Solutions 1-5

Solutions 1-5 - ENGR62/MS&E111 Spring 2008...

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Unformatted text preview: ENGR62/MS&E111 Spring 2008 Introduction to Optimization April 11, 2008 Prof. Ben Van Roy Homework Assignment 1: Solutions Network Routing See the Excel solution worksheet on the course web page. Linear Algebra Review (Chapter 2) Problem 1 Expanding out the first four expressions, we have: a T x = 0 x 2 =- 1 2 x 1 (1) a T x = 1 x 2 =- 1 2 x 1 + 1 2 (2) b T x = 0 x 2 =- 2 x 1 (3) b T x = 1 x 2 =- 2 x 1 + 1 (4) Each of these corresponds to a line in the plane. Expanding out expressions 5 and 6, we have: [ a | b ] x = [0 , 1] T 1 2 2 1 x 1 x 2 = 1 (5) [ a | b ] x = [0 , 2] T 1 2 2 1 x 1 x 2 = 2 (6) Each of these is a system with a unique solution. Using pencil and paper we get that x = 2 3- 1 3 and x = 4 3- 2 3 are their respective solutions. We thus obtain the following graph:- 6 x 1 x 2 1: a T x =0 2: a T x =1 3: b T x =0 4: b T x =1 5 6 where the shaded region represents all x R 2 such that a T x 0. 1 Problem 2 Part 1 Finding a b such that Ax = b has no solution is equivalent to finding a vector, b R 3 , such that b is not in C ( A ). One easy choice is: b = 1 (7) The only way to get the first two b values in some linear combination of the columns of A is to set x 1 = 0 ,x 2 = 0. But, this leads to 0 = 1 in the third row equation. So, Ax = b cannot have a solution in this case. Part 2 Finding a non-zero b such that Ax = b has a solution is equivalent to finding a vector, b R 3 , such that b is in C ( A ). One easy choice is: b = 1 2 (8) which leads to a solution of x = 1 . Problem 5 By definition, S is a subspace of R N if and only if the following are true 1 : 1. For any x S and y S , x + y S 2. For any x S and scalar , x S 3. 0 S Now, given that U and V are both subspaces in R N for some N , lets look at U V . For any x U V and y U V , x and y must each be elements of both U and V by the definition of the intersection operator. Thus, ( x + y ) U and ( x + y ) V since U and V are both subspaces and thus satisfy property 1 above. Thus ( x + y ) U V for any such x and y . Also, for any x U V and any scalar , x must be in both U and V since these are subspaces and thus satisfy property 2 above. Thus, x U V . Finally, we have 0 U V since 0 is an element of both U and V . Thus, by the definition above, U V is a subspace of R N . On the other hand, U V is not necessarily a subspace in R N . We can show this by counterexample. Let U = span ( 1 ) and V = span ( 1 ). Now take x = 1 and y = 1 . Note that x U V and y U V by the definition of the union operator. But, we have x + y = 1 1 which is not an element of U V . Thus, property 1 above does not hold for U V and this cannot not a subspace....
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Solutions 1-5 - ENGR62/MS&E111 Spring 2008...

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