If v1,…,vp are in a vector space V, then Span{v1,…,vp} is a subspace of V.
Proof:
Let v1,…,vp be vectors in a vector space V. The span of these vectors, denoted by Span{v1,…,vp}, is the set of all linear combinations of v1,…,vp:
To show that Span{v1,…,vp} is a subspace of V, we verify the three subspace properties:
1. The zero vector is in Span{v1,…,vp}:
Let c1=c2=⋯=cp=0. Then:
c1v1+c2v2+⋯+cpvp=0.
Hence, 0∈Span{v1,…,vp}.
2. Closed under addition:
Let u,w∈Span{v1,…,vp}. Then:
u=a1v1+a2v2+⋯+apvpw=b1v1+b2v2+⋯+bpvp,
where a1,a2,…,ap∈R and b1,b2,…,bp∈R. Adding u and w gives:
u+w=(a1+b1)v1+(a2+b2)v2+⋯+(ap+bp)vp.
Since ai+bi∈R, u+w∈Span{v1,…,vp}.
3. Closed under scalar multiplication:
Let u∈Span{v1,…,vp} and let c∈R. Then:
u=a1v1+a2v2+⋯+apvp,
where a1,a2,…,ap∈R. Multiplying by c gives:
cu=(ca1)v1+(ca2)v2+⋯+(cap)vp.
Since cai∈R, cu∈Span{v1,…,vp}.
Thus, Span{v1,…,vp} satisfies the conditions for a subspace of V.