Running head: ASSIGNMENT 1 Linear Algebra Student Name Institution ASSIGNMENT 2 Question 1 Question 1(a) Let B= {𝑥 2 , x,1} et S = {𝑥 2 +x,2x-1,x+1} be two basis of 𝑃2 Let 𝑖𝐵 and 𝑖𝑠 be the coordinate map induced on 𝑃2 B= {𝑥 2 , x,1} T: 𝑃2 ⟶ 𝑃2 S= S = {𝑥 2 +x,2x-1,x+1} B and S are two basis of 𝑃2 1 2 [𝑇]𝐵,𝑆 = [−1 3 2 2 0 5 ] = {x𝜖 𝑅 3 | 𝑇𝑥 = 0} −2 𝑥1 1 2 0 0 [−1 3 5 ] [𝑥2 ] = [0] 2 2 −2 𝑥3 0 𝑥1 - 2𝑥2 = 0 𝑥1 = 2𝑥2 ……………………. Equation (1) -𝑥1 +3𝑥2 +5𝑥3 = 0 ……………………… Equation (2) 2𝑥1 + 2𝑥2 -2𝑥3 =0 ……………………….. Equation (3) From Equation (1) and Equation (2) -2𝑥2 +3𝑥2 +5𝑥3 = 0 𝑥2 +5𝑥3 = 0 From Equation (1) and Equation (3) 4𝑥2 +2𝑥2 -2𝑥3 = 0 6𝑥2 -2𝑥3 = 0 3𝑥2 -𝑥3 = 0 𝑥 Therefore, 𝑥3 = 3𝑥2 = 3( 22 ) ASSIGNMENT 𝑥3 = 𝑥2 = 3 3𝑥2 2 𝑥1 2 1 3 ⟶ 𝑥1 (1, 2 , 2 ) 1 3 Null ([𝑇]𝐵,𝜃 ) = {c(1, 2 , 2 ) ; c𝜖ℝ } 1 1 Null[𝑇𝐵𝑆 ] = Span{(2 )} 3 2 To find 𝐼𝐵 (𝑇) , 𝑥1 𝑥1 + 2𝑥2 1 2 0 𝑥 −𝑥 [−1 3 5 ] [ 2 ] = [ 1 + 3𝑥2 + 5𝑥3 ] 2 2 −2 3∗3 𝑥3 3∗1 2𝑥1 + 2𝑥2 − 2𝑥3 1 2 0 = 𝑥1 [−1] + 𝑥2 [3] + 𝑥3 [ 5 ] 2 2 −2 Thus 0 1 2 [ 5 ] = −2 [−1] + 1 [3] −2 2 2 1 2 1 2 = 𝑥1 [−1] + 𝑥2 [3] - 2𝑥3 [−1] + 𝑥3 [3] 2 2 2 2 1 2 = (𝑥1 -2𝑥3 ) [−1] + (𝑥2 +𝑥3 ) [3] 2 2 1 1 𝐼𝐵 (([𝑇])𝐵,𝑠 ) span {(−1 ) , (3 )} 2 2 Therefore, 1 2 0 (𝑇𝐵𝑆 )= Span{(−1 ) , (3 ) , (5 )} 2 2 2 ASSIGNMENT 4 Question 1(b) T:𝑀3∗3 ⟶ P2 𝑎 𝑑 T( 𝑔 𝑏 𝑒 ℎ 𝑐 𝑓 ) = (a-b)𝑥 2 +c(x-1)+d 𝑖 1 0 T(0 0 0 0 0 0) = 𝑥 2 0 0 1 T(0 0 0 0 0 0) = (-1) 𝑥 2 0 0 0 T(0 0 1 0 0 0) = 1(X-1) 0 0 0 T(0 0 0 0 0 0) = 1 1 Now, 𝑎 𝑑 T( 𝑔 𝑏 𝑒 ℎ 𝑐 𝑓 ) = 0, a=1, e=1,i=1 and b,c,d,f,g,h = 0 1 1 0 So the basis of Ker(T) = (0 1 0 0 0 0) 1 And the basis of Image of T (Im(T) ) = {1,X,𝑋 2 } Question 2 We have shall check the following