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?
On
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 $$ (A \otimes B)^{-1} = A^{-1} \otimes B^{-1} $$
On
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. $$ (A\circ B)^\theta = A^\theta\circ B^\theta $$
On
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.
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.