Can we recover all $k$-minors of a square matrix from some of them? This is a cross-post.
Let $k,n$ be natural numbers, $1<k<n$. Suppose we have an "unknown"  invertible $n \times n$ matrix $A$ over a field of characteristic zero. (we do not know the entries of $A$).

Can we recover all the $k$-minors of $A$ from a fixed*, ordered partial list of them?

Explicitly: We are given the values of $r$ of the minors-- a list of $r$ numbers-- and we are told which number corresponds to which minor. Can we recover the other minors?
*The list should be independent of the matrix $A$.
This question is similar to this one, but not identical to it. Here I am talking about a square matrix.
Comment: Knowing of all matrix $k$-minors of $A$  is equivalent to knowing $A$ up to a multiplication by a $k$-th root of unity, since for invertible endomorphisms, $\bigwedge^k A=\bigwedge^k B$ if and only if $A=\lambda B$ where $\lambda^k=1$.

Some non-degeneracy assumptions on $A$ are necessary here: We at least need to assume that $\text{rank}(A)>k$. Otherwise, if $\text{rank}(A)\le k$, then even if we know all the $k$-minors of $A$ except one, we cannot recover the last one. 
Indeed, take $A=\pmatrix{D&0\\ 0&0}$ where $D$ is any diagonal matrix of size $k$. The $k$-minor corresponding to the first $k$ rows and columns (which is $\det D$) cannot be recovered from the other $k$-minors (which are zeroes). 
 A: It depends how you frame the question, but the answer is yes in some sense. Let $A$ be the $n \times n$ generic matrix with linear entries in $\mathbb{k}[x_1, \ldots, x_{n^2}]$. I denote by $I_m$ the ideal generated by the $m \times m$ minors of $A$. 
It has been proved by Bruns that there exists $q=n^2-m^2+1$ homogeneous elements of $I_m$, say $g_1, \dotsc, g_q$, of $I_m$ such that $\sqrt{(g_1,\ldots,g_q)} = I_m$ (and then by Bruns and Schwanzel that the bound $n^2-m^2+1$ is optimal). This proves in particular that for any minor $M_m$ of size $m$, there is an integer $r>0$ such that $M_m^r$ is an algebraic combination of the $g_i$. 
You may have a look at the sections 1 to 5 of the book Determinantal Rings by Bruns and Vetter to see how they construct this "wonderful poset" of generators of $I_m$ which has cardinal $n^2-m^2+1$. I must nevertheless admit that their construction looks a bit intimidating (at least to me) and I would be extremely interested to see a simple construction of this poset.
A: Here is another point of view on the question.
Assume that you are interested in $k$-minors, what you're going to do is focus on submatrices of $A$ of size $k\times n$ by eliminating $n-k$ rows. Such a $k\times n$ submatrix has $n\choose k$ $k$-minors, and these are subject to what is known as Plücker equations:

Theorem: an ordered collection of $n\choose k$ integers is the collection of (lexicographically ordered) maximal minors of some
  $k\times n$ matrix if and only if these numbers satisfy a set of polynomial equations known as
  Plücker equations.
Context: see these lecture notes by Alexander Yong.
Proof: see Schubert Calculus by Kleiman and Lakso.

In practice, it means that there is some maximal number of $k$-minors that can be fixed independently, after which all the others will be uniquely determined by the equations.
Plücker equations for $(n,k)$ can be displayed by typing Grassmannian(k-1,n-1) into Macaulay2 (the $-1$ come from projective reasons).
Here is one of these equations for $n=6, k=3$:
      $$p_{2,3,4} p_{1,3,6} -p_{1,3,4}p_{2,3,6} +p_{1,2,3}p_{3,4,6}=0$$ 
As expected, if all minors are zero except one of them, then the Plücker equations will be of no help to find that one, as the variables always come by pairs.
