Explanation Given: In the vector space P 2 , p ( x ) = a 0 + a 1 x + a 2 x 2 and q ( x ) = b 0 + b 1 x + b 2 x 2 . Approach: An inner product on a real vector space V is a function that assigns a real number, 〈 u , v 〉 , to each pair of vectors u and v , and that satisfies the following properties: 1) 〈 u , u 〉 ≥ 0 and 〈 u , u 〉 = 0 if and only if u = 0 . 2) 〈 u , v 〉 = 〈 v , u 〉 3) 〈 a u , v 〉 = a 〈 u , v 〉 4) 〈 u , v + w 〉 = 〈 u , v 〉 + 〈 u , w 〉 Calculation: (Property-1) For the vector space P 2 , 〈 p , q 〉 = p ( 0 ) q ( 0 ) + p ( 1 ) q ( 1 ) + p ( 2 ) q ( 2 ) Here, p ( x ) = a 0 + a 1 x + a 2 x 2 And, q ( x ) = b 0 + b 1 x + b 2 x 2 Thus, p ( 0 ) = a 0 + a 1 ( 0 ) + a 2 ( 0 ) 2 = a 0 And, q ( 0 ) = b 0 + b 1 ( 0 ) + b 2 ( 0 ) 2 = b 2 Also, p ( 1 ) = a 0 + a 1 ( 1 ) + a 2 ( 1 ) 2 = a 0 + a 1 + a 2 p ( 2 ) = a 0 + a 1 ( 2 ) + a 2 ( 2 ) 2 = a 0 + 2 a 1 + 4 a 2 And, q ( 1 ) = b 0 + b 1 ( 1 ) + b 2 ( 1 ) 2 = b 0 + b 2 + b 2 q ( 2 ) = b 0 + b 1 ( 2 ) + b 2 ( 2 ) 2 = a 0 + 2 b 1 + 4 b 2 So, by property (1) 〈 p , p 〉 = p ( 0 ) p ( 0 ) + p ( 1 ) p ( 1 ) + p ( 2 ) p ( 2 ) = ( a 0 ) 2 + ( a 0 + a 1 + a 2 ) 2 + ( a 0 + 2 a 1 + 4 a 2 ) 2 Since, the square of a real number is always positive, this implies that ( a 0 ) 2 + ( a 0 + a 1 + a 2 ) 2 + ( a 0 + 2 a 1 + 4 a 2 ) 2 > 0 So, 〈 p , p 〉 > 0 Also, 〈 p , p 〉 = 0 if and only if all of the a 0 , a 1 , a 2 are zero that is, p ( x ) = 0 . Thus, 〈 p , p 〉 ≥ 0 Hence, property (1) is satisfied. (Property-2) For the vector space P 2 , 〈 p , q 〉 = a 0 b 0 + a 1 b 1 + a 2 b 2 So, by the property (2) 〈 q , p 〉 = q ( 0 ) p ( 0 ) + q ( 1 ) p ( 1 ) + q ( 2 ) p ( 2 ) = p ( 0 ) q ( 0 ) + p ( 1 ) q ( 1 ) + p ( 2 ) q ( 2 ) = 〈 p , q 〉 Hence, property (2) is satisfied

Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

5th Edition

ISBN: 9780134689531

Author: Lee Johnson, Dean Riess, Jimmy Arnold

Publisher: PEARSON

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Chapter 5.6, Problem 6E

To determine

To prove:

The expression 〈p,q〉=p(0)q(0)+p(1)q(1)+p(2)q(2) is an inner product on P2.

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Chapter 5.6, Problem 5E

Chapter 5.6, Problem 7E

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Chapter 5 Solutions

Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

Ch. 5.2 - For u,v and w given in given Exercises 13,...Ch. 5.2 - For u,v and w given in given Exercises 13,...Ch. 5.2 - For u,v and w given in given Exercises 13,...Ch. 5.2 - Prob. 4E Ch. 5.2 - Prob. 5E Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...Ch. 5.2 - Prob. 7E Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...

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To prove: The expression 〈 p , q 〉 = p ( 0 ) q ( 0 ) + p ( 1 ) q ( 1 ) + p ( 2 ) q ( 2 ) is an inner product on P 2 .

Solution Summary: The author proves that the expression langle p,qrangle =p(0)q

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To prove:

The expression 〈p,q〉=p(0)q(0)+p(1)q(1)+p(2)q(2) is an inner product on P2.

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Chapter 5 Solutions

To prove: The expression 〈 p , q 〉 = p ( 0 ) q ( 0 ) + p ( 1 ) q ( 1 ) + p ( 2 ) q ( 2 ) is an inner product on P 2 .

Solution Summary: The author proves that the expression langle p,qrangle =p(0)q

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To prove: The expression 〈p,q〉=p(0)q(0)+p(1)q(1)+p(2)q(2) is an inner product on P2.

Want to see the full answer?

Chapter 5 Solutions

To prove:

The expression 〈p,q〉=p(0)q(0)+p(1)q(1)+p(2)q(2) is an inner product on P2.