Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given a matrix $A\in \mathbb{R}^{n\times n}$, I am looking for a symmetric matrix $S\in\mathbb{R}^{n\times n}$ such that

$$ S A + A^T S = I $$

$A$ can be assumed to be regular (with positive determinant, if this is of any help).

The difficulty is of course that $S$ must be symmetric, otherwise one could simply take $2S = A^{-T}$. In principle this is a linear equation with $\frac{n(n+1)}{2}$ unknowns and this can be solved for $S$.

Is there a nicer way to find $S$ such as a closed solution formula using some factorization? Has this problem been studied anywhere?

share|improve this question
    
Do you already know (and is it true at all) that the system of linear equations is uniquely solvable? –  darij grinberg Jan 6 '12 at 14:27
    
Yes, this follows from geometric considerations (which are a bit too lengthy for me to reproduce here). –  Philipp Jan 6 '12 at 14:39
2  
Why is there a $S^T$ in the equation if $S$ is symmetric? –  Federico Poloni Jan 6 '12 at 14:59
    
fair enough; I have deleted the transpose. –  Philipp Jan 6 '12 at 15:34
add comment

1 Answer 1

up vote 10 down vote accepted

These matrix equations are called Lyapunov equations and are extensively studied in control theory.

For instance, if $A$ is Hurwitz (all eigenvalues in the left half-plane), then the unique symmetric solution of $A^TX+XA+Q$ is $$ X=\int_0^\infty e^{A^T t } Q e^{At} dt. $$

share|improve this answer
    
thank you!! this is exactly what I was looking for! –  Philipp Jan 6 '12 at 15:19
    
could you please point me to a reference for the above result? thanks! –  Philipp Jan 6 '12 at 15:37
    
Check page 230 of math.rutgers.edu/~sontag/mct.html (there is a full version of the book available online on the page). The proof is not complicated. –  Federico Poloni Jan 6 '12 at 17:32
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.