Explanation Given: The matrix is A = [ 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 ] Approach: For the given ( n × n ) matrix A , find its Hessenberg form H = Q A Q − 1 . Then, find all the linearly independent eigenvectors for the matrix H . If { v 1 , v 2 , ... , v k } are the eigenvectors and λ i ’s; 1 ≤ i ≤ n are the eigenvalues with algebraic multiplicity m i for H such that k < n then find its generalized vector by solving ( H − I ) v = v i to get v i + 1 ; 1 ≤ i ≤ k . Here, v i + 1 is the generalized eigenvector associated with the eigenvalue of H corresponding to the vector v i . Then, the general solution of y ′ = H y is given by y ( t ) = ∑ i = 1 n e λ i t [ c 1 v i + c 2 ( v i + 1 + t v i ) + ... + c m i ( v i + ( m i − 1 ) + t v i + ( m i − 2 ) + ... + t m i − 1 ( m i − 1 ) ! v i ) ] . And the general solution of x ′ = A x is given by x ( t ) = Q − 1 y ( t ) . Calculation: The given matrix is an Hessenberg matrix so, consider H = [ 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 ] The characteristic equation for H is | H − λ I | = 0 | 1 − λ 0 0 0 1 1 − λ 0 0 0 1 1 − λ 0 0 0 1 1 − λ | = 0 ( 1 − λ ) ( 1 − λ ) ( 1 − λ ) ( 1 − λ ) = 0 This gives λ = 1 , 1 , 1 , 1 Let X be the eigenvector for H corresponding to the eigenvalue λ = 1 . Then, ( H − I ) X = 0 That is, [ 1 − 1 0 0 0 1 1 − 1 0 0 0 1 1 − 1 0 0 0 1 1 − 1 ] [ x 1 x 2 x 3 x 4 ] = [ 0 0 0 0 ] [ 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 ] [ x 1 x 2 x 3 x 4 ] = [ 0 0 0 0 ] This implies x 1 = 0 x 2 = 0 x 3 = 0 Thus, X = [ 0 0 0 x 4 ] = x 4 [ 0 0 0 1 ] Let V be the first generalized eigenvector associated with λ = 1 . Then, ( H − I ) V = [ 0 0 0 1 ] [ 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 ] [ v 1 v 2 v 3 v 4 ] = [ 0 0 0 1 ] This implies v 1 = 0 v 2 = 0 v 3 = 1 Thus, V = [ 0 0 1 0 ] Now, let V be the second generalized eigenvector associated with λ = 1

Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

5th Edition

ISBN: 9780134689531

Author: Lee Johnson, Dean Riess, Jimmy Arnold

Publisher: PEARSON

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Chapter 7.8, Problem 4E

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The general solution of x′=Ax.

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Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

Ch. 7.1 - In Exercises 16, find a symmetric matrix A such...Ch. 7.1 - Prob. 2E Ch. 7.1 - In Exercises 16, find a symmetric matrix A such...Ch. 7.1 - Prob. 4E Ch. 7.1 - In Exercises 16, find a symmetric matrix A such...Ch. 7.1 - Prob. 6E Ch. 7.1 - In Exercises 712, find a substitution x=Qy that...Ch. 7.1 - In Exercises 712, find a substitution x=Qy that...Ch. 7.1 - In Exercises 712, find a substitution x=Qy that...Ch. 7.1 - In Exercises 712, find a substitution x=Qy that...

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