Elegant proofs that similar matrices have the same characteristic polynomial?

Question

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).

Answer 1

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

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

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

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

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

$=\det(xI - A)$

Answer 2

(Too long for a comment.)

In lhf's proof, $x$ might be mistaken for a "general element of $R$." (I am taking matrices over a commutative ring $R$.)

Some of the things that are easy to overlook in their proof:

1. Characteristic polynomial is not the "function polynomial", so the equality is really to be checked in $R[x]$. (Note that same polynomial functions can be induced from different polynomials.)

2. But the proof above remains the same. This is because you can again "pull out" $M^{-1}$ and $M$ in $xI - M^{-1}AM$ but this time, in $R^{n\times n}[X]$. (I use $X$, and not $x$, deliberately, for clarity.) But this needs care.

3. Commutativity is required so that the determinant has the usual properties.