# Show that a matrix A may have all leading principal minors greater or equal to zero, yet not be positive semi-definite.

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.

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

## 1 Answer

HINT:

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

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