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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?

tolec
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    Formulas exist for the transpose (T), complex () and hermitian (H) conjugates as well as the element-wise inverse ($\Theta$) of a Hadamard product $$\eqalign{ (A&\circ&B)^T &= A^T&\circ&B^T \ (A&\circ&B)^ &= A^&\circ&B^ \ (A&\circ&B)^H &= A^H&\circ&B^H \ (A&\circ&B)^\Theta &= A^\Theta&\circ&B^\Theta \ }$$ but sadly there is no formula for the standard matrix inverse. – greg Mar 04 '21 at 16:34

4 Answers4

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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 $$ (A \otimes B)^{-1} = A^{-1} \otimes B^{-1} $$

Ben Grossmann
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  • I sort of ask this question because I know Kronecker product has this property and Hadamard product is a submatrix of Kronecker product, but a similar equation is not true. – tolec Jun 11 '15 at 20:52
  • Ah. Well, in general, you can't find the inverse of a submatrix in terms of the inverse of the overall matrix. – Ben Grossmann Jun 11 '15 at 20:56
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    Hi, how can I prove that $A^{-1}\circ B^{-1}\ge (A\circ B)^{-1}$ ? Is it true that proving $A^{-1}\otimes B^{-1}\ge (A\otimes B)^{-1}$ is enough? – Fareed Abi Farraj Jan 06 '20 at 21:06
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    @Fareed I don't know what you mean by $\geq$ in this context and anyway I don't happen to know the answer off the top of my head. Please make a new post about your question if you're interested in an answer. – Ben Grossmann Jan 07 '20 at 01:08
  • I mean by $\ge$, if $A\ge B$ i.e. $A-B\ge 0$ then $A-B$ is positive semidefinite. – Fareed Abi Farraj Jan 07 '20 at 05:07
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    @FareedAbiFarraj I still don't know. We can get $A^{-1} \circ B^{-1}$ as a submatrix of $A^{-1} \otimes B^{-1}$, but I see no way to get $(A \circ B)^{-1}$ by extracting a submatrix. – Ben Grossmann Jan 07 '20 at 07:22
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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. $$ (A\circ B)^\theta = A^\theta\circ B^\theta $$

lynn
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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.

marty cohen
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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.

Daniel
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