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Unformatted text preview: Unit 21 Subspaces of Rfractur m We know that Rfractur m is the set of all mvectors. We refer to Rfractur m as “ m space” , where a “space” is an infinite set (whose members all have the same dimension). In the previous unit, we looked at a small finite subset of a “space” and determined whether or not that set had a certain property, the property of being linearly independent. In this unit, we will study subsets of “spaces” (usually, but not always, infinite subsets) which have the property that any vector which is a linear combination of vectors in the subset is also in the subset. We call these spe cial subsets “subspaces” . Definition 21.1. A subspace of Rfractur m is a nonempty set S of vectors in Rfractur m such that both of the following conditions are met: 1. if vectorv 1 and vectorv 2 are both in S then the vector vectorv 1 + vectorv 2 is also in S (i.e., S is closed under addition); 2. if vectorv 1 is in S and k is any scalar then the vector kvectorv 1 is also in S (i.e., S is closed under scalar multiplication). Notice: The set Rfractur m meets both of these conditions, so Rfractur m can be consid ered as a subspace of Rfractur m . That is, if we say “consider any subspace of Rfractur m ”, that would include Rfractur m itself, and any properties which we prove to be true for all subspaces will also be true for Rfractur m . As well, anything we can do with or to a subspace, we can also do with or to Rfractur m . 1 Also Notice: The set S = { vector } (where vector ∈ Rfractur m ) is also a subspace of Rfractur m . To see this, let vectorv 1 ∈ S and vectorv 2 ∈ S . Then vectorv 1 = vectorv 2 = vector 0, since that is the only vector is S . Thus vectorv 1 + vectorv 2 = vector 0 + vector 0 = vector ∈ S . Also, for any scalar k , kvectorv 1 = k vector 0 = vector 0. We see that S is closed under both addition and scalar mul tiplication, so S is a subspace of Rfractur m . (This is the only finite set which is a subspace of Rfractur m .) Let’s look at examples of other subspaces. Example 1 . Is the set S of all vectors in Rfractur 4 with first and last components being 0 a subspace of Rfractur 4 ? Solution: We need to determine whether or not the set S has the 2 prop erties necessary for it to be a subspace. In order to check this, we need to define 2 “typical” members of the set, vectorv 1 and vectorv 2 . We need to express these vectors in a way which shows the form they have which identifies them as being members of S . To do this, we look at the way that the set S is defined. In this case, S is the set of vectors in Rfractur 4 which have first and fourth components being 0. Their other components may contain any value. So a typical element of S is a 4vector which has components 1 and 4 both being 0, and any nonspecific values in the second and third components, such as vectorv 1 = (0 , a, b, 0). Another typical vector in S also has 0’s in the first and fourth components, and has any other nonspecific values in components 2...
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 Fall '09
 OLDS,VICKY
 Calculus, Linear Algebra, Vectors, Linear Independence, Vector Space, Linear combination

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