inner product for <,∙,> ; let u,v,w 𝜖 v and 𝜆 is a scalar then; <𝑇(𝑢),𝑇(𝑣)> a) 1 =<𝑇(𝑣),𝑇(𝑤)> = ̅̅̅̅̅̅̅̅̅̅̅ < 𝑣, 𝑢 > [∵ <,> is an inner product on V] b) 1 = < 𝑇(𝑢 + 𝑣), 𝑇(𝑤) > ASSIGNMENT 5 = < 𝑇(𝑤) + 𝑇(𝑣), 𝑇(𝑤) > = < 𝑇(𝑤), 𝑇(𝑣) > +< 𝑇(𝑣), 𝑇(𝑤) > =1 =+ 1 c) <𝜆u+v>1 = < 𝑇(𝜆𝑢), 𝑇(𝑣) > = < 𝜆𝑇(𝑢), 𝑇(𝑣) > [∵ T is linear] = < 𝑇(𝑤), 𝑇(𝑣) > +< 𝑇(𝑣), 𝑇(𝑤) > =1 =+ 1 d) 1 = = 0 iff T(u) = 𝜃 where 𝜃is the null element of u= 𝜃 [∵ T is injective ] so, ≥ 0 for all v and 1 for all u𝜖 V and 1 = 0 iff u=0 hence <,>1 satisfies all axioms of inner product and it is a inner product on V If T is positive definite matrix, then T will give n inner product 1 = ∫0 (𝑓(𝑡) − 𝑓 ′ (𝑡))(𝑔(𝑡) − 𝑔′ (𝑡))dt 1 = ∫0 (𝑓(𝑡) − 𝑓 ′ (𝑡))2 dt If = 0 ⟹ 𝑓(𝑡) − 𝑓 ′ (𝑡) = 0 ⟹ 𝑓(𝑡) = 𝑓 ′ (𝑡) ⟹ 𝑓(𝑡) = 𝑒 𝑡 ∀ t and f 𝜖 𝑝𝑛 But 𝑒 𝑡 ∉ 𝑝𝑛 thus not an inner product since = 0 ≇ f = 0 hence false Question 3 Question 3(a) ∞ (f,g) = ∫0 𝑒 −𝑡 f(t)g(t) dt P(x) = 𝑥 3 ASSIGNMENT 6 The basis of 𝑃2 is {1,x,𝑥 2 } We need to determine the orthogonal basis of 𝑃2 Let 𝑣1 = 1 ………………….. Equation 1 𝑣2 = x ………………….. Equation 2 𝑣3 = 𝑥 3 ………………….. Equation 3 Therefore by Gram Schmidt orthogonalization process, we get 𝑤1 =1 …………………………………… Equation 4 𝑣1 = 1 ……………………………….. Equation 5 (𝑣 ,𝑤 ) 𝑤2 = 𝑣2 - (𝑤2 ,𝑤1 )…………………………..Equation 6 1 1 From equation 1 and 2, (𝑥,1) 𝑤2 = 𝑥- (1,1).1 𝑤2 = 𝑥- 1 ∞ =1 0 ∞ (x,1) = ∫0 𝑒 −𝑡 .t.1 dt = [−𝑡𝑒 −𝑡 − 𝑒 −𝑡 ] ∞ ∞ (1,1) = ∫0 𝑒 −𝑡 .1 dt = [−𝑒 −𝑡 ] = 1 0 (𝑣 ,𝑣 ) (𝑣 ,𝑤 ) 𝑤3 = 𝑣3 - (𝑤3,𝑤1 )* 𝑤1 - (𝑤3 ,𝑤2 ) 1 = 𝑥2 – (𝑥 2 ,1) (1,1) 1 - 2 (𝑥 2 ,(𝑥−1) 2 (x-1) (𝑥−1,𝑥−1) Therefore, ∞ (𝑥 2 , 1) = ∫0 𝑒 −𝑡 .𝑡 2 dt= 0+2 = 2 ∞ (𝑥 2 , 𝑥 − 1) = ∫0 𝑒 −𝑡 .