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? 



Prescribing areas of parallelograms (or 2x2 principal minors)
In the case where the matrix $(a^2\_{ij})\\_{i,j=1,\ldots,n}$ is nonsingular, then the problem reduces to the condition that it has a single positive eigenvalue. In fact, we have the following for any $n\times n$ nonzero symmetric matrix $A$ with nonnegative components and zero diagonal (I'll remove the square from $a\_{ij}$, as it doesn't seem to help).



> 
> If there exist $\nu\_i\in \mathbb{R}^n$ such that
>  $$
> \begin{align}
> A\_{ij}=\lVert\nu\_i\rVert^2\lVert\nu\_j\rVert^2-\langle\nu\_i,\nu\_j\rangle^2&&{\rm(1)}
> \end{align}
> $$
>  then $A$ has a single positive eigenvalue (counting multiplicities).
> 
> 
> Conversely, if $A$ is nonsingular and has a single positive eigenvalue, then there exists $\nu\_i\in\mathbb{R}^n$ satisfying (1).
> 
> 
> 


First, suppose that (1) holds. Then, there exist nonnegative reals $\lambda\_i$ and a positive semidefinite matrix $S$ such that $A\_{ij}=\lambda\_i\lambda\_j(1-S\_{ij}^2)$, simply by taking $\lambda\_i=\lVert\nu\_i\rVert^2$ and $S\_{ij}=\langle\hat\nu\_i,\hat\nu\_j\rangle$, where $\hat\nu\_i=1\_{\lbrace\nu\_i\not=0\rbrace}\nu\_i/\lVert\nu\_i\rVert$. Let $\lambda=(\lambda\_i)\\_{i=1,\ldots,n}\in\mathbb{R}^n$. Using the fact that the componentwise square of a positive semidefinite matrix is itself positive semidefinite,
$$
x^{\rm T} A x= \langle x,\lambda\rangle^2-\sum\_{ij}(\lambda\_ix\_i)S\_{ij}^2(\lambda\_jx\_j)\le\langle x,\lambda\rangle^2.
$$
In particular, $x^{\rm T}A x\le0$ for all vectors orthogonal to $\lambda$. So, the space generated by eigenvectors with positive eigenvalues cannot contain any nonzero members orthogonal to $\lambda$, and has dimension at most one. But, as the trace of $A$ is zero, it must have at least one positive eigenvalue. 


Conversely, suppose that $A$ has a single positive eigenvalue and is nonsingular. Diagonalization gives
$$
A=u u^{\rm T}-\sum\_{\alpha=1}^{n-1} v\_{\alpha}v\_{\alpha}^{\rm T}
$$
for nonzero orthogonal $u,\nu\_\alpha\in\mathbb{R}^n$. As the diagonal of $A$ is zero,
$$
u\_i^2=\sum\_\alpha\nu\_{\alpha,i}^2.
$$
Using Cauchy–Schwarz,
$$
A\_{ij}\le u\_iu\_j+\sqrt{\sum\_\alpha v\_{\alpha,i}^2\sum\_\beta v\_{\beta,j}^2}
=u\_iu\_j+\vert u\_iu\_j\vert.
$$
So, the $u\_i$ are all nonzero, otherwise $A$ would have a row with no positive elements. Then, $u\_iu\_j > 0$.Writing $S=\lbrace i=1,2,\ldots,n\colon u\_i > 0\rbrace$ we would have $A\_{ij}=0$ for $i\in S$ and $j\not\in S$. Breaking $A$ down into two blocks on which the row and column indices are respectively in $S$ and not in $S$, we can break the problem down to the case where $u\_i$ are all of the same sign. W.l.o.g., take $u\_i > 0$. For $1 > \epsilon > 0$, define the matrix
$$
\begin{align}
S\_{ij}&=\sqrt{1-\epsilon u\_i^{-1}u\_j^{-1} A\_{ij}}\cr
&=1-\frac12\epsilon u\_i^{-1}u\_j^{-1} A\_{ij}+O(\epsilon^2)\cr
&=1-\epsilon/2+\frac\epsilon2\sum\_{\alpha}u\_i^{-1}u\_j^{-1}v\_{\alpha,i}v\_{\alpha,j}+O(\epsilon^2)
\end{align}
$$
As the vectors $u,v\_{\alpha}$ are linearly independent, the same is true of the vectors $\tilde u=(1,1,\ldots,1)$ and $\tilde v\_\alpha=(u\_1^{-1}v\_{\alpha,1},\ldots,u\_n^{-1}v\_{\alpha,n})$. Then,
$$
x^{\rm T}Sx=(1-\epsilon)\langle\tilde u,x\rangle^2+\frac\epsilon2\left(\langle\tilde u,x\rangle^2+\sum\_\alpha\langle\tilde v\_\alpha,x\rangle^2\right)+O(\epsilon^2\lVert x\rVert^2)
$$
is nonnegative for all $\epsilon$ small enough and $x\in\mathbb{R}^n$. In this case $S$ is positive definite and (by Gram-Schmidt, for example) there are $\hat\nu\_i\in\mathbb{R}^n$ with $S\_{ij}=\langle\hat\nu\_i,\hat\nu\_j\rangle$. Setting $\nu\_i=\epsilon^{-1/4}u\_i^{1/2}\hat\nu\_i$ gives (1).




---


That concludes the case where $A$ is nonsingular. The singular case is, I think, considerably more complicated.


