Math 375 – Homework 4 Oct 26, 2017 Shawn Math 375 No comments yet Math 375 – 9/28 Oct 26, 2017 Shawn Math 375 No comments yet Math 375 – 9/27 Oct 26, 2017 Shawn Math 375 No comments yet Math 375 – 9/26 Oct 26, 2017 Shawn Math 375 No comments yet Math 375 – 9/25 Oct 26, 2017 Shawn Math 375 No comments yet 1 … 3 4 5 6 7 8
![Distance • Definition ○ Distance between two vectors x,y is defined as ○ distance(x,y)=‖x−y‖=√((x−y,x−y) ) • Example 1 ○ Given § V=R2 § (x,y)=x_1 y_1+x_2 y_2 ○ Distance between two vectors is § distance(x,y) § =‖x−y‖ § =√((x−y,x−y) ) § =√((x_1−y_1 )^2−(x_2−y_2 )^2 ) • Example 2 ○ Given § V={all continuous function f:[0,1]→R § (f,g)=∫_0^1▒f(x)g(x)dx ○ Distance between two functions is § distance(f,g) § =‖f−g‖ § =√((f−g,f−g) ) § =∫_0^1▒〖(f(x)−g(x))^2 dx〗 ○ Also known as root mean square distance Triangle Inequality (Version 1) • Statement ○ ‖a+b‖≤‖a‖+‖b‖ • Proof ○ ‖a+b‖≝(a+b,a+b) ○ =(a,a)+(a,b)+(b,a)+(b,b) ○ =(a,a)+2(a,b)+(b,b) ○ ≤‖a‖^2+2‖a‖‖b‖+‖b‖^2 ○ =(‖a‖+‖b‖)^2 ○ Therefore ‖a+b‖≤‖a‖+‖b‖ Triangle Inequality (Version 2) • Statement ○ distance(x,y)≤distance(x,z)+distance(z,y) • Proof ○ Let a=x−z, b=z−y ○ then a+b=x−y ○ ‖x−y‖≤‖x−z‖+‖z−y‖ ○ distance(x,y)≤distance(x,z)+distance(z,y) Orthogonal • Definition ○ {v_1,…,v_n } are orthogonal if (v_k,v_l )=0, ∀k≠l • Theorem ○ If {v_1,…,v_n } are orthogonal ○ and v_k≠0 for all k∈{1,2, …,n} ○ then {v_1,…,v_n } is linearly independent • Proof ○ Suppose § c_1 v_1+…+c_n v_n=0 ○ Then we have to show § c_1=c_2=…=c_n=0 ○ Let k∈{1,2, …,n}, then § (c_1 v_1+…+c_n v_n,v_k )=(0,v_k ) § c_1 (v_1,v_k )+…+c_k (v_k,v_k )+…+c_n (v_n,v_k )=0 ○ Because (v_k,v_l )=0, ∀k≠l, we have § 0+…+0+c_k (v_k,v_k )+0+…+0=0 § c_k (v_k,v_k )=0 ○ Because v_k≠0 § (v_k,v_k )≠0 § c_k=0/((v_k,v_k ) )=0 ○ Therefore § c_1=c_2=…=c_n=0 • Theorem ○ If x=c_1 v_1+…+c_n v_n ○ and {v_1,…,v_n } are non zero and orthogonal ○ then c_k=((x,v_k ))/((v_k,v_k ) ) • Proof ○ (x,v_k ) ○ =(c_1 v_1+…+c_n v_n,v_k ) ○ =c_1 (v_1,v_k )+…+c_k (v_k,v_k )+…+c_n (v_n,v_k ) ○ =0+…+0+c_k (v_k,v_k )+0+…+0 ○ =c_k (v_k,v_k ) ○ ⇒c_k=((x,v_k ))/((v_k,v_k ) ) Gramm-Schmidt Process • Introduction ○ If V has a basis {v_1,…,v_n } ○ then there is an orthogonal basis {w_1,…,w_n } ○ The process to find the orthogonal basis is called ○ Gramm-Schmidt Process • Process ○ w_1=v_1 ○ w_2=v_2−((v_2,w_1 ))/((w_1,w_1 ) ) w_1 ○ w_3=v_3−((v_3,w_1 ))/((w_1,w_1 ) ) w_1−((v_3,w_2 ))/((w_2,w_2 ) ) w_2 ○ ⋮ ○ w_k=v_k−∑_(i=0)^(k−1)▒〖((w_k,w_i ))/((w_i,w_i ) ) w_i 〗 • Proof: (w_3,w_2 )=0 ○ Assume we ve already shown (w_1,w_2 )=(w_1,w_3 )=0 ○ (w_3,w_2 ) ○ =(v_3,w_2 )−((v_3,w_1 ))/((w_1,w_1 ) )⋅(w_1,w_2 )−((v_3,w_1 ))/((w_1,w_1 ) )⋅(w_1,w_2 ) ○ =(v_3,w_2 )−(v_3,w_2 ) ○ =0 • Example 1 ○ Given § V=R2 § (x,y)=x_1 y_1+x_2 y_2 ○ Find the orthogonal basis for v_1=(█(1@1)),v_2=(█(1@2)) § w_1=v_1=(█(1@1)) § w_2=v_2−((v_2,w_1 ))/((w_1,w_1 ) ) w_1=(█(−1/2@1/2)) § {(█(1@1)),(█(−1∕2@1∕2))} • Example 2 ○ Given § V={all continous functions f:[0,1]→R § (f,g)=∫_0^1▒f(x)g(x)dx ○ Find the orthogonal basis for f_1 (x)=1, f_2 (x)=x § g_1 (x)=f_1 (x)=1 § g_2 (x)=f_2 (x)−((f_2,g_1 ))/((g_1,g_1 ) ) g_1 (x)=x−1/2 § {1,x−1/2}](https://shawnzhong.com/wp-content/uploads/2017/10/img_59f255d65119f.png)
![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 ○ cosθ=((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 § cosθ=((x,y))/(‖x‖⋅‖y‖ )=√3/2 § ⇒θ=π/6](https://shawnzhong.com/wp-content/uploads/2017/10/img_59f255bd04121.png)
