For an invertible matrix A, prove that A and A-1 have the same eigenvectors. How are the eigenvalues of A related to the eigenvalues of A-1? Letting x be an eigenvector of A gives Ax = Ax for a corresponding eigenvalue 1. Using matrix operations and the properties of inverse matrices gives which of the following? Ax = Ax Ax = Ax Ax = ix Ax = Ax A/(Ax) = A/(Ax) O(A/A)x = (A/2)x Ix = (A/2)x x = AA-1x A-'Ax = A-1x Ax/A = Ax/A AxA-1 = ixA-1 O (A/A)x = ixA-1 Ix = AA-1x x = AA-1x OXAA-1 XI = AA-1x x = AA-1x Ix = ixA-1 x = AxA-1 A-lx = 1x %3D A-1x = 1x A-1x = 1x A-1x = 1x This shows that ---Select--- is an eigenvector of A-1 with eigenvalue --Select-

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter4: Eigenvalues And Eigenvectors
Section4.3: Eigenvalues And Eigenvectors Of N X N Matrices
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For an invertible matrix A, prove that A and A-1 have the same eigenvectors. How are the eigenvalues of A related to the eigenvalues of A1?
Letting x be an eigenvector of A gives Ax = ìx for a corresponding eigenvalue 1. Using matrix operations and the properties of inverse matrices gives which of the following?
Ax = 1x
Ax = Ax
Ax =
Ax = 1x
A/(Ax)
A/(1x)
Ax/A = 1x/A
AxA-1 =
A-1Ax = A-12x
Ix = AA-1x
= AA-1x
O(A/A)x = (A/1)x
(A/A)x
1xA-1
OXAA-1
XI = AA-1x
x = AA-1x
1x
Ix =
(A/2)x
Ix = ixA-1
X =
x = AxA-1
A-1x = 1x
A-1x = 1x
A-1x = 1x
A-1x
This shows that ---Select--- v is an eigenvector of A
with eigenvalue
--Select--- v
Transcribed Image Text:For an invertible matrix A, prove that A and A-1 have the same eigenvectors. How are the eigenvalues of A related to the eigenvalues of A1? Letting x be an eigenvector of A gives Ax = ìx for a corresponding eigenvalue 1. Using matrix operations and the properties of inverse matrices gives which of the following? Ax = 1x Ax = Ax Ax = Ax = 1x A/(Ax) A/(1x) Ax/A = 1x/A AxA-1 = A-1Ax = A-12x Ix = AA-1x = AA-1x O(A/A)x = (A/1)x (A/A)x 1xA-1 OXAA-1 XI = AA-1x x = AA-1x 1x Ix = (A/2)x Ix = ixA-1 X = x = AxA-1 A-1x = 1x A-1x = 1x A-1x = 1x A-1x This shows that ---Select--- v is an eigenvector of A with eigenvalue --Select--- v
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