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).

  • 2
    $\begingroup$ (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? $\endgroup$ – Gro-Tsen Nov 15 '19 at 9:26
  • $\begingroup$ (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. $\endgroup$ – Kapil Nov 15 '19 at 15:00
  • $\begingroup$ Why do you need the non-degeneracy condition? Aren't you assuming $A$ is invertible? $\endgroup$ – Eric Canton Nov 17 '19 at 3:12
  • $\begingroup$ @EricCanton Yeah, I am assuming that $A$ is invertible. I included the discussion about non-degeneracy conditions as a motivation for why requiring invertibility. $\endgroup$ – Asaf Shachar 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.

  • $\begingroup$ 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? $\endgroup$ – Asaf Shachar Dec 11 '19 at 14:17
  • $\begingroup$ @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. $\endgroup$ – Arnaud Mortier Dec 11 '19 at 14:45
  • $\begingroup$ 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... $\endgroup$ – Asaf Shachar Dec 11 '19 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.