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 = Ax for a corresponding eigenvalue 2. Using matrix operations and the preperties of inverse matrices gives which of the following? Ax = Ax Ax = x Ax = ix AXA-1 = 2xA-1 = 2A-1x xI = JA-1x A/(Ax) = A/(Ax) Ax/A = Ax/A Ax = ix OA/A)x = (A/A)x Ix = (A/A)x O(A/A)x = ixA-1 Ix = 2xA-1 A-Ax = A-12x Ix = A-1x x = 1A-1x x = JA-x x= xA-1 A-x = 1x A-x = 1x A-x = 1x A-x = 1x This shows that-Select-- vis an eigenvector of A with eigenvalue -Select--v Select Need Help? 1/x 1/A Submit Answer

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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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 = 1x for a corresponding eigenvalue 1. Using matrix operations and the properties of inverse matrices gives which of the following?
%3D
Ax = Ax
Ax = ix
%3D
Ax = 1x
%3D
A/(Ax) = A/(1x)
Ax/A = x/A
Ax = Ax
AXA-1 = 1xA-1
= 1A-1x
XI = A-1x
x = A-1x
%3D
%3D
%3D
O(A/A)x = (A/2)x
Ix = (A/2)x
x = AA-1x
A- Ax = A-12x
Ix = A-1x
x = JA-1x
O(A/A)x = ixA-1
%3D
%3D
%3D
%3D
OXAA 1
Ix =
%3D
%3D
X = XA-1
%3D
%3D
A-x = 1x
A-1x = 1x
X.
%3D
A-x = 1x
%3D
%3D
A-1x = 1x
%3D
This shows that -Select-- vis an eigenvector of A with eigenvalue-Select--- v
-Select--
Need Help?
1/x
1/2
Submit Answer
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 = 1x for a corresponding eigenvalue 1. Using matrix operations and the properties of inverse matrices gives which of the following? %3D Ax = Ax Ax = ix %3D Ax = 1x %3D A/(Ax) = A/(1x) Ax/A = x/A Ax = Ax AXA-1 = 1xA-1 = 1A-1x XI = A-1x x = A-1x %3D %3D %3D O(A/A)x = (A/2)x Ix = (A/2)x x = AA-1x A- Ax = A-12x Ix = A-1x x = JA-1x O(A/A)x = ixA-1 %3D %3D %3D %3D OXAA 1 Ix = %3D %3D X = XA-1 %3D %3D A-x = 1x A-1x = 1x X. %3D A-x = 1x %3D %3D A-1x = 1x %3D This shows that -Select-- vis an eigenvector of A with eigenvalue-Select--- v -Select-- Need Help? 1/x 1/2 Submit Answer
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