/*! 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} 17E Let \({\rm{y}} = \left( {\begin{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: 17E (page 331)

Let \({\rm{y}} = \left( {\begin{aligned}{{}}4\\8\\1\end{aligned}} \right)\), \({{\bf{u}}_1} = \left( {\begin{aligned}{{}}{{2 \mathord{\left/ {\vphantom {2 3}} \right.} 3}}\\{{1 \mathord{\left/

{\vphantom {1 3}} \right.} 3}}\\{{2 \mathord{\left/{\vphantom {2 3}} \right.} 3}}\end{aligned}} \right)\),\({{\bf{u}}_1} = \left( {\begin{aligned}{{}}{{{ - 2} \mathord{\left/{\vphantom {{ - 2} 3}} \right.} 3}}\\{{2 \mathord{\left/{\vphantom {2 3}} \right.} 3}}\\{{1 \mathord{\left/{\vphantom {1 3}} \right.} 3}}\end{aligned}} \right)\)and \(W = {\bf{Span}}\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\).

  1. Let\(U = \left( {\begin{aligned}{{}}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}\end{aligned}} \right)\). Compute \({U^T}U\) and \(U{U^T}\).
  2. Compute \({\bf{Pro}}{{\bf{j}}_W}{\bf{y}}\) and \(\left( {U{U^T}} \right){\bf{y}}\).

Short Answer

Expert verified
  1. The matrices are \[{U^T}U = \left[ {\begin{aligned}1&0\\0&1\end{aligned}} \right]\], and \[U{U^T} = \left[ {\begin{aligned}{\frac{8}{9}}&{\frac{{ - 2}}{9}}&{\frac{2}{9}}\\{\frac{{ - 2}}{9}}&{\frac{5}{9}}&{\frac{4}{9}}\\{\frac{2}{9}}&{\frac{4}{9}}&{\frac{5}{9}}\end{aligned}} \right]\].
  2. \[{\rm{Pro}}{{\rm{j}}_W}{\bf{y}} = U{U^T}{\bf{y}} = \left[ {\begin{aligned}2\\4\\5\end{aligned}} \right]\].

Step by step solution

01

Form \[U = \left[ {\begin{aligned}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}\end{aligned}} \right]\]

(a)

Given vectors are,\[{{\bf{u}}_1} = \left[ {\begin{aligned}{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\], and \[{{\bf{u}}_2} = \left[ {\begin{aligned}{{{ - 2} \mathord{\left/

{\vphantom {{ - 2} 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\].

Find \[U = \left[ {\begin{aligned}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}\end{aligned}} \right]\].

\[U = \left[ {\begin{aligned}{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{{ - 2} \mathord{\left/

{\vphantom {{ - 2} 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\]

02

Determine \[{U^T}U\] and \[U{U^T}\]

First, find the transpose of \[U\], that is \[{U^T}\].

\[{U^T} = \left[ {\begin{aligned}{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{{ - 2} \mathord{\left/

{\vphantom {{ - 2} 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\]

Now, compute \[{U^T}U\] and \[U{U^T}\].

\[\begin{aligned}{U^T}U = \left[ {\begin{aligned}{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{{ - 2} \mathord{\left/

{\vphantom {{ - 2} 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\left[ {\begin{aligned}{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{{ - 2} \mathord{\left/

{\vphantom {{ - 2} 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\\ = \left[ {\begin{aligned}{\frac{4}{9} + \frac{1}{9} + \frac{4}{9}}&{\frac{{ - 4}}{9} + \frac{2}{9} + \frac{2}{9}}\\{\frac{{ - 4}}{9} + \frac{2}{9} + \frac{2}{9}}&{\frac{4}{9} + \frac{4}{9} + \frac{1}{9}}\end{aligned}} \right]\\ = \left[ {\begin{aligned}1&0\\0&1\end{aligned}} \right]\end{aligned}\]

\[\begin{aligned}U{U^T} = \left[ {\begin{aligned}{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{{ - 2} \mathord{\left/

{\vphantom {{ - 2} 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\left[ {\begin{aligned}{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{{ - 2} \mathord{\left/

{\vphantom {{ - 2} 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\\ = \left[ {\begin{aligned}{\frac{4}{9} + \frac{4}{9}}&{\frac{2}{9} - \frac{4}{9}}&{\frac{4}{9} - \frac{2}{9}}\\{\frac{2}{9} - \frac{4}{9}}&{\frac{1}{9} + \frac{4}{9}}&{\frac{2}{9} + \frac{2}{9}}\\{\frac{4}{9} - \frac{2}{9}}&{\frac{2}{9} + \frac{2}{9}}&{\frac{4}{9} + \frac{1}{9}}\end{aligned}} \right]\\ = \left[ {\begin{aligned}{\frac{8}{9}}&{\frac{{ - 2}}{9}}&{\frac{2}{9}}\\{\frac{{ - 2}}{9}}&{\frac{5}{9}}&{\frac{4}{9}}\\{\frac{2}{9}}&{\frac{4}{9}}&{\frac{5}{9}}\end{aligned}} \right]\end{aligned}\]

