Explanation Given: The matrix is, A = [ 2 3 2 1 ] 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 ) = | 2 − λ 3 2 1 − λ | = ( 2 − λ ) ( 1 − λ ) − 6 = λ 2 − 3 λ − 4 = ( λ + 1 ) ( λ − 4 ) The characteristic equation of A is, ( λ + 1 ) ( λ − 4 ) = 0 λ = − 1 , 4 The eigenvalues for the matrix A are λ 1 = − 1 and λ 2 = 4 . The polynomial of the eigenvalues can be written as, p ( λ ) = ( λ + 1 ) ( λ − 4 ) Since the polynomial have power 1 then, Thus, the multiplicity of the eigenvalues λ 1 is 1 and λ 2 is 1

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ISBN: 9780321990167

Author: ANNIN

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Question

Chapter 7.2, Problem 17P

To determine

Whether the given matrix is defective or nondefective.

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Chapter 7.2, Problem 16P

Chapter 7.2, Problem 18P

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Chapter 7 Solutions

EBK DIFFERENTIAL EQUATIONS AND LINEAR A

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

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