# 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, $$1. Suppose we have an "unknown" invertible $$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).

• (Just thinking out loud.) I feel it might be worth generalizing the question to the case where you delete $k$ columns and $ℓ$ rows, with the convention that the “determinant” of a non-square matrix is simply the wedge product of its columns (say). So one sub-question becomes: how many wedge products of $n-k$ among $n$ column vectors do you need to recover them all? (The Plücker relations certainly have something to say here.) And another: now what if we also delete some entries? Nov 15 '19 at 9:26
• (Another thinking out loud attempt!) The matrix defines a section of $S^{*}\otimes Q$ on $G_s\times G_q$ where $S$ is the universal sub-bundle on the Grassmannian $G_s$ of rank $k$ sub-spaces and $Q$ is the universal quotient bundle on the Grassmannian $G_q$ of rank $k$ quotient spaces. Such a section is determined by its values at finitely many points. However, you are not asking for the value, rather only the value of its image in $\det(S)^*\otimes\det(Q)$. Moreover, the points are "pre-determined" since you are only using points determined by basis vectors. Nov 15 '19 at 15:00
• Why do you need the non-degeneracy condition? Aren't you assuming $A$ is invertible? Nov 17 '19 at 3:12
• @EricCanton Yeah, I am assuming that $A$ is invertible. I included the discussion about non-degeneracy conditions as a motivation for why requiring invertibility. Nov 18 '19 at 13:32

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.

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.

• Thank you! Can you elaborate on what do you mean "by 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"? Do you have a more precise statement? e.g. can you say how many minors will be needed to determine all the rest? (say for an invertible matrix). Can we describe an explicit choice of a subset of the minors that will suffice? Dec 11 '19 at 14:17
• @AsafShachar To obtain a more explicit statement, one would need to take a closer look at the list of polynomials, which can be a huge task depending on $k$ and $n$. I'll see what this gives in small examples when I have more time. Dec 11 '19 at 14:45
• Thank you. I am merely curious. Perhaps there is a structural way to simplify this "computational search", but I don't see such a thing at the moment... Dec 11 '19 at 14:48