asst1sol

# asst1sol - C&O 350 Linear Optimization Winter 2009...

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Unformatted text preview: C&O 350 Linear Optimization Winter 2009 Solutions to Assignment 1 Question # Max. marks Part marks 1 10 1 3 2 4 2 10 4 2 2 2 3 6 4 3 5 11 2 3 2 4 Total 40 1. (Linear Algebra Review) [10 marks = 1+3+2+4] (a) Let S = { v 1 ,...,v n } be a set of n vectors from the vector space V . Complete the definition that begins S is a linearly independent set . (b) Is the set S = 2 , 2 5 3 , 3- 1 4 linearly independent in R 3 ? Provide convincing evidence. (c) Let S = { x,y,z } be a linearly independent set, where x,y,z R n . Decide whether the following statements are true or false. If true, then give a proof, and if false, then give a counter-example. i. The set { x + y,y + z,z- x } is a linearly independent set. ii. The set { x + y,y + z,z + x } is a linearly independent set. Solution : (a) S is a linearly independent set if and only if for all scalars c 1 ,...,c n , the equation n i =1 c i v i = 0 implies that c 1 = ... = c n = 0 . (b) The set is independent. Let v 1 ,v 2 ,v 3 be the three vectors as listed (in order) in S . Suppose that av 1 + bv 2 + cv 3 = 0 for some scalars a,b,c . Then each component gives a linear equation in the variables a,b,c . The first and third components give the system 2 b + 3 c = 0 3 b + 4 c = 0 . Hence b = c = 0. The equation given by the second component, 2 a + 5 b- c = 0 now becomes 2 a = 0, and therefore a = b = c = 0. Thus the set S is linearly independent. (c) (i) FALSE. ( z- x ) = ( x + y )- ( y + z ) hence, the three new vectors are linearly dependent. For a specific counterexample, we may take x = (1 , , 0) T , y = (0 , 1 , 0) T , z = (0 , , 1) T , then z- x = (- 1 , , 1) T is linearly dependent on x + y = (1 , 1 , 0) T and y + z = (0 , 1 , 1) T . 1 (ii) TRUE. Let a,b,c be scalars such that a ( x + y ) + b ( y + z ) + c ( x + z ) = 0 . It follows that a + c,a + b,b + c are scalars such that ( a + c ) x + ( a + b ) y + ( b + c ) z = 0 . Since { x,y,z } is a linearly independent set of vectors, this implies that a + c = a + b = b + c = 0. Thus a = b = c = 0, and { x + y,y + z,x + z } is linearly independent....
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## This note was uploaded on 02/05/2010 for the course CO 350 taught by Professor S.furino,b.guenin during the Spring '07 term at Waterloo.

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asst1sol - C&O 350 Linear Optimization Winter 2009...

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