MAT1341-L05-Subspaces1

# MAT1341-L05-Subspaces1 - SUBSPACES AND SPANS JOS ´ E MALAG...

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Unformatted text preview: SUBSPACES AND SPANS JOS ´ E MALAG ´ ON-L ´ OPEZ In a vector space we have two basic operations: addition and scalar multiplication. Linear algebra is about the study of the objects that are completely described in terms of such operations. Specifically, we will pay attention to: 1) all the elements in a vector space V that can be written in terms of the two basic operations, which leads to the notion of linear combination; 2) all the subsets of a vector space V that inherit a structure of vector space from the structure given in V , which leads to the notion of subspace. Subspaces A subset W of a vector space V is a subspace if W is a vector space under the operations of addition and scalar multiplication given on V . Example. A vector space V is a subspace of V . Also, if is the zero vector of V , then { } is a subspace of V . We say that { } is the zero subspace . Remark. It is possible that a subset S of a vector space V could have a structure of a vector space under operations different from the op- erations of addition and scalar product of V . In such case, S is not a subspace of V . As an example, take the set S of all the 2 × 2 matrices of the form parenleftbigg a 1 1 b parenrightbigg where a and b are real numbers. We have that S is a vector space under parenleftbigg a 1 1 b parenrightbigg + parenleftbigg a ′ 1 1 b ′ parenrightbigg = parenleftbigg a + a ′ 1 1 b + b ′ parenrightbigg and α⋆ parenleftbigg a 1 1 b parenrightbigg = parenleftbigg αa 1 1 αb parenrightbigg . But we have that S is not a vector space under the standard operations for M 2 × 2 . Since the standard operations for M 2 × 2 are different from the operations defined above, we have that S is not a subspace of M 2 × 2 . 1 If we take a look at the axioms defining a vector space we notice that the only axioms that we need to check for a subset W of a vector space V to see if it is a subspace are: • W is closed under addition; • existence of additive identity; • existence of additive inverse; • W is closed under scalar multiplication. Now, let w be any vector in W ⊆ V . If W is closed under scalar multiplication then − w = ( − 1) w is in W . In other words, we do not need to check the existence of additive inverse in W . Summarizing, we have obtained: Theorem 1. Let V be a vector space. Let W be a non-empty subspace of V . Then W is a subspace of V if and only if (1) The zero vector in V is in W . (2) W is closed under addition. (3) W is closed under scalar multiplication. Example. R is a subspace of C . Example. The set { ( x y ) T | x + y = 0 } is a subspace of R 2 under the standard operations. Indeed, notice that • (0 0) T is in W : we just observe that (0 0) T satisfies the equa- tion defining W , 0 + 0 = 0. Thus (0 0) T ∈ W ....
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MAT1341-L05-Subspaces1 - SUBSPACES AND SPANS JOS ´ E MALAG...

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