[proofplan] We show that $tI - B = P^{-1}(tI - A)P$ when $B = P^{-1}AP$, then take determinants using [Determinant Multiplicativity](/theorems/395) to conclude $\chi_B(t) = \chi_A(t)$. This establishes $\chi_\alpha(t)$ as a well-defined invariant of $\alpha \in \mathrm{End}(V)$. [/proofplan] [step:Express $tI - B$ as a conjugate of $tI - A$] Since $B = P^{-1}AP$, compute \begin{align*} tI - B = tI - P^{-1}AP = P^{-1}(tI)P - P^{-1}AP = P^{-1}(tI - A)P. \end{align*} The first equality uses $tI = P^{-1}(tI)P$, which holds because $P^{-1}IP = I$. [/step] [step:Take determinants to conclude $\chi_B(t) = \chi_A(t)$] By [Determinant Multiplicativity](/theorems/395) applied to the product $P^{-1} \cdot (tI - A) \cdot P$ in $\mathrm{Mat}_n(\mathbb{F}[t])$: \begin{align*} \chi_B(t) = \det(tI - B) = \det(P^{-1}(tI - A)P) = \det(P^{-1}) \cdot \det(tI - A) \cdot \det(P). \end{align*} Since $\det(P^{-1}) \cdot \det(P) = 1$, this gives $\chi_B(t) = \det(tI - A) = \chi_A(t)$. [guided] Multiplicativity of the determinant was proved in [Determinant Multiplicativity](/theorems/395) for matrices over a field, but here we apply it to $P^{-1}(tI - A)P$ where $tI - A$ has entries in the polynomial ring $\mathbb{F}[t]$. This is valid because the proof of multiplicativity uses only the ring axioms for the entries, and $\mathbb{F}[t]$ is a commutative ring. Alternatively, one can view the identity as a polynomial identity: both sides are polynomials in $t$ over $\mathbb{F}$, and they agree for all $t \in \mathbb{F}$ by the field-level determinant multiplicativity, so they agree as polynomials (since a non-zero polynomial of degree $n$ has at most $n$ roots, but equality holds at infinitely many points if $\mathbb{F}$ is infinite; for finite fields, the algebraic argument via $\mathbb{F}[t]$ applies directly). [/guided] [/step]