Does positivity of the n(n-1)/2 principal minors formed from 2 x 2 submatrices ensure positive-definiteness of the n x n matrix itself?

Question

I am interested in conditions under which an $n \times n$ matrix ($\rho$) is positive definite. Of course, one necessary and sufficient set of conditions is that the $n$ leading minors of $\rho$ each be positive.

What if I require that all the $\frac{n (n-1)}{2}$ principal minors constructed by taking the determinants of the $2 \times 2$ submatrices \begin{equation} \left( \begin{array}{cc} \rho _{\text{ii}} & \rho _{\text{ij}} \\ \rho _{\text{ji}} & \rho _{\text{jj}} \\ \end{array} \right) \end{equation} for $i= 1,\ldots,n-1$, $j=i+1,\ldots n$ be themselves positive?

Clearly, for the trivial $n=2$ case, positive-definiteness holds, but is this the case for all/any $n>2$?

Answer 1

The one I put first was $$ \left( \begin{array}{rrr} 3 & -2 & -2 \\ -2 & 3 & -2 \\ -2 & -2 & 3 \\ \end{array} \right) $$

At the cost of possible non-integer value on the main diagonal, we may take real $w> 0$ in

$$ \left( \begin{array}{rrrrr} w & -1 & -1 & -1 & -1 \\ -1 & w & -1 & -1 & -1 \\ -1 & -1 & w & -1& -1 \\ -1 & -1 & -1 & w& -1 \\ -1 & -1 & -1 & -1& w \\ \end{array} \right) $$ where we call the dimension $n.$ The eigenvalues of the matrix of all entries $-1$ are $0,0,0,.., -n \; . \; \; \; $ We have added $(w+1)I,$ so these eigenvalues are $w+1, w+1, ...., w+1, w+1 - n. \; \; $ If $ w > n-1 $ all eigenvalues become positive. So, to get all principal minors (other than the entire matrix) positive we may take $w > n-2; \; \;$ if $w < n-1$ the determinant of the entire matrix is negative. So, take $n-2 < w < n-1.$ If we make the off-diagonal elements $-2$ instead, we may use an integer in $w=2n-3$

There is a pattern we may use for a basis of eigenvectors, the columns of $$ \left( \begin{array}{rrrrr} 1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 \\ 1 & 0 & 0 & 0 & 4 \\ \end{array} \right), $$ are pairwise orthogonal.