Let $v\_1,...,v\_n\in \mathbb{R}^n$ be linearly independent. The parallelepiped defined by these vectors is $P(v\_1,...,v\_n)=\{\sum\_{i=1}^{n}\alpha\_i v\_i|~0\le\alpha\_i\le 1\}$. Observe that while the collection $v\_1,...,v\_n$ is reconstructible from $P(v\_1,...,v\_n)$ as a subset of $\mathbb{R}^n$, it is not reconstructible from $P(v\_1,...,v\_n)$ as a geometric figure. Indeed, if $U$ is an orthogonal operator on $\mathbb{R}^n$, then $P(Uv\_1,...,Uv\_n)=UP(v\_1,...,v\_n)$ is isometric to $P(v\_1,...,v\_n)$. Moreover, $P(v\_1,...,-v\_i,...,v\_n)$ is a translate of $P(v\_1,...,v\_n)$ by $-v\_i$. You can also view this as the shift of origin: instead of looking from the point $0$ we are now looking from the vertex $v\_i$. In fact $P(\pm v\_1,...,\pm v\_n)$ correspond to every of $2^{n}$ vertex of this parallelepiped.


For $A=\{i\_1,...,i\_k\}\subset\{1,...,n\}$ define $V(A)=V\_k(P(v\_{i\_1},...,v\_{i\_k}))$, where $V\_k$ is the $k$-dimensional volume.


Let $w\_1,...,w\_n\in \mathbb{R}^n$ also be linearly independent, and for $A\subset\{1,...,n\}$ define $W(A)$ analogously.


I can show the following 


**Proposition**. If $W(A)=V(A)$ for every $A\subset\{1,...,n\}$, then $P(w\_1,...,w\_n)$ and $P(v\_1,...,v\_n)$ are isometric, i.e. there are $a\_1,...,a\_n=\pm 1$ and an orthogonal operator $U$ on $\mathbb{R}^n,$ such that $w\_i=a\_iUv\_i$, for every $i$.


However, I can only do it using a result about principal minors of a symmetric matrix determining it up to multiplying both $i$-th row and $i$-th column by $\pm 1$ (in our case we consider the Gram matrices, whose principal minors are exactly the squares of the corresponding volumes).



> 
> Is the Proposition known? Is there a geometric proof of it?
> 
> 
> 


I tried to build a proof on the fact that we know the distance from every $v\_i$ to the span of any other combination of $v\_j$ (including the empty one), but geometry kind of gets intertwined with combinatoric of what is orthogonal to what, and I got stuck.


PS In fact the Proposition is equivalent the result that I've mentioned (in the real case), and it is proven e.g. here:


*Rising, Justin; Kulesza, Alex; Taskar, Ben*, [**An efficient algorithm for the symmetric principal minor assignment problem**](http://dx.doi.org/10.1016/j.laa.2014.04.019), Linear Algebra Appl. 473, 126-144 (2015). [ZBL1314.65050](https://zbmath.org/?q=an:1314.65050). 



Parallelepiped is defined by the volumes of its faces
Here is a geometric proof of the result. $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\be}{\boldsymbol{e}}$ As I comment at the end of the proof, this actually proves a stronger fact.


Define two equivalence relations"$\sim\_n$" and "$\approx\_n$" on the set of bases of $\bR^n$.


$$ (v\_1,\dotsc, v\_n)\sim\_n (w\_1,\dotsc, w\_n) $$


$\DeclareMathOperator{\vol}{vol}$ $\newcommand{\eps}{\epsilon}$ if for any subset $I\subset \{1,\dotsc, n\}$ we have
$$
\vol(v\_i, \;\;i\in I)=\vol(w\_i,\;\;i\in I) 
$$
and
$$
(v\_1,\dotsc, v\_n)\approx\_n (w\_1, \dotsc, w\_n) $$
if there exists orthogonal map $T$ and scalars $\eps\_i=\pm 1$ such that


$$
w\_i= \eps\_iTv\_i,\;\;\forall i=1,\dotsc , n.
$$
$\newcommand{\Llra}{\Longleftrightarrow}$ The result states that $\sim\_n \Llra \approx\_n$. Clearly $\approx\_n\implies \sim\_n$.


