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?

That is, suppose that we are given the values of $r$ of the minors- i.e. we are given an indexed list of $r$ numbers, and we are told which number corresponds to which minor. Can we recover the other minors?

Comment:

Some non-degeneracy conditions 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 unknown minor.

Indeed, take $A=\pmatrix{D&0\\ 0&0}$ where $D$ is any diagonal matrix of size $k$. We can't recover the $k$-minor corresponding to the first $k$ rows and columns (which is $\det D$) from the other $k$-minors (which are zeroes). This example was suggested by user1551.

Solution 1:

Yes, you can.

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.

Part of this answer was borrowed from this question/answer of mine, which amazingly was asked just 10 days before yours.