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 A12 Letting x be an eigenvector of A gives Ax = Ax for a corresponding eigenvalue 2. Using matrix operations and the properties of inverse matrices gives which of the following Ax = Ax Ах Ах XY= Xy Ax = Ax Ax/A = Ax/A AXA-1 AXA-1 A/(Ax) = A/(2x) %3D = AxA-1 Ix = AXA1 x = AxA O(A/A)x = (A/A)x Ix = (A/A)x AAx = A-2x Ix = AA-x x = JA-x OXAA-1 - JA-1x x = AA-x x = A-1x Alx = 1x Ax = 1x A1x = 1x A-lx = 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 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 A12
Letting x be an eigenvector of A gives Ax = Ax for a corresponding eigenvalue 2. Using matrix operations and the properties of inverse matrices gives which of the following?
XY = X
Ax/A = Ax/A
Ax = Ax
Ax = Ax
A/(Ax)
O(A/A)x = (A/2)x
Ax = Ax
AXA-1
AXA-1
%3D
A/(2x)
%3D
O(A/A)x = ixA-1
Ix = AxA-
x = AxA-1
Ax = 1x
OXAA-1 = 1A-1x
xI = Ax
x = Ax
AAx = A-lax
O Ix = A-x
x = JAx
%3!
Ix = (A/A)x
x = AA-1x
Ax = 1x
!!
A1x = 1x
A-lx = 1x
This shows that
Selectvis an eigenvector of A
-1
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 A12 Letting x be an eigenvector of A gives Ax = Ax for a corresponding eigenvalue 2. Using matrix operations and the properties of inverse matrices gives which of the following? XY = X Ax/A = Ax/A Ax = Ax Ax = Ax A/(Ax) O(A/A)x = (A/2)x Ax = Ax AXA-1 AXA-1 %3D A/(2x) %3D O(A/A)x = ixA-1 Ix = AxA- x = AxA-1 Ax = 1x OXAA-1 = 1A-1x xI = Ax x = Ax AAx = A-lax O Ix = A-x x = JAx %3! Ix = (A/A)x x = AA-1x Ax = 1x !! A1x = 1x A-lx = 1x This shows that Selectvis an eigenvector of A -1 with eigenvalue -Select- v
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