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The Jacobian matrix of a function $f: \mathbb{R}^n \to \mathbb R^n$, f a $C^1$ function is a matrix of the form $J_f(x) = \left( \frac{\partial f_i}{\partial x_j}(x)\right)_{\substack{1 \leq i \leq m \\ 1 \leq j \leq n}}$. We get the eigenvalues if we solve $\det(J_f - E_n \lambda) = 0$ where $E_n$ is the $n \times n$ identity matrix. Is there any known result or method that gives me conditions when all these eigenvalues have non zero real part?

Thanks!

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    $\begingroup$ I don't think so. All $n\times n$ matrices are Jacobian matrices of some function. Example: given $A$ consider $f(x)=Ax$. The Jacobian of $f$ at $0$ (or at any other point) is exactly $A$. Being a Jacobian is no special property for a matrix and so I do not expect that this information tells you something about the eigenvalues. $\endgroup$ – Giuseppe Negro Jan 17 '17 at 14:29
  • $\begingroup$ Can I generalize my question then. Are there conditions on a matrix $A$ such that all its eigenvalues have non zero real part? $\endgroup$ – JDoe Jan 17 '17 at 14:41
  • $\begingroup$ @GiuseppeNegro I had an idea maybe you can tell if I am correct: Since the Jacobian is symmetric I can transform it into a diagonal matrix (I am not remembering it wrong) and then the eigenvalues can be calculated as $\sum(a_{ii}- \lambda =0)$ ($a_{ii}$ are elements in the diagonal matrix)and these are linear factors. Thus the eigenvalues have nonzero real part if all $a_{ii}$ have nonzero real part. Correct? Its just a quick idea I am not sure if I remembered everything correctly. $\endgroup$ – JDoe Jan 17 '17 at 14:52
  • $\begingroup$ Why should the Jacobian be symmetric? The matrix you have in mind is the Hessian. The Jacobian has no significant properties, AFAIK. $\endgroup$ – Giuseppe Negro Jan 17 '17 at 14:58
  • $\begingroup$ Yeah I realized this too. Thanks! $\endgroup$ – JDoe Jan 17 '17 at 15:36

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