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This question is a little ill-posed, but I've been playing with some equations and am just wondering if this resembles any known problems that have been solved.

Suppose I have two real, positive definite (square) matrices $\mathbf{A}$ and $\mathbf{C}$, and I wish to find another real, positive definite matrix $\mathbf{B}$ such that $\mathbf{A B} + \mathbf{B}^{-1}\mathbf{C}$ is as close as possible to identity.

I'll entertain any reasonable definition of "close" that makes the problem tractable. Maybe minimization of $\| \mathbf{A B} + \mathbf{B}^{-1}\mathbf{C} - \mathbb{I} \|$ for some choice of norm.

Does anyone have any insight or experience with such a problem?

Thanks,

Nick

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  • $\begingroup$ It reminds me of Sylvester's equation. There however we have no inverse. $\endgroup$
    – user35593
    Commented Mar 8, 2016 at 20:48
  • $\begingroup$ I wonder if the minimization problem could be cast in terms of a Frobenius norm. $\endgroup$
    – Nick
    Commented Mar 22, 2016 at 3:28

1 Answer 1

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You are actually looking to solve the continuous algebraic Riccati equation. For convenience, I will write your $B$ as $X=X^T$. Then the equation you're trying to solve is simply

$$ X - XAX + (-C) = 0$$

Or even more explicitly, writing the Cholesky factorization of $A=BB^T$

$$ \left( \frac{1}{2}I \right)^TX + X\left( \frac{1}{2}I \right) - XBB^TX + (-C) = 0$$

The solution of an algebraic Riccati equation using Hamiltonian matrices is a standard topic in control theory. But I would instead refer you to the "care" command in MATLAB.

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    $\begingroup$ But in many cases this algebraic Riccati equation has no real solutions. $\endgroup$
    – Robert Israel
    Commented Mar 11, 2016 at 18:46
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    $\begingroup$ Excellent point. In order to get a real symmetric posdef $X$ "close" to solving the ARE, we would have to reformulate the problem into a semidefinite program. $\endgroup$
    – Richard Zhang
    Commented Mar 11, 2016 at 19:20

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