# Existence of an integer matrix with maximal subdeterminants $a_1, \ldots, a_n$

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:

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 $$\begin{pmatrix}-d & c_2&\cdots&c_n\\ 0 \\ \vdots &&\large M \\ 0\end{pmatrix} \,.$$
• This is where I was headed! I worked it out for $n$ = 4 and planned to get into detail for the general case, so thanks! But I'm confused by the comment "The case $n$ = 2 being trivial." Isn't the starting number for the induction $n$=3; and (for $n$ = 2), how would we interpret the matrix with $n$-2 = 0 rows? I have a feeling you are really starting with $n$ = 3, and going down one to 2, but it's not clear to me how this works. May 7, 2022 at 2:15
• To clarify, the proof is by induction for $n \geq 2$, and $n = 2$ is the base case (where we can take the matrix $(a_2, a_1)$). May 7, 2022 at 8:13
Hint: Assume that the numbers have gcd $$1$$. Let $$A$$ a matrix of determinant $$1$$ with first row $$(a_1,- a_2, \ldots, (-1)^{n-1} a_n)$$. Since $$\det A =1$$ we have $$A= \operatorname{adj}(\operatorname{adj} (A))$$ ( see adjugate).
$$\bf{Added:}$$ We can also prove the following result: Consider a $$k\times n$$ matrix with elements in a PID. Then there exists an $$(n-k)\times n$$ matrix such that the $$n-k$$ minors are the same as the corresponding $$k$$ minors. Use a similar idea and the Jacobi formula for the minors of the adjugate.