Clearly $\sim\_1\Llra \approx\_1$. We then argue by induction. Assume $\sim\_n\Llra \approx\_n$. To prove that $\sim\_{n+1}\Llra \approx\_{n+1}$ it suffices to show that if


$$(v\_0, v\_1,\dotsc, v\_n)\sim \_{n+1} (w\_0, v\_1,\dotsc, v\_n), $$


then


$$(v\_0, v\_1,\dotsc, v\_n)\approx\_{n+1} (w\_0, v\_1,\dotsc, v\_n). $$


$\DeclareMathOperator{\span}{span}$ $\DeclareMathOperator{\Proj}{Proj}$ For $I\subset \{1,\dotsc, n\}$ we set


$$ V\_I:=\span\{v\_i;\;\;i\in I\}. $$


and for any vector $u$ we denote by $[u]\_I$ its orthogonal projection on $V\_I$. 


Since $vol(v\_0, v\_i, i\in I)=\vol(w\_0, v\_i, I\in I)$, $\forall I\subset \{1,\dotsc, n\}$ we deduce


$$\big\Vert\; v\_0-[v\_0]\_I\;\big\Vert =\Big\Vert\; w\_0-[w\_0]\_I\;\big\Vert. $$
Pythagoras' Theorem implies 


$$
\big\Vert\;[v\_0]\_I\;\big\Vert =\big\Vert\;[w\_0]\_I\;\big\Vert.
\tag{$\ast$}
$$


Since 
$$
\vol(v\_0,v\_i)=\vol(w\_0,v\_i),\;\;\forall i=1,\dotsc, n,
$$
there exists $\eps=\pm 1$ such that


$$
(v\_0,v\_i)=\eps(w\_0,v\_i),
$$
where $(-,-)$ denotes the inner product. We set 
$$
S\_i:=\big\{\; \eps=\pm 1;\;\; (v\_0,v\_i)=\eps(w\_0,v\_i)\;\big\}. 
$$


**Remark.** Above, the two inner products are simultaneously nonzero or simultaneously zero. When they are zero $S\_i=\{-1,1\}$. Note that $|S\_i|=1$ iff $(v\_0,v\_i)\neq 0$.



> 
> **Lemma 1.** If $i\neq j$ and $(v\_i,v\_j)\neq 0$, then $S\_i\cap S\_j\neq \emptyset$.
> 
> 
> 


**Proof.** For simplicity assume $i=1, j=2$. If $(v\_0,v\_1)=(v\_0,v\_2)=0$ then $(w\_0,v\_1)=(w\_0,v\_2)=0$ and the result is obviously true. Suppose that $(v\_0,v\_1)\neq 0$. Denote by $\be\_1,\be\_2$ the orthonormal basis of $\span(v\_1,v\_2)$ obtained from the basis $v\_1,v\_2$ via Gram-Schmidt. Then


$$
v\_1=x\be\_1,\;\; v\_2=y\be\_1+z\be\_2 
$$
where $x,z>0$ and $y\neq 0$ since $(v\_1,v\_2)\neq 0$. Write


$$a\_1=(v\_0,\be\_1),\;\;a\_2=(v\_0,\be\_2), $$
$$ b\_1=(w\_0,\be\_1),\;\;b\_2=(w\_0,\be\_2). $$


Then $a\_1=\eps\_1 b\_1=(v\_0,v\_1)\neq 0$. Since $a\_1^2+a\_2^2=b\_1^2+b\_2^2$ we deduce $a\_2=\pm b\_2$. We want to show that $a\_2=\eps\_1 b\_2$. Let $a\_2=\eta\_2b\_2$, $\eta\_2=\pm 1$. We want to prove that $\eps\_1\in S\_2$. We argue by contradiction. We have 


$$
(v\_0,v\_2)=\eps\_2(w\_0,v\_2),\;\;\eps\_2\neq \eps\_1,
$$
so that, given that $a\_1=\eps\_1b\_2$, we get


$$
a\_1y+a\_2 z=\eps\_2\eps\_1a\_1 y+\eps\_2\eta\_2 a\_2z
$$
Since $\eps\_2\neq \eps\_1$ we have $\eps\_2\eps\_1=-1$ and we deduce 


$$2 a\_1y= (\eps\_2\eta\_2-1)a\_2z\neq 0 $$


so $a\_2\neq 0$ and $\eta\_2\eps\_2=-1$, i.e., $\eta\_2=\eps\_1 $ and $b\_2=\eps\_1 a\_2$ and $(w\_0,v\_2)=\eps\_1(v\_0,v\_2)$. $\Box$


