Characteristic polynomial does not depend on the basis

Statement

Let $K$ be a commutative ring, let $A$ be an $n \times n$ matrix over $K$, and let $P$ be an invertible $n \times n$ matrix over $K$. Then the characteristic polynomials of $A$ and $P^{-1} A P$ are equal: $$p_{P^{-1} A P}(t) = p_A(t).$$ So similar matrices have the same characteristic polynomial.

Remarks

The proof is $\det(t I - P^{-1} A P) = \det(P^{-1} (t I - A) P) = \det(t I - A)$. Hence the characteristic polynomial of a linear map does not depend on the basis. The converse fails: $I_2$ and $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ both have $p(t) = (t - 1)^2$, but they are not similar.

Explanations

L1 human

A change of basis does not change the characteristic polynomial. So the eigenvalues, the trace and the determinant, which we can read from it, are the same in every basis. They are properties of the linear map, not of one matrix.