A5_soln - Math 235 Assignment 5 Solutions 1. Use the...

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Unformatted text preview: Math 235 Assignment 5 Solutions 1. Use the Gram-Schmidt procedure to produce an orthonormal set from the given set in R4 under the standard inner product. a) {(1, 0, 0, 1), (1, 1, 0, 2), (0, −1, 1, 1)} Solution: Denote the given basis by z1 = (1, 0, 0, 1), z2 = (1, 1, 0, 2), z3 = (0, −1, 1, 1). Let w1 = z1 . Then, we get w2 = perpw1 (z2 ) = z2 − projw1 (z2 ) = z2 − z2 · w1 w1 w1 2 3 1 = (1, 1, 0, 2) − (1, 0, 0, 1) = − (1, −2, 0, −1) 2 2 To simplify calculations we use w2 = (1, −2, 0, −1) instead. Then, we get w3 = z3 − (z3 · w2 ) z3 · w1 w− w2 21 w1 w2 2 1 1 1 = (0, −1, 1, 1) − (1, 0, 0, 1) − (1, −2, 0, −1) = − (2, 2, −3, −2) 2 6 3 Thus, the set {w1 , w2, w3 } is an orthogonal basis for span{z1 , z2 , z3 }. To obtain an orthonormal basis, we normalize each vector to get 1 1 1 √ (1, 0, 0, 1), √ (1, −2, 0, −1), √ (2, 2, −3, −2) . 2 6 21 b) {(1, −3, −1, 1), (1, 1, 1, 1), (2, 0, 1, 2), (2, 0, 1, 1)} Solution: Denote the given basis by z1 = (1, −3, −1, 1), z2 = (1, 1, 1, 1), z3 = (2, 0, 1, 2), z4 = (2, 0, 1, 1). Let w1 = z1 . Then, we get w2 = perpw1 (z2 ) = z2 − projw1 (z2 ) = z2 − = (1, 1, 1, 1) + z2 · w1 w1 w1 2 2 1 (1, −3, −1, 1) = (7, 3, 5, 7) 12 6 To simplify calculations we use w2 = (7, 3, 5, 7) instead. Then, we get w3 = z3 − (z3 · w2 ) z3 · w1 w− w2 21 w1 w2 2 33 3 (7, 3, 5, 7) = (0, 0, 0, 0) = (2, 0, 1, 2) − (1, −3, −1, 1) − 12 132 2 Thus, z3 was linearly dependent on the first two vectors and so we ignore it. z4 · w1 (z4 · w2 ) w3 = z4 − w1 − w2 w1 2 w2 2 2 26 1 = (2, 0, 1, 1) − (1, −3, −1, 1) − (7, 3, 5, 7) = (5, −1, 2, −6) 12 132 11 Thus, the set {w1 , w2, w3 } is an orthogonal basis for span{z1 , z2 , z3 }. To obtain an orthonormal basis, we normalize each vector to get 1 1 1 √ (1, −3, −1, 1), √ (7, 3, 5, 7), √ (5, −1, 2, −6) . 12 132 66 2. On M (2, 2) define the inner product < A, B >= tr(AT B ). Use the Gram-Schmidt procedure to produce an orthogonal set from the set S= Solution: Let w1 = 11 . Then, 01 < 11 0 −1 , > 11 01 011 11 0 −1 0 −1 = − = 2 01 11 01 11 3 11 01 11 0 −1 30 , , 01 11 11 . w2 = 0 −1 − 11 So, we pick w2 = 0 −1 . 11 < 30 0 −1 30 11 < , > , > 11 11 11 01 11 0 −1 − 2 2 01 11 11 0 −1 01 11 w3 = 30 − 11 = Thus {w1 , w2 , w3 } is an orthogonal set. 411 2 0 −1 1 5 −2 30 − − = 11 301 31 1 3 1 −3 3 3. Find a and b to obtain the best fitting equation of the form y = a + bt for the given data. a) t −2 −1 0 1 2 y3 5 7 9 10 Solution: Let X T = 1 1 111 and y T = (3, 5, 7, 9, 10). Then we get −2 −1 0 1 2 a = (X T X )−1 X T y 50 0 10 −1 = = Thus a = 6.8 and b = 1.8. b) 120 10 0 1 34 6.8 = 18 1.8 3 5 1 1 111 7 −2 −1 0 1 2 9 10 t12345 y34557 11111 and y T = (3, 4, 5, 5, 7). Then we get 12345 3 4 −1 5 15 1 1 1 1 1 5 = 1 11 −3 24 = 2.1 a = (X T X )−1 X T y = 15 35 1 2 3 4 5 10 −3 1 81 .9 5 7 Solution: Let X T = Thus a = 2.1 and b = −.9. 