# HWset7 - Solutions HW 7 5.1.16 Consider the vectors u 1 = 1...

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Unformatted text preview: Solutions HW 7 5.1.16 Consider the vectors u 1 = 1 2 1 2 1 2 1 2 , u 2 = 1 2 1 2- 1 2- 1 2 , u 3 = 1 2- 1 2 1 2- 1 2 in R 4 . Can you find a vector u 4 such that u 1 , u 2 , u 3 , u 4 are orthonormal? If so, how many such vectors are there? Note that u 1 ,u 2 ,u 3 are already orthonormal, so we just need to find u 4 , which must satisfy u 1 · u 4 = u 2 · u 4 = u 3 · u 4 = 0 and u 4 · u 4 = 1. If u 4 = ( a,b,c,d ) T , then the first three conditions give the linear equations 1 2 a + 1 2 b + 1 2 c + 1 2 d = 0 1 2 a + 1 2 b- 1 2 c- 1 2 d = 0 1 2 a- 1 2 b + 1 2 c- 1 2 d = 0 Row reducing these equations leads to the system a + b + c + d = 0 b + d = 0 c + d = 0 which has general solution u 4 = ( a,b,c,d ) T = ( t,- t,- t,t ) T . Any such vector has dot product zero with u 1 ,u 2 ,u 3 . For an orthonormal basis, we want u 4 · u 4 = 1, which gives t = ± 1 2 , so there are 2 choices for u 4 , namely 1 2- 1 2- 1 2 1 2 or - 1 2 1 2 1 2- 1 2 5.1.20 (See the book for the full statement of the problem, which involves finding a least squares estimate for a line y = mx ). Let x be the vector of x coordinates and y be the vector of y coordinates. Geometrically, { m x ) } is a 1 dimension vector space. We which to minimize the distance from y into this space; this is accomplished when m x is the projection of y into this space. Note that x || x || is a unit vector in this space, so the corre- sponding projection should be ( y · x || x || ) x || x || = y · x || x || 2 x . In particular, we should take m = y · x || x || 2 = 4182 . 9 198 . 53 2 = 0 . 106 ... . r = m || x || || y ||...
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HWset7 - Solutions HW 7 5.1.16 Consider the vectors u 1 = 1...

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