Lect15 - Chapter 3 Vector Spaces Math1111 Basis and Dimension Theorem 3.4.1 Theorem 3.4.1 Let v 1 ·· v n be a spanning set for a vector space V

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Unformatted text preview: Chapter 3. Vector Spaces Math1111 Basis and Dimension Theorem 3.4.1 Theorem 3.4.1 Let { v 1 , ··· , v n } be a spanning set for a vector space V . Then any collection { u 1 , ··· , u m } of m vectors in V where m > n is linearly dependent. Chapter 3. Vector Spaces Math1111 Basis and Dimension Theorem 3.4.1 Theorem 3.4.1 Let { v 1 , ··· , v n } be a spanning set for a vector space V . Then any collection { u 1 , ··· , u m } of m vectors in V where m > n is linearly dependent. Proof. Let u 1 , ··· , u m ∈ V where m > n . Want to show: u 1 , ··· , u m is linearly dependent. Chapter 3. Vector Spaces Math1111 Basis and Dimension Theorem 3.4.1 Theorem 3.4.1 Let { v 1 , ··· , v n } be a spanning set for a vector space V . Then any collection { u 1 , ··· , u m } of m vectors in V where m > n is linearly dependent. Proof. Let u 1 , ··· , u m ∈ V where m > n . Want to show: u 1 , ··· , u m is linearly dependent. As V = Span ( v 1 , ··· , v n ) , we have u i = a i 1 v 1 + a i 2 v 2 + ··· + a in v n for i = 1, ··· , m , for some real numbers a ij ( 1 ≤ i ≤ m , 1 ≤ j ≤ n ). Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.1 (Cont’d) Consider c 1 u 1 + ··· + c m u m = . Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.1 (Cont’d) Consider c 1 u 1 + ··· + c m u m = . Want to find nontrival solutions for c 1 , c 2 , ··· , c m Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.1 (Cont’d) Consider c 1 u 1 + ··· + c m u m = . Want to find nontrival solutions for c 1 , c 2 , ··· , c m Note: c 1 u 1 + ··· + c m u m can be written as m X i = 1 a i 1 c i v 1 + m X i = 1 a i 2 c i v 2 + ··· + m X i = 1 a i n c i v n . Want to find nontrivial solution for the linear system ∑ m i = 1 a i 1 c i = , ∑ m i = 1 a i 2 c i = , ··· , ∑ m i = 1 a i n c i = . Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.1 (Cont’d) Consider c 1 u 1 + ··· + c m u m = . Note: c 1 u 1 + ··· + c m u m can be written as m X i = 1 a i 1 c i v 1 + m X i = 1 a i 2 c i v 2 + ··· + m X i = 1 a i n c i v n . Want to find nontrivial solution for the linear system ∑ m i = 1 a i 1 c i = , ∑ m i = 1 a i 2 c i = , ··· , ∑ m i = 1 a i n c i = . i.e. To solve      a 1 1 c 1 + a 2 1 c 2 + ··· + a m 1 c m = a 1 2 c 1 + a 2 2 c 2 + ··· + a m 2 c m = . . . a 1 n c 1 + a 2 n c 2 + ··· + a m n c m = Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.1 (Cont’d) Consider c 1 u 1 + ··· + c m u m = . Note: c 1 u 1 + ··· + c m u m can be written as m X i = 1 a i 1 c i v 1 + m X i = 1 a i 2 c i v 2 + ··· + m X i = 1 a i n c i v n ....
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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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Lect15 - Chapter 3 Vector Spaces Math1111 Basis and Dimension Theorem 3.4.1 Theorem 3.4.1 Let v 1 ·· v n be a spanning set for a vector space V

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