How to solve $AX-XA=C\circ X$ in closed form

Question

Given a hermitian matrix $A$ and a anti-symmetric matrix $C$, consider the matrix equation: \begin{eqnarray} \left[ A. - C\circ\right]X = X.A \end{eqnarray} where $A.B$ represents standard matrix multiplication, but $A\circ B$ is the (elementwise) Hadamard product between two matrices. Further, with $C$ being a real, anti-symmetric matrix: \begin{eqnarray} C_{ij}=-C_{ji}. \end{eqnarray} Is there a way to solve this equation for $X$ in an exact way? By using the eigenbasis of $A$ for instance, I don't arrive to any closed expression for the elements of $X$. It looks related to the Sylvester or Lyapunov equations, but I don't really know how to proceed further.

Answer 1

$ \def\l{\lambda}\def\o{{\tt1}}\def\p{\partial} \def\L{\left}\def\R{\right}\def\LR#1{\L(#1\R)} \def\vecc#1{\operatorname{vec}\LR{#1}} \def\Diag#1{\operatorname{Diag}\LR{#1}} \def\qiq{\quad\implies\quad} \def\c#1{\color{red}{#1}} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} $You can use Kronecker products to create a vector equation which will yield a closed-form solution if a solution exists. $$\eqalign{ &AX - XA = C\circ X \\ &\LR{\LR{I\otimes A} - \LR{A^T\otimes I}}\vecc{X} = \Diag{\vecc{C}}\vecc{X} \\ }$$ Defining some auxiliary variables allows the equation to be written more compactly $$\eqalign{ x &= \vecc{X},\quad B &= \Diag{\vecc{C}},\quad M &= \LR{I\otimes A} - \LR{A^T\otimes I} \\ Mx &= Bx\\ }$$ This is a Generalized Eigenvalue Problem. $\;$This can often be reduced to an ordinary Eigenvalue Problem, if either $M$ or $B$ is invertible.

However, since $C$ is skew symmetric it has $0$s along its diagonal, so $B$ is not invertible. Similarly, $M$ is rank deficient due to the way it is constructed, so it is not invertible either.

In any case, $x$ is the eigenvector associated to the unit eigenvalue $(\l=\o),\,$ if such an eigenvalue exists. The eigenvector can then be reconstituted into the corresponding matrix solution.

Note that if $x$ is a solution, then so is any scalar multiple $\beta x.\;$ This is also true of your matrix equation, i.e. if $X$ satisfies the equation then so does $\beta X$.