$\text{tr}(X)=\text{tr}(A^{-1}B)$ for $AX+XA=2B$ Suppose X is a solution to following equation over positive definite matrices $A$,$B$
$$XA+AX=2B$$
The following seems to hold (numerically)
$$\text{tr}(X)=\text{tr}(A^{-1} B)$$
Can anyone see the way to prove this?

Since $A$ is positive definite, it has an inverse. Hence
$$ +=2 \\ \Rightarrow 
A^{-1}XA +X=2A^{-1}B.$$
Taking $Tr(\cdot)$ on both sides leads to
$$Tr(A^{-1}XA) + Tr(X) = 2Tr(A^{-1}B)\\
\Rightarrow Tr(AA^{-1}X) + Tr(X) = 2Tr(A^{-1}B)\\
\Rightarrow Tr(X) + Tr(X) = 2Tr(A^{-1}B) \\
\Rightarrow  Tr(X) = Tr(A^{-1}B) $$
The second line above is due to the rotation property of the $Tr(\cdot)$ operation.
