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-x = x 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 following? Ax = ix Ax = Ax Ax = ix Ax = ix AXA-1 = ixA-1 OXAA-1 = 2A-x XI = JA-x x = 2A-1x A/(Ax) = A/(ix) Ax/A = ix/A O(A/A)x = (A/2)x Ix = (A/2)x x = JA-1x A-'Ax = A-lax Ix = JA-x x = 1A-x O(A/A)x = ixA-1 Ix = iXA-1 x = ixA-1 = 1x A-1x = 1x A-lx = 1x A-1x = 1x This shows that -Select--- v is an eigenvector of A- 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
Problem 41EQ
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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?
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 following?
Ax = Ax
Ax = ix
Ax = Ax
Ax = ix
A/(Ax) = A/(Ax)
Ax/A = ix/A
AXA-1 = ixA-1
A-'Ax = A-lax
Ix = JA-1x
O(A/A)x = (A/2)x
Ix = (A/1)x
O(A/A)x = iXA-1
Ix = iXA-1
x = iXA-1
OXAA-1 = 1A-1x
XI = AA-x
x = 1A-1x
x = AA-x
x = 1A-1x
A-lx =
1x
A-lx = 1x
A-'x = x
A-x = 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? 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 following? Ax = Ax Ax = ix Ax = Ax Ax = ix A/(Ax) = A/(Ax) Ax/A = ix/A AXA-1 = ixA-1 A-'Ax = A-lax Ix = JA-1x O(A/A)x = (A/2)x Ix = (A/1)x O(A/A)x = iXA-1 Ix = iXA-1 x = iXA-1 OXAA-1 = 1A-1x XI = AA-x x = 1A-1x x = AA-x x = 1A-1x A-lx = 1x A-lx = 1x A-'x = x A-x = 1x This shows that ---Select--- is an eigenvector of A with eigenvalue --Select--- v ...................... ..
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