hwsolB15 - p = V*c = c1*v1 c2*v2 c3*v3 = 11/3-5/3 4 8/3(iv...

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Sheet1 Page 1 HW 15 Solutions _______________ v1 = [ 5 3 -1 1 ].' v2 = [ 5 -1 7 3 ].' v3 = [ 1 1 -10 0 ].' s = [ 3 -1 4 4 ].' (i) We form the matrix V = [v1 v2]: V = [ 5 5 3 -1 -1 7 1 3 ] Then V'*V = [ 36 18 18 84 ] V'*s = [ 12 56 ] Solve (V'*V)*c = V'*s to obtain c1 = 0 and c2 = 2/3 The projection f is given by f = V*c = c1*v1 + c2*v2 = [ 10/3 -2/3 14/3 2 ].' (ii) The error vector is f-s = [ 1/3 1/3 2/3 -2 ].' and its norm equals ||f-s|| = 2.1602 (iii) We form the matrix V = [v1 v2 v3]: V = [ 5 5 1 03/01/01 -1 7 -10 1 3 0 ] Then V'*V = [ 36 18 18 18 84 -66 18 -66 102 ] and V'*s = [ 12 56 -38 ] Solving (V'*V)*c = V'*s, we obtain:
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Sheet1 Page 2 c1 = -1/3 , c2 = 1 , c3 = 1/3 The projection p is given by
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Unformatted text preview: p = V*c = c1*v1 + c2*v2 + c3*v3 = [ 11/3 -5/3 4 8/3 ].' (iv) The error vector is p-s = [ 2/3 -2/3 0 -4/3 ].' and its norm equals ||p-s|| = 1.6330 (v) v4 = [ 14 2 0 8 ].' V = [v1 v2 v3 v4] V = [ 5 5 1 14 3 -1 1 2 -1 7 -10 0 1 3 0 8 ] V is nonsingular. Therefore, the range of the matrix V is R^4 which represents the set of all points in the 4-dimensional space. It can also show that any point x = [x1 x2 x3 x4].' can be uniquely represented in terms of a linear combination of columns of V. This includes x=s, and thus the projection of s on the range of the matrix V is s itself. The resulting error norm is zero. ( Note that s = v4 - v1 - v2 - v3 , i.e., s = V * [-1 -1 -1 1].' )...
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hwsolB15 - p = V*c = c1*v1 c2*v2 c3*v3 = 11/3-5/3 4 8/3(iv...

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