Since the search of the eigenvalues is in general not "simple", equally valid, is the method of reducing with moves of Gauss, that preserve the determinant, (add to multiple rows of other rows, move rows in an even number of times, etc ..) to be traced back to a upper triangular form with zeros below the diagonal. The product of the elements of the diagonal is the determinant, then, if all the elements are greater (resp. lower) than zero then the associated quadratic form is positive definite (resp. negative); if they are greater than or equal (resp. less than or equal) to zero is positive semidefinite (resp. negative semidefinite); undefined if some elements along the diagonal are positive and others negative.

Obviously this is true if the starting matrix is symmetric. What happens if the matrix is not symmetric? Is there a way to study the sign of real part of its eigenvalues not calculate them manually?

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    $\begingroup$ For a non-symmetric matrix, the eigenvalues might not even be real, in which case it isn't clear what you mean by "sign". Is there some property the matrix has which ensures realness? Then maybe you have something to work with. $\endgroup$ – Erick Wong Jul 13 '15 at 15:15
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    $\begingroup$ Excuse me, @ErickWong. I mean the sign of the real part of the eigenvalues. I corrected my post now. $\endgroup$ – Lely Jul 13 '15 at 15:18
  • $\begingroup$ If the eigenvalues are real, you can use Descartes' rule of signs in the characteristic polynomial. $\endgroup$ – Daniel Jul 13 '15 at 17:31
  • $\begingroup$ See: Routh–Hurwitz theorem, Routh–Hurwitz stability criterion $\endgroup$ – user147263 Jul 13 '15 at 20:06

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