Explanation Given: The given matrix A is, A = [ − 2 1 4 − 2 1 4 − 2 1 4 ] Approach: An n × n matrix that is similar to a diagonal matrix is said to be diagonalizable. An n × n matrix A is similar to a diagonal matrix if and only if A is nondefective. In such a case, if v 1 , v 2 , ......... , v n denote n linearly independent eigenvectors of A and S = [ v 1 , v 2 , .......... , v n ] , Then, S − 1 A S = diag ( λ 1 , λ 2 , ......... , λ n ) , Where λ 1 , λ 2 , ........ , λ n are the eigenvalues of A (non necessarily distinct) corresponding in the eigenvectors v 1 , v 2 , ............ , v n . Calculation: Consider the given matrix A , A = [ − 2 1 4 − 2 1 4 − 2 1 4 ] The order the matrix A is n = 3 . The characteristic equation of A is given as det ( A − λ I ) = 0 , det ( A − λ I ) = 0 | [ − 2 1 4 − 2 1 4 − 2 1 4 ] − [ λ 0 0 0 λ 0 0 0 λ ] | = 0 | [ − 2 − λ 1 4 − 2 1 − λ 4 − 2 1 4 − λ ] | = 0 Solve the above equation for the eigenvalues ( λ ) of A , | − 2 − λ 1 4 − 2 1 − λ 4 − 2 1 4 − λ | = 0 ( − 2 − λ ) | 1 − λ 4 1 4 − λ | − 1 | − 2 4 − 2 4 − λ | + 4 | − 2 1 − λ − 2 1 | = 0 { ( − 2 − λ ) [ ( 1 − λ ) ( 4 − λ ) − ( 1 ) ( 4 ) ] − 1 [ ( − 2 ) ( 4 − λ ) − ( − 2 ) ( 4 ) ] + 4 [ ( − 2 ) ( 1 ) − ( − 2 ) ( 1 − λ ) ] } = 0 { ( − 2 − λ ) [ 4 − λ − 4 λ + λ 2 − 4 ] − 1 [ − 8 + 2 λ + 8 ] + 4 [ − 2 + 2 − 2 λ ] } = 0 Further solve the above equation. { ( − 2 − λ ) [ λ 2 − 5 λ ] − [ 2 λ ] + 4 [ − 2 λ ] } = 0 { − 2 λ 2 + 10 λ − λ 3 + 5 λ 2 − 2 λ − 8 λ } = 0 { − λ 3 + 3 λ 2 } = 0 ( λ 2 ) ( − λ + 3 ) = 0 Further solve the above equation. λ = 3 , 0 , 0 Therefore, the eigenvalues of A are 3 , 0 and 0 . The eigenvector corresponding to the eigenvalue λ 1 = 3 is determined based on the linear system given as, ( A − λ I ) v = 0 ( [ − 2 1 4 − 2 1 4 − 2 1 4 ] − [ λ 0 0 0 λ 0 0 0 λ ] ) [ v 1 v 2 v 3 ] = [ 0 0 0 ] [ − 2 − λ 1 4 − 2 1 − λ 4 − 2 1 4 − λ ] [ v 1 v 2 v 3 ] = [ 0 0 0 ] Substitute 3 for λ in the above equation. [ − 2 − 3 1 4 − 2 1 − 3 4 − 2 1 4 − 3 ] [ v 1 v 2 v 3 ] = [ 0 0 0 ] [ − 5 1 4 − 2 − 2 4 − 2 1 1 ] [ v 1 v 2 v 3 ] = [ 0 0 0 ] The augmented matrix associated with the equation is, [ A | 0 ] = [ − 5 1 4 − 2 − 2 4 − 2 1 1 | 0 0 0 ] Apply R 3 → R 3 − R 2 . [ A | 0 ] = [ − 5 1 4 − 2 − 2 4 0 3 − 3 | 0 0 0 ] Apply R 2 → − 1 2 R 2 and R 3 → 1 3 R 3 . [ A | 0 ] = [ − 5 1 4 1 1 2 0 1 − 1 | 0 0 0 ] Apply R 2 → 5 R 2 + R 1 . [ A | 0 ] = [ − 5 1 4 0 6 − 6 0 1 − 1 | 0 0 0 ] Apply R 2 → 1 6 R 2 . [ A | 0 ] = [ − 5 1 4 0 1 − 1 0 1 − 1 | 0 0 0 ] Apply R 3 → R 3 − R 2 . [ A | 0 ] = [ − 5 1 4 0 1 − 1 0 0 0 | 0 0 0 ] Thus, [ − 5 ( v 1 ) + 1 ( v 2 ) + 4 ( v 3 ) 0 ( v 1 ) + 1 ( v 2 ) − 1 ( v 3 ) 0 ( v 1 ) + 0 ( v 2 ) + 0 ( v 3 ) ] = [ 0 0 0 ] [ − 5 v 1 + v 2 + 4 v 3 v 2 − v 3 0 ] = [ 0 0 0 ] − 5 v 1 + v 2 + 4 v 3 = 0 and v 2 − v 3 = 0 Further solve the above equation

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Chapter 7.3, Problem 9P

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Whether the given matrix A is diagonalizable, when possible find the matrix S such that S−1AS=diag(λ1,λ2,........,λn).

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Chapter 7.3, Problem 8P

Chapter 7.3, Problem 10P

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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

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Whether the given matrix A is diagonalizable, when possible find the matrix S such that S − 1 A S = diag ( λ 1 , λ 2 , ........ , λ n ) .

Whether the given matrix A is diagonalizable, when possible find the matrix S such that S−1AS=diag(λ1,λ2,........,λn).

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Chapter 7 Solutions