Title says it all, but I'll rephrase it to be clear.

A is an $n\times n$ matrix whose leading principal minors are all greater than or equal to zero.

A leading principal minor is the determinant of the k-th submatrix of A, consisting of the first k rows and k columns of A.

Show that matrix A can satisfy this conditions yet also not be positive semidefinite.

Thank you.

  • $\begingroup$ Does it includes $det(A)\geq 0$? $\endgroup$ – Snufsan Sep 19 '14 at 0:23
  • $\begingroup$ yes it does include that $\endgroup$ – alan213123 Sep 19 '14 at 0:29
  • $\begingroup$ Then I think its impossible, check en.wikipedia.org/wiki/Sylvester%27s_criterion $\endgroup$ – Snufsan Sep 19 '14 at 0:31
  • $\begingroup$ I will have a look, thank you. $\endgroup$ – alan213123 Sep 19 '14 at 0:33


\begin{equation*} \left( \begin{array}{cc} 0 & 0 \\ 0 & -1 \\ \end{array} \right) \end{equation*}

  • 1
    $\begingroup$ Thank you. A hint is actually a lot more useful to me. $\endgroup$ – alan213123 Sep 19 '14 at 4:03

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