Fix an integer $n$. What can you say about a (not necessarily square) matrix $A$ with integer entries that has the property that for any $k$, every $k\times k$ minor of $A$ is divisible by $n^{k-1}$? Have you seen such matrices before? Do they have a name? Are they in a bijection with some set whose description does not require knowing about determinants?

The question I really care about is the same but with "many-variable Laurent polynomials with integer coefficients" replacing the integers in the above (except $k$), but I suspect that it doesn't really make a difference.

The reason I care is that I have but I don't understand an amusing (I think) generalized Alexander invariant of tangles and virtual tangles with excellent composition properties and with values in such matrices as above, and I would like to understand its target space. A handout and a video are at http://www.math.toronto.edu/~drorbn/Talks/GWU-1203/ but no writeup exists at present.

  • 2
    $\begingroup$ Your condition seems to be preserved under integral elementary row and column operations. So you can probably reduce your matrix to some normal form generalizing the Euclidean algorithm. $\endgroup$
    – Ian Agol
    Commented May 16, 2012 at 22:40
  • $\begingroup$ I've edited the tags, with the desirable side-effect to bump the question back to the front page. $\endgroup$ Commented May 23, 2012 at 1:07

1 Answer 1


If $n$ is prime, we have the following equivalences:

  1. Your condition.

  2. Your condition for $2 \times 2$ minors.

  3. There are integer vectors $v$ and $w$ such that $A \equiv v w^T \mod n$.

The implications $1 \implies 2$ and $2 \implies 3$ are straightforward. For $3 \implies 1$, write $A = v w^T + n B$, and expand out $\det(v w^T + n B)$ as a sum of products of minors from $v w^T$ and from $n B$. All the terms which involve a minor of $v w^T$ larger than $1 \times 1$ are zero, so every term is divisible by $n^{k-1}$.

When $n$ is not prime, we still have $3 \implies 1 \implies 2$, and $2$ does not imply the others; look at $\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}$ with $n=4$. Also, for $n$ nonprime, $1$ does not imply $3$, look at $\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$ for $n=4$.

From your described motivation, it sounds like it would be interesting to know whether your matrices obey $3$.

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    $\begingroup$ My $n$ is not prime, yet I'd better check if I'm lucky and in fact condition 3 holds for my matrices. I was aware of it but never optimistic enough to actually check it. $\endgroup$ Commented May 16, 2012 at 19:57
  • $\begingroup$ Another way of saying this is that over a field, the determinantal rank (size o the largest nonvanishing minor) is equal to the rank of a matrix. $\endgroup$ Commented May 17, 2012 at 5:34
  • $\begingroup$ Long time to check, and probably too late for people to look at my original problem, yet I finally checked and condition 3 definitely does not hold for the matrices I care about... $\endgroup$ Commented May 21, 2012 at 0:25

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