# Solve $XA + A^T = I$ for $X$

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"? – Hurkyl Apr 13 '12 at 19:00
Yes, sorry for the typo; $A$ is invertible – Mahoni Apr 13 '12 at 23:21
@Mahoni, solutions remains the same, false assumption that I made before was that A must be invertible to solve it – Julius Apr 13 '12 at 23:32

To solve it, $A$ must be invertible too $$X\cdot A+A^T=I$$ $$X\cdot A=I-A^T$$ $$X\cdot A\cdot A^{-1}=(I-A^T)A^{-1}$$ $$X=A^{-1}-A^T\cdot A^{-1}$$
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. – Hurkyl Apr 13 '12 at 18:55
multiply both sides by $A^{-1}$ so you get
$X+ A^{T}\cdot A^{-1}=A^{-1}$ and you get your final result.