For a real square matrix $S$ with all eigenvalues on the imaginary axis, does the following hold? Given a real square matrix $S$ which has all the eigenvalues on the imaginary axis, is it true that there exists a positive definite matrix $P=P^{T}>0$ such that 
\begin{equation}
 S^T{P}+PS=0? 
\end{equation}
I found that claim in a research paper (which is not yet available online!) but I do not find a way to prove this. 
While trying to do an example problem, I noticed that for a matrix 
\begin{equation}
S=\begin{bmatrix}\omega & \sigma\\-\sigma & \omega\end{bmatrix}, 
\end{equation}
the claim holds if $P$ is an identity matrix. Additionally, for any block-diagonal matrix $S_1$ with the diagonal matrix being in the same form as $S$, there exists a corresponding unique $P$ which satisfies the claim. 
But in general, the matrix which has all eigenvalues on the imaginary axis, needs not have the same form as $S_1$, and is there a way to show that there exists a unique $P$ for which the claim holds? 

The answer is no. Here $S\in M_n(\mathbb{R})$ where $n$ is even.
Choose $n=4,S=\begin{pmatrix}U&I_2\\0&U\end{pmatrix}$ where $U=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. Let $P=[p_{i,j}]\in M_4(\mathbb{R})$ s.t. $S^TP+PS=0$.
With an easy calculation, we find that $P$ is in the form $\begin{pmatrix}0&0&0&-a\\0&0&a&0\\0&a&b&0\\-a&0&0&b\end{pmatrix}$; thus $P$ cannot be symmetric $>0$.
