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I want to solve the following matrix equation. Could anyone give me a hand? Thanks.

Given an $n \times n$ matrix $\mathbf A$ (diagonally dominant), I need to solve an $n \times n$ symmetric matrix $\mathbf X$ such that

$\mathbf A\mathbf X+\mathbf X\mathbf A^\top =\mathbf I$,

where $\mathbf I$ is an $n \times n$ identity matrix.

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    I don't get the question. Let A=0, then surely no such X exists. Also, please remove those irrelevant tags.2012-04-16
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    We assume that $A$ is a nonzero matrix.2012-04-16
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    Is $A$ *strictly* diagonally dominant? If so, $A$ is of course non-zero.2012-04-16
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    Yeah, it is strictly diagonally dominant.2012-04-16
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    By Gershgorin circle theorem, diagonally dominated matrices are non-singular. Therefore A inverse(which is square) exists2012-04-16

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