^{\mathsf H}R^{\mathsf H} \\ &= SR^{\mathsf H} \end{align} since $S$ is Hermitian. Then, \begin{align} P &= \overline S - RSR^{\mathsf H} \end{align}

I don't see how to isolate $S$ here.

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Notation: $R^{\mathsf H}$ denotes the Hermitian of $R$ (the conjugate transpose) and $\overline S$ denotes the conjugate of $S$.

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My motivation is that I would like to fill in my table:

So what I really need are $S$, $U$ in terms of $H$, $J$.

Answer 1

If $Id−\overline{R}R$ is invertible, then $(Id−\overline{R}R)^{−1}\overline{R}P=U$ and $(Id−\overline{R}R)^{−1}\overline{P}=S$. In fact, since $U$ is symmetric, $\overline{U}=U^H=RS$ and so, from $P=\overline{S}-RU$ we obtain $$ \overline{P}=S-\overline{R}\overline{U}=S-\overline{R}RS=(Id-\overline{R}R)S, $$ and so $(Id−\overline{R}R)^{−1}\overline{P}=S$. On the other hand $\overline{R}\overline{S}=U$, and so, from $P=\overline{S}-RU$ we obtain $$ \overline{R}P=\overline{R}\overline{S}-\overline{R}RU=U-\overline{R}RU=(Id-\overline{R}R)U, $$ and so $(Id−\overline{R}R)^{−1}\overline{R}P=U$.