Prove: similar matrices have the same characteristic polynomial

[ak416]

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 I am not sure how it will be affected...

 

[matt grime]

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.)

 

[HallsofIvy]

det(AB)= det(A)det(B) so $det(Q^{-1}(1-\lambda P Q}=det(Q^{-1})det(1- \lambda P) det(Q)$.

 

[ak416]

thanks i got it now (after a few manipulations of my own).