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acm104hw1

# acm104hw1 - Alex Gittens ACM 104 HW 1 Exercise 1 Show that...

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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 = { 0 } : W is then a subspace of V and span( W ) = W . Assume W = { 0 } 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 = { 0 } is a subset of V such that span( W ) = W , then if x, y W and α is a scalar, then αx + 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 ) = { 0 } = 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 β = { 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 i =1 a i v i = n i =1 a i Tx i = T n i =1 a i x i = 0 n i =1 a i x i = 0 , since T is one-to-one. By the linear independence of S , this means a i = 0, for all i , which shows T ( S ) is linearly independent.

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acm104hw1 - Alex Gittens ACM 104 HW 1 Exercise 1 Show that...

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