𝑡 2 (𝑡 − 1) dt ∞ ∞ = ∫0 𝑡 3 𝑒 −𝑡 dt - ∫0 𝑡 2 𝑒 −𝑡 dt ASSIGNMENT 7 = 0+(3*2)-2 =4 ∞ (x-1,x-1) = ∫0 𝑒 −𝑡 .(𝑡 − 1) (𝑡 − 1) dt ∞ = ∫0 𝑒 −𝑡 .(𝑡 2 − 2𝑡 + 1) dt ∞ 0 = 2-2+[𝑒 −𝑡 ] =2-2+1 =1 Therefore, 2 4 𝑤3 = 𝑥 2 - 1- 1 (x-1) = 𝑥 2 - 2 − 4(𝑥 − 1) = 𝑥 2 -2-4x+4 = 𝑥 2 4x+2 Thus, orthogonal basis of 𝑃2 IS {1,x-1, 𝑥 2 4x+2} Orthogonal projection of P(x) = 𝑥 3 onto 𝑃2 = 𝑃𝑟𝑜𝑗𝑤1 P(x) 𝑃𝑟𝑜𝑗𝑤2 P(x)+ 𝑃𝑟𝑜𝑗𝑤3 P(x) = (𝑃(𝑥),𝑤1 ) (𝑤1 ,𝑤1 ) 𝑤1 + (𝑃(𝑥),𝑤2 ) (𝑤2 ,𝑤2 ) (𝑃(𝑥),𝑤3 ) 𝑤2+ (𝑤3 ,𝑤3 ) 𝑤3 (𝑃(𝑥), 𝑤1 ) = (𝑥 3 , 1) ∞ = ∫0 𝑒 −𝑡 .𝑡 3 dt = 0+(3*2) =6 (𝑃(𝑥), 𝑤2 ) = (𝑥 3 , x-1) ∞ = ∫0 𝑒 −𝑡 .𝑡 3 (𝑡 − 1) dt ∞ ∞ = ∫0 𝑒 −𝑡 𝑡 4 dt - ∫0 𝑒 −𝑡 𝑡 3 dt ASSIGNMENT 8 = 0+(24-6) = 18 (𝑃(𝑥), 𝑤3 ) = (𝑥 3 , 𝑥 2 − 4𝑥+2) ∞ = ∫0 𝑒 −𝑡 .𝑡 3 (𝑡 2 − 4𝑡 + 2) dt = (0 + 5 ∗ 18)-(4*18+12) = 0+(24-6) = 18+12 = 30 ∞ (𝑤3 , 𝑤3 ) = ∫0 𝑒 −𝑡 .(𝑡 2 − 4𝑡 + 2)( 𝑡 2 − 4𝑡 + 2)dt ∞ = ∫0 𝑒 −𝑡 .(𝑡 4 − 4𝑡 3 + 2𝑡 2 − 4𝑡 3 + +16𝑡 2 -8t +2𝑡 2 -8t +4t) dt ∞ = ∫0 𝑒 −𝑡 .(𝑡 4 − 4𝑡 3 + 2𝑡 2 − 4𝑡 3 + 16𝑡 2 -8t +2𝑡 2 -8t +4t) dt ∞ = ∫0 𝑒 −𝑡 .(𝑡 4 − 8𝑡 3 + 20𝑡 2 − 16𝑡 + 4) dt = 24-8*6+20*2-16*1+4 =4 Therefore, orthogonal projection of P(x) = 𝑥 3 onto 𝑃2 is 30 = 6*1+18𝑤2 + 4 𝑤3 = 6+18(x-1)+ 15 2 (𝑥 2 -4x+2) 15 = 6+18x-18+ 2 𝑥 2 -30x+15 = 15 2 𝑥 2 -12x+3 Question 3(b) ∈ = {f*(𝑃2 |f(0) = 0} Let a+bx+(𝑥 2 +a𝑥 3 ∈ E) ASSIGNMENT 9 ⟹ a+b(0) +c(0)2 +d(0)3 -E a+b(0) +c(0)2 +d(0)3 -E = 0 a=0 Therefore, the elements of the form +(𝑥 2 +a𝑥 3 ∈ E) and it is spanned by x, 𝑥 2 , 𝑥 3 . Therefore basis of E is (x, 𝑥 2 , 𝑥 3 ) . Question 4 Question 4 In order to prove 𝑢0 is unique in T(v) = < 𝑢0 , v> ∀ r 𝜖 v Let 𝑢0 is not unique ⟹ T(v) = < 𝑢0 , v> T(v) = < 𝑢1 , v> ⟹ < 𝑢0 , v> = < 𝑢1 , v> ⟹ < 𝑢0 -𝑢1 , 0> = 0 So by property of inner product ⟹ 𝑢0 -𝑢1 (𝑢0 =𝑢1 ) hence 𝑢0 is unique Now to find q 𝜖 𝑝2 ∀ P 𝜖 𝑃2 1 P(0) = ∫−1 𝑞(𝑡)p(t) dt Let q(t) = A+Bt+C𝑡 2 1 P(0) = ∫−1(A + Bt + C𝑡 2 )p(t) dt 1 Let P(t) = 1 ⟹ 𝑃2 (𝑥) ⟹ 1 = ∫−1(A + Bt + C𝑡 2 ) dt 2𝐶 1= 2A+ 3 ……………………………………………. Equation 1 1 P(t) = t ⟹ 0 = ∫−1(At + Bt 2 + C𝑡 3 )dt 2𝐵 P(0) = 0 ⟹ 0 = 0+ 3 +0 ⟹ B=0 ……………………. Equation 2 Let P(t) = 𝑡 2 ⟹ 𝑃2 ASSIGNMENT 10 P(0) = 0 1 0 = ∫−1(At 2 + Bt 3 + C𝑡 4 )dt t3 ′ t5 ′ 0 = 2A( 3 ) +0 +2C( 5 ) 0= 2𝐴 2𝐶 3 + 3 = 0 ⟹ 5A = -3C ………………. Equation 3 From equation 1, 2𝐶 1=2𝐴+ 3 −8 ⟹ 1= 15 𝐶 C= −15 8 9 A= 8 Therefore, 9 q(t) = + 0.t8 9 q(t) = 8 - 15 2 𝑡 8 15 2 𝑡 8 Question 5 Question 5(a) 3 2 Given, A= [2 3 2 2 2 2] 3 3−𝑥 The characteristic equation of A is, | 2 2 2 3−𝑥 2 2 2 | 3 − 𝑥 𝑥=0 i.e (3-x)*[(3 − 𝑥)2 − 4] – 2*[6-2x-4] +2*[4-+2x] = 0 (3-x)*[5 − 6𝑥 + 𝑥 2 ] + 8x -8 =0 ASSIGNMENT 11 𝑥 3 - 9𝑥 2 +15x -7 =0 (𝑥 − 1)2 (X-7) = 0 X= 1, 1, 7 Thus the Eigenvalues of A are 1, 1, and 7 then the Eigenvectors are given as: −1 [1] 𝑐 0 −1 [0] +𝑑 1 Where c, d ≠ 0 The corresponding Eigenvector to the above Eigenvalue 7 is : 1 [1] 𝑐1 −1 −1 1 Let P = [ 1 0 1] and a non-singular matrix 0 1 1 Then, 3 2 AP = [2 3 2 2 2 −1 −1 1 −1 −1 2] [ 1 0 1] = [ 1 0 3 0 1 1 0 1 −1 −1 1 1 0 PD = [ 1 0 1 ] [0 1 0 1 1 0 0 0 −1 −1 0] = [ 1 0 1 0 1 7 7] 7 7 7] 7 Thus, AP = PD i.e A= PDPT where P-1= PT Question 5(b) The maximum value of G(u) 𝑢1 3(𝑢1 2 +𝑢2 2 +𝑢3 2 )+3(𝑢1 𝑢2 +𝑢2 𝑢3 +𝑢3 𝑢1 ) G[𝑢2 ] = 𝑢1 2 +𝑢2 2 +𝑢3 2 𝑢3 Then the maximum value of G(u) is at (-1,0,1), (0,-1,1), (-1,1,0),(1,0,-1),(1,-1,0),and (0,1,-1) ASSIGNMENT 12 Name: Description: ...