1.2. Vecto r spaces 11 is the set {(v l5 ..., v^) : v G ^} i n which addition and multiplication by scalars are defined by: (vv...,vk) + (uv...,uk) = (v x +uv...,vk + u k ), a(vv...,vk) = (av v ...,avk). The direct sum of vector spaces is clearly a vector space. In th e cas e tha t Y x ,..., Y k ar e al l subspace s o f th e sam e vecto r space Y, w e defined thei r sum, £ ^. = {v : v = Y!J=\ V J, V J ^ Yj), i n example/of 1.2.3 . The "natural" map of Yx 0 0 Y k int o Yx + + Y k , defined by (1.2.7) *((v 1 ,...,v i k ))=£v 7 . 5 l is clearly linea r an d surjective . I t is a n isomorphism whe n th e sub- spaces are independent DEFINITION: Th e subspaces Yj, j = 1,.. . ,£, o f a vector space Y are independent if £ v = 0 with v G ^ implie s that v = 0 for all j . 2 Proposition. Let Yj be subspaces of Y. The map & defined by (1.2.7) w aw isomorphism if and only if the subspaces are independent. PROOF:f is clearly linear and surjective. I t is injective if and only if every vector in the range has a unique preimage, that is, if (1.2.8) V p vj G Yj an d V[ + + = v[ + + Vk implies that Vj v' fo r every j. Subtracting and writing v = Vj Vj, (1.2.8) i s equivalen t to : £v . = 0 wit h v G Yj. Th e subspace s ar e independent if and only if this implies that v, = 0 for all j. In view of the proposition, we refer to the sum £ Wj of independent subspaces of a vector space as their direct sum, an d write 0 # } instead of £ 30}. If Y = ^ © 5^, we refer to ^ a s a complement of 3r i n y, an d vice versa. 2 Properly speaking : the set {y.} i s independent.
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