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

Info iconThis preview shows pages 1–3. 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: 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....
View Full Document

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.

Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

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