Hence, \[{U^T}U = \left[ {\begin{aligned}1&0\\0&1\end{aligned}} \right]\] and \[U{U^T} = \left[ {\begin{aligned}{\frac{8}{9}}&{\frac{{ - 2}}{9}}&{\frac{2}{9}}\\{\frac{{ - 2}}{9}}&{\frac{5}{9}}&{\frac{4}{9}}\\{\frac{2}{9}}&{\frac{4}{9}}&{\frac{5}{9}}\end{aligned}} \right]\].

03

Write the Theorem

Let \[{\bf{y}} \in {\mathbb{R}^n}\], then \[{\rm{Pro}}{{\rm{j}}_W}{\bf{y}} = U{U^T}{\bf{y}}\], if \[U = \left[ {\begin{aligned}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}& \cdots &{{{\bf{u}}_p}}\end{aligned}} \right]\].

04

Find \[{\bf{Pro}}{{\bf{j}}_W}{\rm{y}}\] and \[\left( {{U^T}U} \right){\rm{y}}\] 

(b).

Since \[U = \left[ {\begin{aligned}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}\end{aligned}} \right]\] from part (a). then \[{\rm{Pro}}{{\rm{j}}_W}{\bf{y}} = U{U^T}{\bf{y}}\].

From part (a) \[U{U^T} = \left[ {\begin{aligned}{\frac{8}{9}}&{\frac{{ - 2}}{9}}&{\frac{2}{9}}\\{\frac{{ - 2}}{9}}&{\frac{5}{9}}&{\frac{4}{9}}\\{\frac{2}{9}}&{\frac{4}{9}}&{\frac{5}{9}}\end{aligned}} \right]\], then find \[U{U^T}{\bf{y}}\], where \[{\bf{y}} = \left[ {\begin{aligned}4\\8\\1\end{aligned}} \right]\].

\[\begin{aligned}U{U^T}{\bf{y}} = \left[ {\begin{aligned}{\frac{8}{9}}&{\frac{{ - 2}}{9}}&{\frac{2}{9}}\\{\frac{{ - 2}}{9}}&{\frac{5}{9}}&{\frac{4}{9}}\\{\frac{2}{9}}&{\frac{4}{9}}&{\frac{5}{9}}\end{aligned}} \right]\left[ {\begin{aligned}4\\8\\1\end{aligned}} \right]\\ = \left[ {\begin{aligned}{\frac{{32}}{9} - \frac{{16}}{9} + \frac{2}{9}}\\{\frac{{ - 8}}{9} + \frac{{40}}{9} + \frac{4}{9}}\\{\frac{8}{9} + \frac{{32}}{9} + \frac{5}{9}}\end{aligned}} \right]\\ = \left[ {\begin{aligned}2\\4\\5\end{aligned}} \right]\end{aligned}\]

So, \[{\rm{Pro}}{{\rm{j}}_W}{\bf{y}} = \left[ {\begin{aligned}2\\4\\5\end{aligned}} \right]\], as \[{\rm{Pro}}{{\rm{j}}_W}{\bf{y}} = U{U^T}{\bf{y}}\].

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 11 and 12, find the closest point to \[{\bf{y}}\] in the subspace \[W\] spanned by \[{{\bf{v}}_1}\], and \[{{\bf{v}}_2}\].

12. \[y = \left[ {\begin{aligned}3\\{ - 1}\\1\\{13}\end{aligned}} \right]\], \[{{\bf{v}}_1} = \left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right]\], \[{{\bf{v}}_2} = \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right]\]

a. Rewrite the data in Example 1 with new \(x\)-coordinates in mean deviation form. Let \(X\) be the associated design matrix. Why are the columns of \(X\) orthogonal?

b. Write the normal equations for the data in part (a), and solve them to find the least-squares line, \(y = {\beta _0} + {\beta _1}x*\), where \(x* = x - 5.5\).

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

Compute the quantities in Exercises 1-8 using the vectors

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

7. \(\left\| {\mathop{\rm w}\nolimits} \right\|\)

Let \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) be an orthonormal set in \({\mathbb{R}^n}\). Verify the following inequality, called Bessel鈥檚 inequality, which is true for each x in \({\mathbb{R}^n}\):

\({\left\| {\bf{x}} \right\|^2} \ge {\left| {{\bf{x}} \cdot {{\bf{v}}_1}} \right|^2} + {\left| {{\bf{x}} \cdot {{\bf{v}}_2}} \right|^2} + \ldots + {\left| {{\bf{x}} \cdot {{\bf{v}}_p}} \right|^2}\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

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.