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sol1 - CO 350 Linear Optimization(Spring 2009 Solution to...

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CO 350 Linear Optimization (Spring 2009) Solution to Assignment 1 [10] 1. (Exercise 2 in page 22 of the course notes.) The dimension of R 3 coincides with the number of vectors we have (3). Therefore, { A 1 , A 2 , A 3 } is a basis for R 3 iff it is linearly independent. Let α 1 , α 2 , α 3 R . Then α 1 A 1 + α 2 A 2 + α 3 A 3 = 0 implies α 1 + α 2 = α 3 , 2 α 2 = α 3 , α 1 + α 2 = - α 3 . First and the last equations imply α 3 = 0, then the second equation implies α 2 = 0, finally the first or the last equation gives α 1 = 0. Therefore, { A 1 , A 2 , A 3 } is a basis for R 3 . A 4 = 3 4 A 1 - 1 4 A 2 - 1 2 A 3 , A 5 = - 1 2 A 1 + 1 2 A 2 , A 6 = 1 4 A 1 + 1 4 A 2 + 1 2 A 3 . [10] 2. (Exercise 3 in page 22 of the course notes.) Suppose to the contrary that { A 1 , A 2 , . . . , A k , A k +1 } is linearly dependent. Then there are real numbers α 1 , α 2 , . . . , α k , α k +1 , at least one of which is nonzero, such that α 1 A 1 + α 2 A 2 + · · · + α k A k + α k +1 A k +1 = 0 . (1) Case 1: α k +1 = 0. Then α 1 A 1 + α 2 A 2 + · · · + α k A k = 0 and at least one of the coefficients, α i , is nonzero. But this contradicts the linear independence of the set { A 1 , A 2 , . . . , A k } . Case 2: α k +1 6 = 0. In this case, we can solve (1) for A k +1 : A k +1 = k X i =1 - α i α k +1 A i . Now this contradicts our assumption that A k +1 is not in the subspace generated by the vectors A 1 , A 2 , . . . , A k . Hence in both possible cases, we encountered a contradiction and the proof is complete. [10] 3. (Exercise 4 in page 22 of the course notes.) Consider any subset S of linearly independent vectors in R m . If S is not a basis, then | S | = k < m . Therefore, there exists a vector A k +1 R m such that A k +1 is not in the subspace spanned by vectors in S . Now, applying the statement proved in the solution of the previous exercise, we see that S ∪{ A k +1 } is linearly independent. We can apply this argument ( m - k ) times to reach a basis. Therefore, given a linearly independent subset of vectors (say k of them) in R m , we can always find m - k vectors to

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