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 $\displaystyle x = \sum_{k=1}^n \lambda_k a_k$.

Necessary Condition
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 $\displaystyle \sum_{k=1}^n \lambda_k a_k = \sum_{k=1}^n \mu_k a_k$.

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.

Sufficient Condition
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 $G$.

Suppose $\displaystyle \sum_{k=1}^n \lambda_k a_k = 0$.

Then, since also $\displaystyle \sum_{k=1}^n 0 a_k = 0$, 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.