October 26, 2017 - Page 2 of 6 - Shawn Zhong

Math 375 – 10/11

$Question • Given ○ V=C([−1,1]) ○ ⟨v,w⟩=∫_(−1)^1▒v(x)w(x)dx • Find the linear polynomial closest to f(x)=e^x • Answer ○ Let S=span{1,x} ○ Projection of f onto S is ○ ⟨1,e^x ⟩/⟨1,1⟩ ⋅1+⟨x,e^x ⟩/⟨x,x⟩ ⋅x ○ Therefore the linear polynomial closest to f(x)=e^x is ○ g(x)=3/e x+(e−e^(−1))/2$

Math 375 – 10/10

$Linear Transformations • Definition ○ Let V and W be two vector spaces ○ Then a map/function/transformation/mapping ○ T:V→W is called linear if ○ {■8(T(x+y)=T(x)+T(y)&∀x,y∈V@T(c⋅x)=c⋅T(x)&∀x∈V,c∈R┤ • Mapping notation ○ In the mapping T:V→W ○ V is called domain ○ W is called codomain or target set ○ T(v) must be defined ∀v∈V ○ T(v) always belongs to W • Example 1 ○ Let V,W be any vector space ○ Define Tx=0, ∀x∈V ○ {█(T(x+y)=0@T(x)+T(y)=0+0=0)┤⇒T(x+y)=Tx+Ty ○ {█(T(c⋅x)=0@c⋅T(x)=c⋅0=0)┤⇒T(c⋅x)=c⋅T(x) ○ Therefore this mapping is a linear transformation • Example 2 ○ Let V,W be any vector space ○ Define Tv=w≠0, ∀v∈V ○ T(x)+T(y)=2w≠w=T(x+y) ○ Therefore this mapping is not a linear transformation • Example 3 ○ Let V=W be the same vector space ○ Define Tx=x, ∀v∈V ○ Then T is a linear transformation ○ T is called the identity map from V to V ○ Common notations: id, id_V, 1_V • Example 4 ○ Let V=W=R2 be the same vector space ○ Define T(x,y)=(2x,2y) ○ T(u)+T(v)=2u+2v=2(u+v)=T(u+v) ○ T(c⋅u)=2c⋅u=c⋅(2u)=c⋅T(u) ○ Therefore T is a linear transformation • Example 5 ○ Let V=W=R2 be the same vector space ○ Define T(a,b)=(b,a) ○ It s reflection in the diagonal • Example 6 ○ Let V=W=R2 be the same vector space ○ Define Tu=u rotated by 30° counter-clockwise ○ Proof by graph T(u+v)=T(u)+T(v) ○ We can also prove that T(c⋅v)=c⋅T(v) ○ Therefore T is a linear transformation Linear Transformation on Basis • Theorem ○ Suppose T:V→W is a linear transformation ○ Let {e_1,…,e_n } be a basis for V ○ Then T is completely defermined by ○ {Te_1,,Te_2…,Te_n } ○ Suppose we known Te_1,Te_2,…,Te_n, ○ and let x∈V be given ○ Then there are c_1,c_2,…,c_n∈R ○ such that x=c_1 e_1+c_2 e_2+…+c_n e_n, then ○ T(x)=T(c_1 e_1+c_2 e_2+…+c_n e_n ) ○ =T(c_1 e_1 )+T(c_2 e_2 )+…+T(c_n e_n ) ○ =c_1 Te_1+c_2 Te_2+…c_n Te_n • Example (Rotation) ○ Let V=W=R2 be the same vector space ○ Define T rotate by θ counter-clockwise ○ Pick a basis {e_1,e_2 }, where § e_1=(█(1@0)) § e_2=(█(0@1)) ○ Compute Te_1, Te_2 § Te_1=(█(cos⁡θ@sin⁡θ )) § Te_2=(█(−sin⁡θ@cos⁡θ )) ○ Compute T(ae_1+be_2 ) § T(ae_1+be_2 ) § =aTe_1+bTe_2 § =a(█(cos⁡θ@sin⁡θ ))+b(█(−sin⁡θ@cos⁡θ )) § =(█(a cos⁡θ−b sin⁡θ@a sin⁡θ+b cos⁡θ )) Solving System of Equations • Setup ○ {█(a_11 x_1+a_12 x_2+…+a_1n x_n=y_1@a_21 x_1+a_22 x_2+…+a_2n x_n=y_2@⋮@a_n1 x_1+a_n2 x_2+…+a_nn x_n=y_n )┤ ○ Define a transformation T: Rn→Rn ○ Let x=(█(x_1@⋮@x_n )), y=(█(y_1@⋮@y_n )) ○ Then Tx=y is a linear transformation • Property of one-to-one map ○ A linear map T:V→W is a one-to-one map ○ if for all u,v∈V ○ Tu=Tv⇒u=v ○ i.e. The equation Tx=y has at most one solution • Example of one-to-one map ○ Let V=R2, W=R3 ○ T(x_1,x_2 )=(x_1,x_2,0)=(y_1,y_2,y_3 ) ○ {█(1x_1+0x_2=y_1@0x_1+1x_2=y_2@0x_1+0x_2=y_3 )┤⇒{█(y_1=x_1@y_2=x_2@y_3=0)┤ ○ Three equations, two unknowns • Theorem ○ A linear map T:V→W is injective ○ if for all x∈V ○ Tx=0⇒x=0$

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

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

Shawn Zhong

Shawn Zhong

Math 375 – 10/11

Math 375 – 10/10

Math 375 – 10/9

Math 375 – Homework 5

Math 375 – 10/5

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