/*! 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} Q6.2-17E Question: In Exercises 17-22, de... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Q6.2-17E (page 331)

Question: In Exercises 17-22, determine which sets of vectors are orthonormal. If a set is only orthogonal, normalize the vectors to produce an orthonormal set.

17. \(\left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{1}{3}}\\{\frac{1}{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - \frac{1}{2}}\\0\\{\frac{1}{2}}\end{array}} \right)\)

Short Answer

Expert verified

The set of vectors \(\left\{ {{\bf{u}},{\bf{v}}} \right\}\) is not orthonormal.

The orthonormal set is \(\left\{ {\frac{{\bf{u}}}{{\left\| {\bf{u}} \right\|}},\frac{{\bf{v}}}{{\left\| {\bf{v}} \right\|}}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}{\frac{{\sqrt 3 }}{3}}\\{\frac{{\sqrt 3 }}{3}}\\{\frac{{\sqrt 3 }}{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - \frac{{\sqrt 2 }}{2}}\\0\\{\frac{{\sqrt 2 }}{2}}\end{array}} \right)} \right\}\).

Step by step solution

01

Definition of orthonormal sets

A set \(\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}, \ldots ,{{\bf{u}}_p}} \right\}\) is called anorthonormal setwhen it is an orthogonal set of unit vectors.

02

Check whether the set of vectors are orthogonal

Consider that, \({\bf{u}} = \left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{1}{3}}\\{\frac{1}{3}}\end{array}} \right),{\rm{ }}{\bf{v}} = \left( {\begin{array}{*{20}{c}}{ - \frac{1}{2}}\\0\\{\frac{1}{2}}\end{array}} \right)\).

Check whether the set of the vectors is orthogonal, as shown below:

\(\begin{array}{c}{\bf{u}} \cdot {\bf{v}} = \left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{1}{3}}\\{\frac{1}{3}}\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}{ - \frac{1}{2}}\\0\\{\frac{1}{2}}\end{array}} \right)\\ = \frac{1}{3}\left( { - \frac{1}{2}} \right) + \frac{1}{3}\left( 0 \right) + \frac{1}{3}\left( {\frac{1}{2}} \right)\\ = - \frac{1}{6} + 0 + \frac{1}{6}\\ = 0\end{array}\)

It is observed that \(\left\{ {{\bf{u}},{\bf{v}}} \right\}\) is an orthogonal set because \({\bf{u}} \cdot {\bf{v}} = 0\).

03

Check whether the set of vectors is orthonormal

Check whether the set of vectors is orthonormal as shown below:

\(\begin{array}{c}{\left\| {\bf{u}} \right\|^2} = {\bf{u}} \cdot {\bf{u}}\\ = {\left( {\frac{1}{3}} \right)^2} + {\left( {\frac{1}{3}} \right)^2} + {\left( {\frac{1}{3}} \right)^2}\\ = \left( {\frac{1}{9}} \right) + \left( {\frac{1}{9}} \right) + \left( {\frac{1}{9}} \right)\\ = \frac{3}{9}\\ = \frac{1}{3}\\{\left\| {\bf{v}} \right\|^2} = {\bf{v}} \cdot {\bf{v}}\\ = {\left( { - \frac{1}{2}} \right)^2} + {\left( 0 \right)^2} + {\left( {\frac{1}{2}} \right)^2}\\ = \left( {\frac{1}{4}} \right) + 0 + \left( {\frac{1}{4}} \right)\\ = \frac{2}{4}\\ = \frac{1}{2}\end{array}\)

It is observed that \(\left\{ {{\bf{u}},{\bf{v}}} \right\}\) is not an orthonormal set.

04

Normalize the vectors to produce an orthonormal set

Normalize the vectors to produce the orthonormal set as shown below:

\(\begin{array}{c}\left\{ {\frac{{\bf{u}}}{{\left\| {\bf{u}} \right\|}},\frac{{\bf{v}}}{{\left\| {\bf{v}} \right\|}}} \right\} = \left\{ {\frac{1}{{\sqrt {\frac{1}{3}} }}\left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{1}{3}}\\{\frac{1}{3}}\end{array}} \right),\frac{1}{{\sqrt {\frac{1}{2}} }}\left( {\begin{array}{*{20}{c}}{ - \frac{1}{2}}\\0\\{\frac{1}{2}}\end{array}} \right)} \right\}\\ = \left\{ {\left( {\begin{array}{*{20}{c}}{\frac{{\sqrt 3 }}{3}}\\{\frac{{\sqrt 3 }}{3}}\\{\frac{{\sqrt 3 }}{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - \frac{{\sqrt 2 }}{2}}\\0\\{\frac{{\sqrt 2 }}{2}}\end{array}} \right)} \right\}\end{array}\)

