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.



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 $$ (-1)^m. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.