Theorem

Proof:

Let $\vec{v}_1, \dots, \vec{v}_p$ be vectors in a vector space $V$. The span of these vectors, denoted by $\text{Span}\{\vec{v}_1, \dots, \vec{v}_p\}$, is the set of all linear combinations of $\vec{v}_1, \dots, \vec{v}_p$:

$\text{Span}\{\vec{v}_1, \dots, \vec{v}_p\} = \{ c_1 \vec{v}_1 + c_2 \vec{v}_2 + \dots + c_p \vec{v}_p \mid c_1, c_2, \dots, c_p \in \mathbb{R} \}.$

To show that $\text{Span}\{\vec{v}_1, \dots, \vec{v}_p\}$ is a subspace of $V$, we verify the three subspace properties:

1. The zero vector is in $\text{Span}\{\vec{v}_1, \dots, \vec{v}_p\}$:
Let $c_1 = c_2 = \dots = c_p = 0$. Then: $$c_1 \vec{v}_1 + c_2 \vec{v}_2 + \dots + c_p \vec{v}_p = \vec{0}.$$ Hence, $\vec{0} \in \text{Span}\{\vec{v}_1, \dots, \vec{v}_p\}$.

2. Closed under addition:
Let $\vec{u}, \vec{w} \in \text{Span}\{\vec{v}_1, \dots, \vec{v}_p\}$. Then: $$\vec{u} = a_1 \vec{v}_1 + a_2 \vec{v}_2 + \dots + a_p \vec{v}_p \quad \\ \quad \vec{w} = b_1 \vec{v}_1 + b_2 \vec{v}_2 + \dots + b_p \vec{v}_p,$$ where $a_1, a_2, \dots, a_p \in \mathbb{R}$ and $b_1, b_2, \dots, b_p \in \mathbb{R}$. Adding $\vec{u}$ and $\vec{w}$ gives: $$\vec{u} + \vec{w} = (a_1 + b_1)\vec{v}_1 + (a_2 + b_2)\vec{v}_2 + \dots + (a_p + b_p)\vec{v}_p.$$ Since $a_i + b_i \in \mathbb{R}$, $\vec{u} + \vec{w} \in \text{Span}\{\vec{v}_1, \dots, \vec{v}_p\}$.

3. Closed under scalar multiplication:
Let $\vec{u} \in \text{Span}\{\vec{v}_1, \dots, \vec{v}_p\}$ and let $c \in \mathbb{R}$. Then: $$\vec{u} = a_1 \vec{v}_1 + a_2 \vec{v}_2 + \dots + a_p \vec{v}_p,$$ where $a_1, a_2, \dots, a_p \in \mathbb{R}$. Multiplying by $c$ gives: $$c \vec{u} = (c a_1)\vec{v}_1 + (c a_2)\vec{v}_2 + \dots + (c a_p)\vec{v}_p.$$ Since $c a_i \in \mathbb{R}$, $c \vec{u} \in \text{Span}\{\vec{v}_1, \dots, \vec{v}_p\}$.

Thus, $\text{Span}\{\vec{v}_1, \dots, \vec{v}_p\}$ satisfies the conditions for a subspace of $V$.