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I wonder if it's possible to find positive-definite matrix $X$ such that $$XB + CX^{-1} = aI$$

$a$ is known non-negative scalar, matrices $X$, $B$ and $C$ are symmetric and have the same size

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Is the matrix $B$ non-negative? If it is the case then there is such an algorithm, and even solutions to some extent. You can make a search with the keywords "algebraic Riccati equation". – Sebastien B Jun 17 '13 at 9:48
@SebastienB, unfortunately no – sbos Jun 17 '13 at 9:55

If at least one of $B$ or $C$ is positive definite, the problem is easy. Suppose $B$ is positive definite. Let $Y=B^{1/2}XB^{1/2}$ and $M=B^{1/2}CB^{1/2}$. Then the equation $XB + CX^{-1} = aI$ is equivalent to $Y^2 - aY + M = 0$. If we orthogonally diagonalise $M$, we see that last equation has a positive definite solution $Y$ if and only if $x^2-ax+\lambda_\max(M)=0$ has a positive root. A similar argument applies if $C$ is positive definite.

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Wow, very neat solution. Could you please explain how to get the expression for $Y$ for the last equation? – sbos Jun 17 '13 at 12:29
@sbos If $Y^2-aY+M=0$ has a solution, then $M=aY-Y^2$ is a polynomial in $Y$ and hence it commutes with $Y$. Yet, $M$ and $Y$ are also real symmetric. Therefore they are simultaneously orthogonally diagonalisable. So, if we orthogonally diagonalise $M$ as $QDQ^T$, i.e. $D=Q^TMQ$, we may assume that $Z=Q^TYQ$ is a diagonal matrix too, and the equation further reduces to $Z^2-aZ+D=0$, which is essentially the $1\times1$ case because both $Z$ and $D$ are diagonal. – user1551 Jun 17 '13 at 13:14
I've just checked: 1) I can formulate my equation as $BX + CX^{-1} = I$ as well. 2) Matrix B is difference of two positive semidefinite matrices and matrix C is positive semidefinite too. Unfortunately I have no guarantee that either B or C is positive definite – sbos Jun 18 '13 at 15:12
how did you use the fact that $B$ is strictly positive definite? Isn't positive semi-definiteness sufficient here? BTW how to show that $(B^{1/2}XB^{1/2})^2 = BX^2B$? – sbos Jun 18 '13 at 17:24
@sbos 1) That $XB+CX^{-1}=I$ implies that $BX+\color{red}{X^{-1}C}=X^{-1}(XB+CX^{-1})X=I$. Now you said that it can be reformulated as $BX+\color{green}{CX^{-1}}=I$. Are they the same $X$? If so, do you mean $C$ and $X^{-1}$ commute? 2) That $B$ can be written as such a difference should not be surprising, as every real symmetric matrix can be written as the difference of two positive semidefinite matrices. 3) If $B$ is singular, we may not be able to infer from $Y^2-aY+M=0$ that $XBX-aX+C=0$. In other words, you get a solution $Y$, but you cannot recover $X$ from it. – user1551 Jun 18 '13 at 17:58

There's no solution in the general case, of course, since you've got about twice as many constraints as free variables.

For a numerical algorithm, I would try gradient descent on $||XB+CY-\alpha I||_F + ||XY-I||_F$, with $Y$ approximating $X^{-1}$.

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It's really the same number of constraints as variables: after multiplying on the right by $X$, you have $XBX + C = aX$, and notice that this is symmetric if $X,B,C$ are symmetric. – Robert Israel Jun 17 '13 at 10:00

Mutliplying on the right by $X$, you have $XBX + C = aX$.
Consider e.g. the case $B = I$, $C = c I$, so the equation says $X^2 - a X + c I = 0$. If $a^2 < 4 c$, there is no hermitian solution.

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Thank you for your answer, Robert. In my case B and C are not even diagonal matrices, probably I should mention this in my question – sbos Jun 17 '13 at 10:09
That's not the point. As I read the question, you wanted to know if there is always a solution. My example shows that sometimes there isn't. – Robert Israel Jun 17 '13 at 13:22

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