**Proof plan.** The [Rational Canonical Form](/theorems/863) decomposes $V$ into $\alpha$-invariant cyclic summands $V_i \cong \mathbb{F}[X]/(f_i)$ with $f_i \mid f_s$ (the minimal polynomial). The characteristic polynomial is $\chi_\alpha = f_1 \cdots f_s$. Since $f_i \mid f_s$ for all $i$, every $f_i(\alpha)$ kills $V_i$; and since $\chi_\alpha = f_1 \cdots f_s$ and each factor kills the corresponding summand, $\chi_\alpha(\alpha) = 0$. **Step 1: Decompose using rational canonical form.** By the [Rational Canonical Form](/theorems/863), \begin{align*} V = V_1 \oplus V_2 \oplus \cdots \oplus V_s \end{align*} with each $V_i \cong \mathbb{F}[X]/(f_i)$ an $\alpha$-invariant subspace, $f_1 \mid f_2 \mid \cdots \mid f_s$. The minimal polynomial is $f_s$ and the characteristic polynomial is $\chi_\alpha = f_1 f_2 \cdots f_s$. **Step 2: $\chi_\alpha(\alpha)$ kills each summand.** [claim: Kills Each Summand] For each $i$, $\chi_\alpha(\alpha)$ kills $V_i$. [/claim] [proof] Since $f_i \mid \chi_\alpha$ (as $\chi_\alpha = f_1 \cdots f_s$ and $f_i$ is one factor), we can write $\chi_\alpha = f_i \cdot g$ for some $g \in \mathbb{F}[X]$. For any $v \in V_i$, $f_i(\alpha)(v) = 0$ (since $V_i \cong \mathbb{F}[X]/(f_i)$ and $f_i$ is the annihilator polynomial of $V_i$). Therefore \begin{align*} \chi_\alpha(\alpha)(v) = (f_i \cdot g)(\alpha)(v) = g(\alpha)(f_i(\alpha)(v)) = g(\alpha)(0) = 0. \end{align*} [/proof] **Step 3: Conclusion.** Since $V = V_1 \oplus \cdots \oplus V_s$ and $\chi_\alpha(\alpha)$ kills each $V_i$, we conclude $\chi_\alpha(\alpha)(v) = 0$ for all $v \in V$, i.e. $\chi_\alpha(\alpha) = 0$. $\square$