Similar linear transformations have the same characteristic and minimal polynomials

Let V be an n-dimensional vector space over a field F. Prove that similar linear transformations on V have the same characteristic and the same minimal polynomial.

Suppose A and B are similar matrices (i.e. linear transformations) on V. Then we have an invertible matrix P such that B = P^{-1}AP.

Now \mathsf{char}(B) = \mathsf{det}(xI - B) = \mathsf{det}(xI - P^{-1}AP) = \mathsf{det}(P^{-1}xP - P^{-1}AP) = \mathsf{det}(P^{-1})\mathsf{det}(xI - A)\mathsf{det}(P) = \mathsf{det}(xI - P) = \mathsf{char}(A). So A and B have the same characteristic polynomial.

Recall that the minimal polynomial of a transformation T is the unique monomial generator of \mathsf{Ann}(V) in F[x] (under the usual action of F[x] on V). Thus, to show that A and B have the same minimal polynomial, it suffices to show that they have the same annihilator in F[x] under their respective actions on V.

To this end, suppose p(x) \in \mathsf{Ann}_A(V) (where \mathsf{Ann}_A denotes the annihilator induced by A). Then as a linear transformation, we have p(A) = 0. Say p(x) = \sum r_ix^i; then \sum r_iA^i = 0. But A^i = PB^iP^{-1}, so that \sum r_iPB^iP^{-1} = 0, and thus (left- and right-multiplying by P^{-1} and P, respectively) \sum r_iB^i = 0. So p(B) = 0 as a linear transformation, and we have p(x) \in \mathsf{Ann}_B(V). The reverse inclusion is similar, so that \mathsf{Ann}_A(V) = \mathsf{Ann}_B(V) as desired.

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