Lect17 - Chapter 3. Vector Spaces Math1111 Basis and...

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Unformatted text preview: Chapter 3. Vector Spaces Math1111 Basis and Dimension Examples (Cont’d) Example . Find dim N ( A ) if A = 1 2 1 2 1 2 1 1 1 1 1 . Chapter 3. Vector Spaces Math1111 Basis and Dimension Examples (Cont’d) Example . Find dim N ( A ) if A = 1 2 1 2 1 2 1 1 1 1 1 . Ans . In this case, N ( A ) = { x ∈ R 5 : A x = } The solutions of A x = are x = α u + β v , where u = (- 1 1 ) T and v = (- 1 1 1 ) T . Chapter 3. Vector Spaces Math1111 Basis and Dimension Examples (Cont’d) Example . Find dim N ( A ) if A = 1 2 1 2 1 2 1 1 1 1 1 . Ans . In this case, N ( A ) = { x ∈ R 5 : A x = } The solutions of A x = are x = α u + β v , where u = (- 1 1 ) T and v = (- 1 1 1 ) T . Apply elementary row operations: 1 2 1 2 1 2 1 1 1 1 1 -→ ······ -→ 1 1 1 1 1- 1 Chapter 3. Vector Spaces Math1111 Basis and Dimension Examples (Cont’d) Example . Find dim N ( A ) if A = 1 2 1 2 1 2 1 1 1 1 1 . Ans . In this case, N ( A ) = { x ∈ R 5 : A x = } The solutions of A x = are x = α u + β v , where u = (- 1 1 ) T and v = (- 1 1 1 ) T . ∴ N ( A ) = Span ( u , v ) . Chapter 3. Vector Spaces Math1111 Basis and Dimension Examples (Cont’d) Example . Find dim N ( A ) if A = 1 2 1 2 1 2 1 1 1 1 1 . Ans . In this case, N ( A ) = { x ∈ R 5 : A x = } The solutions of A x = are x = α u + β v , where u = (- 1 1 ) T and v = (- 1 1 1 ) T . ∴ N ( A ) = Span ( u , v ) . Note: u , v are linearly independent. Exercise Chapter 3. Vector Spaces Math1111 Basis and Dimension Examples (Cont’d) Example . Find dim N ( A ) if A = 1 2 1 2 1 2 1 1 1 1 1 . Ans . In this case, N ( A ) = { x ∈ R 5 : A x = } The solutions of A x = are x = α u + β v , where u = (- 1 1 ) T and v = (- 1 1 1 ) T . ∴ N ( A ) = Span ( u , v ) . Note: u , v are linearly independent. Exercise ∴ { u , v } is a basis for N ( A ) , so dim N ( A ) = 2 . Chapter 3. Vector Spaces Math1111 Basis and Dimension Examples (Cont’d) Example . Find dim N ( A ) if A = 1 2 1 2 1 2 1 1 1 1 1 . Ans . In this case, N ( A ) = { x ∈ R 5 : A x = } The solutions of A x = are x = α u + β v , where u = (- 1 1 ) T and v = (- 1 1 1 ) T . ∴ N ( A ) = Span ( u , v ) . Note: u , v are linearly independent. Exercise ∴ { u , v } is a basis for N ( A ) , so dim N ( A ) = 2 . Observation dim N ( A ) = number of free variables in the solution of A x = Chapter 3. Vector Spaces Math1111 Vector Spaces Homework 9 Reading Textbook - p.148-150 Homework 9 Chapter 3 - Section 4 Exercises: Qn. 9, 10, 11, 12, 14, 16, 17, 18 Chapter 3. Vector Spaces Math1111 Change of Basis Motivation Question What is coordinate? For example, what is ( 4,3 ) in R 2 meant? Chapter 3. Vector SpacesChapter 3....
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This note was uploaded on 04/30/2010 for the course MATH 1111 taught by Professor Dr,li during the Spring '10 term at HKU.

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Lect17 - Chapter 3. Vector Spaces Math1111 Basis and...

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