I stumbled upon an interesting problem in my calculations, which I can't figure out how to solve directly. It goes as follows: show that
$$ -2\left(I\otimes (Z-Q)C^{-1} + (Z-Q)C^{-1} \otimes I \right)^{-1}vec(Q) = vec(C) $$
where $C, Q$ are any square symmetric matrices and $Z$ is skew-symmetric; $I$ is identity matrix, and $vec(\cdot):\mathbb{R}^{n\times n} \to \mathbb{R}^{n^2}$ is a vectorization opertor (vertical column stacking); $\otimes$ is a kronecker product.
One solution is to multiply both sides by $\left(I\otimes (Z-Q)C^{-1} + (Z-Q)C^{-1} \otimes I \right)$ and turning it to the Sylvester equation of the form $-2Q=(Z-Q)C^{-1}C + C\left((Z-Q)C^{-1}\right)^T \implies -2Q=-2Q$.
Q: how to do this directly, without getting rid of vectorization, just by shuffling around the LHS of the first equation?