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Question1: Find all the eigenvalues for the matrix A below 2 -1 0 A=-1 2 -1 0 -1 2

Question1: Find all the eigenvalues for the matrix A below 2 -1 0 A=-1 2 -1 0 -1 2

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter4: Eigenvalues And Eigenvectors
Section4.5: Iterative Methods For Computing Eigenvalues
Problem 15EQ
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Chapter 7: Eigenvalues Question1: Find all the eigenvalues for the matrix A below Solution: -1 Power Method: max A =-1 2 -1 [A VjT absolute value No error 0 -1 2 0.0000 1.0000 0.0000 -1.0000 2.0000 -1.0000 2.0000 Solution: The characteristic equation A-A1=0 is 2. |2-2 -1 3 -1 2-2 -il=0 4 -1 2-a 5 6. -0.7071 1.0000 0.7071-24141 3.4141 -2.4141 3.4141 0.0003 Cubic polynomial: (2-4)[(2-4)(2-4)-(-19(-1)] -(-1D[(-1)(2-2)-(-1)(0)] +o[(-1)(-1)-(2-2)(0)]=0 -12' +62-_A+_=0 0.7071 1.0000 -0.7071 The dominant eigenvalue is 3.4142 And its corresponding eigenvector is (-0,7071 1 -0.7071)' Solution: Inverse Power Method: "EQN - DEGREE → 3" . Eigenvalue 1=3.4142 Eigenvalue 2=-2 Eigenvalue 3= (0.75 0.5 0.25 A= 0.51 0.25 0.5 0.75 0.5 max No [A VT absolute value error Question 2: Given the matrix 0.0000 1.0000 0.0000 0.5000 1.0000 0.5000 1.0000 2 -1 0 A= -1 2 -1 0 -1 2 3. Use the power method to find the dominant eigenvalues, A / the smallest eigenvalues, 2 and its corresponding cigenvector, v, for the matrix A using V° = (0 1 0). Stop the iteration until |4-A|s0.005 6. 0.7071 1.0000 0.7071 1.2071 L7071 1.2071 1.7071 0.0002 0.7071 L0000 0.7071 The dominant cigenvalue is 1/1.7071. And its corre sponding cigenvector is (0.7071 1 07071) * * * * x>>
Transcribed Image Text:Chapter 7: Eigenvalues Question1: Find all the eigenvalues for the matrix A below Solution: -1 Power Method: max A =-1 2 -1 [A VjT absolute value No error 0 -1 2 0.0000 1.0000 0.0000 -1.0000 2.0000 -1.0000 2.0000 Solution: The characteristic equation A-A1=0 is 2. |2-2 -1 3 -1 2-2 -il=0 4 -1 2-a 5 6. -0.7071 1.0000 0.7071-24141 3.4141 -2.4141 3.4141 0.0003 Cubic polynomial: (2-4)[(2-4)(2-4)-(-19(-1)] -(-1D[(-1)(2-2)-(-1)(0)] +o[(-1)(-1)-(2-2)(0)]=0 -12' +62-_A+_=0 0.7071 1.0000 -0.7071 The dominant eigenvalue is 3.4142 And its corresponding eigenvector is (-0,7071 1 -0.7071)' Solution: Inverse Power Method: "EQN - DEGREE → 3" . Eigenvalue 1=3.4142 Eigenvalue 2=-2 Eigenvalue 3= (0.75 0.5 0.25 A= 0.51 0.25 0.5 0.75 0.5 max No [A VT absolute value error Question 2: Given the matrix 0.0000 1.0000 0.0000 0.5000 1.0000 0.5000 1.0000 2 -1 0 A= -1 2 -1 0 -1 2 3. Use the power method to find the dominant eigenvalues, A / the smallest eigenvalues, 2 and its corresponding cigenvector, v, for the matrix A using V° = (0 1 0). Stop the iteration until |4-A|s0.005 6. 0.7071 1.0000 0.7071 1.2071 L7071 1.2071 1.7071 0.0002 0.7071 L0000 0.7071 The dominant cigenvalue is 1/1.7071. And its corre sponding cigenvector is (0.7071 1 07071) * * * * x>>
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