Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Q4E (page 267)

In Exercises 1–4, the matrix A is followed by a sequence \(\left\{ {{{\bf{x}}_k}} \right\}\) produced by the power method. Use these data to estimate the largest eigenvalue of A, and give a corresponding eigenvector.

4.\(A = \left( {\begin{aligned}{ {20}{r}}{4.1}&{ - 6}\\3&{ - 4.4}\end{aligned}} \right)\)

\(\left( {\begin{aligned}{ {20}{l}}1\\1\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{c}}1\\{.7368}\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{c}}1\\{.7541}\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{c}}1\\{.7490}\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{c}}1\\{.7502}\end{aligned}} \right)\)

Short Answer

Expert verified

The largest eigenvalue of A is \( - .4012\) , and the corresponding eigenvector is \(\left( {\begin{aligned}{ {20}{c}}1\\{.7502}\end{aligned}} \right)\).

Step by step solution

01

Given information

A matrix \(A = \left( {\begin{aligned}{ {20}{l}}{4.1}&{ - 6}\\3&{ - 4.4}\end{aligned}} \right)\). A sequence \(\left\{ {{x_k}} \right\}\).

02

Find the Eigenvalue

Compute the value of\(A{x_k}\)and identify the largest entry as follows:

\(A{x_0} = \left( {\begin{aligned}{ {20}{c}}{4.1}&{ - 6}\\3&{ - 4.4}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{l}}1\\1\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}{ - 1.9}\\{ - 1.4}\end{aligned}} \right)\),\({\mu _0} = - 1.9\)

\(A{x_1} = \left( {\begin{aligned}{ {20}{c}}{4.1}&{ - 6}\\3&{ - 4.4}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}1\\{.7368}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}{ - .3208}\\{ - .24192}\end{aligned}} \right)\),\({\mu _1} = - .3208\)

\(A{x_2} = \left( {\begin{aligned}{ {20}{c}}{4.1}&{ - 6}\\3&{ - 4.4}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}1\\{.7541}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}{ - .4246}\\{ - .31804}\end{aligned}} \right)\),\({\mu _2} = - .4296\)

\(A{x_3} = \left( {\begin{aligned}{ {20}{c}}{4.1}&{ - 6}\\3&{ - 4.4}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}1\\{.7490}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}{ - .394}\\{.2956}\end{aligned}} \right)\),\({\mu _3} = - .394\)

\(A{x_4} = \left( {\begin{aligned}{ {20}{c}}{4.1}&{ - 6}\\3&{ - 4.4}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}1\\{.7502}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}{ - .4012}\\{ - .30088}\end{aligned}} \right)\),\({\mu _4} = - .4012\)

Hence, the largest absolute entry is 0.4012. So, the eigenvalue is equal to 0.4012. The corresponding eigenvector is \(\left( {\begin{aligned}{ {20}{c}}1\\{.7502}\end{aligned}} \right)\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Question: Let \(A = \left( {\begin{array}{*{20}{c}}{ - 6}&{28}&{21}\\4&{ - 15}&{ - 12}\\{ - 8}&a&{25}\end{array}} \right)\). For each value of \(a\) in the set \(\left\{ {32,31.9,31.8,32.1,32.2} \right\}\), compute the characteristic polynomial of \(A\) and the eigenvalues. In each case, create a graph of the characteristic polynomial \(p\left( t \right) = \det \left( {A - tI} \right)\) for \(0 \le t \le 3\). If possible, construct all graphs on one coordinate system. Describe how the graphs reveal the changes in the eigenvalues of \(a\) changes.

Question: Construct a random integer-valued \(4 \times 4\) matrix \(A\).

  1. Reduce \(A\) to echelon form \(U\) with no row scaling, and use \(U\) in formula (1) (before Example 2) to compute \(\det A\). (If \(A\) happens to be singular, start over with a new random matrix.)
  2. Compute the eigenvalues of \(A\) and the product of these eigenvalues (as accurately as possible).
  3. List the matrix \(A\), and, to four decimal places, list the pivots in \(U\) and the eigenvalues of \(A\). Compute \(\det A\) with your matrix program, and compare it with the products you found in (a) and (b).

Suppose \({\bf{x}}\) is an eigenvector of \(A\) corresponding to an eigenvalue \(\lambda \).

a. Show that \(x\) is an eigenvector of \(5I - A\). What is the corresponding eigenvalue?

b. Show that \(x\) is an eigenvector of \(5I - 3A + {A^2}\). What is the corresponding eigenvalue?

M] In Exercises 19 and 20, find (a) the largest eigenvalue and (b) the eigenvalue closest to zero. In each case, set \[{{\bf{x}}_{\bf{0}}}{\bf{ = }}\left( {{\bf{1,0,0,0}}} \right)\] and carry out approximations until the approximating sequence seems accurate to four decimal places. Include the approximate eigenvector.

19.\[A{\bf{=}}\left[{\begin{array}{*{20}{c}}{{\bf{10}}}&{\bf{7}}&{\bf{8}}&{\bf{7}}\\{\bf{7}}&{\bf{5}}&{\bf{6}}&{\bf{5}}\\{\bf{8}}&{\bf{6}}&{{\bf{10}}}&{\bf{9}}\\{\bf{7}}&{\bf{5}}&{\bf{9}}&{{\bf{10}}}\end{array}} \right]\]

Let\(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\) be a basis for a vector space\(V\). Find \(T\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right)\) when \(T\) isa linear transformation from \(V\) to \(V\) whose matrix relative to \(B\) is

\({\left( T \right)_B} = \left( {\begin{aligned}0&{}&{ - 6}&{}&1\\0&{}&5&{}&{ - 1}\\1&{}&{ - 2}&{}&7\end{aligned}} \right)\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

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.

Sign-up for free