# A Linear Algebra Problem

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?

-
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
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

## 1 Answer

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 $$X=\int_0^\infty e^{A^T t } Q e^{At} dt.$$

-
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