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I have the following quadratic matrix equation:

$$ XAX+X = B $$

where both $A$ and $B$ are given positive definite matrices, and $X$ is a covariance matrix and, hence, positive definite.

When there is no constraint, the equation can be solved via Bernoulli iteration in the following form:

$$X_{k+1} = -A^{-1}(I-BX_k^{-1})$$

However, this does not seems to preserve positive semidefinite.

Any guidance would be appreciated. Thank you.

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  • $\begingroup$ 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 $B$. $\endgroup$
    – Markus Sprecher
    Commented Nov 7, 2019 at 10:04
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    $\begingroup$ 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. $\endgroup$
    – Markus Sprecher
    Commented Nov 7, 2019 at 10:09

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Turning my previous comments into an answer.

What you have is an algebraic Riccati equation: indeed, setting $F=-\frac12 I$, you get $F^TX+XF + B = XAX$. Since $A$ and $B$ are positive definite, the pairs $(F,A)$ and $(F^T,B)$ are controllable and observable, so the classical solvability criteria are satisfied and the equation has a unique stabilizing, positive definite solution. You will find solution algorithms implemented in most numerical software; among them, Matlab's care and icare, or scipy.linalg.solve_continuous_are.

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  • $\begingroup$ Thank you Poloni! I have just tried it and it works very well! $\endgroup$
    – lisi
    Commented Nov 10, 2019 at 1:09

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