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Jul 20, 2020 at 17:36 history edited Rodrigo de Azevedo CC BY-SA 4.0
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Nov 9, 2019 at 8:02 answer added Federico Poloni timeline score: 3
Nov 7, 2019 at 10:09 comment added Markus Sprecher Let $f(X)=XAX+X$ we have $f(X+dX)=f(X)+dX\cdot A\cdot X+X\cdot A\cdot dX+dX+O(|dX|^2)$. We could define an iteration by $X_{k+1}=X_k+dX$ where $dX$ is the solution to $f(X_k)+dX\cdot A\cdot X_k+X_k\cdot A\cdot dX+dX=B$. This is a continuous Lyapunov equation.
Nov 7, 2019 at 10:04 comment added Markus Sprecher another iteration would be $X_{k+1}=B-X_k A X_k$. This would preserve the symmetry of the matrix and, if you start close enough to a solution, also the positive definiteness. If it converges depends however on $A$ and $B$.
Nov 7, 2019 at 8:14 history edited lisi CC BY-SA 4.0
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Nov 7, 2019 at 8:09 history edited lisi CC BY-SA 4.0
edited body
Nov 7, 2019 at 7:55 history asked lisi CC BY-SA 4.0