Shawn Zhong

Shawn Zhong

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Shawn Zhong

钟万祥
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Home / 2017 / October / Page 6

Math 375 – Homework 3

  • Oct 26, 2017
  • Shawn
  • Math 375
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Math 375 – 9/21

  • Oct 26, 2017
  • Shawn
  • Math 375
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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
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Math 375 – 9/20

  • Oct 26, 2017
  • Shawn
  • Math 375
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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}
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Math 375 – 9/19

  • Oct 26, 2017
  • Shawn
  • Math 375
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Math 375 – 9/18

  • Oct 26, 2017
  • Shawn
  • Math 375
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Question 1 • Let V be a vector space,S⊆T⊆V be subsets • Prove or disprove: • S independence⇒T independence ○ False ○ Counterexample 1 § V={0} § T={0} § S=∅ ○ Counterexample 2 § V=R2 § T={(0,1),(1,0),(1,1)} § S={(0,1),(1,0)} • T independence⇒S independence ○ True • span(S)=V⇒span(T)=V ○ True • span(T)=V⇒span(S)=V ○ False ○ Counterexample § V=R3 § T={(1,0,0),(0,1,0),(0,0,1)} § S={(1,0,0)} Question 2 • For which functions f:R→R is {f,f^′ } linear dependent • f(x)=〖Ae〗^tx where k≠1 and A≠0
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