Answer :
Answer:
See proof below
Step-by-step explanation:
Assume that V is a vector space over the field F (take F=R,C if you prefer).
Let [tex]x\in span(S_1)[/tex]. Then, we can write x as a linear combination of elements of s1, that is, there exist [tex]v_1,v_2,\cdots,v_k \in S_1[/tex] and [tex]a_1,a_2,\cdots,a_k\in F[/tex] such that [tex]x=a_1v_1+a_2v_2+\cdots+a_kv_k[/tex]. Now, [tex]S_1\subseteq S_2[/tex] then for all [tex]y\in S_1[/tex] we have that [tex]y\in S_2[/tex]. In particular, taking [tex]y=v_j[/tex] with [tex]j=1,2,\cdots,k[/tex] we have that [tex]v_j\in S_2[/tex]. Then, x is a linear combination of vectors in S2, therefore [tex]x\in span(S_2)[/tex]. We conclude that [tex]span(S_1)\subseteq span(S_2)[/tex].
If, additionally [tex]S_2\subseteq S_1[/tex] then reversing the roles of S1 and S2 in the previous proof, [tex]span(S_2)\subseteq span(S_1)[/tex]. Then [tex]span(S_1)\subseteq span(S_2)\subseteq span(S_1)[/tex], therefore [tex]span(S_1)=span(S_2)[/tex].