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I want to solve

$$X \cdot A + A^T = I$$

for $X$, $A$ and $X$ are arbitrary matrices and $A$ is invertible. I know that $A \cdot A^{-1} = I$, this helps, but I don't know how to deal with the additional $+A^T$.

How can I approach this?

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    Did you mean "$A$ is invertible" in your hypothesis instead of "$X$ is invertible"?2012-04-13
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    Yes, sorry for the typo; $A$ is invertible2012-04-13
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    @Mahoni, solutions remains the same, false assumption that I made before was that A must be invertible to solve it2012-04-13

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Rewrite $X\cdot A+A^T=I$ as $X\cdot A=I-A^T$ and premultiply, since $A$ is invertible, by $A^{-1}$ on he right to get $$X\cdot A\cdot A^{-1}=(I-A^T)A^{-1}.$$ Thus, this gives $$X=(I-A^T)A^{-1}=A^{-1}-A^T\cdot A^{-1}.$$

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    Not true. e.g. $$A = \left(\begin{array}{cc}1 & i\\0 & 0 \end{array}\right)$$ and $$X = \left(\begin{array}{cc}0 & 1 \\ -i & 0\end{array}\right)$$. Every 2x2 counter-example is complex, though.2012-04-13
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    @Hurkyl, thank you for the note2012-04-13
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multiply both sides by $ A^{-1} $ so you get

$ X+ A^{T}\cdot A^{-1}=A^{-1} $ and you get your final result.