Prove: similar matrices have the same characteristic polynomial

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Prove: Similar matrices have the same characteristic polynomial.

By characteristic polynomial of A i mean det(A-tI) where t is a scalar.
A is similar to B if A = Q^-1 B Q for some invertible matrix Q. (i.e. B is the matrix representation of the same linear transformation as A but under a different basis.)

I do know that similar matrices have the same determinant. This can be easily proved using det(AB) = det(A)det(B). But when you change the diagonal entries of the determinant im not sure how it will be affected...
 

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  • #2
matt grime
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You've just said you know the answer. (A and B similar implies they have the same char poly, which is what you were asked to prove.)
 
  • #3
HallsofIvy
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det(AB)= det(A)det(B) so [itex]det(Q^{-1}(1-\lambda P Q}=det(Q^{-1})det(1- \lambda P) det(Q)[/itex].
 
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thanks i got it now (after a few manipulations of my own).
 

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