/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q21E Use the transpose definition of ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Q21E (page 331)

Use the transpose definition of the inner product to verify parts (b) and (c) of Theorem 1. Mention the appropriate facts from Chapter 2.

Short Answer

Expert verified

Using the definition of inner product, verified that:

  1. \(\left( {{\bf{u}} + {\bf{v}}} \right) \cdot {\bf{w}} = {\bf{u}} \cdot {\bf{w}} + {\bf{v}} \cdot {\bf{w}}\)
  2. \(\left( {c{\bf{u}}} \right) \cdot {\bf{v}} = c\left( {{\bf{u}} \cdot {\bf{v}}} \right)\)

Step by step solution

01

 Verification of theorem 1(b).

Part (b) of theorem 1 is,\(\left( {{\bf{u}} + {\bf{v}}} \right) \cdot {\bf{w}} = {\bf{u}} \cdot {\bf{w}} + {\bf{v}} \cdot {\bf{w}}\).

Prove it by using the definition of Inner Product.

\(\begin{aligned}{c}\left( {{\bf{u}} + {\bf{v}}} \right) \cdot {\bf{w}} = {\left( {{\bf{u}} + {\bf{v}}} \right)^T} \cdot {\bf{w}}\\ = \left( {{{\bf{u}}^T} + {{\bf{v}}^T}} \right) \cdot {\bf{w}}\\ = {{\bf{u}}^T} \cdot {\bf{w}} + {{\bf{v}}^T} \cdot {\bf{w}}\\ = {\bf{u}} \cdot {\bf{w}} + {\bf{v}} \cdot {\bf{w}}\end{aligned}\)

Hence, part (b) of theorem 1 has been verified.

02

 Verification of theorem 1(c).

Part (c) of theorem 1 is,\(\left( {c{\bf{u}}} \right) \cdot {\bf{v}} = c\left( {{\bf{u}} \cdot {\bf{v}}} \right)\).

Prove it by using the definition of Inner Product.

\(\begin{aligned}{c}\left( {c{\bf{u}}} \right) \cdot {\bf{v}} = {\left( {c{\bf{u}}} \right)^T} \cdot {\bf{v}}\\ = c\left( {{{\bf{u}}^T}{\bf{v}}} \right)\\ = c\left( {{\bf{u}} \cdot {\bf{v}}} \right)\end{aligned}\)

Hence, part (c) of theorem 1 has been verified.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

To measure the take-off performance of an airplane, the horizontal position of the plane was measured every second, from \(t = 0\) to \(t = 12\). The positions (in feet) were: 0, 8.8, 29.9, 62.0, 104.7, 159.1, 222.0, 294.5, 380.4, 471.1, 571.7, 686.8, 809.2.

a. Find the least-squares cubic curve \(y = {\beta _0} + {\beta _1}t + {\beta _2}{t^2} + {\beta _3}{t^3}\) for these data.

b. Use the result of part (a) to estimate the velocity of the plane when \(t = 4.5\) seconds.

(M) Use the method in this section to produce a \(QR\) factorization of the matrix in Exercise 24.

Let \({\mathbb{R}^{\bf{2}}}\) have the inner product of Example 1. Show that the Cauchy-Schwarz inequality holds for \({\bf{x}} = \left( {{\bf{3}}, - {\bf{2}}} \right)\) and \({\bf{y}} = \left( { - {\bf{2}},{\bf{1}}} \right)\). (Suggestion: Study \({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^{\bf{2}}}\).)

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

6. \(\left( {\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm x}\nolimits} }}} \right){\mathop{\rm x}\nolimits} \)

Find the distance between \({\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{10}\\{ - 3}\end{aligned}} \right)\) and \({\mathop{\rm y}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\{ - 5}\end{aligned}} \right)\).
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.