Let $J$ be a real valued $2\times 2$ matrix with determinant equal to 1. The eigenvalues of $J$ are complex in general. Let $Q$ be a $2\times 2$ matrix satisfying $$J^T Q J = Q$$
Question: Does there exist a $Q$ which is symmetric positive-definite?
My work so far:
I have found a solution for $Q$ by using the vec operator and the kronecker product. Taking the $\text{vec}$ of both sides of the above equations yields $$(J\otimes J^T) \text{vec}(Q) = \text{vec}(Q)$$
Notice $\text{vec}(Q)$ is an eigenvector of $(J\otimes J^T)$ with eigenvalue 1. This eigenvector exists because $J$ has determinant one, so $(J\otimes J^T)$ has at least two eigenvalues equal to one. The multiplicity 2 on the eigenvalue should give us two degrees of freedom to ensure that $Q$ is symmetric positive definite.