Math 375 Archives - Page 12 of 15 - Shawn Zhong

Math 375 – 9/27

$Theorem • Statement ○ Let W_1,W_2⊆V be subspace ○ W_1∪W_2 is a subspace ⇔W_1⊆W_2 or W_2⊆W_1 • Proof: W_1⊆W_2 or W_2⊆W_1⇒W_1∪W_2 is a subspace ○ Obvious • Proof: W_1∪W_2 is a subspace ⇒W_1⊆W_2 or W_2⊆W_1 ○ Suppose § ∃v_1∈W_1, s.t. v_1∉W_2 § ∃v_2∈W_2, s.t. v_2∉W_1 ○ Then § v_1+v_2∉W_1 ○ Indeed, if § v_1+v_2=w∈W_1 ○ Then § v_2=w−v_1∈W_1 § Contradiction ○ Likewise § v_1+v_2∉W_2 ○ Therefore § v_1+v_2∉W_1∪W_2 Question 1 • Let V be a vector space, ⟨⋅,⋅⟩ is an inner product on V • Prove ○ ∀ v,w∈V ○ ⟨u,v⟩=0⇔‖v+c⋅w‖≥‖v‖, ∀c∈R • Proof: ⟨u,v⟩=0⇒‖v+c⋅w‖≥‖v‖ ○ c^2 ‖w‖^2≥0 ○ ‖v‖^2+c^2 ‖w‖^2≥‖v‖^2 ○ ‖v‖^2+2c⟨u,v⟩+c^2 ‖w‖^2≥‖v‖^2 ○ ‖v+c⋅w‖^2≥‖v‖^2 ○ ‖v+c⋅w‖≥‖v‖ • Proof: ‖v+c⋅w‖≥‖v‖⇒⟨u,v⟩=0 ○ ‖v+c⋅w‖≥‖v‖ ○ ‖v‖^2+2c⟨u,v⟩+c^2 ‖w‖^2≥‖v‖^2 ○ In order for the inequality to be true for all c ○ ⟨u,v⟩=0 Question 2 • Let V be a finite-dimensional vector space • ⟨⋅,⋅⟩ is an inner product on V • Let W⊆V be a subspace • Define W^⊥={v∈V│⟨v,w⟩=0, ∀w∈W} • Prove that ○ W^⊥ is a subspace ○ dim⁡W+dim⁡〖W^⊥=dim⁡V 〗$

Math 375 – 9/26

$Inner Product • Definition (on real vector space) ○ An inner product on a real vector space V ○ is a real-valued function (x,y) with x,y∈V ○ for which: § (x+y,z)=(x,z)+(y,z), \ ∀x,y,z∈V § (tx,y)=t(x,y), ∀x,y∈V,and t∈R § (x,y)=(y,x), ∀x,y∈V § (x,x)≥0, ∀x∈V § (x,x)=0⇒x=0 • Definition (on complex vector space) ○ An inner product on a real vector space V ○ is a real-valued function (x,y) with x,y∈V ○ for which: § (x+y,z)=(x,z)+(y,z), \ ∀x,y,z∈V § (tx,y)=t(x,y), ∀x,y∈V,and t∈R § (x,y)=((y,x) ) ̅, ∀x,y∈V § (x,x)≥0, ∀x∈V § (x,x)=0⇒x=0 ○ Note: (x,ty)=((ty,x) ) ̅=t ̅(x,y) • Example in R2 ○ Let V=R2 ○ The following is an inner product for V § (x,y)=x_1 y_1+x_2 y_2+…+x_n y_n ○ Proof: (tx,y)=t(x,y) § (tx,y) § =(tx_1 ) y_1+(tx_2 ) y_2+…+(tx_n ) y_n § =t(x_1 y_1 )+t(x_2 y_2 )+…+t(x_n y_n ) § =t(x_1 y_1+x_2 y_2+…+x_n y_n ) § =t(x,y) • Example in ℂ^n ○ Let V=ℂ^n ○ The following is an inner product for V § (x,y)=x_1 (y_1 ) ̅+x_2 (y_2 ) ̅+…+x_n (y_n ) ̅ ○ Proof § (x+y,z)=(x,z)+(y,z) § (tx,y)=t(x,y) § (x,y)=((y,x) ) ̅ § (x,x)≥0 § (x,x)=0⇒x=0 • Counterexample in Rn ○ Let V=Rn ○ Whether the following is an inner product for V § (x,y)=x_1 y_1−x_2 y_2 ○ We need to check § (x+y,z)=(x,z)+(y,z) § (tx,y)=t(x,y) § (x,y)=((y,x) ) ̅ § (x,x)≥0 § (x,x)=0⇒x=0 • Counterexample in Rn ○ Let V=Rn ○ Whether the following is an inner product for V § (x,y)=x_1 y_1 ○ We need to check § (x+y,z)=(x,z)+(y,z) § (tx,y)=t(x,y) § (x,y)=((y,x) ) ̅ § (x,x)≥0 § (x,x)=0⇒x=0 • Example in Rn ○ Let V=Rn ○ The following is an inner product for V § (x,y)=(x_1+x_2 )(y_1+y_2 )+x_2 y_2 • Example in function space ○ V=C([a,b])={all continuous function on [a,b]} ○ The following is an inner product for V § (f,g)=∫_a^b▒f(x)g(x)dx, where a<b ○ We need to check § (f+g,h=(f,h+(g,h § (t⋅f,g)=t(f,g) § (f,g)=(g,f) § (f,f)≥0 § (f,f)=0⇒f=0 Length of Vector • Definition ○ √((x,x) )=‖x‖ is called the length of x ○ Note: (x,x)=‖x‖^2 • Cauchy Schwarz Inequality ○ (x,y)≤|x||y|, for all x,y∈V ○ Proof on page 16 Angle • Definition ○ If x,y∈V (x≠0,y≠0) ○ Then the angle between x,y is θ where ○ cos⁡θ=((x,y))/(‖x‖⋅‖y‖ ) • Note ○ Cauchy Schwarz Inequality implies ○ −1≤((x,y))/(‖x‖⋅‖y‖ )≤1 • Orthogonal ○ Vectors x,y are called orthogonal or perpendicular if ○ (x,y)=0 • Example ○ Given § V={all polynomials} § (f,g)=∫_0^1▒f(x)g(x)dx ○ Find the angle θ between f(x)=1 and g(x)=1 § ‖f‖=√(∫_0^1▒f(x)f(x)dx)=√(∫_0^1▒〖1^2 dx〗)=1 § ‖g‖=√(∫_0^1▒g(x)g(x)dx)=√(∫_0^1▒〖x^2 dx〗)=√3/3 § (f,g)=∫_0^1▒f(x)g(x)dx=∫_0^1▒xdx=1/2 § cos⁡θ=((x,y))/(‖x‖⋅‖y‖ )=√3/2 § ⇒θ=π/6$

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

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

Shawn Zhong

Shawn Zhong

Math 375

Math 375 – 9/27

Math 375 – 9/26

Math 375 – 9/25

Math 375 – Homework 3

Math 375 – 9/21

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