psLinAlgebra2Ans

psLinAlgebra2Ans - Fall 2007 ARE211 Problem Set #07- Answer...

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Unformatted text preview: Fall 2007 ARE211 Problem Set #07- Answer key Second Lin Algebra Problem Set Problem 1 What is the difference between a minimum spanning set for a vector space and a basis for a vector space. Provide an example highlighting this difference. Ans: The minimum spanning set does not have to be part of the vector space V , while a basis has to be an element of the vector set. For example, consider the vector space V = { x | x = α parenleftbigg 1 1 parenrightbigg ,α ∈ R } which is a straight line through the origin. A minimal spanning set for V are the two vectors: v 1 = parenleftbigg 1 parenrightbigg , v 2 = parenleftbigg 1 parenrightbigg Take away any of the two vectors and the remaining vector does not span V . However they are not a basis as neither v 1 nor v 2 are an element of V . 1 2 Problem 2 Simon & Blume question 11.2 (page 243). Explain your answer. Ans: There are several ways to test for linear dependence of the vectors v 1 , v 2 , ..., v n . The definition of linear independence was Show that the linear equation system bracketleftbig v 1 v 2 ··· v n bracketrightbig x = has the only soulution x = 0 . (Where x is a (nx1) column vector). a) (i) Given: v 1 = parenleftbigg 2 1 parenrightbigg , v 2 = parenleftbigg 1 2 parenrightbigg (ii) Let’s consider the linear equation system bracketleftbig v 1 v 2 bracketrightbig x = 2 1 . . . 1 2 . . . Add (-2) times the second row to the first row: ⇔ - 3 . . . 1 2 . . . Divide the first row by (-3): ⇔ 0 1 . . . 1 2 . . . Add (-2) times the first row to the second: ⇔ 0 1 . . . 1 0 . . . Hence we know from the first row that x 1 = 0 and from the second row that x 2 = 0 (iii) Since is the only solution to the linear equation syatem in (ii), we know that v 1 and v 2 are linearly independent. b) (i) Given: v 1 = parenleftbigg 2 1 parenrightbigg , v 2 = parenleftbigg- 4- 2 parenrightbigg (ii) Let’s consider the linear equation system bracketleftbig v 1 v 2 bracketrightbig x = 2- 4 . . . 1- 2 . . . 3 Add (-2) times the second row to the first row: ⇔ . . . 1- 2 . . . Clearly, x 1 = 2 and x 2 = 1 is a nonzero solution to the system (iii) Since the linear equation system in (ii) has a nonzero solution, we know that v 1 and v 2 are linearly dependent. c) (i) Given: v 1 = 1 1 , v 2 = 1 1 (ii) Let’s consider the linear equation system bracketleftbig v 1 v 2 bracketrightbig x = 1 0 ....
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This note was uploaded on 08/01/2008 for the course ARE 211 taught by Professor Simon during the Fall '07 term at Berkeley.

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psLinAlgebra2Ans - Fall 2007 ARE211 Problem Set #07- Answer...

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