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It's a simple exercise to show that two similar matrices has the same eigenvalues and eigenvectors (my favorite way is noting that they represent the same linear transformation in different bases).

However, to show that two matrices has the same characteristic polynomial it does not suffice to show that they have the same eigenvalues and eigenvectors - one needs to say something smart about the algebraic multiplicities of the eigenvalues. Moreover, we might be working over a field which is not algebraically closed and hence simply "don't have" all the eigenvalues. This can be overcome, of course, by working in the algebraic closure of the field, but it complicates the explanation.

I'm looking for a proof that is simple and stand-alone as much as possible (the goal is writing an expository article about the subject, so clarity is the most important thing, not efficiency).

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    Ind$e$ed, this is just plain wrong; what is correct is that two similar matrices can be viewed as representing the same linear transformation in different bases, and then their eigenvectors are "the same" in the sense that they are two representations (in the different bases) of the coordinates of the eigenvectors of the transformation.2011-12-03

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If you define the characteristic polynomial of a matrix $A$ to be $\det(xI - A)$, then for $M$ invertible we have:

$\det(xI - M^{-1} A M)$

$= \det(M^{-1} xI M - M^{-1} A M)$

$= \det(M^{-1} (xI-A) M)$

$= \det (M^{-1}) \det(xI-A) \det(M)$

$=\det(xI - A)$

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    @MarcvanLeeuwen Won't the simplest construction work? K naturally embeds in K[X], and any K-module can be enlarged to a K[X]-module (by taking the tensor product, I'm not sure), and an endomorphism A induces an endomorphism of this K[X]-module, and then the determinant of $Ix-A$ will be the characteristic polynomial.2016-10-21