Lect16

# Lect16 - Chapter 3. Vector Spaces Math1111 Basis and...

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Unformatted text preview: Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.4 iii) Theorem 3.4.4 iii) Let V be a vector space, dim V = n ≥ 1 . iii) Suppose v 1 , ··· , v m where m > n span V . We may drop suitably m- n vectors to get a basis. Proof. Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.4 iii) Theorem 3.4.4 iii) Let V be a vector space, dim V = n ≥ 1 . iii) Suppose v 1 , ··· , v m where m > n span V . We may drop suitably m- n vectors to get a basis. Proof. Suppose v 1 , ··· , v m ( m > n ) span V . Claim 1: We can drop m- n vectors from v 1 , ··· , v m such that the remaining n vectors span V . Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.4 iii) Theorem 3.4.4 iii) Let V be a vector space, dim V = n ≥ 1 . iii) Suppose v 1 , ··· , v m where m > n span V . We may drop suitably m- n vectors to get a basis. Proof. Suppose v 1 , ··· , v m ( m > n ) span V . Claim 1: We can drop m- n vectors from v 1 , ··· , v m such that the remaining n vectors span V . Proof. By Thm 3.4.1, v 1 , ··· , v m are linearly dependent. Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.4 iii) Theorem 3.4.4 iii) Let V be a vector space, dim V = n ≥ 1 . iii) Suppose v 1 , ··· , v m where m > n span V . We may drop suitably m- n vectors to get a basis. Proof. Suppose v 1 , ··· , v m ( m > n ) span V . Claim 1: We can drop m- n vectors from v 1 , ··· , v m such that the remaining n vectors span V . Proof. By Thm 3.4.1, v 1 , ··· , v m are linearly dependent. One of v 1 , ··· , v m is a linear combination of other m- 1 vectors. W.L.O.G, assume v m ∈ Span ( v 1 , ··· , v m- 1 ) . Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.4 iii) Theorem 3.4.4 iii) Let V be a vector space, dim V = n ≥ 1 . iii) Suppose v 1 , ··· , v m where m > n span V . We may drop suitably m- n vectors to get a basis. Proof. Suppose v 1 , ··· , v m ( m > n ) span V . Claim 1: We can drop m- n vectors from v 1 , ··· , v m such that the remaining n vectors span V . Proof. By Thm 3.4.1, v 1 , ··· , v m are linearly dependent. One of v 1 , ··· , v m is a linear combination of other m- 1 vectors. W.L.O.G, assume v m ∈ Span ( v 1 , ··· , v m- 1 ) ....
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## This note was uploaded on 09/06/2010 for the course MATH MATH1111 taught by Professor Forgot during the Fall '08 term at HKU.

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Lect16 - Chapter 3. Vector Spaces Math1111 Basis and...

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