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I am trying to find two sets of two linear independent row vectors in $\mathbb Z^2$ satisfies certain properties, I made a program in Matlab to generate such vectors, however, it still hasn't found the vectors I was looking for. I am hoping some expert here can provide some insight.

Suppose we have row vectors $x_1$, $x_2$ , $y_1$ , $y_2 \in \mathbb{Z}^2$, where {$x_1$, $x_2$} and { $y_1$, $y_2$} are two linear independent sets.

Now, consider the $4$ by $4$ matrix $\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \\ x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix}$, let $M_1 = \begin{bmatrix} x_1 & y_1 & (0, 0) \\ -y_1 & x_1 & y_1 \\ x_2 & y_2 &(0, 0) \\ -y_2 & x_2 & y_2 \end{bmatrix}$ and $M_2 =\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \\ x_2 & y_2 \\ -y_2 & x_2 \\ (0, 0) &y_1 \\ (0, 0) &y_2 \end{bmatrix}$.

I am trying to find $x_1$, $x_2$, $y_1$, $y_2$ such that the last determinant divisor (i.e. the $\gcd$ of the determinants of all $4$ by $4$ minors of the matrix) of $M_1$ is not $1$, while the last determinant divisor of $M_2$ is $1$.

Thank you for reading, any idea would be really appreciated. I also posted this question on mathoverflow.

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  • $\begingroup$ I have the following simple idea which can be helpful. The vectors satisfy the required condition iff there exists a prime number $p$ (supposed to be any prime factor of the last determinant divisor of $M_1$) such that when we consider the matrices modulo $p$ then the last determinant divisor of $M_1$ is $0$, but the last determinant divisor of $M_2$ is $1$. What about small $p$, such as $2$ or $3$? $\endgroup$ Aug 18, 2023 at 12:59
  • $\begingroup$ Moreover, it seems when we deal with the vectors modulo $p$ we can relax the independence condition, because it seems we can always assure it by some lift of the vectors modulo $p$ to the vectors over $\mathbb Z$. $\endgroup$ Aug 18, 2023 at 12:59
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    $\begingroup$ @AlexRavsky I think in that case the last determinant divisor of $M_1$ is $1$ as well $\endgroup$
    – ghc1997
    Aug 19, 2023 at 11:03
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    $\begingroup$ I used a program to compute its Smith normal form, the last determinant divisor is the "last diagonal" entry $\endgroup$
    – ghc1997
    Aug 19, 2023 at 11:05
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    $\begingroup$ @AlexRavsky Thank you for your time. I did it by hand initially as well and the computation was a bit messy. I found a package in Matlab that compute the Smith normal form of a matrix really helpful $\endgroup$
    – ghc1997
    Aug 19, 2023 at 13:47

1 Answer 1

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There are no matrices $M_1$ and $M_2$ which you are trying to find because of the following proposition.

Proposition. Let $p$ be any prime. Then $p$ divides the last determinant divisor of $M_1$ iff $p$ divides the last determinant divisor of $M_2$. Note that we can relax the independence condition.

Proof. By elementary transformations which do not change the last determinant divisors, we can transform the matrices $M_1$ to and $M_2$, to $M_1’= \begin{bmatrix} x_1 & y_1 & (0, 0) \\ x_2 & y_2 &(0, 0) \\ (0, 0) & x_1 & y_1 \\ (0, 0) & x_2 & y_2 \end{bmatrix}$ and $M_2’= \begin{bmatrix} x_1 & (0, 0) \\ x_2 & (0, 0) \\ y_1 & x_1 \\ y_2 & x_2 \\ (0, 0) &y_1 \\ (0, 0) &y_2 \end{bmatrix}$, respectively.

For any matrix $M$ over $\mathbb Z$ let $\overline{M}$ be the matrix $M$ with its entries replaced by their residues modulo $p$. We consider the following matrices over the field $\mathbb Z_p$ of residues modulo $p$. Put $X=\overline{\begin{bmatrix} x_1 \\ x_2\end{bmatrix}}$, and $Y=\overline{\begin{bmatrix} y_1 \\ y_2\end{bmatrix}}$, $N_1=\overline{M’_1}=\begin{bmatrix} X & Y & 0\\ 0 & X & Y\end{bmatrix}$, and $N_2=\overline{M’_2}=\begin{bmatrix} X & 0\\ Y & X \\ 0 & Y\end{bmatrix}$. Then $p$ divides the last determinant divisor of $M_1$ (resp. $M_2$) iff the rank of $N_1$ (resp. $N_2$) is at most $3$. If any of matrices $X$ and $Y$ is nonsingular then both $N_1$ and $N_2$ have rank $4$ and we are done. If any of matrices $X$ and $Y$ is zero then the rank $N_1$ equals the rank of $N_2$ and we are done. So it remains to consider the case when both ranks of $X$ and $Y$ are $1$. Simultaneously applying to $X$ and $Y$ the elementary transformations which do not change the ranks of $N_1$ and $N_2$, we can transform $X$ to $\begin{bmatrix} \overline{1} & 0 \\ 0 & 0\end{bmatrix}$ and $Y$ to some matrix $Z$. If the last row of $Z$ is zero then both matrices $N_1$ and $N_2$ are singular and we ore done. Otherwise by the elementary row transformations which do not change the ranks of $N_1$ and $N_2$ we can keep the matrix $X$ and annulate the first row of the matrix $Z$. Then we can easily see that matrices $N_1$ and $N_2$ have equal rank, so and we are done. $\square$

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