Explanation Given: The given matrix is, A = [ − 1 0 0 − 2 ] S − 1 A S = J ... ( 4 ) Approach: Write the equation of characteristics polynomial for the matrix A . det ( A − λ I ) = 0 ... ( 1 ) Write the condition of generalized vector for a non zero vector v to the eigenvalue λ ( A − λ I ) v = 0 ... ( 2 ) The inverse of the matrix S is given by, S − 1 = adj ( S ) | S | ... ( 3 ) Here, p is positive integer and v is null space matrix. Calculation: Consider the matrix, A = [ − 1 0 0 − 2 ] Substitute [ − 1 0 0 − 2 ] for A and [ 1 0 0 1 ] for I in equation ( 1 ) . det ( [ − 1 0 0 − 2 ] − λ [ 1 0 0 1 ] ) = 0 det ( [ − 1 0 0 − 2 ] − [ λ 0 0 λ ] ) = 0 det ( [ − 1 − λ 0 0 − 2 − λ ] ) = 0 Solve further, | − 1 − λ 0 0 − 2 − λ = | 0 ( − 1 − λ ) ( − 2 − λ ) = 0 λ = − 1 , − 2 So, the Jordan canonical for is a diagonal matrix with its eigenvalue as diagonal matrix elements. JCF ( A ) = [ − 1 0 0 − 2 ] For λ = − 1 , [ 0 0 0 − 1 ] [ v 1 v 2 ] = [ 0 0 ] Here, − v 2 = 0 v 1 = 0 Substitute k for v 1 in above equation. v 1 = k 1 So the solution is, [ v 1 v 2 ] = [ k 1 0 ] = k 1 [ 1 0 ] So, the eigenvector corresponding to the eigenvalue λ = − 1 is [ 1 0 ] . The solution set is { ( r , 0 ) : r ∈ R } , and the corresponding eigenvector is ( 1 , 0 ) . For λ = − 2 , [ 0 0 0 − 1 ] [ v 1 v 2 ] = [ 0 0 ] Here, v 2 = k 2 v 1 = 0 Substitute k 2 for v 2 in 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 7.6, Problem 18P

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The Jordan canonical form $J$ for the matrix $A$ and an invertible matrix $S$ , such that $S−1AS=J$ .

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Chapter 7.6, Problem 17P

Chapter 7.6, Problem 19P

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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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The Jordan canonical form J for the matrix A and an invertible matrix S , such that S − 1 A S = J .

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The Jordan canonical form $J$ for the matrix $A$ and an invertible matrix $S$ , such that $S−1AS=J$ .

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Chapter 7 Solutions