Routh-Hurwitz for eigenvalues

The Routh-Hurwitz 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 Routh-Hurwitz 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?

• What does Gantmacher says about that ? His second volume is dedicated to such questions. – Denis Serre Mar 17 '11 at 11:44
• I have often wondered about this question as well. @Denis Serre - I don't believe Gantmacher addresses the matrix version of this question. – 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 $P>0$.
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.