Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

[M] In Exercises 37–40, use a matrix program to find the eigenvalues of the matrix. Then use the method of Example 4 with a row reduction routine to produce a basis for each eigenspace.

37. \(\left( {\begin{array}{*{20}{c}}8&{ - 10}&{ - 5}\\2&{17}&2\\{ - 9}&{ - 18}&4\end{array}} \right)\)

Short Answer

Expert verified

The eigenvalues are \(\lambda = \left( {3,13,13} \right)\). The basis for eigenspace of \(\lambda = 3\) is \(N = \left( {\begin{array}{*{20}{c}}5\\{ - 2}\\9\end{array}} \right)\). The basis for eigenspace of \(\lambda = 13\) is \(N = \left\{ {\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)} \right\}\).

Step by step solution

01

Command for input the matrix in MATLAB

Write the command given below to input the matrix

\( > > {\rm{ A }} = {\rm{ }}\left( {{\rm{8 }} - 10{\rm{ }} - {\rm{5}};{\rm{ 2 17 2}};{\rm{ }} - {\rm{9 }} - 18{\rm{ 4}}} \right)\)

02

Command for finding the eigenvalues of the matrix

\( > > {\rm{ ev}} = {\rm{eig}}\left( {\rm{A}} \right)\)

The output will be\({\rm{ev}} = \left( {3,13,13} \right)\).

Hence, the eigenvalues are \(\lambda = \left( {3,13,13} \right)\).

03

Command for finding the null basis for each eigenvalue

Write the command given below to find the null basis corresponding to \(\lambda = 3\):

\( > > {\rm{ N}} = {\rm{A}} - {\rm{ev}}\left( 1 \right)*{\rm{eye}}\left( 3 \right)\)

The output is given below:

Hence, the basis for eigenspace of \(\lambda = 3\) is \(N = \left( {\begin{array}{*{20}{c}}5\\{ - 2}\\9\end{array}} \right)\).

Write the command given below to find the null basis corresponding to \(\lambda = 3\):

\( > > {\rm{ N}} = {\rm{A}} - {\rm{ev}}\left( 2 \right)*{\rm{eye}}\left( 3 \right)\)

The output is given below:

\(N = \left( {\begin{array}{*{20}{c}}{ - 2}&{ - 1}\\1&0\\0&1\end{array}} \right)\)

Hence, the basis for eigenspace of \(\lambda = 13\) is \(N = \left\{ {\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)} \right\}\).

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 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 \(A = \left( {\begin{array}{*{20}{c}}{.5}&{.2}&{.3}\\{.3}&{.8}&{.3}\\{.2}&0&{.4}\end{array}} \right)\), \({{\rm{v}}_1} = \left( {\begin{array}{*{20}{c}}{.3}\\{.6}\\{.1}\end{array}} \right)\), \({{\rm{v}}_2} = \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\2\end{array}} \right)\), \({{\rm{v}}_3} = \left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\) and \({\rm{w}} = \left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right)\).

  1. Show that \({{\rm{v}}_1}\), \({{\rm{v}}_2}\), and \({{\rm{v}}_3}\) are eigenvectors of \(A\). (Note: \(A\) is the stochastic matrix studied in Example 3 of Section 4.9.)
  2. Let \({{\rm{x}}_0}\) be any vector in \({\mathbb{R}^3}\) with non-negative entries whose sum is 1. (In section 4.9, \({{\rm{x}}_0}\) was called a probability vector.) Explain why there are constants \({c_1}\), \({c_2}\), and \({c_3}\) such that \({{\rm{x}}_0} = {c_1}{{\rm{v}}_1} + {c_2}{{\rm{v}}_2} + {c_3}{{\rm{v}}_3}\). Compute \({{\rm{w}}^T}{{\rm{x}}_0}\), and deduce that \({c_1} = 1\).
  3. For \(k = 1,2, \ldots ,\) define \({{\rm{x}}_k} = {A^k}{{\rm{x}}_0}\), with \({{\rm{x}}_0}\) as in part (b). Show that \({{\rm{x}}_k} \to {{\rm{v}}_1}\) as \(k\) increases.

Question 19: Let \(A\) be an \(n \times n\) matrix, and suppose A has \(n\) real eigenvalues, \({\lambda _1},...,{\lambda _n}\), repeated according to multiplicities, so that \(\det \left( {A - \lambda I} \right) = \left( {{\lambda _1} - \lambda } \right)\left( {{\lambda _2} - \lambda } \right) \ldots \left( {{\lambda _n} - \lambda } \right)\) . Explain why \(\det A\) is the product of the n eigenvalues of A. (This result is true for any square matrix when complex eigenvalues are considered.)

Question: Construct a random integer-valued \(4 \times 4\) matrix \(A\), and verify that \(A\) and \({A^T}\) have the same characteristic polynomial (the same eigenvalues with the same multiplicities). Do \(A\) and \({A^T}\) have the same eigenvectors? Make the same analysis of a \(5 \times 5\) matrix. Report the matrices and your conclusions.

(M)Use a matrix program to diagonalize

\(A = \left( {\begin{aligned}{*{20}{c}}{ - 3}&{ - 2}&0\\{14}&7&{ - 1}\\{ - 6}&{ - 3}&1\end{aligned}} \right)\)

If possible. Use the eigenvalue command to create the diagonal matrix \(D\). If the program has a command that produces eigenvectors, use it to create an invertible matrix \(P\). Then compute \(AP - PD\) and \(PD{P^{{\bf{ - 1}}}}\). Discuss your results.

For the Matrices A find real closed formulas for the trajectory x(t+1)=Ax(t)where x(0)=[01]

A=[43-34]

See all solutions

Recommended explanations on Math Textbooks

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