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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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Recurrence relation for matrices

This is an instance of the Riccati equation, which can be solved using the Matrix Sign function. Relevant section from Higham's "Functions of Matrices" book:
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