Sylvester's equation with Hadamard product So I've been struggling with this problem for a very, very long time. For $W, Y \in \mathbb{R}^{T \times N}, U \in \mathbb{R}^{N \times K}$ and $R\in \mathbb{R}^{T \times T}$, find $V \in \mathbb{R}^{T \times K}$ such that
$$
W \circ (V U^T)U + V + R^T R V = C 
$$
where $\circ$ is the Hadamard (or element-wise) product. I know that in the absence of $W$ this would be Sylvesters's equation and thus (under condition ...) be solvable by
$$
vec(V) = (I_K \otimes R^T R + (U^T U + I_K) \otimes  I_T)^{-1}vec(C)
$$
where $\otimes$ is the Kronecker-product. Does anyone know how to solve this when the Hadamard-product $W \circ \dots$ is inserted into the mix?
EDIT: The solution posed by Omnomnomnom was very good and with a minor fix turned out to solve my problem. For future reference I post the full solution. We first need to vectorize the equation
$$
\begin{align}
\text{vec}\left[W \circ (V U^T)U + V + R^T R V \right] = \text{vec}[C] \\
\text{vec}\left[W \circ (V U^T)U\right] + \text{vec}(V) + \text{vec}\left[R^T R V \right] = \text{vec}[C]
\end{align}
$$
If we consider the first term as $\text{vec}\left(A B\right)$ where $A = W \circ (U V^T)$ and $B = U$ we can use that $\text{vec}(A B) = (B^T \otimes I_T)\text{vec}(A)$ to compute
$$
\text{vec}\left[W \circ (VU^T) U\right] = (U^T \otimes I_T)\text{vec}\left[W\circ(VU^T)\right]
$$
Having done this the remaining steps are exactly those of Omnomnomnom's. We compute
$$
\begin{align}
\text{vec}\left[W\circ (VU^T)\right] &= \text{diag}(\text{vec}(W))\text{vec}[VU^T] \\
&= \text{diag}(\text{vec}(W))(U \otimes I_T)\text{vec}(V)
\end{align}
$$
and 
$$
\text{vec}(I_K V) = (I_K \otimes I_T)\text{vec}(V)
$$ 
and 
$$
\text{vec}[R^T R V] = (I_K \otimes R^T R) \text{vec}(V)
$$ to arrive at the fully vectorized equation
$$
\left[(U^T \otimes I_T)\text{diag}(\text{vec}(W))(U \otimes I_T) + (I_K \otimes I_T) + (I_K \otimes R^T R) \right]\text{vec}(V) = \text{vec}(C)
$$
which we can guarantee to be solvable if the entries of $W$ are non-negative. Once again thank you Omnomnomnom.
 A: The trick here is to get a formula for $\operatorname{vec}(W \circ X)$ in terms of $\operatorname{vec}(X)$.  In particular, we have
$$
\operatorname{vec}(W \circ X) = \operatorname{diag}(\operatorname{vec}(W))\operatorname{vec}(X).
$$
With that, we now have
$$
\operatorname{vec}(W \circ (V U^T)U + V + R^T R V) = \\
\operatorname{vec}(W \circ (V U^T)U) + 
\operatorname{vec}(V) + 
\operatorname{vec}(R^T R V) =\\
\operatorname{diag}(\operatorname{vec}(W))\operatorname{vec}((V U^T)U) + 
\operatorname{vec}(V) + 
\operatorname{vec}(R^T R V) =\\
\operatorname{diag}(\operatorname{vec}(W))(U^TU \otimes I_T)\operatorname{vec}(V) + 
\operatorname{vec}(V) + 
(I_K \otimes R^TR)\operatorname{vec}(V) =\\
[\operatorname{diag}(\operatorname{vec}(W))(U^TU \otimes I_T) + 
I_K \otimes I_T + 
(I_K \otimes R^TR)]\operatorname{vec}(V).
$$
So, we have
$$
[\operatorname{diag}(\operatorname{vec}(W))(U^TU \otimes I_T) + 
I_K \otimes I_T + 
(I_K \otimes R^TR)]\operatorname{vec}(V) = \operatorname{vec}(C).
$$
When the matrix on the left is invertible, we can say that
$$
\operatorname{vec}(V) = [\operatorname{diag}(\operatorname{vec}(W))(U^TU \otimes I_T) + 
I_K \otimes I_T + 
(I_K \otimes R^TR)]^{-1} \operatorname{vec}(C).
$$
If you prefer, this can be slightly condensed into 
$$
\operatorname{vec}(V) = [\operatorname{diag}(\operatorname{vec}(W))(U^TU \otimes I_T) + 
(I_K \otimes (I_T + R^TR))]^{-1} \operatorname{vec}(C).
$$
If the entries of $W$ are non-negative, then we can guarantee that the matrix will be invertible, which is to say that your original equation has a unique solution.
