practiceproblems2sol

# practiceproblems2sol - SOLUTIONS FOR SELECTED PRACTICE...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: SOLUTIONS FOR SELECTED PRACTICE PROBLEMS Section 1.1 3. (b) The plane in question is the set of all points of the form (3 , − 6 , 7) + t ( (3 , − 6 , 7) − ( − 2 , , − 4) ) + s ( (3 , − 6 , 7) − (5 , − 9 , − 2) ) = (3 + 5 t − 2 s, − 6 − 6 t + 3 s, 7 + 11 t + 9 s ) , with t,s ∈ R . Section 1.3 31. (a) Suppose first that v ∈ W , and show that, in this case, v + W = W (in particular, v + W would be a subset of V ). On the one hand, for every w ∈ W , we have v + w ∈ W , hence v + W ⊂ W . On the other hand, any w ∈ W can be written as w = v + u , with u = w − v ∈ W . Thus, v + W ⊃ W . Now suppose v / ∈ W , and show that W ′ = v + W is not a subspace of V . To this end, it suffices to prove that v ∈ W ′ , but 2 v / ∈ W ′ (subspaces are closed under multiplication by a scalar). Note first that 0 ∈ W , hence v +0 ∈ W ′ . Suppose, for the sake of contradiction, that 2 v ∈ W ′ . That is, there exists w ∈ W for which v + w = 2 v . Solving for w , we obtain w = v , which is impossible. Section 1.4 5. (g) Yes . We have to show that there exist x 1 ,x 2 ,x 3 ∈ R such that parenleftbigg 1 2 − 3 4 parenrightbigg = x 1 parenleftbigg 1 − 1 0 parenrightbigg + x 2 parenleftbigg 0 1 0 1 parenrightbigg + x 3 parenleftbigg 1 1 0 0 parenrightbigg . In other words, we need to know if the system of 4 linear equations x 1 + x 3 = 1 x 2 + x 3 = 2 − x 1 = − 3 x 2 = 4 has a solution. It is straightforward that the triple x 1 = 3, x 2 = 4, x 3 = − 2 solves this system. 1 2 SOLUTIONS FOR SELECTED PRACTICE PROBLEMS No , parenleftbigg 1 0 0 1 parenrightbigg does not belong to span( S ). If it did, there would exist a,b,c ∈ R satisfying parenleftbigg 1 0 0 1 parenrightbigg = a parenleftbigg 1 − 1 0 parenrightbigg + b parenleftbigg 0 1 0 1 parenrightbigg + c parenleftbigg 1 1 0 0 parenrightbigg = parenleftbigg a + c b + c − a b parenrightbigg Thus, the triple ( a,b,c ) should be a solution for the following system of four equations with three unknowns: a + c = 1 b + c = 0 − a = 0 b = 1 Equations 3 and 4 lead us to conclude that a = 0 and b = 1. By Equation 1, c = 1. Plugging these values into Equation 3, we obtain 2 = 0, which is false. Section 1.5 13. (b) Assume the set { u,v,w } is linearly independent, and prove that the set { u + v,u + w,v + w } is linearly independent – that is a 1 ( u + v ) + a 2 ( u + w ) + a 3 ( v + w ) = 0 implies a 1 = a 2 = a 3 = 0. Note that a 1 ( u + v ) + a 2 ( u + w ) + a 3 ( v + w ) = ( a 1 + a 2 ) u + ( a 1 + a 3 ) v + ( a 2 + a 3 ) w, hence, by the linear independence of { u,v,w } , a 1 ( u + v ) + a 2 ( u + w ) + a 3 ( v + w ) = 0 iff a 1 + a 2 = 0, a 1 + a 3 = 0, and a 2 + a 3 = 0. Solving this system of linear equations, we obtain a 2 = a 3 = − a 1 . Furthermore, 0 = a 2 + a 3 = − 2 a 1 , hence a 1 = a 2 = a 3 = 0....
View Full Document

## This note was uploaded on 02/02/2012 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.

### Page1 / 10

practiceproblems2sol - SOLUTIONS FOR SELECTED PRACTICE...

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

View Full Document
Ask a homework question - tutors are online