Matrix obtained by sequence of elementary transformations is equal to the original coordinate matrix multiplied by some matrices in $GL(n,\mathbb{Z})$

Question

Let $X = \{ x_{1}, x_{2}, \cdots, x_{n}\}$ be a set of elements of the free abelian group $A_{n}$. Let $M$ be the $(n\times n)$-matrix of coordinates of elements $x_{i}$ in terms of the basis $B=\{b_{1},b_{2},\cdots, b_{n}\}$ (where $x_{i}=r_{i1}b_{1}+r_{i2}b_{2}+\cdots + r_{in}b_{n}$, and the coordinates $r_{ij}$ define the matrix $M=(r_{ij})$)

Assume $M^{\prime}$ is a matrix obtained from $M$ by a sequence of elementary transformations.

I need to prove that $M^{\prime}=ZMY$, where $Z$, $Y$ are some matrices in $GL(n, \mathbb{Z})$.

I know of a similar result from linear algebra that says that if a matrix is obtained as a result of an elementary transformation, then it is equal to the original matrix multiplied by a corresponding elementary matrix. But here, we're talking about a sequence of transformations, as in multiple transformations, and we only have two matrices in $GL$ that we're multiplying the original one by, which may or may not be elementary.

I don't know how, therefore, to even approach this. Can someone please help me out (and be willing to answer follow-up questions and be very, very patient?)?

Thank you.

Answer 1

You've said that you know that if a matrix $M_1$ is obtained from a matrix $M_0$ by an elementary transformation, then there is an elementary matrix $E_1$ such that $M_1=M_0E_1$ or $M_1=E_1M_0$. Now suppose $M_2$ is obtained from $M_1$ by another elementary transformation, so it is obtained from $M_0$ by a sequence of two elementary transformations. Then there is an elementary matrix $E_2$ such that $M_2=E_2M_1$ or $M_2=M_1E_2$. If $M_1=M_0E_1$, then $M_2=E_2M_0E_1$ or $M_2=E_2E_1M_0$. If $M_1=E_1M_0$, then $M_2=E_2E_1M_0$ or $M_2=E_1M_0E_2$. In all cases, we can write $M_2=XM_0Y$ for some $X,Y\in GL(n,\mathbb{Z})$. For instance, in the case $M_2=E_2E_1M_0$, we have $X=E_2E_1$ and $Y=I$.

So that handles the case that your sequence has length $2$. The general case is similar: you obtain $M'$ from $M$ by repeatedly multiplying on the left or right by elementary matrices, so you have $M'=XMY$ where $X$ is the product of all the elementary matrices you multiplied by on the left and $Y$ is the product of all the elementary matrices you multiplied by on the right.

More formally, you can prove that $M'$ has the form $XMY$ by induction on the length of your sequence. The base case is that your sequence has length $0$, in which case $M'=M$ and you can take $X=Y=I$. Given the result for sequences of length $m$, suppose $M'$ is obtained from $M$ by a sequence of $m+1$ elementary transformations. Let $N$ be the matrix you obtain by performing just the first $m$ elementary transformations. Then by the induction hypothesis, $N=X_0MY_0$ for some $X_0,Y_0\in GL(n,\mathbb{Z})$. Since $M'$ is obtained from $N$ by just one more elementary transformation, there is an elementary matrix $E$ such that $M'=EN$ or $M'=NE$. We then have either $M'=EX_0MY_0$ or $M'=X_0MY_0E$. In the first case we can take $X=EX_0$ and $Y=Y_0$, and in the second case we can take $X=X_0$ and $Y=Y_0E$. Either way, we get $M'=XMY$.