# Help to find two sets of two linear independent vectors satisfies certain properties

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.

• 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 $3$? Aug 18, 2023 at 12:59
• 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 $\mathbb Z$. Aug 18, 2023 at 12:59
• @AlexRavsky I think in that case the last determinant divisor of $M_1$ is $1$ as well Aug 19, 2023 at 11:03
• I used a program to compute its Smith normal form, the last determinant divisor is the "last diagonal" entry Aug 19, 2023 at 11:05
• @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 Aug 19, 2023 at 13:47

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. $$\square$$