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

Proof

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

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{k \mathop = 1}^n \left({\lambda_k - \mu_k}\right) a_k\)	\(=\)	\(\displaystyle \sum_{k \mathop = 1}^n \left({\lambda_k a_k - \mu_k a_k}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k \mathop = 1}^n \lambda_k a_k - \sum_{k=1}^n \mu_k a_k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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.

$\Box$

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

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

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 27$: Theorem $27.4$