Is there a formula for the inverse of Hadamard product? Say $A$ and $B$ are two square, positive-semidefinite matrices. Is there an expression in terms of matrix product, transpose, and inverse for the Hadamard product $A∘B$? 
For example, "$(A∘B)^{-1} = A^{-1} ∘ B^{-1}$" (which is not true).
Edit: I understand that $A∘B$ may not be invertible, but is there any expression if invertibility is given?
 A: There won't be such a formula. In fact, if $A$ and $B$ are invertible, we can't guarantee that $A\circ B$ will even have an inverse.  In particular, consider
$$
A = \pmatrix{1&0\\0&1}, \quad
B = \pmatrix{0&1\\1&0}
$$
we note that $A=A^{-1}$ and $B = B^{-1}$, but $A\circ B = 0$.

It is often, however, useful to consider the Hadamard product as a submatrix of the Kronecker product.  For the Kronecker product, we have
$$
(A \otimes B)^{-1} = A^{-1} \otimes B^{-1}
$$
A: The Hadamard (aka element-wise) inverse $A^{\theta}$ is such that $A\circ A^\theta = 1$.
If any element of $A$ is zero, then the Hadamard inverse is undefined. 
It satisfies a product rule similar to that of the Kronecker product, i.e. 
$$ (A\circ B)^\theta = A^\theta\circ B^\theta $$
A: I don't think so.
If $A$ is invertible,
and
$B_{i, j}
=1/A_{i,j}
$,
then
$B$ is also invertible
but
$A∘B$
is all ones
and therefore
not invertible.
A: Yes, I think there is a formula for that, in terms of the Hadamard inverse of the first matrix and the inverse of the second one. For example, if:
$\Sigma = A \circ B$
then the inverse of $\Sigma$ can be written as:
$\Sigma^{-1} = A^{\circ (-1)} \circ B^{-1}$
where $A^{\circ (-1)}$ is the Hadamard inverse of $A$, which is defined as
$[A^{\circ (-1)}]_{i,j} = 1/[A]_{i,j}$
Thus B must be invertible, and if $\Sigma$ is invertible as well (as in your question), it follows that $A^{\circ (-1)}$ is well defined. 
Also, a sufficient though not necessary condition is then that $B$ and $A$ be invertible, as Reams proofs that if $A$ is invertible then $A^{\circ (-1)}$ is positive definite.
