pset5sol

pset5sol - 18.06 Problem Set 5 Solutions Problem 1 Do...

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18.06 Problem Set 5 Solutions Problem 1: Do problems 5 and 6 from section 4.2. Solution (5) For a 1 = ( - 1 , 2 , 2), the projection matrix is P 1 = 1 9 1 - 2 - 2 - 2 4 4 - 2 4 4 . For a 2 = (2 , 2 , - 1), the projection matrix is P 2 = 1 9 4 4 - 2 4 4 - 2 - 2 - 2 1 . We compute to see P 1 P 2 = 0. This is because a 1 and a 2 are perpendicular. Precisely, if you project a vector b to a 1 , then it results a vector p on the same line as a 1 , whose projection to a 2 is zero. (6) In this case, A = - 1 2 2 2 2 - 1 . We compute A T A = ± 10 0 0 10 ² , and p = A ( A T A ) - 1 A T b = 1 2 1 5 0 . The projection of b onto a 3 is p 3 = P 3 b = 4 9 - 2 9 4 9 . The projection p 1 = P 1 b = 1 9 - 2 9 - 2 9 . The projection p 2 = P 2 b = 4 9 4 9 - 2 9 . So p 1 + p 1 + p 3 = 1 0 0 , which is b ! The reason is that a 3 is perpendicular to a 1 and a 2 , hence when you compute the three projections of a vector and add them
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This note was uploaded on 05/06/2010 for the course 18 18.06 taught by Professor Strang during the Fall '08 term at MIT.

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pset5sol - 18.06 Problem Set 5 Solutions Problem 1 Do...

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