This is a [cross-post](https://math.stackexchange.com/questions/3274711/can-we-recover-all-matrix-minors-from-some-of-them).


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](https://mathoverflow.net/questions/43965/how-many-minors-i-need-to-check-to-conclude-all-minors-will-vanish), 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). 



Can we recover all $k$-minors of a square matrix from some of them?
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](http://www.home.uni-osnabrueck.de/wbruns/brunsw/pdf-article/NumbEqDet.published.pdf) 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](http://www.home.uni-osnabrueck.de/wbruns/brunsw/detrings.pdf) 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.


