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, 2019 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, 2019 at 15:00
  • $\begingroup$ Why do you need the non-degeneracy condition? Aren't you assuming $A$ is invertible? $\endgroup$ Nov 17, 2019 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. $\endgroup$ Nov 18, 2019 at 13:32

2 Answers 2


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? $\endgroup$ Dec 11, 2019 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. $\endgroup$ Dec 11, 2019 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... $\endgroup$ Dec 11, 2019 at 14:48

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