# Smith normal form and last invariant factor of certain matrices

I've copied over this question from what I asked on Mathematics Stack Exchange, in the hope that some experts can provide some relevant insight.

Suppose we have row vectors $$x_1$$, $$x_2$$ , $$y_1$$ , $$y_2 \in \mathbb{Z}^2$$, 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 $$M_2 =\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \\ x_2 & y_2 \\ -y_2 & x_2 \\ (0, 0) &y_1 \\ (0, 0) &y_2 \end{bmatrix}$$.

The last invariant factor of $$M_i$$ is defined to be the $$\gcd$$ of the determinants of all $$4$$ by $$4$$ minors of $$M_i$$.

Question:If the last invariant factor of $$M_2$$ is $$1$$, does this implies that the last invariant factor of $$M_1$$ is $$1$$?

My idea was to apply some elementary row and column operations on $$M_1$$ and $$M_2$$, call the resulting matrix $$M_1'$$ and $$M_2'$$, and show that all of the $$4$$ by $$4$$ minors in $$M_2'$$ can be found in $$M_1'$$. However, it is not obvious to me how to find $$M_1'$$ and $$M_2'$$.

There is also another potential idea, let $$r_1, \ldots, r_4$$ denotes the rows of $$M_1$$, and $$r'_1, \ldots r'_6$$ denote the rows of $$M_2$$, then an eqvaiblent statement to the question would be: $$\text{ \frac{\mathbb{Z}^4 }{ \langle r'_1, \ldots, r'_6 \rangle} is torsion free implies that \frac{\mathbb{Z}^6 }{ \langle r_1, \ldots, r_4 \rangle} is torsion free. }$$

Thank you for reading, any idea for this would be really appreciated.

• It appears at first sight to be a completely different question from the one in your link to MSE. Aug 16 at 15:55

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. $$\square$$