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This question is related to another problem I asked a question about here. In fact, it is part (b) of a problem whose part (a) was

Let $X = \{ x_{1}, x_{2}, \dots, x_{n}\}$ 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 $B=\{b_{1},b_{2},\dots, b_{n}\}$ (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.

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

For the part of the problem I'm asking about now, I need to show that $X$ is a basis for $A_{n}$ if and only if $\det (M) = \pm 1$.

For the direction $(\Longleftarrow)$, I was given the hint: "How is $\det(M)$ changed under transformations of $M$?" Using this hint, I was able to get to the following point:

Suppose $\det M = \pm 1$. By Part (a), if $M^{\prime}$ is a matrix obtained by a sequence of elementary transformations on $M$, then $M^{\prime}=ZMY$, where $Z$, $Y$ $\in GL(n,\mathbb{Z})$. Since $Z, Y \in GL(n, \mathbb{Z})$, $\det(Z) = \pm 1$ and $\det(Y) = \pm 1$. From linear algebra, we know that the determinant of a product of matrices is equal to the product of the determinants, so $\det(Z) \cdot \det (M) \cdot \det(Y) = (\pm 1) (\pm 1) (\pm 1) = \pm 1 = \det(M^{\prime})$.

This was as far as I was able to get though, because I am having trouble understanding how to reconcile this with $X$ being a basis for $A_{n}$.

For the $(\Longrightarrow)$ direction, I was able to get even less, primarily because of the fact that I don't understand how $X$ being a basis for $A_{n}$ has anything to do with what the determinant of $M$ is.

I started by saying that if $X$ is a basis for $A_{n}$, then $\forall a_{i} \in A_{n}$, $\exists c_{i1}, c_{i2}, \cdots , c_{in} \in \mathbb{Z}$ such that $a_{i}=c_{i1}x_{1} + c_{i2}x_{2}+\cdots + c_{in}x_{n}$, where the $c_{ij}$ are the entries in the matrix $M$. And wasn't sure where to go from there.

Could someone please tell me where to go from where I've stopped (in both directions)? I'm very much stuck and this point. (And be prepared to be patient and answer lots of follow-up questions!)

Thank you! :)

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    $A_n$ is $\mathbb{Z}^n$. So you want to show that if $M \in \mathbb{Z}^{n \times n}, \det(M) = \pm 1$ then $M^{-1} \in \mathbb{Z}^{n \times n}$ – 2017-01-30
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    It is not clear what is your teacher's approach. Did you prove some properties about the possible basis $B$, in particular about $b_1$ ? – 2017-01-30
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    @user1952009 not that I'm aware of. – 2017-01-30
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    What you wrote isn't enough for proving the claim, so I think something else is proven before, about the elementary transformations and the basis. And a solution is to express $M^{-1}$ in term of the [adjugate matrix](https://en.wikipedia.org/wiki/Adjugate_matrix#Definition) – 2017-01-30
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    @user1952009 there was this: https://math.stackexchange.com/questions/2100821/basis-for-free-abelian-group-a-n-of-rank-n – 2017-01-30
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    My apologies for this marginal comment, but you are using $X$ for two fairly different objects here. – 2017-01-30
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    @JessyunBourne It is not useful for the claim : a matrix $M \in \mathbb{Z}^{n \times n}$ has its inverse $M^{-1} \in \mathbb{Z}^{n \times n}$ if and only if $\det(M) = \pm 1$ – 2017-01-30
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    @AndreasCaranti yeah, that's my prof. And it's confusing me too. – 2017-01-30
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    @user1952009 I realized there was a typo in the original problem and have edited my question to reflect that. – 2017-02-01

1 Answers 1

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In the following $X = \{ x_{1}, x_{2}, \dots, x_{n}\}$.

Write $$ [X] = \begin{bmatrix}x_{1}\\x_{2}\\ \vdots\\x_{n}\end{bmatrix} $$ and similarly for $[B]$.

Then you have $$ [X] = M [B]. $$

Now $\det(M) = \pm 1$ iff $M$ is invertible, i.e., it has an inverse which is itself an integer matrix. Thus if $\det(M) = \pm1$ you have $$ [B] = M^{-1} [X]. $$

Now since $B$ is a basis, given any $v \in A_{n}$ there is a unique row vector $c$ such that $v = c [B]$. If $\det(M) = \pm 1$ we have $$ v = c [B] = (c M^{-1}) [X], $$ so $v$ can be written as a linear combination of the elements of $X$. This is unique, since if $v = d [X]$ for some row vector $d$, then $$ v = d [X] = d M [B], $$ so $d M$ is uniquely determined, as $B$ is a basis, and thus so is $d = (d M) M^{-1}$. Therefore $X$ is a basis of $A_{n}$.

Conversely, if $X$ is a basis, every element of $B$ can be written as a linear combination of the elements of $X$, so there is a matrix $N$ such that $[B] = N [X] = N M [B]$. Since $B$ is a basis, we have that $N M = I$, the $n \times n$ identity matrix, and taking determinants ones sees that $\det(M) = \pm 1$.

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    You are assuming that $M^{-1}$ is an integer matrix ? – 2017-01-30
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    Only when this is the case, that is when $\det(M) = \pm 1$ is an invertible integer, yes. – 2017-01-30
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    Yes but this is what we need to prove – 2017-01-30
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    @AndreasCaranti if $M$ is the coordinate matrix of $X$ in terms of the basis elements $B$, then why do we need $N$? – 2017-02-01
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    @AndreasCaranti also realized there were a couple of typos in the original problem and have now edited my question to reflect that. Does that change your answer at all? – 2017-02-01
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    @JessyunBourne, I am not sure I fully understand the suggestion you were given - my bad. My answer assumes a fact in matrix theory. Let $A$ be a square matrix with coefficients in a commutative ring $R$. If $A$ is invertible, then $\det(A)$ is an invertible element of $A$, This is because from $A B = I$ you get $\det(A) \det(B) = \det(I) = 1$. Conversely, let $B$ be the [adjugate](https://en.wikipedia.org/wiki/Adjugate_matrix) of $A$, Then $A B = \det(A) I = B A$. It follows that if $\det(A)$ is invertible in $R$, then $A$ is invertible. – 2017-02-02
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    @AndreasCaranti Re: "$[B] = N [X] = N M [B]$. Since $B$ is a basis, we have that $N M = I$, the $n \times n$ identity matrix, and taking determinants ones sees that $\det(M) = \pm 1$." How is $N[X] = NM[B]$? How do we know that $NM = I$? – 2017-02-02
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    @AndreasCaranti I also posted a +50 bounty on this question. If you can fill in these details that I'm having trouble understanding where they come from, I would be glad to award you the bounty. – 2017-02-02
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    $N[X]=NM[B]$ follows immediately from $[X]=M[B]$. Since $B$ is a basis, the rows of the matrix $[B]$ span ${\mathbb Z}^n$, and so $[B]$ is an invertible matrix. So $[B]=NM[B] \Rightarrow NM=I$ (multiply the equation on the right by $[B]^{-1})$. – 2017-02-02
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    @AndreasCaranti n/m I figured it out! So as soon as I can, the +50 bounty will be yours. – 2017-02-03
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    @DerekHolt, in my intention, the group $A_{n}$ is an abstract group, not (already) realized as $\mathbb{Z}^{n}$, and $[B]$ is a formal column vector, whose entries are elements of $A_{n}$. So from my point of view $[B]$ is *not* an $n \times n$ integer matrix. – 2017-02-03
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    @JessyunBourne, thx a lot! – 2017-02-03