# Hadamard product of two matrices as a matrix multiplication

I have encountered the following problem:

I have two $$N$$-by-$$N$$ complex valued matrices $$A, B$$, and then I form a third matrix as a Hamadard product of the previous two: $$C = A \odot B$$, so that for each element we have $$C_{i,j} = A_{i,j} \cdot B_{i,j}$$. My question is whether it is possible to express matrix $$C$$ as a regular matrix multiplication (dot products) of $$A, B$$, with a finite number of additional matrices? For instance $$C = T^{(L)} \cdot A \cdot T^{(M)} \cdot B \cdot T^{(R)}$$, where $$T$$ are just some auxiliary matrices.

I know that some people have asked this question already, however, the common solutions involve representing matrices in a vector form or using things like SVD. All these as well as using the sum of multiplications on projector operators I want to avoid as my final goal is to find $$C^{-1}$$, which is why I restricted the desired form of $$C$$ in terms of dot products.

• This post is related. As I state on that post, I don't believe that there is a nice formula (or algorithm) for the inverse of $A \odot B$ that makes any use of $A^{-1}$ or $B^{-1}$. Commented Mar 25, 2021 at 14:01
• Nice to meet you again, Ben! I did not find this thread at first, thank you for providing the link. So, probably, I need to study specific symmetry properties of $A, B$ I encounter for my problem in order to use this knowledge when constructing the inverse. Commented Mar 25, 2021 at 14:05
• That said, it is potentially possible to use the inverse of $A \otimes B$ to get the inverse of the submatrix $A \odot B$ if one uses the approach described here, where one uses the Woodbury formula instead of the Sherman Morrison formula Commented Mar 25, 2021 at 14:05
• I do think that your alternative approach of studying the symmetry properties of $A$ and $B$ is more likely to work out. Perhaps you should post a question that's a bit more specific about the matrices $A$ and $B$. Commented Mar 25, 2021 at 14:09

Notice that if $$C$$ is a product of matrices and $$A,B$$ are part of this product then $$\det(C)=x\det(A)\det(B)$$.
Now, choose $$A=\begin{pmatrix}1 & 2\\ 1 & 2\end{pmatrix}$$ and $$B=\begin{pmatrix}1 & 1\\ 1 & -1\end{pmatrix}$$ then $$C=\begin{pmatrix}1 & 2\\ 1 & -2\end{pmatrix}$$ and $$\det(C)=-4$$, but $$\det(A)=0$$.
So $$\det(C)\neq x\det(A)\det(B)$$.