4 4. Verify that the following systems are inconsistent, then determine for each system the x that minimizes Ax − b . a) x1 + 2x2 = 5 2x1 − 3x2 = 6 x1 − 12x2 = −4 Solution: Writing the augmented matrix and row-reducing gives 12 5 100 2 −3 6 ∼ 0 1 0 , 1 −12 −4 001 so it is inconsistent. The x that minimizes Ax − b must satisfy AT Ax = AT b. Solve for x in this system gives −1 5 1 157 16 13 6 −16 383/98 12 1 T −1 T 6= x = (A A) A b = = 16 6 40 −16 157 32/49 2 −3 −12 686 −4 Thus x = (383/98, 32/49) minimizes Ax − b . b) x1 + x2 = 7 x1 − x2 = 4 x1 + 3x2 = 14 Solution: Writing the augmented matrix and row-reducing gives 117 100 1 −1 4 ∼ 0 1 0 , 1 3 14 001 so it is inconsistent. The x that minimizes Ax − b must satisfy AT Ax = AT b. Solve for x in this system gives −1 7 1 11 −3 25 33 1 1 1 3 5/6 T −1 T 4= x = (A A) A b = = 3 11 1 −1 3 −3 3 45 5/2 24 14 Thus x = (35/6, 5/2) minimizes Ax − b . 5 5. Prove that if S is a k -dimensional subspace of Rn , then S ⊥ is an (n − k ) dimensional subspace of Rn . · · · vk · x = 0. v1 v2 Since {v1 , . . . , vk } is linearly independent, the matrix . has rank k and hence the solution . . vk ⊥ Solution: Let {v1 , . . . , vk } be a basis for S . Then S ⊥ is the solution space of the system v1 · x = 0, v2 · x = 0, space S has dimension n − k . 6. Prove that perpS = projS ⊥ for any subspace S of Rn . Solution: Let {v1 , . . . , vk } be an orthonormal basis for S and let {vk+1 , . . . , vn } be an orthonormal basis for S ⊥ . Then, {v1 , . . . , vk , vk+1, . . . , vn } is an orthonormal basis for Rn and so any x ∈ Rn can be written as Then x =< x, v1 > v1 + · · · + < x, vk > vk + < x, vk+1 > vk+1 + · · · + < x, vn > vn . perpS x = x − projS (x) = x − < x, v1 > v1 + · · · + < x, vk > vk =< x, vk+1 > vk+1 + · · · + < x, vn > vn = projS ⊥ (x) 7. Let A be an m × n real matrix. Prove that the nullspace of AT is the orthogonal complement of the column space of A. T v1 T . . Solution: Let A = v1 · · · vn , so A = . . T vn T v1 x ⊥ T . If x ∈ col(A) , then vi · x = 0. But then A x = . = 0 hence x ∈ nul(AT ). . T vn x On the other hand, let x ∈ nul(AT ), then AT x = 0 so vi · x = 0 for all i. Now, pick any b ∈ col(A). Then b = c1 v1 + · · · + cn vn . But then hence x ∈ col(A) . ⊥ b · x = (c1 v1 + · · · + cn vn ) · x = 0, 6 MATLAB −0.4566 −0.3307 −0.3156 −0.1205 0.8719 −0.3796 1) a) N = 0.4994 0.0819 0.5149 0.4493 −0.3338 −0.6441 form the orthonormal basis. −0.6679 0.3693 −0.2477 −0.1406 . The columns of the matrix −0.6838 −0.1058 −0.1183 −0.5078 b) We verify that N T N = I using the command N ′ ∗ N 0 10 20 30 2) b) The design matrix is X = 40 50 60 70 80 d) 0 100 400 900 1600 2500 3600 4900 6400 0 1000 8000 27000 64000 125000 216000 343000 512000 c) The least squares solution is beta = (1.6288, −0.0366). ...
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This note was uploaded on 04/18/2010 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.

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