Math 375 - 9/28

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}