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 1. Using matrix operations and the properties of inverse matrices gives which of the following? Ax = ix Ax = ix Ax = ix AxA-1 = ixA-1 A/ (Ax) = A / (àx) Ax /A = ix /A Ax = ix 4-1 = AA-1x A-'Ax = A-1¼x O(A/ A)x = (A / 2)x Ix = (A/ A)x O(A/ A)x = ixA-1 Ix = ixA-1 OXAA- XI = AA-1x Ix = AA-1x x = A-1x x = AA-lx 1. x = 1A-1x x = iXA-1 1x A-1x = 1x A-1x = A-1x = A-1x =

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- have the same eigenvectors. How are the eigenvalues of A related to the eigenvalues of A¯?
A-lx = x
Letting x be an eigenvector of A gives Ax = 1x for a corresponding eigenvalue 1. Using matrix operations and the properties of inverse matrices gives which of the following?
Ax =
Ax = 1x
Ax = 1x
Ax = 1x
AxA-1 = ixA-1
A/ (Ax) = A / (1x)
Ax / A = Ax / A
Ах / A
A-lAx = A¬1^x
(A/ A)x = (A / 1)x
(A / 1)x
XAA¬1 = 1A¬1×
(A/ A)x
= ixA-1
X.
Ix =
XI =
Ix =
Ix
iXA-1
x = AA-1x
X,
AA-1x
X.
x = 1XA-1
X =
X.
X =
X =
A-1x
1x
=
A-lx = Ix
A-1x = x
A-1x
-1
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- have the same eigenvectors. How are the eigenvalues of A related to the eigenvalues of A¯? A-lx = x Letting x be an eigenvector of A gives Ax = 1x for a corresponding eigenvalue 1. Using matrix operations and the properties of inverse matrices gives which of the following? Ax = Ax = 1x Ax = 1x Ax = 1x AxA-1 = ixA-1 A/ (Ax) = A / (1x) Ax / A = Ax / A Ах / A A-lAx = A¬1^x (A/ A)x = (A / 1)x (A / 1)x XAA¬1 = 1A¬1× (A/ A)x = ixA-1 X. Ix = XI = Ix = Ix iXA-1 x = AA-1x X, AA-1x X. x = 1XA-1 X = X. X = X = A-1x 1x = A-lx = Ix A-1x = x A-1x -1 This shows that ---Select--- v is an eigenvector of A with eigenvalue ---Select--- v
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