$\text{tr}(X)=\text{tr}(A^{-1}B)$ 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?

$XA+AX=2B$
$A^{-1}XA+A^{-1}AX=A^{-1}2B=2A^{-1}B$
$\text{tr}(A^{-1}XA)+\text{tr}(A^{-1}AX)=\text{tr}(A^{-1}2B)=2\text{tr}(A^{-1}B)$
Since $\text{tr}(A^{-1}XA) = \text{tr}(X) $
We get $\text{tr}(X)=\text{tr}(A^{-1} B)$
