Let $(a_{ij})$ be a $n\times n$ symmetric matrix such that $a_{ij}\geq 0$ for all $i,j$ and $a_{ii}=0$ for all $i$. Under which conditions on the $a_{ij}$'s can one find $n$ vectors $v_1,\ldots,v_n\in{\mathbb R}^n$ such that for all $i,j$ the area of the parallelogram spanned by $v_i$ and $v_j$ equals $a_{ij}$:

$\forall i,j:\quad\|v_i\|^2\|v_j\|^2-\langle v_i,v_j\rangle^2=a_{ij}^2$ ?

Here is the only and obvious necessary condition I know about: if $a_{ij}=0$ for some $i\neq j$, then  $a_{ik}a_{jl}=a_{il}a_{jk}$ for all $k,l$. 

What if $a_{ij}>0$ for all $i\neq j$ ?

Thank you. 

**Edit.** As Noah Stein suggested, a useful reformulation of the question is: can one prescribe the $2\times2$ principal minors of a symmetric positive semidefinite matrix?     

**Edit 2.** See also George Lowther comment. It is always possible (and easy!) to prescribe the $2\times 2$ principal minors of a symmetric $n\times n$ matrix. If the $a_{ij}$'s, $1 \leq i < j \leq n $,
 are to be those minors, we simply need to choose $n$ numbers $g_{ii}$ such that $g_{ii}g_{jj}\geq a_{ij}$ for all $i\neq j$. Then we are done with the symmetric matrix $G=(g_{ij})$ whose off diagonal entries are given by $g_{ij}=\epsilon_{ij}\sqrt{g_{ii}g_{jj}-a_{ij}}$, where $\epsilon_{ij}=\pm 1$.

So the initial question becomes : under what conditions on the $a_{ij}$'s, can one find $n$ real numbers $g_{ii}\geq 0$, and $\epsilon_{ij}=\pm 1$, so that the matrix $G$ defined above is positive semidefinite?