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Unformatted text preview: M341 Homework Solution 10 Section 4.4: 12: If S is linearly dependent then by Theorem 4.8, there is a vector v in S such that v can be expressed as a linear combination of other vectors in S . Therefore v ∈ span( S − { v } ) . We also have ( S − { v } ) ⊆ span( S − { v } ). If follows that S ⊆ span( S − { v } ) and by Theorem 4.5, span ( S ) ⊆ span( S − { v } ). Obviously we have span ( S − { v } ) ⊆ Span ( S ) and therefore span ( S − { v } ) = span ( S ) for some v ∈ S . Conversely, if there is a vector v ∈ S such that span ( S −{ v } ) = span ( S ) . Then v ∈ span ( S −{ v } ) since v ∈ S ⊆ span ( S ) . Therefore we can nd vectors v 1 , v 2 , . . . , v n in S −{ v } such that v is a linear combination of v 1 , v 2 , . . . , v n . By Theorem 4.8, S is linearly dependent. 13(c): Let S = { [ x 1 , x 2 , x 3 , x 4 ] ∈ R 4  x i = ± 1 , for each i } . Then by simpli ed span method we can see that span ( S ) = R 4 since 1 1 1 1 1 1 1 − 1 1 1 − 1 1 1 1 − 1 − 1 1 − 1 1 1 1 − 1 1 − 1 1 − 1 − 1 1 1 − 1 − 1 − 1 − 1 1 1 1 − 1 1 1 − 1 − 1 1 − 1 1 − 1 1 − 1 − 1 − 1 − 1 1 1 − 1 − 1 1 − 1 − 1 − 1 − 1 1 − 1 − 1 − 1 − 1 is reduced to...
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 Spring '08
 HIETMANN
 Linear Algebra, Vector Space, −1, vk

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