2017 - Page 17 of 43 - Shawn Zhong

Math 375 – 10/9

$Question • Let V be a finite-dimensional inner product space • S⊆V is a subspace of V • Let S^⊥={v∈V│∀s∈S,⟨v,s⟩=0} Prove (S^⊥ )^⊥=S Answer: First, (S⊥)⊥ is the orthogonal complement of S ⊥, which is itself the orthogonal complement of S, so (S⊥)⊥ = S means that S is the orthogonal complement of its orthogonal complement. To show that it is true, we want to show that S is contained in (S⊥)⊥ and, conversely, that (S⊥)⊥ is contained in S; if we can show both containments, then the only possible conclusion is that (S⊥)⊥ = S. To show the first containment, suppose v ∈ S and w ∈ S ⊥. Then hv, wi = 0 1 by the definition of S ⊥. Thus, S is certainly contained in (S⊥)⊥ (which consists of all vectors in R n which are orthogonal to S ⊥) To show the other containment, suppose v ∈ (S⊥)⊥ (meaning that v is orthogonal to all vectors in S ⊥) then we want to show that v ∈ S. I’m sure there must be a better way to see this, but here’s one that works. Let {u1, . . . , up} be a basis for S and let {w1, . . . , wq} be a basis for S ⊥. If v ∈/ S, then {u1, . . . , up, v} is a linearly independent set. Since each vector in that set is orthogonal to all of S ⊥, the set {u1, . . . , up, v, w1, . . . , wq} is linearly independent. Since there are p+q+1 vectors in this set, this means that p+q+1 ≤ n or, equivalently, p + q ≤ n − 1. On the other hand, if A is the matrix whose ith row is u T i , then the row space of A is S and the nullspace of A is S ⊥. Since S is p-dimensional, the rank of A is p, meaning that the dimension of nul(A) = S ⊥ is q = n − p. Therefore, p + q = p + (n − p) = n, contradicting the fact that p + q ≤ n − 1. From this contradiction, then, we see that, if v ∈ (S⊥)⊥, it must be the case that v ∈ S$

Math 375 – Homework 5

Math 375 – 10/5

$Best Approximation of Elements • Theorem ○ V: vector space with inner product ○ L⊆V: finite dimensional linear subspace ○ If x∈V then there exists excatly one z∈L ○ that minimizes the distance to x ○ i.e. ∀y∈L, ‖y−x‖≥‖z−x‖ and ○ If y≠z then ‖y−x‖>‖z−x‖ • Solution ○ L is finite dimensional therefore it has a basis ○ Gram-Schmidt says that we can assume the basis is orthonormal ○ i.e. L has a basis {e_1,e_2,…,e_n } where {■8((e_k,e_l )=0&k≠l@(e_k,e_k )=1&∀k)┤ ○ Then z is given by z=(x,e_1 ) e_1+(x,e_2 ) e_2+…+(x,e_n ) e_n ○ Since z is a linear combination of {e_1,…,e_n }, z∈L • Claim ○ x−z is perpendicular to all u∈L ○ i.e. if u∈L then u⊥x−z ○ i.e. (u,x−z)=0 ○ i.e. (u,x)=(u,z) • Proof: (u,x)=(u,z) ○ Let u∈L be given ○ Then {e_1,…e_n } is a basis for L ○ So for certain u_1,…,u_n∈R ○ Calculate (u,x) § (u,x)=(u_1 e_1+…+u_n e_n,x) § =u_1 (e_1,x)+…+u_n (e_n,x) ○ Calculate (u,z) § (u,z)=(u_1 e_1+…+u_n e_n,(x,e_1 ) e_1+…+(x,e_n ) e_n ) § =[u_1 (x_1,e_1 )(e_1,e_1 )+…+u_1 (x_1,e_n )(e_1,e_n )]+… +[u_n (x_1,e_1 )(e_n,e_1 )+…+u_n (x_n,e_n )(e_n,e_n )] § =u_1 (x,e_1 )+u_n (x,e_2 )+…+u_n (x,e_n ) ○ Therefore (u,x)=(u,z) ○ i.e. u⊥x−z, ∀u∈L • Proof: ∀y∈L, ‖y−x‖≥‖z−x‖ ○ Let y∈L be given ○ {█(y−x=(y−z)+(z−x)@y−z⊥z−x)┤ ○ ⇒‖y−x‖^2=‖y−z‖^2+‖z−x‖^2 ○ ⇒‖y−x‖^2≥‖z−x‖^2 ○ ⇒‖y−x‖≥‖z−x‖ ○ Also if y≠z then ‖y−x‖>‖z−x‖ Foorier Series • V={all continuous function f:[0,π]→R • (f,g)\=∫_0^π▒f(x)g(x)dx • Let f_n (x)=sin⁡(nx) • ⇒(f_n,f_m )=∫_0^π▒〖sin⁡(nx) sin⁡(mx)dx〗 x Z........_T as *U x 11×-21 My ex I TZ _ - _ - -- _ -- 80 s$

Math 375 – 10/4

$Proof Writing • Question ○ Let V be a vector space ○ Let x,y∈V such that {x,y} is independent ○ Prove that {2x+y,3x+2y} is independent • Proof ○ Let c_1,c_2∈R be arbitrary constant ○ c_1 (2x+y)+c_2 (3x+2y)=0 ○ (2c_1+3c_2 )x+(c_1+2c_2 )y=0 ○ Let {█(d_1=2c_1+3c_2@d_2=c_1+2c_2 )┤, d_1,d_2∈R ○ d_1 x+d_2 y=0 ○ Because {x,y} is independent ○ d_1=d_2=0 ○ {█(d_1=2c_1+3c_2=0@d_2=c_1+2c_2=0)┤⇒c_1=c_2=0 ○ Therefore {2x+y,3x+2y} is independent • Prompt ○ Exchange proofs with someone else. In a different color of pen or pencil, give them written feedback on their proof. ○ The main things to be looking for are: § Is the proof logically valid? § Is the proof understandable and clearly written? § Is the proof well-organized? ○ Here are some more questions it might be useful to ask (but don$

Math 375 – 10/2

Question 1 • Is S={(█(1@2@−1@0)),(█(−1@−2@−1@0)),(█(π@√2@−1@1/2)),(█(−3@2@2@1)),(█(1@2@0@3))} independent? • Claim ○ If S is linearly dependent • Proof ○ If S is linearly independent, then § dim⁡(span(S))=|S|=5 ○ But because span(S) is a subspace of R4 § dim⁡(span(S))≤dim⁡〖R4 〗=4 ○ So S is linearly dependent Question 2 • Prove ○ 1,sin⁡x,sin⁡2x is linearly independent • Claim ○ ∀ a,b,c∈R ○ if a+b⋅sin⁡x+sin⁡2x=0, ∀x∈[0,1] ○ then a=b=c=0 • Proof ○ Set x=0 ⇒a=0 ○ Set x=π/6 ⇒ 1/2 b+√3/2 c=0 ○ Set x=π/4 ⇒b=c=0 ○ Therefore a=b=c=0

Shawn Zhong

Shawn Zhong

Math 375 – 10/9

Math 375 – Homework 5

Math 375 – 10/5

Math 375 – 10/4

Math 375 – 10/2

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