Linear Combination of Sequence is Linear Combination of Set

Theorem
Let $G$ be an $R$-module.

Let $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ be a sequence of elements of $G$.

Let $b$ be an element of $G$.

Then:
 * $b$ is a linear combination of the sequence $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$


 * $b$ is a linear combination of the set $\left \{{a_k: 1 \mathop \le k \mathop \le n} \right\}$
 * $b$ is a linear combination of the set $\left \{{a_k: 1 \mathop \le k \mathop \le n} \right\}$

Necessary Condition
By definition of linear combination of subset:


 * Every linear combination of $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ is a linear combination of $\left \{{a_k: 1 \mathop \le k \mathop \le n} \right\}$.

Sufficient Condition
Let $b$ be a linear combination of $\left \{{a_k: 1 \mathop \le k \mathop \le n} \right\} = \left\{{a_1, a_2, \ldots, a_n}\right\}$.

Then there exists:
 * a sequence $\left \langle {c_j} \right \rangle_{1 \mathop \le j \mathop \le m}$ of elements of $\left\{{a_1, a_2, \ldots, a_n}\right\}$

and:
 * a sequence $\left \langle {\mu_j} \right \rangle_{1 \mathop \le j \mathop \le m}$ of scalars such that:


 * $\displaystyle b = \sum_{j \mathop = 1}^m \mu_j c_j$

For each $k \in \left[{1 \,.\,.\, n}\right]$, let $\lambda_k$ be defined as follows.

If:
 * $a_k \in \left\{{c_1, c_2, \ldots, c_m}\right\}$

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:
 * $a_k \notin \left\{{c_1, c_2, \ldots, c_m}\right\}$

or:
 * $a_i = a_j$ for some index $i$ such that $1 \le i < k$

let $\lambda_k = 0$.

It follows that:
 * $\displaystyle b = \sum_{j \mathop = 1}^m \mu_j c_j = \sum_{k \mathop = 1}^n \lambda_k a_k$

Let $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ and $\left \langle {b_j} \right \rangle_{1 \mathop \le j \mathop \le m}$ be sequences of elements of $G$ such that $\left\{{a_1, a_2, \ldots, a_n}\right\}$ and $\left\{{b_1, b_2, \ldots, b_m}\right\}$ are identical.

Then as a consequence of the above:
 * an element is a linear combination of $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$


 * it is a linear combination of $\left \{{a_k: 1 \mathop \le k \mathop \le n} \right\}$
 * it is a linear combination of $\left \{{a_k: 1 \mathop \le k \mathop \le n} \right\}$