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Let $A = DB$, where $D$ is any diagonal matrix with positive diagonal entries and $B = [b_{ij}]$ is any symmetric matrix s.t. $B^{\circ -1} = \left[\frac{1}{b_{ij}}\right]$ is positive definite. Can we find the inverse $(A+A^T)^{-1}$ in terms of $B$ and $B^{-1}?$ Here $A^T$ denotes the transpose of $A.$

Any suggestions will be really helpful. Thanks.

VSP
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    The inverse may not exists. Do you have any other hypothesis about A that guarantees the existence of the inverse? – jjagmath Jun 24 '21 at 16:20
  • Related. –  Jun 24 '21 at 16:21
  • For example, $A = \begin{pmatrix} 0 & 1 \ -1 & 0 \end{pmatrix}$ is invertible, but $A + A^T = 0$ is certainly not. Unless your definition of non-symmetric also prohibits skew-symmetric matrices. – Matthew Buck Jun 24 '21 at 16:25
  • @hardmath, i have edited. – VSP Jun 25 '21 at 04:40
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    Do you have any reason to think this task is possible? From only the description given, it seems somewhat random to me. Have you tried the case $n=2$, where all inverses can easily be found by hand? – Greg Martin Jun 25 '21 at 04:55

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