Theorem. For any set of vectors \(\displaystyle{S}={\left\lbrace{v}{1},…,{v}{n}\right\rbrace}\) in a vector space V, span (S) is a subspace of V.

Proof. Let \(\displaystyle{u},{w}∈{s}{p}{a}{n}{\left({S}\right)},{k}∈{F}.\) Then there exist \(\displaystyle{c}{1},…,{c}{n}∈{R},…,{c}{n}∈{R}{\quad\text{and}\quad}{k}{1},…,{k}{n}∈{F}{k}∈{F}\) such that

\(\displaystyle{u}={c}{1}{v}{1}+{c}{2}{v}{2}+\ldots+{c}{n}{v}{n}\)

and

\(\displaystyle{w}={k}{1}{v}{1}+{k}{2}{v}{2}+\ldots+{k}{n}{v}{n}\)

Note that

\(\displaystyle{u}+{w}={\left({c}{1}{v}{1}+{c}{2}{v}{2}+\ldots+{c}{n}{v}{n}\right)}+{\left({k}{1}{v}{1}+{k}{2}{v}{2}+\ldots+{k}{n}{v}{n}\right)}={\left({c}{1}+{k}{1}\right)}{v}{1}+{\left({c}{2}+{k}{2}\right)}{v}{2}+\ldots+{\left({c}{n}+{k}{n}\right)}{v}{n}∈{s}{p}{a}{n}{\left({S}\right)}\)

Thus span (S) is closed under addition. M1)

\(ku=k(c_1v_1+c_2v_2+\dots+c_nv_n) =k(c_1v_1)+k(c_2v_2)+\dots+k(c_nv_n) =(kc_1)v_1+(kc_2)v_2+\dots+(kc_n)v_n \in span(S)\)

This hosws that span (S) is closed under scalar multiplication. Hence , span (S) is a subspace of V.