Unique Representation by Ordered Basis

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Theorem

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


Then $\sequence {a_n}$ is an ordered basis of $G$ if and only if:

For every $x \in G$ there exists one and only one sequence $\sequence {\lambda_n}$ of scalars such that $\displaystyle x = \sum_{k \mathop = 1}^n \lambda_k a_k$.


Proof

Necessary Condition

Let $\sequence {a_n}$ be an ordered basis of $G$.

Then every element of $G$ is a linear combination of $\set {a_1, \ldots, a_n}$, 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: $\sequence {\lambda_n}$ and $\sequence {\mu_n}$.

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

Then:

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


So $\lambda_k = \mu_k$ for all $k \in \closedint 1 n$ as $\sequence {a_n}$ is a linearly independent sequence.

$\Box$


Sufficient Condition

Now suppose there is one and only one sequence $\sequence {\lambda_n}$ such that the condition holds.

It is clear that $\set {a_1, \ldots, a_n}$ 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:

$\forall k \in \closedint 1 n: \lambda_k = 0$


Therefore $\sequence {a_n}$ is a linearly independent sequence.

$\blacksquare$


Also see


Sources