Unique Representation by Ordered Basis
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 $\ds x = \sum_{k \mathop = 1}^n \lambda_k a_k$.
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 $\ds \sum_{k \mathop = 1}^n \lambda_k a_k = \sum_{k \mathop = 1}^n \mu_k a_k$.
Then:
\(\ds \sum_{k \mathop = 1}^n \paren {\lambda_k - \mu_k} a_k\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \paren {\lambda_k a_k - \mu_k a_k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \lambda_k a_k - \sum_{k \mathop = 1}^n \mu_k a_k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
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 $\ds \sum_{k \mathop = 1}^n \lambda_k a_k = 0$.
Then, since also $\ds \sum_{k \mathop = 1}^n 0 a_k = 0$, 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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.4$