We now define a graph $\Gamma$ with vertices $\{1,\dotsc, n\}$ where two distinct vertices $i,j$ are connected by an edge if $(v\_i,v\_j)\neq 0$. Denote by $\Gamma\_1,\dotsc, \Gamma\_c$ its connected components. Let $I\_\alpha$ denote the set of vertices of $\Gamma\_\alpha$. We obtain an orthogonal decomposition 


$$ V:=\span(v\_1,\dotsc, v\_n)=\bigoplus\_{\alpha} V\_{I\_\alpha}. $$


For any vector $u$ we set 
$$
[u]\_\alpha:=[u]\_{I\_\alpha}.
$$



> 
> **Lemma 2.** Suppose that $i,j$ belong to the same component $\Gamma\_\alpha$ and $|S\_i|=1$. Then $S\_i\subset S\_j$.
> 
> 
> 


**Proof.** $\DeclareMathOperator{\dist}{dist}$ We argue by induction on the distance $\dist\_{\Gamma\_\alpha}(i,j)$. The case $\dist\_{\Gamma\_\alpha}(i,j)=1$ is covered by **Lemma 1**. 


We assume the result is true whenever $\dist(i,j)<d$ and we prove that it is true when $\dist(i,j)=d$. 


For simplicity we assume that $i=1$ and $v\_1,v\_2,\dotsc, v\_{d+1}=v\_j$ is a path of length $d$ in $\Gamma\_\alpha$ connecting $i$ to $j$.


If there exists $k$, $1<k<d+1$ such that $(v\_0,v\_k)\neq 0$ then $|S\_k|=1$, $\dist(v\_i,v\_k), \dist(v\_k,v\_j)<d$ and the induction assumption implies 


$$ S\_i\subset S\_k\subset S\_j.$$


Since $v\_1,\dotsc, v\_{d+1}$ is a minimal path connecting $i$ to $j$ we deduce that for any $1\leq k<\ell \leq d+1$, $\ell-k\geq 2$, the vertices $v\_j,v\_\ell$ are not adjacent so $(v\_k, v\_\ell)=0$. 


Consider the orthonormal basis $\be\_1,\dotsc,\be\_{d+1}$ of $\span\{v\_1,\dotsc, v\_{d+1}\}$ obtained from $v\_1,\dotsc, v\_{d+1}$ via Gram-Schmidt. Then


$$v\_1=c\_{01}\be\_1,\;\;v\_2=c\_{12}\be\_1+c\_{02}\be\_2,\;\;v\_k=c\_{1k}\be\_{k-1}+c\_{0k}\be\_k,\;\;k=2,\dotsc, d+1, $$


$$c\_{0k}>0,\;\;c\_{1k}\neq 0. $$


We denote by $v\_0'$ and $w\_0'$ the orthogonal projections of $v\_0$ and respectively $w\_0$ on $\span\{v\_1,\dotsc, v\_{d+1}\}$.


Then


$$v\_0'=\sum\_{k=1}^{d+1} a\_k\be\_k,\;\; w\_0'=\sum\_{k=1}^{d+1} b\_k\be\_k. $$


From ($\ast$) we deduce $a\_k=\pm b\_k$, $\forall k$. Moreover, $a\_1=(v\_0,v\_1)\neq 0$. For simplicity we assume that $a\_1=b\_i$, i.e. $S\_i=S\_1=\{1\}$. We have to prove that $1\in S\_{d+1}$ i.e.,


$$(v\_0, v\_{d+1})= (w\_0, v\_{d+1}). $$ 


For $k=2,\dotsc, d$ we have


$$ a\_{k-1}c{1k}+a\_k c\_{0k}=(v\_0,v\_k)=0=((w\_0,v\_k)=b\_{k-1}c\_{1k}+b\_kc\_{0k} $$


