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My question is the following

Is there a characterisation of $n\times n$ matrices with real entries whose real eigenvalues are positive?

I am interested in this question because I am analysing some matrices where I know the determinant up to a sign, and I want to show that the sign must be positive. These matrices often have complex eigenvalues, but because the entries are real these eigenvalues will appear in conjugate pairs. So when computing the determinant as the product of the eigenvalues, these complex conjugate pairs will multiply together to make positive quantities (zero is never an eigenvalue of the matrices I am studying). So I want to show that the remaining real eigenvalues must be positive. I understand that there is a possibility that some of the real eigenvalues could be negative and the determinant would still be positive, but I suspect that they are all going to be positive. In order to see if this is true, I am looking for a characterisation of such matrices (as in my question).

Any ideas or references to known results about this question are much appreciated.

Edit: I found a sufficient condition (although, I'm fairly certain it is not necessary):

If for each $0\leq k\leq n$ the sum of the determinants of the principal $k\times k$ minors is positive, then the matrix has no negative eigenvalues.

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  • $\begingroup$ Positive definite. $\endgroup$ – Doug M Oct 23 '18 at 19:07
  • $\begingroup$ @DougM I'm not always considering symmetric matrices, and I am not looking for the real parts of all eigenvalues to be positive (just the strictly real eigenvalues to be positive). $\endgroup$ – Dave Oct 23 '18 at 19:10
  • $\begingroup$ An observation: your condition at the end could also be stated as "the coefficients of the characteristic polynomial are all positive". But given that you are trying to determine one of those coefficients, I'm not sure it helps. $\endgroup$ – Joppy Oct 24 '18 at 1:41
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    $\begingroup$ @Joppy not quite: the coefficients would be alternating signs. But yes, that sufficient condition I added could be phrased in terms of the characteristic polynomial, and that was actually the motivation I had for it (application of Descartes' rule of signs). $\endgroup$ – Dave Oct 24 '18 at 1:44

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