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Unformatted text preview: Direct Sum of Subspaces Paul Skoufranis September 29, 2011 The purpose of this document is to demonstrate that vector spaces can have ’nice’ decompositions. In particular, we will present some proofs relating to spaces of vector spaces. We begin with a definition for the sum of two sets of vectors. Definition) Let V be a vector space. Suppose that S 1 and S 2 are nonempty subsets of V . The sum of S 1 and S 2 , denoted S 1 + S 2 , is { ~x + ~ y  ~x ∈ S 1 ,~ y ∈ S 2 } (that is, S 1 + S 2 is the set of vectors of V that can be obtained by adding a vector in S 1 to a vector in S 2 ). Example) Let S 1 be the xaxis in R 3 and let S 2 be the yaxis in R 3 . Then S 1 + S 2 is the xyplane in R 3 . To see this, we notice that if we add a vector in the direction of the xaxis to a vector in the direction of the yaxis, we obtain a vector in the xyplane (that is ( x, , 0) + (0 ,y, 0) = ( x,y, 0) is in the xyplane). Moreover, every vector in the xyplane can be written as the sum of a vector in the direction of the xaxis and a vector in the direction of the yaxis (that is, ( x,y, 0) = ( x, , 0) + (0 ,y, 0)). Example) Let S 1 = { ( x, 0)  x ∈ R } ⊆ R 2 and S 2 = { ( y,y )  y ∈ R } ⊆ R 2 } . Then S 1 + S 2 ∈ R 2 (as span { (1 , 0) , (1 , 1) } = R 2 ). The following is our first result about the sum of two subspaces of a vector space. Proposition) Let V be a vector space and suppose W 1 and W 2 are subspaces of V . Then W 1 + W 2 is a subspace of V that contains W 1 and W 2 . Proof : Clearly W 1 + W 2 ⊆ V . To show that W 1 + W 2 is a subspace of V , we need to demonstrate the three properties of being a subspace. 1. We need to demonstrate that ~ ∈ W 1 + W 2 . We recall that since W 1 and W 2 are subspaces of V , ~ ∈ W 1 and ~ ∈ W 2 . Therefore, since ~ 0 = ~ 0 + ~ 0, ~ ∈ W 1 , and ~ ∈ W 2 , ~ ∈ W 1 + W 2 by the definition of W 1 + W 2 . 2. We need to demonstrate that W 1 + W 2 is closed under addition. Suppose ~x,~ y ∈ W 1 + W 2 are arbitrary vectors. Therefore, by the definition of W 1 + W 2 , there exists vectors ~x 1 ,~ y 1 ∈ W 1 and ~x 2 ,~ y 2 ∈ W 2 such that ~x = ~x 1 + ~x 2 and ~ y = ~ y 1 + ~ y 2 . Therefore ~x + ~ y = ( ~x 1 + ~x 2 ) + ( ~ y 1 + ~ y 2 ) = ( ~x 1 + ~ y 1 ) + ( ~x 2 + ~ y 2 ) Since ~x 1 ,~ y 1 ∈ W 1 and W 1 is a subspace of V , ~x 1 + ~ y 1 ∈ W 1 . Since ~x 2 ,~ y 2 ∈ W 2 and W 2 is a subspace of V , ~x 2 + ~ y 2 ∈ W 2 . Therefore, we have shown that....
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This note was uploaded on 04/09/2012 for the course MATH 172a taught by Professor Kong,l during the Fall '08 term at UCLA.
 Fall '08
 Kong,L
 Math, Vector Space

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