If A is invertible, show that AB is similar to BA for all B.
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Similar matrices have the same characteristic polynomials if A or B is invertible
A= PBP^(-1)
B =P^-(1)BP
Characteristic polynomial of A = det(xI-A) = det(xI-PBP^(-1))
= det(xPIP^(-1) - PBP^(-1))
=det(P(xI-B)P^(-1))
=det(P) det(xI-B) det (P^(-1))
=characteristic polynomial of B
A= PBP^(-1)
B =P^-(1)BP
Characteristic polynomial of A = det(xI-A) = det(xI-PBP^(-1))
= det(xPIP^(-1) - PBP^(-1))
=det(P(xI-B)P^(-1))
=det(P) det(xI-B) det (P^(-1))
=characteristic polynomial of B
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How would you go about showing that Tr(A) = Tr(B) or that A and B have the same eigenvalues?
In order to show that AB~BA, shouldn't I prove all the properties of similar matrices?
In order to show that AB~BA, shouldn't I prove all the properties of similar matrices?
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THEY HAVE THE SAME LETTERS
DUH
DUH