571_3.1 - 3.1/3.2 DEFINITONS AND EXAMPLES OF VECTOR SPACES...

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3.1/3.2 DEFINITONS AND EXAMPLES OF VECTOR SPACES AND SUBSPACES KIAM HEONG KWA p. 113 A (real) linear space or a (real) vector space is an ordered triple ( V, , ± ) , where V is a nonempty set and : V × V V and ± : R × V V are functions, such that the following axioms hold: A1. p q = q p for any p , q V . A2. ( p q ) r = p ( q r ) for any p , q , r V . (In view of this, we write p q r = p ( q r ) .) A3. There is an element 0 V such that p 0 = p for any p V . A4. For each p V , there is an element - p V such that p ( - p ) = 0 . A5. α ± ( p q ) = ( α ± p ) ( α ± q ) for any α R and any p , q V . A6. ( α + β ) ± p = ( α ± p ) ( β ± p ) for any α,β R and any p V . A7. ( αβ ) ± p = α ± ( β ± p ) for any α,β R and any p V . A8. 1 ± p = p for any p V . Notation: Here we have written p q for ( p , q ) and α ± p for ± ( α, p ) for any α R and any p , q V . Implicit in the definition of the vector space ( V, , ± ) are the clo- sure properties : C1. For any scalar α and p V , α ± p V . C2. For any p , q V , p q V . The functions and ± are called addition and scalar multiplica- tion respectively. The elements of V are called vectors . In this context, the elements of R are called scalars . It is customary to re- fer to V as the linear space instead of the ordered triple ( V, , ± ) on the assumption that the addition and scalar multiplication are tacitly given. Notation: It is not uncommon to write + for , so that p q is written as p + q for any p , q V . It is also not uncommon to represent ± by juxtaposition, so that α ± p is written as α p for any α R and any p V . Date : June 26, 2011. 1
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2 KIAM HEONG KWA p. 115 If V is a linear space and p V , then Theorem 3.1.1 (1) 0 p = 0 ; (2) if p + q = 0 for a q V , then q = - p ; (3) (-1) p =- p . Proof of (1). By axioms A6 and A8, p A 8 = 1 p = (0 + 1) p A 6 = 0 p + 1 p A 8 = 0 p + p . Then by axioms A2, A3, and A4, 0 A 4 = p + ( - p ) = (0 p + p ) + ( - p ) A 2 = 0 p + ( p + ( - p )) A 4 = 0 p + 0 A 3 = 0 p . ± Proof of (2). By axioms A1, A3, and A4, ( - r ) + r A 1 = r + ( - r ) A 4 = 0 and 0 + r A 1 = r + 0 A 3 = r for any r V . Hence the assumption p + q = 0 and axiom A2 imply that q = 0 + q = (( - p ) + p ) + q A 2 = ( - p ) + ( p + q ) = ( - p ) + 0 = - p . ± Proof of (3). By axioms A6 and A8 and part (1), p + ( - 1) p A 8 = 1 p + ( - 1) p A 6 = (1 + ( - 1)) p = 0 p (1) = 0 . Hence ( - 1) p = - p by part (2). ± p. 116 The element 0 in a linear space V is unique. Exercise 3.1.7 Proof. Let ˜ 0 V be such that p + ˜ 0 = p for all p V . In particular, 0 + ˜ 0 = 0 . This, together with axioms A1 and A3, yields ˜ 0 A 3 = ˜ 0 + 0 A 1 = 0 + ˜ 0 = 0 . ± p. 116
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This note was uploaded on 11/11/2011 for the course MATH 571 taught by Professor Kim during the Summer '08 term at Ohio State.

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571_3.1 - 3.1/3.2 DEFINITONS AND EXAMPLES OF VECTOR SPACES...

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