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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 1 , 2 , 3 R . Then 1 A 1 + 2 A 2 + 3 A 3 = 0 implies 1 + 2 = 3 , 2 2 = 3 , 1 + 2 =- 3 . First and the last equations imply 3 = 0, then the second equation implies 2 = 0, finally the first or the last equation gives 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 1 , 2 ,..., k , k +1 , at least one of which is nonzero, such that 1 A 1 + 2 A 2 + + k A k + k +1 A k +1 = 0 . (1) Case 1: k +1 = 0. Then 1 A 1 + 2 A 2 + + k A k = 0 and at least one of the coefficients, i , is nonzero. But this contradicts the linear independence of the set { A 1 ,A 2 ,...,A k } . Case 2: k +1 6 = 0. In this case, we can solve (1) for A k +1 : A k +1 = k X i =1- i 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) times to reach a basis....
View Full Document

This note was uploaded on 04/11/2010 for the course CO 350 taught by Professor S.furino,b.guenin during the Fall '07 term at Waterloo.

Page1 / 5

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online