# Integer matrices with a strange divisibility property

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.

• 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. May 16 '12 at 22:40
• I've edited the tags, with the desirable side-effect to bump the question back to the front page. May 23 '12 at 1:07

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

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 $3$.

• 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. May 16 '12 at 19:57
• 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. May 17 '12 at 5:34
• 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... May 21 '12 at 0:25