Explanation Given: The vector space ℝ 2 mapping is given as, 〈 v , w 〉 = v 1 w 1 − v 1 w 2 − v 2 w 1 + 4 v 2 w 2 Approach: Let, v , w and u be nonzero vectors in ℝ 2 as, v = ( v 1 , v 2 ) And, w = ( w 1 , w 2 ) And, u = ( u 1 , u 2 ) All the properties which need to be verified on inner product on ℝ 2 are given as, 1. 〈 v , v 〉 ≥ 0 , furthermore 〈 v , v 〉 = 0 if and only if v = 0 . 2. 〈 w , v 〉 = 〈 v , w 〉 . 3. 〈 k v , w 〉 = k 〈 v , w 〉 . 4. 〈 v + w , u 〉 = 〈 v , u 〉 + 〈 w , u 〉 . Here, the vectors v , w and u are in ℝ 2 and k is any real number. Calculation: Consider the first property, 〈 v , v 〉 ≥ 0 , furthermore 〈 v , v 〉 = 0 if and only if v = 0 . For vector v in ℝ 2 as v = ( v 1 , v 2 ) . Check whether the given vector space ℝ 2 mapping 〈 v , w 〉 = v 1 w 1 − v 1 w 2 − v 2 w 1 + 4 v 2 w 2 satisfies 〈 v , v 〉 ≥ 0 , furthermore 〈 v . v 〉 = 0 if and only if v = 0 . As, 〈 v , v 〉 = 〈 ( v 1 , v 2 ) , ( v 1 , v 2 ) 〉 = v 1 v 1 − v 1 v 2 − v 2 v 1 + 4 v 2 v 2 = v 1 2 − v 1 v 2 − v 2 v 1 + 4 v 2 2 = v 1 2 − 2 v 1 v 2 + v 2 2 + 3 v 2 2 Further solve the above equation. 〈 v , v 〉 = ( v 1 − v 2 ) 2 + 3 v 2 2 As, square of any real number is always positive. 〈 v , v 〉 = ( v 1 − v 2 ) 2 + 3 v 2 2 ≥ 0 If 〈 v , v 〉 = 0 , ( v 1 − v 2 ) 2 + 3 v 2 2 = 0 The above expression will be zero if and only if, v 1 = v 2 = 0 That is if v = 0 . Thus, first property that is 〈 v , v 〉 ≥ 0 , furthermore 〈 v , v 〉 = 0 if and only if v = 0 is satisfied by the mapping 〈 v , w 〉 = v 1 w 1 − v 1 w 2 − v 2 w 1 + 4 v 2 w 2 . Consider the second property. 〈 w , v 〉 = 〈 v , w 〉 For vector v and w in ℝ 2 as v = ( v 1 , v 2 ) and w = ( w 1 , w 2 ) . Check whether the given vector space ℝ 2 mapping 〈 v , w 〉 = v 1 w 1 − v 1 w 2 − v 2 w 1 + 4 v 2 w 2 satisfies 〈 w , v 〉 = 〈 v , w 〉 . As, 〈 v , w 〉 = 〈 ( v 1 , v 2 ) , ( w 1 , w 2 ) 〉 = v 1 w 1 − v 1 w 2 − v 2 w 1 + 4 v 2 w 2 … ( 1 ) Now, 〈 w , v 〉 = 〈 ( w 1 , w 2 ) , ( v 1 , v 2 ) 〉 = w 1 v 1 − w 2 v 1 − w 1 v 2 + 4 w 2 v 2 … ( 2 ) From equation ( 1 ) and ( 2 ) . 〈 w , v 〉 = 〈 v , w 〉 Thus, the second property that is 〈 w , v 〉 = 〈 v , w 〉 satisfied by the mapping 〈 v , w 〉 = v 1 w 1 − v 1 w 2 − v 2 w 1 + 4 v 2 w 2 . Consider the third property. 〈 k v , w 〉 = k 〈 v , w 〉 For vector v and w in ℝ 2 as v = ( v 1 , v 2 ) and w = ( w 1 , w 2 ) , k is all real number

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Chapter 5.1, Problem 27P

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To verify:

That the 〈v,w〉=v1w1−v1w2−v2w1+4v2w2 defines a valid inner product on ℝ2.

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Chapter 5.1, Problem 26P

Chapter 5.1, Problem 28P

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Chapter 5 Solutions

Differential Equations and Linear Algebra (4th Edition)

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To verify: That the 〈 v , w 〉 = v 1 w 1 − v 1 w 2 − v 2 w 1 + 4 v 2 w 2 defines a valid inner product on ℝ 2 .

Solution Summary: The author explains how the langle v,wrangle =v_1 satisfies all the four properties of inner product on R2.

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To verify:

That the 〈v,w〉=v1w1−v1w2−v2w1+4v2w2 defines a valid inner product on ℝ2.

Want to see the full answer?

Chapter 5 Solutions

To verify: That the 〈 v , w 〉 = v 1 w 1 − v 1 w 2 − v 2 w 1 + 4 v 2 w 2 defines a valid inner product on ℝ 2 .

Solution Summary: The author explains how the langle v,wrangle =v_1 satisfies all the four properties of inner product on R2.

Videos

To verify: That the 〈v,w〉=v1w1−v1w2−v2w1+4v2w2 defines a valid inner product on ℝ2.

Want to see the full answer?

Chapter 5 Solutions

To verify:

That the 〈v,w〉=v1w1−v1w2−v2w1+4v2w2 defines a valid inner product on ℝ2.