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Theorem

If v1,,vp \vec{v}_1, \dots, \vec{v}_p are in a vector space V V , then Span{v1,,vp} \{ \vec{v}_1, \dots, \vec{v}_p \} is a subspace of V V .

Proof:

Let v1,,vp \vec{v}_1, \dots, \vec{v}_p be vectors in a vector space V V . The span of these vectors, denoted by Span{v1,,vp} \text{Span}\{\vec{v}_1, \dots, \vec{v}_p\} , is the set of all linear combinations of v1,,vp \vec{v}_1, \dots, \vec{v}_p :

Span{v1,,vp}={c1v1+c2v2++cpvpc1,c2,,cpR}. \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 Span{v1,,vp} \text{Span}\{\vec{v}_1, \dots, \vec{v}_p\} is a subspace of V V , we verify the three subspace properties:

1. The zero vector is in Span{v1,,vp} \text{Span}\{\vec{v}_1, \dots, \vec{v}_p\} :
Let c1=c2==cp=0 c_1 = c_2 = \dots = c_p = 0 . Then: c1v1+c2v2++cpvp=0. c_1 \vec{v}_1 + c_2 \vec{v}_2 + \dots + c_p \vec{v}_p = \vec{0}. Hence, 0Span{v1,,vp} \vec{0} \in \text{Span}\{\vec{v}_1, \dots, \vec{v}_p\} .

2. Closed under addition:
Let u,wSpan{v1,,vp} \vec{u}, \vec{w} \in \text{Span}\{\vec{v}_1, \dots, \vec{v}_p\} . Then: u=a1v1+a2v2++apvpw=b1v1+b2v2++bpvp, \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 a1,a2,,apR a_1, a_2, \dots, a_p \in \mathbb{R} and b1,b2,,bpR b_1, b_2, \dots, b_p \in \mathbb{R} . Adding u \vec{u} and w \vec{w} gives: u+w=(a1+b1)v1+(a2+b2)v2++(ap+bp)vp. \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 ai+biR a_i + b_i \in \mathbb{R} , u+wSpan{v1,,vp} \vec{u} + \vec{w} \in \text{Span}\{\vec{v}_1, \dots, \vec{v}_p\} .

3. Closed under scalar multiplication:
Let uSpan{v1,,vp} \vec{u} \in \text{Span}\{\vec{v}_1, \dots, \vec{v}_p\} and let cR c \in \mathbb{R} . Then: u=a1v1+a2v2++apvp, \vec{u} = a_1 \vec{v}_1 + a_2 \vec{v}_2 + \dots + a_p \vec{v}_p, where a1,a2,,apR a_1, a_2, \dots, a_p \in \mathbb{R} . Multiplying by c c gives: cu=(ca1)v1+(ca2)v2++(cap)vp. c \vec{u} = (c a_1)\vec{v}_1 + (c a_2)\vec{v}_2 + \dots + (c a_p)\vec{v}_p. Since caiR c a_i \in \mathbb{R} , cuSpan{v1,,vp} c \vec{u} \in \text{Span}\{\vec{v}_1, \dots, \vec{v}_p\} .

Thus, Span{v1,,vp} \text{Span}\{\vec{v}_1, \dots, \vec{v}_p\} satisfies the conditions for a subspace of V V .