Suppose $A$ is $d\times d$ positive definite matrix, I'd like to find a matrix $X$ such that the following is true for all positive definite matrices $d\times d$ matrices $B$
$$AB+BA=XBX^T$$
When does solution exist? Is there a name for this equation?
On
First of all, we prove the following properties:
Property 1:
If $\;M\;$ is a real symmetric matrix, there exists a symmetric matrix $\;M^{\frac{1}{2}}\;$ such that $\;M^{\frac{1}{2}}\cdot M^{\frac{1}{2}}=M.\;$ Moreover, if $\;M\;$ is positive definite, then $\;M^{\frac{1}{2}}\;$ is real.
Proof:
By applying the spectral theorem for real symmetric matrices, we get that there exists an orthogonal matrix $\;U\;$ such that $\;U^T\cdot M\cdot U=F\;$ is diagonal, hence
$M=U\cdot F\cdot U^T\;$.
Let $\;G\;$ be the diagonal matrix whose entries are the square roots of the entries of $\;F\;,\;$ so it results that $\;F=G\cdot G\;.$
The matrix $\;M^{\frac{1}{2}}=U\cdot G\cdot U^T\;$ is symmetric and
$\;M^{\frac{1}{2}}\cdot M^{\frac{1}{2}}=U\cdot G\cdot U^T\cdot U\cdot G\cdot U^T=U\cdot G\cdot G\cdot U^T=\\\qquad\qquad\;\;=U\cdot F\cdot U^T=M\;.$
Moreover, if $\;M\;$ is positive definite, then all its eigenvalues are positive, therefore all the diagonal entries of the matrix $\;F\;$ are positive too, consequently $\;G\;$ and $\;M^{\frac{1}{2}}\;$ are real matrices.
Property 2:
If $\;M\;$ is an invertible real symmetric matrix, then
$Y^T\cdot Y=M\implies\exists\;U$ orthogonal matrix such that $\;Y=U\cdot M^{\frac{1}{2}}\;.$
Proof:
By applying the Property 1, it follows that there exists a symmetric matrix $\;M^{\frac{1}{2}}\;$ such that $\;M^{\frac{1}{2}}\cdot M^{\frac{1}{2}}=M\;.$
Since $\;M\;$ is invertible, then $\;\det M\ne0\;,\;$ hence $\;\det\left(M^{\frac{1}{2}}\right)\ne0\;,\;$ consequently $\;M^{\frac{1}{2}}\;$ is invertible too.
Given that $M^{\frac{1}{2}}$ is symmetric, then $\left(M^{\frac{1}{2}}\right)^{-1}$ is symmetric too.
If $\;Y^T\cdot Y=M\;,\;$ then
$Y^T\cdot Y=M^{\frac{1}{2}}\cdot M^{\frac{1}{2}}\;,$
$\left(M^{\frac{1}{2}}\right)^{-1}\cdot Y^T\cdot Y\cdot\left(M^{\frac{1}{2}}\right)^{-1}=I\;,$
$\left[Y\cdot\left(M^{\frac{1}{2}}\right)^{-1}\right]^T\cdot\left[Y\cdot\left(M^{\frac{1}{2}}\right)^{-1}\right]=I\;,\;$ hence
$Y\cdot\left(M^{\frac{1}{2}}\right)^{-1}=U\;$ is an othogonal matrix and
$Y=U\cdot M^{\frac{1}{2}}\;.$
Property 3:
If $\;M\;$ is a real symmetric matrix, then
$Y=U\cdot M^{\frac{1}{2}}\;$ where $\;U\;$ is any orthogonal matrix $\implies Y^T\cdot Y=M\;.$
Proof:
$Y^T\cdot Y=\left(U\cdot M^{\frac{1}{2}}\right)^T\cdot\left(U\cdot M^{\frac{1}{2}}\right)=M^{\frac{1}{2}}\cdot U^T\cdot U\cdot M^{\frac{1}{2}}=\\\qquad\quad=M^{\frac{1}{2}}\cdot M^{\frac{1}{2}}=M\;.$
Now we are going to solve the matrix equation:
$X\cdot B\cdot X^T=AB+BA\;,\quad\color{blue}{(*)}$
where $\;A\;$ and $\;B\;$ are positive definite matrices.
By applying the spectral theorem for real symmetric matrices, we get that there exists an orthogonal matrix $\;P\;$ such that $\;P^T\cdot B\cdot P=D\;$ is diagonal, hence
$B=P\cdot D\cdot P^T\;$.
Let $\;E\;$ be the diagonal matrix whose entries are the square roots of the entries of $\;D\;,\;$ so it results that $\;D=E\cdot E\;.$
Given that $\;B\;$ is a positive definite matrix, all its eigenvalues are positive, hence all the diagonal entries of $\;D\;$ and $\;E\;$ are positive too, consequently $\;D\;$ and $\;E\;$ are invertible matrices.
$X\cdot B\cdot X^T=AB+BA\;,$
$X\cdot P\cdot E\cdot E\cdot P^T\cdot X^T=AB+BA\;,$
$\left(E\cdot P^T\cdot X^T\right)^T\cdot E\cdot P^T\cdot X^T=AB+BA\;.$
By letting $\;Y=E\cdot P^T\cdot X^T\;,\;$ we get that
$Y^T\cdot Y=AB+BA\;.\quad\color{blue}{(**)}$
Since $\;AB+BA\;$ is a real symmetric matrix, by applying the Property 3, it follows that
$Y=Q\cdot\left(AB+BA\right)^{\frac{1}{2}}$
is a solution of the matrix equation $(**)$ whatever orthogonal matrix $\;Q\;$ is.
$Y=Q\cdot\left(AB+BA\right)^{\frac{1}{2}}\;,\;$
$E\cdot P^T\cdot X^T=Q\cdot\left(AB+BA\right)^{\frac{1}{2}}\;,\;$
$X^T=P\cdot E^{-1}\cdot Q\cdot\left(AB+BA\right)^{\frac{1}{2}}\;,\;$
$X=\left(AB+BA\right)^{\frac{1}{2}}\cdot Q^T\cdot E^{-1}\cdot P^T\;.$
Therefore,
$X=\left(AB+BA\right)^{\frac{1}{2}}\cdot Q^T\cdot E^{-1}\cdot P^T\;,$
where $\;E,\;P\;$ are matrices such that $\;B=P\cdot E\cdot E\cdot P^T\;,$
is a solution of the matrix equation $(*)$ whatever orthogonal matrix $\;Q\;$ is.
As it stands, the equation is solvable if and only if $A$ is a positive scalar matrix.
Since $AB+BA=XBX^T$ for all positive definite matrices $B$, if we pass $B$ to a limit, the equation is still satisfied when $B$ is positive semidefinite. In particular, $Auu^T+uu^TA=(Xu)(Xu)^T$ for every nonzero vector $u$. Since the rank of the RHS is at most one, $Au$ must be a scalar multiple of $u$. As $u$ is arbitrary, this means every nonzero vector is an eigenvector of $A$ and hence $A$ is a scalar matrix. Thus $A=cI$ for some $c>0$ because $A$ is positive definite.
Conversely, if $A=cI$ for some $c>0$, we have $2cuu^T=(Xu)(Xu)^T$ for every nonzero vector $u$. Hence $Xu$ is identically equal to $\pm\sqrt{2c}u$, meaning that the only solutions are $X=\pm\sqrt{2c}I$.