Lecture2

Lecture2 - Lecture 2 Subspaces In most applications we will...

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Subspaces In most applications we will be working with a subset W of a vector space V such that W itself is a vector space. Question: Do we have to test all the axioms to find out if W is a vector space? The answer is NO. Theorem. Let W 6 = be a subset of a vector space V . Then W , with the same addition and scalar multiplication as V , is a vector space if and only if the following two conditions hold: 1. u + v W for all u,v W (or W + W W ) 2. r · u W for all r R and all u W (or R W W ). In this case we say that W is a subspace of V . Proof. Assume that W + W W and R W W . To show that W is a vector space we have to show that all the 10 axioms of Definition 1.1 hold for W . But that follows because the axioms hold for V and W is a subset of V : A1 (Commutativity of addition) For u,v W , we have u + v = v + u . This is because u,v are also in V and commutativity holds in V . 7
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Lecture2 - Lecture 2 Subspaces In most applications we will...

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