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### Recurrence relation for matrices

Another different way to obtain an equivalent formula: let $Y = A^{1/2}XA^{1/2}$. Then, multiplying your equation from both sides by $A^{1/2}$, we have $Y^2+Y = A^{1/2}BA^{1/2} = C$. The matrix $C$ ...
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### Libraries for solving Lyapunov's equation

There's a remarkably simple way to express solution in terms of eigenvalues/eigenvectors. If found this in Matlab implementation, even though none of the standard textbooks talk about this. Here's a ...
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