Solutions 1-5

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

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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 α , αx ∈ S 3. 0 ∈ S Now, given that U and V are both subspaces in R N for some N , let’s 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 α , αx must be in both U and V since these are subspaces and thus satisfy property 2 above. Thus, αx ∈ 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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## This note was uploaded on 10/12/2009 for the course ENGR 62 taught by Professor Unknown during the Spring '06 term at Stanford.

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

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