Span • L(S)={x∈V│■8(∃n∈N∃c_1,…,c_n∈R∃x_1,…,x_n∈S@x=c_1 x_1+…+c_n x_n )} Theorem • Statement ○ S⊆V is a subspace ⇔S=L(S) • Proof: S=L(S)⇒S⊆V is a subspace ○ Let s,t∈S, k∈R ○ Then s+k⋅t∈L(S) ○ L(S)=S⇒s+k⋅t∈S ○ ⇒S is closed under addition and scalar multiplication ○ Therefore S is a subspace of V • Proof: S⊆V is a subspace⇒S=L(S) ○ If T⊆V and T is a subspace, then L(S)⊆T ○ Setting T=S, we have L(S)⊆S ○ We also know that S⊆L(S) ○ So S=L(S) by definition of set equality Question 1 • Example of L(S∩T)≠L(S)∩L(T), where S,T⊆V ○ V=R2 ○ S={v_1,v_2 }, T={w_1,w_2 } ○ L(S∩T)=L(∅)={0} ○ L(S)=L(R)=R2 Question 2 • Let S_1,…,S_n be subsets of V • When is L(S_1 )∪…∪L(S_n ) a subspace? • L(S_1 )∪L(S_2 ) is a subspace ⇔L(S_1 )⊆L(S_2 ) or L(S_2 )⊆L(S_1 ) .02 ore I U. • Tu,
