The Subspace Span is the Set of all Linear Combinations

Last updated Nov 1, 2022

Let $V$ be a Vector Space on Field $F$. Let $S \subset V$. Then $${c_{1} \mathbf{x}{1} + \cdots + c{n} \mathbf{x}{n} \in V : n \in \mathbb{Z}{\geq 0}; \mathbf{x}{1}, \dots, \mathbf{x}{n} \in S; c_{1}, \dots, c_{n} \in F} = \text{span} S$$

Let $W = {c_{1} \mathbf{x}{1} + \cdots + c{n} \mathbf{x}{n} \in V : n \in \mathbb{Z}{\geq 0}; \mathbf{x}{1}, \dots, \mathbf{x}{n} \in S; c_{1}, \dots, c_{n} \in F}$.

First recall that the Subspace Span is a Vector Subspace. For any $n \in \mathbb{N}$, $\mathbf{x}{1}, \dots, \mathbf{x}{n} \in S$, $c_{1}, \dots, c_{n} \in F$, we have that $\mathbf{x}{1}, \dots, \mathbf{x}{n} \in \text{span}S$ since $S \subset \text{span}S$. Since A Subset of a Vector Space is a Subspace iff it contains all its Linear Combinations, we see that $c_{1} \mathbf{x}{1} + \cdots + c{n} \mathbf{x}_{n} \in \text{span} S$ and $W \subset \text{span} S$. $\checkmark$

On the other hand, we will show that $W$ is a Vector Subspace, in which case $W \supset S$ ($1 * \mathbf{x}$ is a Linear Combination for $\mathbf{x} \in S$). Thus, by definition of Subspace Span, $\text{span} S \subset W$. Together with above, we get $\text{span} S = W$. To that end, let $\mathbf{u}, \mathbf{v} \in W$ and let $a \in F$. Then there exist $n, m \in \mathbb{Z}{\geq 0}$, $\mathbf{x}{1}, \dots, \mathbf{x}{n}, \mathbf{y}{1}, \dots, \mathbf{y}{m} \in S; c{1}, \dots, c_{n}, d_{1}, \dots, d_{m} \in F$ so that $$\begin{align*} a \mathbf{u} + \mathbf{v} &= a \sum\limits_{i=1}^{n} c_{i} \mathbf{x}{i} + \sum\limits{j=1}^{m} d_{j} \mathbf{y}{j}\\ &=ac{1}\mathbf{x}{1} + \dots + ac{n}\mathbf{x}{n} + d{1} \mathbf{y}{1} + \dots + d{m} \mathbf{y}_{m}\\ &\in W. \end{align*}$$ A Subset of a Vector Space is a Subspace iff it is closed under scaling and addition, so $W$ is a Vector Subspace, completing the proof. $\checkmark$ $\blacksquare$

1. This works as a general alternate definition to Subspace Span.