MIT18_024S11_soln_pset1

# MIT18_024S11_soln_pset1 - Solutions for PSet 1 1(1.10:22(a...

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Unformatted text preview: Solutions for PSet 1 1. (1.10:22) (a) Let S = { x 1 ,...,x k } ⊂ V . As L ( S ) = span ( S ), we can write: k L ( S ) = { y : y ∈ V where y = c i x i } c i is scalar i =1 For c j = 1 ,c i = 0 ,i = j , we have y = i c i x i = x j ∈ L ( S ). Thus x j ∈ S implies that x j ∈ L ( S ) and S ⊆ L ( S ). (b) As T is a subspace of linear space V , T is a non-empty subset of V and T satisfies all closure axioms. Since S ⊆ T , we know (using the notation above) that { x 1 ,...,x k } ⊆ T . Now let y ∈ L ( S ). Then by definition there exist c i ∈ R , for i = 1 ,...,k , such that y = c i x i . By the closure i axioms, c i x i ∈ T and thus L ( S ) ⊆ T . i (c) Since L ( S ) is a subspace of V , one direction is obvious. Now, suppose by contradiction that S is a subspace of V but S = L ( S ). Since S ⊂ L ( S ), this implies there exists y ∈ L ( S ) − S . As y ∈ L ( S ), there exist c i ∈ R , i = 1 ,...,k , such that y = As S is a subset i c i x i . and thus closed under addition and scalar multiplication, y ∈ S . This implies a contradiction and proves the result. (d) Assume S = { x 1 ,...,x k } ,T = { x 1 ,...,x n } where n ≥ k . Let y ∈ L ( S ). Then y = k c i x i for some c i ∈ R . For c j = for all j = k + 1 ,...,n , k i =1 n y = c i x i + c j x j . Thus, y ∈ L ( T ). i =1 j = k +1 (e) As S and T are subspaces of V , they are both closed under addition and scalar multiplication. Let x, y ∈ S ∩ T and c ∈ R . As cx + y ∈ S and cx + y ∈ T we...
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## This note was uploaded on 01/18/2012 for the course MATH 18.024 taught by Professor Christinebreiner during the Spring '11 term at MIT.

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MIT18_024S11_soln_pset1 - Solutions for PSet 1 1(1.10:22(a...

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