M115A-DirectSumOfSubspaces

# M115A-DirectSumOfSubspaces - Direct Sum of Subspaces Paul...

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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 non-empty 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 x -axis in R 3 and let S 2 be the y -axis in R 3 . Then S 1 + S 2 is the xy -plane in R 3 . To see this, we notice that if we add a vector in the direction of the x -axis to a vector in the direction of the y -axis, we obtain a vector in the xy -plane (that is ( x, 0 , 0) + (0 , y, 0) = ( x, y, 0) is in the xy -plane). Moreover, every vector in the xy -plane can be written as the sum of a vector in the direction of the x -axis and a vector in the direction of the y -axis (that is, ( x, y, 0) = ( x, 0 , 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 ~ 0 W 1 + W 2 . We recall that since W 1 and W 2 are subspaces of V , ~ 0 W 1 and ~ 0 W 2 . Therefore, since ~ 0 = ~ 0 + ~ 0, ~ 0 W 1 , and ~ 0 W 2 , ~ 0 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 ~x + ~ y is the sum of a vector in W 1 and a vector in W 2 . Hence ~x + ~ y W 1 + W 2 by the definition of W 1 + W 2 . Therefore, since ~x, ~ y W 1 + W 2 were arbitrary, W 1 + W 2 is closed under addition.

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