Unique Representation by Ordered Basis

Theorem
Let $$G$$ be a unitary $R$-module.

Then $$\left \langle {a_n} \right \rangle$$ is an ordered basis of $$G$$ iff for every $$x \in G$$ there exists one and only one sequence $$\left \langle {\lambda_n} \right \rangle$$ of scalars such that $$x = \sum_{k=1}^n \lambda_k a_k$$.

Proof

 * Let $$\left \langle {a_n} \right \rangle$$ be an ordered basis of $$G$$.

Then every element of $$G$$ is a linear combination of $$\left\{{a_1, \ldots, a_n}\right\}$$, which is a generator of $$G$$, by Generated Submodule is Linear Combinations.

Thus there exists at least one such sequence of scalar.

Now suppose there were two such sequences of scalars: $$\left \langle {\lambda_n} \right \rangle$$ and $$\left \langle {\mu_n} \right \rangle$$.

That is, suppose $$\sum_{k=1}^n \lambda_k a_k = \sum_{k=1}^n \mu_k a_k$$.

Then:

$$ $$ $$

So $$\lambda_k = \mu_k$$ for all $$k \in \left[{1 \,. \, . \, n}\right]$$ as $$\left \langle {a_n} \right \rangle$$ is a linearly independent sequence.


 * Now suppose there is one and only one sequence $$\left \langle {\lambda_n} \right \rangle$$ such that the condition holds.

It is clear that $$\left\{{a_1, \ldots, a_n}\right\}$$ generates $$G$$.

If $$\sum_{k=1}^n \lambda_k a_k = 0$$, then, since also $$\sum_{k=1}^n 0 a_k = 0$$, we have, by hypothesis, $$\forall k \in \left[{1 \,. \, . \, n}\right]: \lambda_k = 0$$.

Therefore $$\left \langle {a_n} \right \rangle$$ is a linearly independent sequence.