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Lect17

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

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Chapter 3. Vector Spaces Math1111 Basis and Dimension Examples (Cont’d) Example . Find dim N ( A ) if A = 1 2 1 0 2 1 2 1 1 1 0 1 0 0 1 .

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Chapter 3. Vector Spaces Math1111 Basis and Dimension Examples (Cont’d) Example . Find dim N ( A ) if A = 1 2 1 0 2 1 2 1 1 1 0 1 0 0 1 . Ans . In this case, N ( A ) = { x 5 : A x = 0 } The solutions of A x = 0 are x = α u + β v , where u = ( - 1 0 1 0 0 ) T and v = ( 0 - 1 0 1 1 ) T .
Chapter 3. Vector Spaces Math1111 Basis and Dimension Examples (Cont’d) Example . Find dim N ( A ) if A = 1 2 1 0 2 1 2 1 1 1 0 1 0 0 1 . Ans . In this case, N ( A ) = { x 5 : A x = 0 } The solutions of A x = 0 are x = α u + β v , where u = ( - 1 0 1 0 0 ) T and v = ( 0 - 1 0 1 1 ) T . Apply elementary row operations: 1 2 1 0 2 1 2 1 1 1 0 1 0 0 1 -→ ······ -→ 1 0 1 0 0 0 1 0 0 1 0 0 0 1 - 1

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Chapter 3. Vector Spaces Math1111 Basis and Dimension Examples (Cont’d) Example . Find dim N ( A ) if A = 1 2 1 0 2 1 2 1 1 1 0 1 0 0 1 . Ans . In this case, N ( A ) = { x 5 : A x = 0 } The solutions of A x = 0 are x = α u + β v , where u = ( - 1 0 1 0 0 ) T and v = ( 0 - 1 0 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 0 2 1 2 1 1 1 0 1 0 0 1 . Ans . In this case, N ( A ) = { x 5 : A x = 0 } The solutions of A x = 0 are x = α u + β v , where u = ( - 1 0 1 0 0 ) T and v = ( 0 - 1 0 1 1 ) T . N ( A ) = Span ( u , v ) . Note: u , v are linearly independent. Exercise

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Chapter 3. Vector Spaces Math1111 Basis and Dimension Examples (Cont’d) Example . Find dim N ( A ) if A = 1 2 1 0 2 1 2 1 1 1 0 1 0 0 1 . Ans . In this case, N ( A ) = { x 5 : A x = 0 } The solutions of A x = 0 are x = α u + β v , where u = ( - 1 0 1 0 0 ) T and v = ( 0 - 1 0 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 0 2 1 2 1 1 1 0 1 0 0 1 . Ans . In this case, N ( A ) = { x 5 : A x = 0 } The solutions of A x = 0 are x = α u + β v , where u = ( - 1 0 1 0 0 ) T and v = ( 0 - 1 0 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 = 0

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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 2 meant?

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Chapter 3. Vector Spaces Math1111 Change of Basis Motivation Question What is coordinate? For example, what is ( 4,3 ) in 2 meant? As { e 1 , e 2 } is a basis of 2 , we may represent it as 4 e 1 + 3 e 2 .
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