Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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
add comment

3 Answers

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.

share|improve this answer
    
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
1  
@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
show 2 more comments

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 $X^{-1}$.

share|improve this answer
1  
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
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.