When is $A^TR + RA = -2RA$

Question

I am working through some exercises on stability theory and Lyapunov analysis.

The equation in question is

\begin{equation} A^TR+RA = -I. \end{equation}

The matrix $A_{n\times n}$ is symmetric, Hurwitz (all eigenvalues have a negative real part) and $R_{n\times n}$ is a symmetric positive definite matrix. The matrix $I_{n\times n}$ is denoting the identity matrix.

Because $A$ is symmetric it is claimed that the previous equation can be restated as

$$2RA = -I.$$

It is easy to come up with counterexamples to show that symmetry alone is not sufficient.

It is obvious that the equation can be rewritten as

$$AR + RA = -I.$$

If both matrices would commute than the claim would be easy to verify.

Questions

1. Is there a criterion to check if two matrices / symmetric matrices commute?

2. What additional conditions (including symmetry of $A$) would be sufficient to guarantee that $A^TR + RA = -I \implies 2RA = -I$?

Answer 1

1. Is there a criterion to check if two matrices / symmetric matrices commute?

It is a hard problem to find a criterion that works all the time and is easier than just checking both products. However, here is one:

Theorem. If $A$, $B$ are symmetric then $AB=BA$ if and only if $AB$ is also symmetric.

A proof here for instance.

2. What additional conditions (including symmetry of $A$) would be sufficient to guarantee that $A^TR + RA = -I \implies 2RA = -I$?

If $A$ is symmetric then the implication $$A^TR + RA = -I \implies 2RA = -I$$ is equivalent to $$AR=RA$$ therefore I don't think there can be more to be said in this direction.