Rank of a hollow matrix

Question

$A$ is a $d \times d$ square and symmetric matrix.

$A(i,i)=0$ for all $i$,

$A(i,j)>0$ and distinct for $i \neq j$.

Is $A$ full-rank, or invertible? This is true for $d=2,3,4$.

Answer 1

If the matrix is symmetric, $a_{i,j} = a_{j,i}$, so they can't all be distinct for $i \ne j$ (if $n \ge 2$). I assume you mean distinct for $i > j$.

In the $4 \times 4$ case, $\det(A)=0$ if $$ a_{1,2} = {\frac {a_{{1,3}}a_{{2,4}}+a_{{1,4}}a_{{2,3}}+2\,\sqrt {a_{{1,3}}a_{{1 ,4}}a_{{2,3}}a_{{2,4}}}}{a_{{3,4}}}} $$ For example, try $$ \pmatrix{0 & t & 1 & 2\cr t & 0 & 3 & 4\cr 1 & 3 & 0 & 5\cr 2 & 4 & 5 & 0\cr}$$ where $t = 2 + 4 \sqrt{6}/5$.

Answer 2

A counterexample is given as follows:

\begin{pmatrix} 0 & 2 & 3 & 15 \\ 2 & 0 & 5 & 9 \\ 3 & 5 & 0 & 6 \\ 15 & 9 & 6 & 0 \end{pmatrix}

This matrix has determinant zero.

Reference: Determinant of symmetric matrix with the main diagonal elements zero