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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?

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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
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4 Answers

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 $P>0$.

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No similar algebraic stability test for matices, it is well-known.

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Would you mind editing your answer and providing references? This might be well-known for you, but not for us, and some insights would be great. –  András Bátkai Aug 28 '11 at 20:14
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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.

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