# Solution of matrix equation or matrix inequality

When there exist a positive definite solution $S$ of the following matrix equation or matrix inequality: $SA^{T}+AS+\alpha S-\beta BB^{T}=0$, that is, what condition on $A,B, \alpha, \beta$ can guarantee the existence of a positive definite matrix solution $S$. The matrix $A$ is a n-by-n square matrix, the matrix $B$ is a n-by-m matrix with $m<n$, and it is full column rank, $\alpha, \beta$ are positive scalars.

This matrix equation form is similar to the continuous time algebraic Riccati equation. It can be some special form of Sylvester equations.

Any help or references are appreciated.

I'm assuming the matrices are real. We may as well absorb the $\alpha$ into $A$ and take $\beta = 1$. So we want a positive definite $S$ with $S A^T + A S = P$, where $P = B^T B$ is positive definite. Multiplying on left and right by $P^{-1/2}$ (the positive definite square root of $P^{-1}$), write this as $T C^T + C T = I$, where $T = P^{-1/2} S P^{-1/2}$ (which is positive definite iff $S$ is) and $C = P^{-1/2} A P^{1/2}$. Now this implies for every real vector $v$, $v^T C T v = v^T T C^T v = v^T v/2$. Of course if $C^T v = 0$ the left side is $0$, so this can't happen if $C$ is singular. If $C$ has a real eigenvalue $\lambda$, there is a nonzero vector $v$ such that $C^T v = \lambda v$, and then $\lambda v^T T v = v^T v/2$, so we get a contradiction if $\lambda < 0$. I suspect these are the only restrictions, i.e. if $C$ (and thus $A$) has no nonpositive real eigenvalues, there is a solution.
EDIT: My suspicion was false. Consider the case $$C = \pmatrix{1 & -3\cr 2 & -2\cr}$$ which has two complex eigenvalues. The unique symmetric solution to $T C^T + C T= I$ is $$T = \left( \begin {array}{cc} -{\frac {17}{8}}&-{\frac {7}{8}}\\ -{\frac {7}{8}}&-{\frac {9}{8}}\end {array} \right)$$ which of course is not positive definite.
• We can write this equation as $SA^{T}+AS+S=P$. For your answers, I think the $S$ is missing. – Xiangyu Meng Oct 6 '12 at 22:07
• As I said, I absorbed the $\alpha$ into $A$. $SA^T + AS + S = S (A+I/2)^T + (A+I/2) S$. – Robert Israel Oct 7 '12 at 4:20