Therefore, the orthonormal set is \(\left\{ {\frac{{\bf{u}}}{{\left\| {\bf{u}} \right\|}},\frac{{\bf{v}}}{{\left\| {\bf{v}} \right\|}}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}{\frac{{\sqrt 3 }}{3}}\\{\frac{{\sqrt 3 }}{3}}\\{\frac{{\sqrt 3 }}{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - \frac{{\sqrt 2 }}{2}}\\0\\{\frac{{\sqrt 2 }}{2}}\end{array}} \right)} \right\}\).

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

In Exercises 9-12 find (a) the orthogonal projection of b onto \({\bf{Col}}A\) and (b) a least-squares solution of \(A{\bf{x}} = {\bf{b}}\).

12. \(A = \left[ {\begin{array}{{}{}}{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{ - {\bf{1}}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}\\{ - {\bf{1}}}&{\bf{1}}&{ - {\bf{1}}}\end{array}} \right]\), \({\bf{b}} = \left( {\begin{array}{{}{}}{\bf{2}}\\{\bf{5}}\\{\bf{6}}\\{\bf{6}}\end{array}} \right)\)

In Exercises 5-14, the space is \(C\left[ {0,2\pi } \right]\) with inner product (6).

7. Show that \({\left\| {\cos kt} \right\|^2} = \pi \) and \({\left\| {\sin kt} \right\|^2} = \pi \) for \(k > 0\).

Exercises 19 and 20 involve a design matrix \(X\) with two or more columns and a least-squares solution \(\hat \beta \) of \({\bf{y}} = X\beta \). Consider the following numbers.

(i) \({\left\| {X\hat \beta } \right\|^2}\)鈥攖he sum of the squares of the 鈥渞egression term.鈥� Denote this number by .

(ii) \({\left\| {{\bf{y}} - X\hat \beta } \right\|^2}\)鈥攖he sum of the squares for error term. Denote this number by \(SS\left( E \right)\).

(iii) \({\left\| {\bf{y}} \right\|^2}\)鈥攖he 鈥渢otal鈥� sum of the squares of the \(y\)-values. Denote this number by \(SS\left( T \right)\).

Every statistics text that discusses regression and the linear model \(y = X\beta + \in \) introduces these numbers, though terminology and notation vary somewhat. To simplify matters, assume that the mean of the -values is zero. In this case, \(SS\left( T \right)\) is proportional to what is called the variance of the set of -values.

19. Justify the equation \(SS\left( T \right) = SS\left( R \right) + SS\left( E \right)\). (Hint: Use a theorem, and explain why the hypotheses of the theorem are satisfied.) This equation is extremely important in statistics, both in regression theory and in the analysis of variance.

Let \(\overline x = \frac{1}{n}\left( {{x_1} + \cdots + {x_n}} \right)\), and \(\overline y = \frac{1}{n}\left( {{y_1} + \cdots + {y_n}} \right)\). Show that the least-squares line for the data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\) must pass through \(\left( {\overline x ,\overline y } \right)\). That is, show that \(\overline x \) and \(\overline y \) satisfies the linear equation \(\overline y = {\hat \beta _0} + {\hat \beta _1}\overline x \). (Hint: Derive this equation from the vector equation \({\bf{y}} = X{\bf{\hat \beta }} + \in \). Denote the first column of \(X\) by 1. Use the fact that the residual vector \( \in \) is orthogonal to the column space of \(X\) and hence is orthogonal to 1.)

Question: In Exercises 3-6, verify that\(\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\)is an orthogonal set, and then find the orthogonal projection of y onto\({\bf{Span}}\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\).

4.\(y = \left[ {\begin{aligned}{\bf{6}}\\{\bf{3}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{3}}\\{\bf{4}}\\{\bf{0}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{4}}}\\{\bf{3}}\\{\bf{0}}\end{aligned}} \right]\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

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.