For an invertible matrix A, prove that A and A 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 = ix for a corresponding eigenvalue 2. Using matrix operations and the properties of inverse matrices gives which of the follo Ax = Ax AXA-1 = ixA-1 Ax = 1x Ax = 1x Ax = 1x Ax/A = Ax/A A/(Ax) = A/(2x) O(A/A)x = (A/2)x Ix = (A/2)x x = 1A-1x A-'Ax = A-1ix o(A/A)x = ixA-1 Ix = 1xA-1 x = 1xA-1 OXAA-1 = 1A-1x xI = 1A-1x x = 1A-1x A-1x = 1x Ix = JA-1x x = JA-1x A-1x = 1x A-1x = 1x A-1x = 1x This shows that ---Select--- v is an eigenvector of A- with eigenvalue ---Select-- v

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
Problem 24EQ
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For an invertible matrix A, prove that A and A 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 a. Using matrix operations and the properties of inverse matrices gives which of the following?
Ax = Ax
AXA-1 = ixA-1
Ax = 1x
Ax = 1x
Ax = 1x
Ax/A = Ax/A
A/(Ax) = A/(2x)
O(A/A)x = (A/2)x
Ix = (A/2)x
x = 1A-1x
A-'Ax = A-1ix
o(A/A)x = ixA-1
Ix = 1xA-1
x = 1xA-1
OXAA-1 = 14-x
xI = 1A-1x
x = 1A-1x
A-1x = 1x
Ix = 1A-1x
x = JA-1x
A-1x = 1x
A-1x = 1x
A-1x = 1x
This shows that --Select--
is an eigenvector of A- with eigenvalue --Select-- v
Transcribed Image Text:For an invertible matrix A, prove that A and A 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 a. Using matrix operations and the properties of inverse matrices gives which of the following? Ax = Ax AXA-1 = ixA-1 Ax = 1x Ax = 1x Ax = 1x Ax/A = Ax/A A/(Ax) = A/(2x) O(A/A)x = (A/2)x Ix = (A/2)x x = 1A-1x A-'Ax = A-1ix o(A/A)x = ixA-1 Ix = 1xA-1 x = 1xA-1 OXAA-1 = 14-x xI = 1A-1x x = 1A-1x A-1x = 1x Ix = 1A-1x x = JA-1x A-1x = 1x A-1x = 1x A-1x = 1x This shows that --Select-- is an eigenvector of A- with eigenvalue --Select-- v
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