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Prove: similar matrices have the same characteristic polynomial

  1. Jul 10, 2006 #1
    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...
     
  2. jcsd
  3. Jul 10, 2006 #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.)
     
  4. Jul 10, 2006 #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].
     
  5. Jul 11, 2006 #4
    thanks i got it now (after a few manipulations of my own).
     
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