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Lect13

# Lect13 - Chapter 3 Vector Spaces Math1111 Span of Vectors...

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Chapter 3. Vector Spaces Math1111 Span of Vectors Spanning set - Example Example . Let P 3 be the vector space consisting of all polynomials of degree < 3 . Let v 1 = 1 - x 2 , v 2 = x + 2 , v 3 = x 2 . Show that { v 1 , v 2 , v 3 } is a spanning set for P 3 .

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Chapter 3. Vector Spaces Math1111 Span of Vectors Spanning set - Example Example . Let P 3 be the vector space consisting of all polynomials of degree < 3 . Let v 1 = 1 - x 2 , v 2 = x + 2 , v 3 = x 2 . Show that { v 1 , v 2 , v 3 } is a spanning set for P 3 . Proof. Any element in P 3 is of the form p ( x ) = ax 2 + bx + c .
Chapter 3. Vector Spaces Math1111 Span of Vectors Spanning set - Example Example . Let P 3 be the vector space consisting of all polynomials of degree < 3 . Let v 1 = 1 - x 2 , v 2 = x + 2 , v 3 = x 2 . Show that { v 1 , v 2 , v 3 } is a spanning set for P 3 . Proof. For any p ( x ) = ax 2 + bx + c , want to find α , β , γ such that p ( x ) = α v 1 + β v 2 + γ v 3 . Any element in P 3 is of the form p ( x ) = ax 2 + bx + c .

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Chapter 3. Vector Spaces Math1111 Span of Vectors Spanning set - Example Example . Let P 3 be the vector space consisting of all polynomials of degree < 3 . Let v 1 = 1 - x 2 , v 2 = x + 2 , v 3 = x 2 . Show that { v 1 , v 2 , v 3 } is a spanning set for P 3 . Proof. For any p ( x ) = ax 2 + bx + c , want to find α , β , γ such that p ( x ) = α v 1 + β v 2 + γ v 3 . As α v 1 + β v 2 + γ v 3 = α ( 1 - x 2 ) + β ( x + 2 ) + γ x 2 = ( α + 2 β ) + β x + ( γ - α ) x 2 , it suffices to solve the system α + 2 β = c , β = b , - α + γ = a .
Chapter 3. Vector Spaces Math1111 Vector Spaces Homework 5 Reading Textbook - p.128-131 Homework 5 Chapter 3 - Section 2 Exercises: Qn. 9, 10, 11, 12, 13, 14.

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Chapter 3. Vector Spaces Math1111 Linear Independence Motivation Recall that both { e 1 , e 2 , e 3 , ( 1 2 3 ) T } and { e 1 , e 2 , e 3 } are spanning sets for 3 . The set { e 1 , e 2 , e 3 } is smaller in the sense that { e 1 , e 2 , e 3 } is a subset of { e 1 , e 2 , e 3 , ( 1 2 3 ) T } .
Chapter 3. Vector Spaces Math1111 Linear Independence Motivation Recall that both { e 1 , e 2 , e 3 , ( 1 2 3 ) T } and { e 1 , e 2 , e 3 } are spanning sets for 3 . The set { e 1 , e 2 , e 3 } is smaller in the sense that { e 1 , e 2 , e 3 } is a subset of { e 1 , e 2 , e 3 , ( 1 2 3 ) T } . Question Can we find a spanning set smaller than { e 1 , e 2 , e 3 } ?

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Chapter 3. Vector Spaces Math1111 Linear Independence Motivation Recall that both { e 1 , e 2 , e 3 , ( 1 2 3 ) T } and { e 1 , e 2 , e 3 } are spanning sets for 3 . The set { e 1 , e 2 , e 3 } is smaller in the sense that { e 1 , e 2 , e 3 } is a subset of { e 1 , e 2 , e 3 , ( 1 2 3 ) T } . Question Can we find a spanning set smaller than { e 1 , e 2 , e 3 } ? Ans. NO . i.e. { e 1 , e 2 , e 3 } is a minimal spanning set .
Chapter 3. Vector Spaces Math1111 Linear Independence Motivation Recall that both { e 1 , e 2 , e 3 , ( 1 2 3 ) T } and { e 1 , e 2 , e 3 } are spanning sets for 3 . The set { e 1 , e 2 , e 3 } is smaller in the sense that { e 1 , e 2 , e 3 } is a subset of { e 1 , e 2 , e 3 , ( 1 2 3 ) T } . Question Can we find a spanning set smaller than { e 1 , e 2 , e 3 } ? Ans. NO . i.e. { e 1 , e 2 , e 3 } is a minimal spanning set . Question When is a spanning set minimal?

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Chapter 3. Vector Spaces Math1111 Linear Independence Motivation (Cont’d) Revisit the example: { e 1 , e 2 , e 3 , ( 1 2 3 ) T } is a spanning set for 3 .
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