Linear Combination of Sequence is Linear Combination of Set
Theorem
Let $G$ be an $R$-module.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of elements of $G$.
Let $b$ be an element of $G$.
Then:
- $b$ is a linear combination of the sequence $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$
- $b$ is a linear combination of the set $\set {a_k: 1 \mathop \le k \mathop \le n}$
Proof
Necessary Condition
By definition of linear combination of subset:
- Every linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is a linear combination of $\set {a_k: 1 \mathop \le k \mathop \le n}$.
$\Box$
Sufficient Condition
Let $b$ be a linear combination of $\set {a_k: 1 \mathop \le k \mathop \le n} = \set {a_1, a_2, \ldots, a_n}$.
Then there exists:
- a sequence $\sequence {c_j}_{1 \mathop \le j \mathop \le m}$ of elements of $\set {a_1, a_2, \ldots, a_n}$
and:
- $\ds b = \sum_{j \mathop = 1}^m \mu_j c_j$
For each $k \in \closedint 1 n$, let $\lambda_k$ be defined as follows.
If:
- $a_k \in \set {c_1, c_2, \ldots, c_m}$
and:
- $a_i \ne a_j$ for all indices $i$ such that $1 \le i < k$
let $\lambda_k$ be the sum of all scalars $\mu_j$ such that $c_j = a_k$.
If:
- $a_k \notin \set {c_1, c_2, \ldots, c_m}$
or:
- $a_i = a_j$ for some index $i$ such that $1 \le i < k$
let $\lambda_k = 0$.
It follows that:
- $\ds b = \sum_{j \mathop = 1}^m \mu_j c_j = \sum_{k \mathop = 1}^n \lambda_k a_k$
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ and $\sequence {b_j}_{1 \mathop \le j \mathop \le m}$ be sequences of elements of $G$ such that $\set {a_1, a_2, \ldots, a_n}$ and $\set {b_1, b_2, \ldots, b_m}$ are identical.
Then as a consequence of the above:
- an element is a linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$
- it is a linear combination of $\set {a_k: 1 \mathop \le k \mathop \le n}$
$\blacksquare$
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Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases