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 A1? 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 Ax = Ax Ax = Ax Ax = Ax A / (Ax) = A / (Ax) O(A / A)x = (A / A)x Ix = (A / A)x x = AA-1x 1x AxA-1 = AxA-1 Ax / A = Ax / A A-Ax = A-1Ax OXAA-1 = NA-1x xl = NA-1x x = NA-x O(A / A)x = AxA-1 Ix = AxA-1 Ix = AA-1x x = AA-1x x = AxA-1 A-x = x A-x = 1x A-1x = A-x = 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-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 A. Using matrix operations and the properties of inverse matrices gives
which of the following?
Ax = Ax
Ax = Ax
Ax = Ax
Ax = Ax
AxA-1 = AxA-1
A / (Ax) = A / (Ax)
Ax / A = Ax / A
A-1Ax = A-1Ax
XAA-1 = AA-x
xl = NA-1x
x = AAx
(A/ A)x = (A / A)x
Ix = (A / A)x
x = AA-1x
O(A / A)x = AXXA-1
Ix = AxA-1
Ix = AA-1x
x = AA-1x
1,
x = AxA-1
A-x = x
A-1x =x
A-1x = 1x
A-1x = 1x
This shows that --Select- v is an eigenvector of A1 with eigenvalue -Select--
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 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 Ax = Ax Ax = Ax Ax = Ax AxA-1 = AxA-1 A / (Ax) = A / (Ax) Ax / A = Ax / A A-1Ax = A-1Ax XAA-1 = AA-x xl = NA-1x x = AAx (A/ A)x = (A / A)x Ix = (A / A)x x = AA-1x O(A / A)x = AXXA-1 Ix = AxA-1 Ix = AA-1x x = AA-1x 1, x = AxA-1 A-x = x A-1x =x A-1x = 1x A-1x = 1x This shows that --Select- v is an eigenvector of A1 with eigenvalue -Select--
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