Explanation Given: The matrix is, A = [ 4 1 6 − 4 0 − 7 0 0 − 3 ] Approach: For a given n × n matrix A , the polynomial p ( λ ) defined by p ( λ ) = det ( A − λ I ) is called the characteristic polynomial of A , and the equation p ( λ ) = 0 is called the characteristic equation of A . The polynomial of the form, p ( λ ) = ( − 1 ) n ( λ − λ 1 ) m 1 ( λ − λ 2 ) m 2 ⋯ ( λ − λ k ) m k Thus, associated with each eigenvalue λ i is a number m i , called the multiplicity of λ i . Let A be an n × n matrix. For a given eigenvalue λ i , let E i denotes the set of all vectors v satisfying A v = λ i v . Then E i is called the eigenspace of a A corresponding to the eigenvalue λ i . Thus, E i is the solution set to the linear system ( A − λ i I ) v = 0 . If n i denotes the dimension of E i , then 1 ≤ n i ≤ m i for each i . In words, the dimension of the eigenspace corresponding to λ i is at most the multiplicity of λ i . An n × n matrix A that has n linearly independent eigenvectors is called nondefective. In such a case, we say that A has a complete set of eigenvectors. If A has less than n linearly independent eigenvectors, it is called defective. Calculation: Calculate the eigenvalues for the matrix A . The characteristic polynomial of A can be written as, p ( λ ) = | 4 − λ 1 6 − 4 − λ − 7 0 0 − 3 − λ | = ( 4 − λ ) [ − λ ( − 3 − λ ) ] − 1 [ − 4 ( − 3 − λ ) ] = ( 3 + λ ) ( λ 2 − 4 λ + 4 ) = ( λ + 3 ) ( λ − 2 ) 2 The characteristic equation of A is, ( λ + 3 ) ( λ − 2 ) 2 = 0 λ = − 3 , 2 , 2 The eigenvalues for the matrix A are λ 1 = − 3 and λ 2 = 2 . The polynomial of the eigenvalues can be written as, p ( λ ) = ( λ + 3 ) ( λ − 2 ) 2 Since the polynomial have power 1 and 2 then, Thus, the multiplicity of the eigenvalues λ 1 is 1 and λ 2 is 2 . Calculate the eigenvector corresponding to eigenvalue λ 1 = − 3 . ( A − λ i I ) v = 0 The augmented matrix for the system is, [ 7 1 6 0 − 4 3 − 7 0 0 0 0 0 ] Applying the elementary row operation, 7 R 2 + 4 R 1 → R 2 . [ 7 1 6 0 0 25 − 25 0 0 0 0 0 ] Applying the elementary row operation, 1 25 R 2 → R 2 . [ 7 1 6 0 0 1 − 1 0 0 0 0 0 ] Applying the elementary row operation, R 1 + 6 R 2 → R 1 . [ 7 7 0 0 0 1 − 1 0 0 0 0 0 ] Applying the elementary row operation, 1 7 R 1 → R 1

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

See similar textbooks

Question

Chapter 7.2, Problem 10P

To determine

The multiplicity of each eigenvalue, basis for each eigenspace, dimension of each eigenspace of the given matrix A and whether the matrix is defective or nondefective.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 7.2, Problem 9P

Chapter 7.2, Problem 11P

Students have asked these similar questions

Graph without using the calculator y-1 = | x+4 |

9:43 AS く Akbar © Printed in the United States 15) Scale: 1 cmal unit on both axes .ill 64% The graph above shows a straight line QT intersecting the y-axis at T. i State the co-ordinates of T. ii Calculate the gradient of QT 16) iii Determine the equation of QT. A (-1, 9) ||| i L Г (5 marks)

Pls help.

Chapter 7 Solutions

Differential Equations and Linear Algebra (4th Edition)

Ch. 7.1 - Prob. 1P Ch. 7.1 - Prob. 2P Ch. 7.1 - Prob. 3P Ch. 7.1 - Prob. 4P Ch. 7.1 - Prob. 5P Ch. 7.1 - Given that v1=(2,1) and v2=(1,1) are eigenvectors...Ch. 7.1 - Prob. 7P Ch. 7.1 - Prob. 8P Ch. 7.1 - Prob. 9P Ch. 7.1 - Prob. 11P

Knowledge Booster