acm104hw1 - Alex Gittens January 19, 2007 ACM 104: HW 1...

Info iconThis preview shows pages 1–2. 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: Alex Gittens January 19, 2007 ACM 104: HW 1 Exercise 1. Show that a subset W of a vector space V is a subspace of V if and only if span( W ) = W . Proof. This statement is trivially true if W = { } : W is then a subspace of V and span( W ) = W . Assume W 6 = { } is a subspace of V , then clearly W span( W ), because if x W , then x = 1 x span( W ). If y span( W ), that is, there are x i W such that y = m i =1 a i x i , then since W is closed under addition and scalar multiplication, y W . This shows span( W ) W , so span( W ) = W . Assume W 6 = { } is a subset of V such that span( W ) = W , then if x, y W and is a scalar, then x + y span( W ) = W , so W is closed under addition and scalar multiplication, therefore W is a subspace of V . Exercise 2. Let S 1 and S 2 be two subsets of a vector space V . Prove that span( S 1 S 2 ) span( S 1 ) span( S 2 ) . Give one example to show equality does not hold in general. Proof. If x span( S 1 S 2 ), then there are y 1 , . . . , y m S 1 S 2 such that x = m i =1 a i y i for some a i R . But each y i is in S 1 and in S 2 , so x span( S 1 ) and x span( S 2 ). This shows span( S 1 S 2 ) span( S 1 ) span( S 2 ). To see equality does not hold in general, take V = R 2 , S 1 = { (0 , 1) , (1 , 0) } and S 2 = { (1 , 1) , (1 ,- 1) } . Then both S 1 and S 2 are bases for V, so span( S 1 ) span( S 2 ) = R 2 , but S 1 S 2 = , so span( S 1 S 2 ) = { } 6 = R 2 . Exercise 3. Let V be a vector space having dimension n , and let S be a subset of V such that span( S ) = V . Prove that there is a subset of S that is a basis for V . (Note: S may be infinite.) Proof. Let = { v 1 , . . . , v n } be a basis for V , then v 1 , . . . , v n are in the span of S , so there are s 1 , 1 , , . . . , s 1 ,m 1 , s 2 , 1 , . . . , s 2 ,m 2 , . . . , s n, 1 , . . . , s n,m n S such that v j = m j k =1 a k s j,k for j = 1 , . . . , n . Therefore V = span( s 1 , 1 , . . . , s n,m n ). Since this is a finite set, there is a subset S of { s 1 , 1 , . . . , s n,m n } that is a basis for V (by Thm 1.18). Since S S , S contains a basis for V . Exercise 4. Let V and W be vector spaces and let T : V W be a linear transformation. Suppose T is one-to-one and S is a subset of V . Prove that S is linearly independent if and only if T ( S ) is linearly independent. Proof. Let S be linearly independent. Assume there are v 1 , . . . , v n T ( S ) and scalars a 1 , . . . , a n such that n i =1 a i v i = 0. Let x i be such that v i = Tx i , then n X i =1 a i v i = n X i =1 a i Tx i = T n X i =1 a i x i !...
View Full Document

This document was uploaded on 01/21/2011.

Page1 / 6

acm104hw1 - Alex Gittens January 19, 2007 ACM 104: HW 1...

This preview shows document pages 1 - 2. Sign up to view the full document.

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