Given $n \geq 2$ and integers $a_1, \ldots, a_n$, does there exist an integer $(n-1) \times n$ matrix whose maximal subdeterminants are $a_1, \ldots, a_n$ (with fixed ordering)?
Example: $n = 3$, $(a_1, a_2, a_3) = (19, 4, 22)$. The matrix
$$\begin{pmatrix}0 & 11 & 2 \\
-2 &95&19\end{pmatrix}$$
has $i$th subdeterminants (with $i$th column removed) equal to $(19, 4, 22)$.
Context:
- The $n=3$ case is precisely this question from the newsletter: Is every vector in $\mathbb Z^3$ a cross product?. (This is where the example comes from.)
- The general case would give an alternative proof for this question: Can the determinant of an integer matrix with a given row be any multiple of the gcd of that row? by taking $(a_1, \ldots, a_n)$ to be coefficients in Bézout's theorem.
Best Answer
This can be proven inductively as in the second linked question, and it generalizes the construction in the first linked question (which is for $n=3$).
When $n = 2$ we can take the matrix $\begin{pmatrix}a_2 & a_1\end{pmatrix}$. Now take $n \geq 3$ and $a_1, \ldots, a_n$ integers, and let $d = \gcd(a_2, \ldots, a_n)$. Construct (using the induction hypothesis) a working $(n-2) \times (n-1)$ matrix for $a_2/d, \ldots, a_n/d$ and call it $M$. The maximal subdeterminants of $M$ are coprime, so there exist integers $c_2, \ldots, c_n$ with $$\det\begin{pmatrix}c_2&\cdots&c_n\\\\ &\large M\end{pmatrix}=a_1 \,.$$ We can then take the matrix $$\begin{pmatrix}-d & c_2&\cdots&c_n\\ 0 \\ \vdots &&\large M \\ 0\end{pmatrix} \,.$$