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 = 1x for a corresponding eigenvalue 2. Using matrix operations and the properties of inverse matrices gives which of the following? Ax = Ax Ax = Ax Ax = Ax AxA-1 = AxA-1 = A-1x Ax = ix A-1Ax = A-1x Ix = AA-1x Ax/A = Ax/A A/(Ax) = A/(Ax) (A/A)x = (A/2)x Ix = (A/A)x x = AA-1x OXAA-1 xI = 1A-1x x = AA-1x O(A/A)x = ixA-1 Ix = ixA-1 x = AA-1x x = AxA-1 1x A-1x = 1x A-1x = 1x A-1x = 1x A-1x = This shows that ---Select--- vis an eigenvector of A-1 with eigenvalue --Select--- v Select-- Need Help? 1/x 1/2 ed Work Revert to Last Response

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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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 = 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 = Ax
Ax = Ax
AxA-1 = 1xA-1
Ax/A = Ax/A
A/(Ax) = A/(1x)
(A/A)x = (A/1)x
A-lAx = A-12x
Ix = AA-1x
x = 1A-1x
= AA-1x
(A/A)x
= 1xA-1
OXAA-1
Ix = (A/1)x
xI = AA-1x
Ix = AxA-1
x = 1A-1x
1x
x = 1A-1x
x = 1xA-1
A-1x =
1x
1x
A-1x = 1x
A-1x =
A-1x
This shows that
-Select--- v is an eigenvector of A-1 with eigenvalue
Select--- v
Select-
Need Help? 1/x
1/1
d Work Revert to Last Response
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 = 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 = Ax Ax = Ax AxA-1 = 1xA-1 Ax/A = Ax/A A/(Ax) = A/(1x) (A/A)x = (A/1)x A-lAx = A-12x Ix = AA-1x x = 1A-1x = AA-1x (A/A)x = 1xA-1 OXAA-1 Ix = (A/1)x xI = AA-1x Ix = AxA-1 x = 1A-1x 1x x = 1A-1x x = 1xA-1 A-1x = 1x 1x A-1x = 1x A-1x = A-1x This shows that -Select--- v is an eigenvector of A-1 with eigenvalue Select--- v Select- Need Help? 1/x 1/1 d Work Revert to Last Response
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