I need to show that a complicated $n \times n$ Hermitian matrix is positive semidefinite. I'm wondering if there are simple sufficient conditions that can be used to show this. For instance, if a matrix with positive diagonal is diagonally dominant, then it is positive semidefinite. Other conditions like this would be helpful.

  • $\begingroup$ It is pos. semidefinite iff all its eigenvalues are (real) non-negative, iff all its principal minors are non-negative. $\endgroup$ – DonAntonio May 26 '17 at 21:28
  • $\begingroup$ Right, I guess I should have been more precise with my definition of "simple". Computing the eigenvalues or the determinants of large matrices is not very simple. $\endgroup$ – user_lambda May 26 '17 at 23:33
  • 2
    $\begingroup$ Anything else would imply, imo, either much more information regarding a particular matrix, or else magic... $\endgroup$ – DonAntonio May 26 '17 at 23:39

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