and we deduce that


$$a\_k=b\_k\neq 0,\;\;\forall k=1,\dotsc, d. $$


If $1\in S\_{d+1}$ we are done. If $-1\in S\_{d+1}$, then


$$ a\_dc\_{1,d+1}+a\_{d+1} c\_{0,d+1}=(v\_0,v\_{d+1})=-(w\_0, v\_{d+1})=-a\_dc\_{1,d+1}-b\_{d+1} c\_{0,d+1}, $$


so that


$$0\neq 2a\_dc\_{1, d+1}=-(a\_{d+1} +b\_{d+1}) c\_{0,d+1}. $$


Hence $a\_{d+1}+b\_{d+1}\neq0$ so $a\_{d+1}=b\_{d+1}$ and thus


$$(v\_0, v\_{d+1})=(w\_0, v\_{d+1}), $$


i.e., $1\in S\_{d+1}$. $\Box$



> 
> **Corollary 3.** For any $\alpha=1,\dotsc, c$ there exists $\eps\_\alpha=\pm 1$ such that
> 
> 
> $$[v\_0]\_\alpha=\eps\_\alpha[w\_0]\_\alpha. $$
> 
> 
> 


**Proof.** If $v\_0\perp v\_i$, $\forall i\in I\_\alpha$ the result is obvious since in this case $[v\_0]\_\alpha=[w\_0]\_\alpha=0$.


Suppose that $(v\_0,v\_i)\neq 0$ so that $S\_i=\{\eps\_i\}$, $\eps\_i=\pm 1$. Using **Lemma 2** we deduce that $S\_i\subset S\_j$, $\forall j\in I\_\alpha$. 


We can take $\eps\_\alpha=\eps\_i$. $\Box$


Choose a unit vector $\be\_0\in\bR^{n+1}$ such that $\be\_0\perp V=\span\{v\_1,\dotsc, v\_n)$. Then we have an orthogonal decomposition


$$ v\_0=a\_0\be\_0+\sum\_\alpha[v\_0]\_\alpha,\;\;w\_0=b\_0\be\_0+\sum\_\alpha [w\_0]\_\alpha. $$


There exists $\eps\_0=\pm 1$ such that $a\_0=\eps\_0b\_0$. Define the orthogonal map $T:\bR^{n+1}\to\bR^{n+1}$ $\newcommand{\bone}{\boldsymbol{1}}$


$$T=\eps\_0\bone\_{\span\be\_0}\oplus \bigoplus\_\alpha\eps\_\alpha\bone\_{V\_{I\_\alpha}}. $$


Then $Tv\_0=w\_0$ and $Tv\_i=\eps\_\alpha v\_i$ for $i\in V\_\alpha$. $\Box$


**Remark.** The connected components $\Gamma\_\alpha$ used in the above proof have a nice geometric interpretation. 


For any $I\subset \{1,\dotsc, n\}$ we denote by $G\_I$ the group of orthogonal transformations $T$of $V\_I$ such that $Tv\_i=\pm v\_i$, $\forall i\in I$. Clearly $\pm \bone\_{V\_I}\subset G\_I$. The set $I$ is called *irreducible* if $G\_I=\{\pm \bone\_{V\_I}\}$.


Note that if two irreducible sets $I,J$ are not disjoint, then their union is also irreducible. Thus, every $i=1,\dotsc, n$ is contained in a unique maximal irreducible subset and we obtain a partition of $\{1,\dotsc, n\}$ into maximal irreducible sets. These maximal irreducible sets are precisely the vertex sets of the components $\Gamma\_\alpha$.


**Comment.** It seems to me that requiring the knowledges of the volumes of all faces is too strong a condition. There are $2^n-1$ faces whereas a basis is determined by $n^2$ numbers. Note that the set of bases of $\bR^n$ modulo the action of $O(n)$ is a space of dimension 


$$ n^2-\frac{ n(n-1)}{2}=\frac{n(n+1)}{2}=n+{n \choose 2}. $$


This suggests that, the result ought to be true under weaker assumptions. 


