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# C find the matrices p and p whose application result

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Unformatted text preview: stating the dimension of V ⊥ . 1 1 1 0 1 1 ￿ 1 = , ￿ 2 = , ￿ 3 = . v 0 v 1 v 0 1 0 0 (b) Give a basis for V and state the dimension of V . Justify, why the vectors in the basis provided as your answer indeed form a basis of V . (c) Find the matrices P and P ⊥ whose application result in the orthogonal projections on V and on V ⊥ , respectively. (d) For what value of a parameter p, p being a real number, the projection of a vector 2 3p 0 w= ￿ 1 , onto space V is the shortest? What is the length of such shortest 0 vector? Solution: (a) V ⊥ is equal to the left null space of A = [￿1 ￿2 ￿3 ], that is, all vectors ￿ vvv y satisfying AT ￿ = 0. To solve it, note that reduction of AT is as follows y 10 0 1 100 1 1001 100 1 1 −1 ￿3 → 0 1 0 −1 +r −r AT = 1 1 1 0 ￿1 → 0 1 1 −1 → 0 1 0 1 0 −1 −￿2 r 0 0 −1 0 ×(−1) 001 0 1 1 0 0 −￿1 r y 100 1 1 y Therefore, system AT ￿ = 0 is equivalent to 0 1 0 −1 2 = 0, so y4 is a free variy y3 001 0 y4 able (non-pivot column) and y1 = −y4 , y2 = y4 , y3 = 0. This gives the solution of AT ￿ = 0 y...
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## This document was uploaded on 02/19/2014 for the course MATH 441 at WVU.

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