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Lect16

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

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Chapter 3. Vector Spaces Math1111 Basis and Dimension Examples (Cont’d) Example . Let U and V be subspaces of a vector space. If U V = { 0 } , show that dim ( U + V ) = dim U + dim V . Proof . Let dim U = m and { u 1 , ··· , u m } be a basis of U . Let dim V = n and { v 1 , ··· , v n } be a basis of V . Want to show: u 1 , ··· , u m , v 1 , ··· , v n form a basis for U + V . Definition When U V = { 0 } , the subspace U + V is called the direct sum of U and V , denoted by U V .

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Chapter 3. Vector Spaces Math1111 Basis and Dimension Examples (Cont’d) Theorem ( Dimension Theorem ) Let U and V be subspaces of a vector space. Then dim ( U + V ) = dim U + dim V - dim ( U V ) .
Chapter 3. Vector Spaces Math1111 Basis and Dimension Examples (Cont’d) Theorem ( Dimension Theorem ) Let U and V be subspaces of a vector space. Then dim ( U + V ) = dim U + dim V - dim ( U V ) . Remark dim ( U V ) = dim U + dim V

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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 .
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 R 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 .

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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 R 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
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 R 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 ) .

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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 R 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
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 R 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 .

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

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