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?
Best Answer
That $S^TP+PS=0$ means $PS$ is skew-symmetric. In turn, $P^{-1/2}(PS)P^{-1/2}=P^{1/2}SP^{-1/2}$ is skew-symmetric too. Thus a necessary condition for $P$ to exist is that $S$ is diagonalisable over $\mathbb C$.
This condition is also sufficient: if $S$ is diagonalisable over $\mathbb C$, it admits a real Jordan form $S=M^{-1}KM$. As $S$ has a purely imaginary spectrum and non-real eigenvalues of a real matrix must occur in conjugate pairs, $K$ is skew-symmetric. Hence $PS=M^TKM$ is skew-symmetric for $P=M^TM>0$.