Reconstruct vectors from determinant in subspace If we have $n$ vectors in $\mathbb{R}^m$ (with $m>n$), we can project these vectors onto $\mathbb{R}^n$ and calculate the determinant of the matrix formed by these $n$ n-dimensional (projected) vectors. Suppose we know all the $C^m_n$ determinants (and the basis they are projected onto), is there a method to calculate the set of vectors that can produce these determinants?
You can assume for our choice of $m,n$, $C^m_n > mn$. Brute force solving $mn$ equations of degree $n$ does not seems practical.
 A: Here's an answer to a closely-related question, which might be the one you should have asked:
Let $P$ be the parallelipiped generated by $v_1, \ldots, v_n \in R^k$, and $S$ the subspace generated by the $v$s.
Consider the projections $u_1, \ldots, u_n$ of the $v_i$s (in order) into some coordinate $n$-plane, $H$, and define
$$
d(H) = det(u_1, \ldots, u_n).
$$
Do this for all $k \choose n$ coordinate $n$-planes. Also compute the volume $D$ of the parallelipiped $P$.
Now the question is: given, for each coordinate $n$-plane, $H$, the number $\frac{d(H)}{D}$,

*

*Is there any relation that must hold among these numbers?


*Can we recover the subspace $S$ from these numbers?
The answer to both is "yes"; the things that must hold are the Plucker relations, and the numbers $\frac{d(H)}{D}$ are called the Plucker coordinates of $S$. The name Plucker should have an umlaut over the "u", but I don't recall how to produce that in MSE, alas.
What's even better is that there's a general formula (involving a lot of determinant-like things) for finding $S$ from the Plucker coordinates. As an example, in the case of a 2-plane through the origin in 3-space, if you compute the (signed) areas of projections of a unit square in your plane onto the $xy$, $yz$, and $zx$ planes, and call them $C, B, A$ respectively, then the plane containing your square is given by
$$
Ax + By + Cz = 0.
$$
A: Absolutely not. Suppose all the vectors $v_1, \ldots, v_n$ lie in the coordinate subspace defined by $x_{n+1} = x_{n+2} = \ldots = x_m= 0$. Then all the determinants will be zero except for the one corresponding to the first coordinate $n$-plane.
Let $M$ be any orthonormal $n \times n$ matrix,
$$
w_i = \pmatrix{M & 0 \\ 0 & I_{m-n}}v_i 
$$
for $i = 1, \ldots, n$.
Then the determinants for the $w$-vectors will be the same as those for the $v$-vectors/
More explicitly, in $R^{10}$, let
$$
v_1 = e_1, v_2 = e_2, w_1 = 2e_1, w_2 = 0.5e_2.
$$
Then the determinants associated to the vs and ws are the same (even though they are NOT orthogonally related!)
