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Unfortunately I haven't received any response for my previous question, so I'm trying to solve it in a different way.

  1. I know that iff matrix $H$ is negative definite, its leading principal minors alternate in sign (starting with < 0 for the $H_{(1,1)}$ element).

  2. I know that for a negative semi-definite matrix, a similar statement (without strict inequalities) can be made about all of the principal minors -- not just the leading ones.

But does $H$ being negative definite tell us anything about all of its principal minors?

e.g. Does it imply that they alternate in sign, like for negative semi-definite matrices, but with strict inequalities?

Matrices aren't in my area of expertise, so apologies if this is blindingly obvious, but I haven't found it stated anywhere, as #1 and #2 above are.

Thanks

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If $H$ is Hermitian and negative definite, then all principal minors (the submatrices) are negative definite.

Let $\tilde H$ be such a negative definite minor with size $m\times m$, then its determinant (as product of its $m$ negative eigenvalues) has sign $$ (-1)^m. $$

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