**Proof plan.** Apply the prime decomposition of the [Structure Theorem for Finitely Generated Modules over Euclidean Domains](/theorems/857) to $V_\alpha$ over $\mathbb{C}[X]$. Since $\mathbb{C}$ is algebraically closed, the only monic irreducibles in $\mathbb{C}[X]$ are linear factors $X - \lambda$. Each summand $\mathbb{C}[X]/((X-\lambda)^k)$ corresponds to a Jordan block via an explicit basis computation. **Step 1: Prime decomposition over $\mathbb{C}[X]$.** By the [Rational Canonical Form](/theorems/863), $V_\alpha \cong \bigoplus_{i=1}^s \mathbb{C}[X]/(f_i)$ with $f_i \mid f_{i+1}$. Since $\mathbb{C}$ is algebraically closed, the [fundamental theorem of algebra](/theorems/347) guarantees that every monic irreducible in $\mathbb{C}[X]$ has the form $X - \lambda$ for some $\lambda \in \mathbb{C}$. The prime decomposition theorem for modules ([Chinese remainder theorem](/theorems/734) applied repeatedly to each $f_i = \prod_j (X - \lambda_j)^{a_{ij}}$) then gives \begin{align*} V_\alpha \cong \bigoplus_{i=1}^t \frac{\mathbb{C}[X]}{((X - \lambda_i)^{a_i})} \end{align*} for some $\lambda_i \in \mathbb{C}$ (not necessarily distinct) and positive integers $a_i$. **Step 2: The basis for each summand.** [claim: Jordan Block] For the summand $W_i = \mathbb{C}[X]/((X-\lambda_i)^{a_i})$, the ordered basis \begin{align*} e_1 = 1, \quad e_2 = X - \lambda_i, \quad \ldots, \quad e_{a_i} = (X-\lambda_i)^{a_i - 1} \end{align*} gives the matrix of $\alpha$ restricted to the corresponding $\alpha$-invariant subspace as the Jordan block \begin{align*} J_{a_i}(\lambda_i) = \begin{pmatrix} \lambda_i & 0 & \cdots & 0 \\ 1 & \lambda_i & \cdots & 0 \\ 0 & 1 & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_i \end{pmatrix}. \end{align*} [/claim] [proof] The action of $\alpha$ on $W_i$ is multiplication by $X = (X - \lambda_i) + \lambda_i$. In the basis above: $e_j \mapsto \lambda_i e_j + e_{j+1}$ for $j < a_i$, and $e_{a_i} \mapsto \lambda_i e_{a_i}$ (since $(X-\lambda_i)^{a_i} = 0$ in $W_i$). The column vectors of this map in the ordered basis give exactly $J_{a_i}(\lambda_i)$. [/proof] **Step 3: Assemble the Jordan form.** The direct sum decomposition of $V_\alpha$ corresponds to a decomposition of $V$ into $\alpha$-invariant subspaces. Concatenating the bases from Step 2 gives a basis of $V$ in which $\alpha$ is represented by the block-diagonal matrix \begin{align*} \begin{pmatrix} J_{a_1}(\lambda_1) & & \\ & \ddots & \\ & & J_{a_t}(\lambda_t) \end{pmatrix}. \qquad \square \end{align*}