# Can the determinant of an integer matrix with $k$ given rows equal the gcd of the $k\times k$ minors of those rows?

I'm interested if the following is true:

Let $n> k\geq1$ be integers, let $A\in\mathbb Z^{k\times n}$ and denote the $\binom nk$ $k\times k$ minors of $A$ by $A_1,\ldots,A_N$. Then the equation $$\left|\begin{array}{c}A\\\hline X\end{array}\right|=\gcd(A_1,\ldots,A_N)$$ always has a solution $X\in\mathbb Z^{n-k\times n}$.

This is a generalisation of Can the determinant of an integer matrix with a given row be any multiple of the gcd of that row? (which is the case $k=1$). For $\style{text-decoration:line-through}{k=n}$ and $k=n-1$ the above conjecture is also (obviously) true.

Note that, as in the linked question, finding the minimal positive value of that determinant is equivalent to finding the minimal hypervolume of an $n$-dimensional simplex whose vertices are in an integer lattice and $k+1$ vertices are fixed.

Idea: The case $\gcd(A_1,\ldots,A_N)=1$ may follow from Elementary proof that if $A$ is a matrix map from $\mathbb{Z}^m$ to $\mathbb Z^n$, then the map is surjective iff the gcd of maximal minors is $1$

This follows from the Smith normal form and so holds for any PID: Smith-normal form preserves the gcd of minors of size $k$ for all $k\leq n$ (in particular, the determinant). This can be seen from Cauchy-Binet's formula.

It suffices to go to Smith normal form to obtain a matrix of the form $$QAP=\begin{pmatrix}D&0\end{pmatrix}$$

with $D$ diagonal of size $k\times k$ and determinant equal to the gcd of $k\times k$ minors of $A$. Extend it in a trivial way:

$$\begin{pmatrix}D&0\\0&I_{n-k}\end{pmatrix}$$

and apply the inverse change of bases that gave the smith normal form to obtain

$$\begin{pmatrix}Q^{-1}&0\\0&I_{n-k}\end{pmatrix} \cdot \begin{pmatrix}D&0\\0&I_{n-k}\end{pmatrix} \cdot P^{-1} = \begin{pmatrix}A\\X\end{pmatrix}$$

for some $X$ (equal to the last $n-k$ rows of $P$).

• Nitpick: gcd in a PID is defined only up to scaling by invertible elements, whereas gcd in $\mathbb {Z}$ is nonnegative by definition. So, be careful with $k=n$ in the original question. – darij grinberg Oct 22 '17 at 8:15
• You're right. If $k<n$ we can multiple the last row by any invertible, if $k=n$ we can't. Edited the question. – punctured dusk Oct 22 '17 at 8:34