Prove that A and A have the same eigenvalues. To prove that A and A' have the same eigenvalues it is sufficient to prove that their characteristic equations are the same. Start with the characteristic equation of A then use the properties of martix transposes and determinants to show it is equivalent to the characteristic equation of A. O 1a1 + ATI = 1(a) - Al = |(21 - A)I = 1ar + Al O jar - ATI = I(an)- ATI = I(a1 - A)TI = 121 - Al O jar - ATI = I(21)" Al = 1(A1 + A)TI = 1a1 - Al O jar + ATI = 1(a)7- ATI = |(a1 + A)I = jaI + Al Are the eigenspaces the same? O yes O no
Prove that A and A have the same eigenvalues. To prove that A and A' have the same eigenvalues it is sufficient to prove that their characteristic equations are the same. Start with the characteristic equation of A then use the properties of martix transposes and determinants to show it is equivalent to the characteristic equation of A. O 1a1 + ATI = 1(a) - Al = |(21 - A)I = 1ar + Al O jar - ATI = I(an)- ATI = I(a1 - A)TI = 121 - Al O jar - ATI = I(21)" Al = 1(A1 + A)TI = 1a1 - Al O jar + ATI = 1(a)7- ATI = |(a1 + A)I = jaI + Al Are the eigenspaces the same? O yes O no
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter7: Distance And Approximation
Section7.1: Inner Product Spaces
Problem 11AEXP
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![Prove that A and A' have the same eigenvalues.
To prove that A and A' have the same eigenvalues it is sufficient to prove that their characteristic equations are the same. Start with the characteristic equation of A' then use the properties of martix transposes and determinants to show it is equivalent to the
characteristic equation of A.
O 1a1 + AT| = |(ar)T - Al = 1(AI – A)'I = 1a1 + A|
O ja1 - ATI = 1(ar)T - ATI = 1(A1 - A)| = |A1 – A|
O 1A1 – A"| = 1(a1)" + Al = 1(AI + A)'I = 1A1 – A|
O jaI + ATI = |(ar)" + ATI = 1(A1 + A)TI = |A1 + A|
Are the eigenspaces the same?
O yes
O no](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F47d3e04e-8ca1-4941-9ce0-116d95754f88%2Fa6a0ec22-0748-4078-a3e8-5ac374f62c1a%2Fuxmev9om_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Prove that A and A' have the same eigenvalues.
To prove that A and A' have the same eigenvalues it is sufficient to prove that their characteristic equations are the same. Start with the characteristic equation of A' then use the properties of martix transposes and determinants to show it is equivalent to the
characteristic equation of A.
O 1a1 + AT| = |(ar)T - Al = 1(AI – A)'I = 1a1 + A|
O ja1 - ATI = 1(ar)T - ATI = 1(A1 - A)| = |A1 – A|
O 1A1 – A"| = 1(a1)" + Al = 1(AI + A)'I = 1A1 – A|
O jaI + ATI = |(ar)" + ATI = 1(A1 + A)TI = |A1 + A|
Are the eigenspaces the same?
O yes
O no
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