TITLE: Are solutions of $\frac{1}{2}(A^T+A)x=b$ and $Ax=b$ related?
QUESTION [7 upvotes]: I saw some statements about these 2 systems while I was reading something about linear algebra. So I am curious if the solutions of these 2 systems are related. If it is, how are they related?
Thanks for any suggestion!

REPLY [3 votes]: We define the equations and vectors $\boldsymbol{x}_{1}$ and $\boldsymbol{x}_{2}$: 
$$A\boldsymbol{x}_{1}=\boldsymbol{b}$$
$$\cfrac{1}{2}(A+A^{T})\boldsymbol{x_{2}}=\boldsymbol{b}$$
Then, suppose to describe $\boldsymbol{x_{2}}$ using $A$ , $\boldsymbol{b}$ and $\boldsymbol{x_{1}}$. According to the formula as Woodbury matrix identity, an inverse matrix can be transformed as follows:
$$
\begin{align}
\boldsymbol{x_{2}}
&=2(A+A^{T})^{-1}\boldsymbol{b} \\
&=2\{A^{-1}-A^{-1}A^{T}(I+A^{-1}A^{T})^{-1}A^{-1}\}\boldsymbol{b} \\
&=2\{I- \ U(I+U)^{-1}\}\boldsymbol{x_{1}}
\end{align}
$$
where, $U=A^{-1}A^{T}$ and $I$ is an identity matrix. Particularly, if $A$ is a symmentric matrix, $U=I$ is established. Therefore:
$$\boldsymbol{x_{2}}=\boldsymbol{x}_{1}$$
In conclusion, there is a relation between $\boldsymbol{x_{1}}$ and $\boldsymbol{x_{2}}$.