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# c3s1 - ~v(3 if ~u ~v = ~u ~w then ~v = ~w(4 ~v = ~ 0 for...

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3.1 Vector Spaces 1 Chapter 3. Vector Spaces 3.1 Vector Spaces Definition 3.1. A vector space is a set V of vectors along with an operation of addition + of vectors and multiplication of a vector by a scalar (real number), which satisfies the following. For all u, v, w V and for all r, s R : (A1) ( u + v ) + w = u + ( v + w ) (A2) v + w = w + v (A3) There exists 0 V such that 0 + v = v (A4) v + ( v ) = 0 (S1) r ( v + w ) = rv + rw (S2) ( r + s ) v = rv + sv (S3) r ( sv ) = ( rs ) v (S4) 1 v = v Definition. 0 is the additive identity . v is the additive inverse of v .

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3.1 Vector Spaces 2 Example. Some examples of vector spaces are: (1) The set of all polynomials of degree n or less, denoted P n . (2) All m × n matrices. (3) The set of all functions integrable f with domain [0 , 1] such that 1 0 | f ( x ) | 2 dx < . This vector space is denoted L 2 [0 , 1]: L 2 [0 , 1] = f 1 0 | f ( x ) | 2 dx < . Theorem 3.1. Elementary Properties of Vector Spaces. Every vector space V satisfies: (1) the vector 0 is the unique additive identity in a vector space, (2) for each
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Unformatted text preview: ~v , (3) if ~u + ~v = ~u + ~w then ~v = ~w , (4) ~v = ~ 0 for all ~v ∈ V , (5) r ~ 0 = ~ 0 for all scalars r ∈ R , (6) ( − r ) ~v = r ( − ~v ) = − ( r~v ) for all r ∈ R and for all ~v ∈ V . 3.1 Vector Spaces 3 Proof of (1) and (3). Suppose that there are two additive identities, ~ 0 and ~ . Then consider: ~ 0 = ~ 0 + ~ (since ~ is an additive identity) = ~ (since ~ 0 is an additive identity). Therefore, ~ 0 = ~ and the additive identity is unique. Suppose ~u + ~v = ~u + ~w . Then we add − ~u to both sides of the equation and we get: ~u + ~v + ( − ~u ) = ~u + ~w + ( − ~u ) ~v + ( ~u − ~u ) = ~w + ( ~u − ~u ) ~v + ~ 0 = ~w + ~ ~v = ~w The conclusion holds. QED Example. Page 189 number 14 and page 190 number 24....
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