Inverse of matrix with hadamard product Let $A$ and $B$, $X$ be matrices with $\mathbb{R}^{n \times n}$ where $A$, $B$ are a dense and sparse matrix, i.e., the almost elements of $B$ are zeros, respectively. I'm looking for a way to solve the equation below with respect to $ X $:
$I \circ B - A(X \circ B) \circ B = 0 $ where $ \circ $ denotes the Hadamard product.
I tried to solve it as follows:
$ A^{-1}(I \circ B) = X \circ B \circ B $
$ X = A^{-1}(I \circ B) ⊘ B ⊘ B$ where ⊘ denotes element-wise division.
But, there are two suspicious parts.
The first is whether $ A $ can be invertible when the hadamard product exists on the right side of matrix multiplication.
The second is validity of element-wise division. $ X $ has a constraint that only the part of $ X $ where the value of matrix $ B $ is filled has a value, otherwise zero. To compute element of $ X $, it is just needed to calculate the part filled, so I think it is valid to use element-wise division, but I don't have confidence.
I hope I can get an answers to either of the questions. Please help, thank you.
 A: $
\def\b{{\cal B}}
$The standard basis vectors $\{e_k\}$ can be used to extract the columns of a matrix.
For example, the $k^{th}$ column of $X$ is given by
$$x_k = Xe_k$$
Unlike all other vectors, the $\{e_k\}$ distribute over Hadamard products, i.e.
$$\big(A\circ B\big)\,e_k \;=\; \big(Ae_k\circ Be_k\big)$$
This property can be used to turn the matrix equation into a sequence of vector equations, i.e.
$$\eqalign{
\Big(B\circ A(B\circ X)\Big)\,e_k &= \Big(B\circ I\Big)\,e_k \\
Be_k\circ A(Be_k\circ Xe_k) &= Be_k\circ Ie_k \\
b_k\circ A(b_k\circ x_k) &= b_k\circ e_k \\
}$$
Hadamard products of vectors can be replaced by normal matrix products with diagonal matrices
$$\eqalign{
\b = {\rm Diag}(b) \quad\implies\quad \b w \;=\; b\circ w \\
}$$
This creates a simple matrix equation for each column of $X$
$$\eqalign{
(\b_k A\b_k)\,x_k &= \b_k e_k \\
x_k &= (\b_k A\b_k)^{-1}\b_k e_k 
   \;=\; \b_k^{-1} A^{-1} e_k \\
}$$
Finally, any matrix can be written as the sum of its columns $\times$ basis vectors, therefore
$$\eqalign{
X\;=\;\sum_{k=1}^n x_k\,e_k^T\;=\;\sum_{k=1}^n\b_k^{-1}A^{-1}e_k\,e_k^T \\
\\
}$$

Since $B$ is very sparse,
$\,\b_k\;\big({\rm and}\;\b_kA\b_k\big)$ will be singular for most values of $k$.
In those cases, the best you can do is use the pseudoinverse to solve for $x_k$
$$\eqalign{
x_k &= \big(\b_k A\b_k\big)^{+}\b_k e_k
  + \Big(I-\big(\b_k A\b_k\big)^{+}\big(\b_k A\b_k\big)\Big)\,w \\
}$$
where $w$ is an arbitrary vector and its coefficient matrix is the nullspace projector.
In particular, if $\,\b_k=0\;$ then $\,x_k=w={\rm arbitrary}.$
Obviously with all of those random $w$-vectors,
the solution isn't unique.
However, if you always choose the same vector, e.g.
$\,w=0,\,$ then you can generate a unique $X$ matrix.
