Theorems - Theorem 1.1 (Cancellation Law for Vector...

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Theorem 1.1 (Cancellation Law for Vector Addition). If x , y , and z are vectors in a vector space V such that x + z = y + z , then x = y . Proof. There exists a vector v in V such that z + v = 0 (VS 4). Thus x = x + 0 = x + (z + v) = (x + z) + v = (y + z) + v = y + (z + v) = y + 0 = y by (VS 2) and (VS 3). Corollary 1. The vector 0 described in (VS 3) is unique. Corollary 2. The vector y described in (VS 4) is unique. 1
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Theorem 1.2. In any vector space V the following statements are true: (a) 0 x = 0 for each x V . (b) ( - a ) x = - ( ax ) = a ( - x ) for each a F and each x V . (c) a 0 = 0 for each a F . Proof. a. By (VS 8), (VS 3), and (VS 1) it follows that 0 x + 0 x = (0 + 0) x = 0 x = 0 x + 0 . Hence 0 x = 0 by Theorem 1.1. b. The vector - ( ax ) is the unique element of V such that ax +[ - ( ax )] = 0 . Thus if ax +( - a ) x = 0 , Corollary 2 to Theorem 1.1 implies that ( - a ) x = - ( ax ). But by (VS 8). ax + ( - a ) x = [ a + ( - a )] x = 0 x = 0 by (a). Consequently ( - a ) x = - ( ax ). In particular ( - 1) x = - x . So by (VS 6), a ( - x ) = a [( - 1) x ] = [ a ( - 1)] x = ( - a ) x. c. By (VS 3) and (VS 7) it follows that a 0 + a 0 = a( 0 + 0 ) = a 0 = a 0 + 0 . Hence a 0 = 0 by Theorem 1.1. 2
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Theorem 1.3. Let V be a vector space and W be a subset of V . Then W is a subspace of V if and only if the following three conditions hold for the operations defined in V : (a) 0 W . (b) ( x + y ) W whenever x,y W . (c) cx W whenever c F and x W . Proof. We begin by assuming that W is a subspace of V ; then W is a vector space with the operations of addition and scalar multiplication defined on V . Hence conditions (b) and (c) hold, and there exists a vector 0 w W such that 0 w + x = x for each x W . But also x + 0 = x , and thus 0 w = 0 by Theorem 1.1. So condition (a) holds. Next suppose that conditions (a), (b), and (c) hold, then by the discussion on pg 17, W is a subspace of V if the additive inverse of each vector in W lies in W . But if x W , then ( - 1) x W by condition (c), and - x = ( - 1) x by Theorem 1.2. Hence W is a subspace of V . 3
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Theorem 1.4. Any intersection of subspaces of a vector space V is a subspace of V . Proof. Let C be a collection of subspaces of V and let W denote the intersection of the subspaces of C . Since every subspace contains the zero vector, 0 W . Let a F and x,y W . Then x and y are contained in each subspace of C. Because each subspace is closed under addition and scalar multiplication, it follows that x + y and ax are contained in each subspace of C . Hence x + y and ax are also contained in W , so that W is a subspace of V by Theorem 1.3. 4
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Theorem 1.5. The span of any subset S of a vector space V is a sub- space of V . Moreover, any subspace of V that contains S must also contain the span of S. Proof. The result is immediate if S = φ because span( φ ) = { 0 } , which is a subspace contained in any subspace of V . If S 6 = φ, then S contains a vector z . So 0 z = 0 is in span( S ). Let x,y span( S ). Then there exists vectors u 1 ,u 1 ,...,u m ,v 1 2 ,...,v n in S and scalars a 1 ,a 2 ,...,a m ,b 1 2 ,...,b n such that x = a 1 u 1 + a 2 u 2 + ··· + a m u m , y = b 1 v 1 + b 2 v 2 + + b n v n . Then x + y = a 1 u 1 + a 2 u 2 + + a m u m + b 1 v 1 + b 2 v 2 + + b n v n and for any scalar c , cx = ( ca 1 ) u 1 + ( ca 2 ) u 2 + + ( ca m ) u m are clearly linear combinations of vectors in S ; so x + y and cx are in span( S ).
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Theorems - Theorem 1.1 (Cancellation Law for Vector...

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