How to solve the matrix equations: $\frac{1}{2}(\Omega Q - Q\Omega)=F$ for $\Omega$ How to solve the following matrix equation: $$\frac{1}{2}(\Omega Q + Q\Omega^T)=F$$

*

*$Q$ is of rank $1$, $\text{tr}(Q)=1$, $Q\succeq0$ (positive semidefinite).  So we can werite $Q = qq^T$ for some $q$ with $\|q\|=1$.

*$\Omega$ is skew-symmetric, i.e., $\Omega^{T} = -\Omega$. So diagonal entries are all $0$.

*$\Omega$, $Q$, $F$ are $4\times 4$ square matrices. $\Omega$ is the variable I want to solve.

I cannot find any method to solve it. Can anyone suggest me how to solve $\Omega$ in terms of $Q$ and $F$?
Even though we cannot obtain the closed form, any approximation or trick can we use?
Thanks!
Note:  $\Omega$ should look like
$$\begin{bmatrix}0 & \omega_3 & -\omega_2 & \omega_1 \\
  -\omega_3 & 0 & \omega_1 & \omega_2 \\
   \omega_2 & -\omega_1 & 0 & \omega_3 \\
   -\omega_1 & -\omega_2 & -\omega_3 & 0 \end{bmatrix}$$
 A: Presumably the matrices are real. Since $Q$ is a rank-one positive semidefinite matrix with unit trace, we can write $Q=uu^T$ for some unit vector $u$. Let $S=2F$. The equation can then be rewritten as
$$
\Omega uu^T+uu^T\Omega^T=S.\tag{1}
$$
When $S=0$, the equation is clearly solvable and its solutions are given by those skew-symmetric matrices $\Omega$ such that $\Omega u=0$. So, if we pick an orthogonal matrix $U$ whose first column is $u$, the general solution in this case is given by $\Omega=U(0\oplus K)U^T$ where $K$ is an arbitrary skew-symmetric matrix.
When $S\ne0$, we will prove that the equation is solvable if and only if $S$ is a traceless rank-$2$ symmetric matrix such that $\{u,Su\}$ is an orthogonal basis of $\operatorname{range}(S)$. When this condition is satisfied, the solutions to $(1)$ are given by those skew-symmetric matrices $\Omega$ such that $\Omega u=Su$.
For necessity, note that when $S\ne0$, the LHS of $(1)$ is a traceless symmetric matrix whose rank is at most two. Since skew-symmetric matrices have no nonzero real eigenvalues, the LHS of $(1)$ cannot have rank one. Therefore the equation is solvable only when $S$ is a traceless symmetric matrix of rank $2$. In this case, if we right-multiply both sides of $(1)$ by $u$, we get $\Omega u=Su$. Since $S$ has rank two, $(1)$ shows that $\Omega u$ and $u$ must be linearly independent and they span the range of $S$. As we also have $u^TSu=0$ (this can be obtained directly from $(1)$ or from $u^TSu=u^T\Omega u=0$), $\{u,Su\}$ is also an orthogonal basis of $\operatorname{range}(S)$.
Conversely, suppose $S$ is a traceless rank-$2$ symmetric matrix such that $\{u,Su\}$ is an orthogonal basis of $\operatorname{range}(S)$. By a change of orthonormal basis, we may assume that $u=e_1$ and $Su$ is parallel to $e_2$. Hence $S=\pmatrix{0&a\\ a&0}\oplus0$ and it is straightforward to verify that $Suu^T+uu^TS=S$. It follows that $(1)$ is satisfied when $\Omega u=Su$. In other words, if we put $a=\|Su\|$ and pick an orthogonal matrix $U$ whose first two columns are $u$ and $\frac{Su}{\|Su\|}$ respectively, then
$$
S=U\left[\begin{array}{cc|c}0&a&\mathbf0\\ a&0&\mathbf0\\ \hline \mathbf0&\mathbf0&\mathbf0\end{array}\right]U^T
$$
and the solutions to $(1)$ are given by skew-symmetric matrices in the form of
$$
\Omega=U\left[\begin{array}{cc|c}0&-a&\mathbf0\\ a&0&-y^T\\ \hline \mathbf0&y&K\end{array}\right]U^T.
$$
A: As a "brute-force" approach, it suffices to solve a $32 \times 16$ linear system.
Since it is given that $\Omega^T = -\Omega$, we can rewrite the equation as
$$
\frac 12 (\Omega Q - Q\Omega) = F.
$$
Note that this is a Sylvester equation. As such, this equation can be rewritten in terms of the vectorization operator and Kronecker product
$$
\frac 12 (Q^T\otimes I_4- I_4 \otimes Q) \operatorname{vec}(\Omega) = \operatorname{vec}(F) \implies\\
(Q^T\otimes I_4- I_4 \otimes Q) \operatorname{vec}(\Omega) = 2\operatorname{vec}(F).
$$
Note that this equation will not have a unique solution (if it has any solution at all), so we must impose the constraint that $\Omega$ is skew-symmetric. This constraint can be expressed with the help of the commutation matrix as
$$
(I_4 + K^{(4,4)})\operatorname{vec}(\Omega) = 0.
$$
Putting everything together, we can express the system to be solved as the block-matrix equation
$$
\pmatrix{Q^T \otimes I_4 - I_4 \otimes Q\\ I_4 + K^{(4,4)}} \operatorname{vec}(\Omega) = \pmatrix{2\operatorname{vec}(F)\\0}.
$$
