**Proof plan.** View $V$ as a finitely generated $\mathbb{F}[X]$-module $V_\alpha$ via the [linear map](/page/Linear%20Map) $\alpha$. Apply the [Structure Theorem for Finitely Generated Modules over Euclidean Domains](/theorems/857) (with $R = \mathbb{F}[X]$, which is Euclidean). No free summands arise since $V$ is finite-dimensional. Reading the companion matrices off the cyclic summands gives the block form. **Step 1: $V_\alpha$ is a finitely generated $\mathbb{F}[X]$-module with no free part.** A basis $v_1, \ldots, v_n$ of $V$ over $\mathbb{F}$ also generates $V_\alpha$ over $\mathbb{F}[X]$ (since $\mathbb{F} \subseteq \mathbb{F}[X]$). The Cayley-Hamilton theorem guarantees that every $v \in V$ is annihilated by the characteristic polynomial, so $\operatorname{Ann}(V_\alpha) \neq 0$. Thus $V_\alpha$ has no free $\mathbb{F}[X]$-summand (a free summand $\mathbb{F}[X]$ is infinite-dimensional over $\mathbb{F}$). **Step 2: Apply the structure theorem.** By the [Structure Theorem for Finitely Generated Modules over Euclidean Domains](/theorems/857), there exist monic polynomials $f_1, \ldots, f_s \in \mathbb{F}[X]$ with $f_1 \mid f_2 \mid \cdots \mid f_s$ such that \begin{align*} V_\alpha \cong \frac{\mathbb{F}[X]}{(f_1)} \oplus \frac{\mathbb{F}[X]}{(f_2)} \oplus \cdots \oplus \frac{\mathbb{F}[X]}{(f_s)}. \end{align*} (We normalise the $d_i$ to be monic, which is possible since $\mathbb{F}[X]$ is a field ring — units in $\mathbb{F}[X]$ are non-zero scalars.) **Step 3: The companion matrix of each summand.** [claim: Companion Matrix Representation] For each cyclic summand $\mathbb{F}[X]/(f_i)$ with $f_i = a_0 + a_1 X + \cdots + a_{r-1}X^{r-1} + X^r$, the action of $\alpha$ restricted to the corresponding invariant subspace, in the basis $\{1, X, X^2, \ldots, X^{r-1}\}$ mod $(f_i)$, is represented by the companion matrix \begin{align*} c(f_i) = \begin{pmatrix} 0 & 0 & \cdots & 0 & -a_0 \\ 1 & 0 & \cdots & 0 & -a_1 \\ 0 & 1 & \cdots & 0 & -a_2 \\ \vdots & & \ddots & & \vdots \\ 0 & 0 & \cdots & 1 & -a_{r-1} \end{pmatrix}. \end{align*} [/claim] [proof] The action of multiplication by $X$ (which corresponds to $\alpha$) maps $X^j \mapsto X^{j+1}$ for $j < r-1$, and $X^{r-1} \mapsto X^r \equiv -(a_0 + a_1 X + \cdots + a_{r-1}X^{r-1})$ modulo $(f_i)$. Reading the images as column vectors in the ordered basis gives exactly $c(f_i)$. [/proof] **Step 4: Block diagonal form.** The direct sum decomposition $V_\alpha \cong \bigoplus_{i=1}^s \mathbb{F}[X]/(f_i)$ corresponds (via the isomorphism) to a decomposition of $V$ into $\alpha$-invariant subspaces $V_1, \ldots, V_s$. The basis of each $V_i$ coming from Step 3, concatenated, gives a basis of $V$ in which the matrix of $\alpha$ is \begin{align*} \begin{pmatrix} c(f_1) & & \\ & \ddots & \\ & & c(f_s) \end{pmatrix}. \end{align*} **Step 5: Reading off the minimal and characteristic polynomials.** The minimal polynomial of $\alpha$ is $f_s$: every $f_i$ annihilates $\mathbb{F}[X]/(f_i)$, and since $f_s$ is the largest factor (with $f_i \mid f_s$ for all $i$), $f_s(\alpha) = 0$; no polynomial of smaller degree annihilates $\mathbb{F}[X]/(f_s)$. The characteristic polynomial is $f_1 f_2 \cdots f_s$ (the product of the degrees matches $\dim V$). $\square$