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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.

Sl0wp0k3
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    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 $B^{-1}$. – Ben Grossmann Mar 25 '21 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. – Sl0wp0k3 Mar 25 '21 at 14:05
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    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 – Ben Grossmann Mar 25 '21 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 $B$. – Ben Grossmann Mar 25 '21 at 14:09

1 Answers1

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In general this is impossible. The reason is very simple.

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 $\det(C)\neq x\det(A)\det(B)$.

Daniel
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