I want to find one solution to the Lyapunov equation, and I use LyapunovSolve[a,c]. But when the solution is not unique, Mathematica did not return any value instead, it shows
The matrix equation has no unique solution.
Is there some method that I can get one solution even if the solution is not unique?
Edit I try to solve equation $$\left( \begin{matrix} 1& 0\\ 0& -1\\ \end{matrix} \right) =\left( \begin{matrix} 1+z& x-iy\\ x+iy& 1-z\\ \end{matrix} \right) X+X\left( \begin{matrix} 1+z& x-iy\\ x+iy& 1-z\\ \end{matrix} \right)$$
$Assumptions = Element[x, Reals] && Element[y, Reals] && Element[z, Reals]; Rho = { {1 + z, x - I*y}, {x + I*y, 1 - z} } ; c = { {1, 0}, {0, -1} } ; LyapunovSolve[Rho, c]
LyapunovSolve[rho, c]returns a solution for me (V13.1). This has a nicer simplified form:xmat = LyapunovSolve[rho, c] // ComplexExpand-- I changedRhotorhoin accordance with best practices. $\endgroup$$Assumptions. (ComplexExpandwill take care of it after the solution.) $\endgroup$