The RouthHurwitz criterion provides a convenient test, even for hand calculation, of whether a polynomial with real coefficients has all its roots in the left half plane. I'm wondering about a similar test for whether the eigenvalues of a real matrix lie in the LHP. In principle, one could find the characteristic polynomial of the matrix and apply the RouthHurwitz test to this, but finding the characteristic polynomial of a matrix, even one of modest size, is not a convenient hand calculation. Does anyone know of a more conveniently applied test for all the eigenvalues to lie in the LHP?

$\begingroup$ What does Gantmacher says about that ? His second volume is dedicated to such questions. $\endgroup$ – Denis Serre Mar 17 '11 at 11:44

$\begingroup$ I have often wondered about this question as well. @Denis Serre  I don't believe Gantmacher addresses the matrix version of this question. $\endgroup$ – alex Aug 28 '11 at 17:37
The most convenient algebraic test for stability of a matrix $A$ known to humankind is to pick an arbitrary $Q>0$ and solve the matrix Lyapunov equation $A^T P + P A + Q = 0$. All eigenvalues of $A$ have negative real part if and only if $P>0$.
No similar algebraic stability test for matices, it is wellknown.

5$\begingroup$ Would you mind editing your answer and providing references? This might be wellknown for you, but not for us, and some insights would be great. $\endgroup$ – András Bátkai Aug 28 '11 at 20:14
From Gershgorin's circle theorem (see Wikipedia) I believe it follows that a matrix $A = [A_{ij}]$ with real entries has eigenvalues with negative real part if \begin{equation} A_{ii} + \sum_{j\neq i} A_{ij} < 0, \quad \forall i. \end{equation} This is of course just a sufficient criterion.
I think this fits the bill:
http://books.google.ie/books?id=LeuNXB2bl5EC&q=routhhurwitz#v=onepage&q=routhhurwitz&f=true