2017 - Page 19 of 43 - Shawn Zhong

Math 375 – 9/25

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

Math 375 – Homework 3

Math 375 – 9/21

$Theorem 1.5 Theorem 1.6 • Statement ○ If {v_1…v_n} and {w_1,…,w_m} are bases for V, then n=m • Proof ○ Suppose n<m ○ w_1,…,w_m,w_(m+1)∈span{v_1…v_n } ○ ⇒{w_1,…,w_n,w_(n+1) } are linearly dependent by previous therom ○ ⇒{w_1,…,w_n,w_(n+1),…,w_m } are also linearly dependent ○ But {w_1,…,w_m} is linearly independet, because it a basis for V ○ So n<m is not true ○ Similarly the assumption n>m also leads to contradiction ○ Therefore n=m • Example ○ Given § f(x)=1+2x+x^2 § g(x)=x^2−4 § h(x)=2x−x^2 § k(x)=x−3 ○ Claim § There exist c_1,c_2,c_3,c_4∈R § such that c_1 f(x)+c_2 g(x)+c_3 h(x)+c_4 k(x)=0 § And at least one of c_1,c_2,c_3,c_4 is not 0 § V ={all polynomials of degree≤2} has basis {1,x,x^2} § f,g,h,k∈span{1,x,x^2 } § ⇒ f,g,h,k are linearly dependent Theorem • Statement ○ If V is a n-dimensional vector space ○ And v_1,…,v_m∈V are linearly independence with m<n ○ Then there exist v_(m+1),…,v_n∈V ○ Such that {v_1,…,v_n} is a basis for V • Outline of proof ○ span{v_1,…,v_m }≠V by the previous theorem ○ Choose v_(m+1)∈V such that v_m∉span{v_1,…,v_m } ○ Then {v_1,…,v_m,v_(m+1) } is also linearly independent ○ If m+1=n, then {v_1,…,v_m,v_(m+1) } is a basis for V ○ Or m+1<n, then repeat the previous steps$

Math 375 – 9/20

$Question 1 • Why is span(∅)={0}? • 0 is the additive identity Question 2 • The basis of V={f∈P_n│f(0)+f^′ (0)=0}? • f(x)=a_0+a_1 x+a_2 x^2+…+a_n x^n • f(0)=a_0, f^′ (0)=a_1 • f(0)+f^′ (0)=0 • ⇒a_0=−a_1 • f(x)=a_1 (x−1)+a_2 x^2+…+a_n x^n • Therefore the basis of V is {x−1,x^2,…,x^n}$

Math 375 – 9/19

Shawn Zhong

Shawn Zhong

Math 375 – 9/25

Math 375 – Homework 3

Math 375 – 9/21

Math 375 – 9/20

Math 375 – 9/19

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