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Let $B = \{ b_{1},b_{2},\cdots , b_{n} \}$ be a basis of the free abelian group $A_{n}$ of rank $n$.

If $d_{1}$, $d_{2}$, $\cdots$, $d_{n}$ are $n$ nonzero integers, then I need to prove that the set $\{ d_{1}b_{1},\, d_{2}b_{2}, \cdots , d_{n}b_{n} \}$ is a basis for $A_{n}$ if and only if $d_{i} \in \{ -1, 1\}$ for all $i$.

Here is my attempt thus far:

$(\implies)$ Suppose that $D = \{ d_{1}b_{1},\, d_{2}b_{2}, \cdots , d_{n}b_{n} \}$ is a basis for $A_{n}$. Then, for any $x \in A_{n}$, $\exists$ a unique representation $x = a_{1}(d_{1}b_{1}) + a_{2}(d_{2}b_{2}) + \cdots + a_{n}(d_{n}b_{n})$, where $d_{i}b_{i} \in D $, $a_{i} \in \mathbb{Z}$, $a_{i} \neq 0$ and $d_{i}b_{i} \neq d_{j}b_{j}$ for $i \neq j$.

Since $B$ is also a basis for $A_{n}$, for that same $x$, we have the unique representation in that basis $x = c_{1}b_{1} + c_{2}b_{2}+\cdots + c_{n}b_{n}$, where $b_{i} \in B$, $c_{i} \in \mathbb{Z}$, $c_{i} \neq 0$, and $b_{i} \neq b_{j}$ for $i \neq j$.

Thus, $a_{1}(d_{1}b_{1}) + a_{2}(d_{2}b_{2}) + \cdots + a_{n}(d_{n}b_{n}) = c_{1}b_{1} + c_{2}b_{2}+\cdots + c_{n}b_{n}$, which implies that $a_{i}d_{i} = c_{i}$ $\forall i$ from $i=1$ to $n$.

That's as far as I got on that side - there must be something I'm missing here that isn't allowing me to see how $d_{i} = \pm 1$ only. Therefore, any help would be very much appreciated!

$(\Longleftarrow)$ In this direction, I really have no idea. Just that if $d_{i} \in \{ -1,1\}$, then the set $D$ becomes $\{\pm b_{1}, \pm b_{2}, \cdots, \pm b_{n} \}$. If I were to express any $x \in A_{n}$ as $x =c_{1}b_{1} + c_{2}b_{2}+\cdots + c_{n}b_{n}$, how could I use that to show that $\pm b_{i}$ would also work?

Thank you.

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    In ($\Longrightarrow$) you don't require $a_i\ne 0$. Hint for ($\Longrightarrow$) - in particular $a_id_i=c_i$ for some $a_i$ when $c_i=1$. Hint for ($\Longleftarrow$) - consider $(\pm c_i)(\pm b_i)$ – 2017-01-16
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    @RobertChamberlain I'm afraid I don't understand your hints. Why would $c_{i}=1$? and for $(\Longleftarrow)$, you mean let $x = (\pm c_{1}) (\pm b_{1}) + (\pm c_{2})(\pm b_{2})+\cdots + (\pm c_{n}) (b_{n})$? How are we justified in doing this? Then, I imagine since you've just written $x$ as a linear combination of $\pm b_{i}$, you've just shown the set $D$ is a basis? Or is there more to it than that? – 2017-01-16
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    No worries! The $a_i$ have to exist for any $x$, so they certainly exist for the $x$ defined by setting each $c_i=1$. For the second part, you want to show that arbitrary $x$ are in the span of $\pm b_i$, you choose the $c_i$ for the original basis, then notice $x=(\pm c_1)(\pm b_1)+...+(\pm c_n)(\pm b_n)$ – 2017-01-16
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    @RobertChamberlain I'm still working on this, and still struggling. I've put out a 50 point bounty if you could answer w/details. I'd appreciate it very much and be forever in your debt :) – 2017-01-20

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$\{d_1b_1,\ldots,d_nb_n\}$ is a basis if and only if each $x\in A_n$ can be written uniquely as $\sum_{i=1}^na_id_ib_i$ for some $a_i\in\mathbb{Z}$. Now if $x$ can be written as $\sum_{i=1}^na_id_ib_i$ then, since $\{b_1,\ldots,b_n\}$ is a basis, $a_id_i$ is uniquely determined, so $a_i=a_i^{-1}a_id_i$ is uniquely determined, hence if such a representation exists then it is unique.

Now, suppose $d_i=\pm 1$ for each $i$, then $d_i^2=1$. As $\{b_1,\ldots,b_n\}$ is a basis, each $x\in A_n$ can be written $\sum_{i=1}^na_ib_i=\sum_{i=1}^n(a_id_i)(d_ib_i)$, thus $\{d_1b_1,\ldots,d_nb_n\}$ is a basis.

Conversely suppose $\{d_1b_1,\ldots,d_nb_n\}$ is a basis then each $x\in A_n$ can be written $\sum_{i=1}^na_id_ib_i$. In particular, for each $i$, $b_i$ can be written in the form $\sum_{i=1}^na_id_ib_i$. As $\{b_1,\ldots,b_n\}$ is a basis, the uniqueness of this representation means $a_id_i=1$. The only integer values for which this is possible are $a_i=d_i=\pm1$.

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    when you say "for each $i$, $b_{i}$ can be written in the form $\sum_{i=1}^{n}a_{i}d_{i}b_{i}$", that sounds a little weird. Did you mean $\exists x_{b}\in A_{n}$ where $x_{b}=\sum_{i=1}^{n}b_{i}$? What you said doesn't make sense because it implies a linear dependency and by definition, a basis is not supposed to have that. – 2017-01-23
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    see comment above. – 2017-01-24
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    No, I mean exactly what I said. $b_i\in A_n$ and the $d_jb_j$ span $A_n$ so there exists integers $a_j$ with $b_i=\sum_{j=1}^na_jd_jb_j$. This doesn't contradict linear dependence, rather we then use linear dependence with $(\sum_{j=1}^na_jd_jb_j)-b_i=0$ or $(a_id_i-1)b_i=\sum_{j\ne i}a_jd_jb_j$ (depending on how you have defined linear dependence, they're both equivalent) to imply $a_id_i=1$. – 2017-01-24