**Proof plan.** By the [Generalised Eigenspace Decomposition](/theorems/411), it suffices to find Jordan bases for nilpotent endomorphisms (since on each generalised eigenspace $V_i$, $\alpha - \lambda_i\,\mathrm{id}$ is nilpotent). We prove existence for nilpotent endomorphisms by induction on dimension, then assemble the full Jordan form. Uniqueness follows from the [Uniqueness of Jordan Form via Nullities](/theorems/413). **Step 1: Reduction to the nilpotent case.** By the [Generalised Eigenspace Decomposition](/theorems/411), $V = \bigoplus_{i=1}^k V_i$ where $V_i = \ker((\alpha - \lambda_i\,\mathrm{id})^{c_i})$. Each $V_i$ is $\alpha$-invariant, and the restriction $\alpha|_{V_i}$ has only eigenvalue $\lambda_i$. The endomorphism $\nu_i = (\alpha - \lambda_i\,\mathrm{id})|_{V_i}$ is nilpotent (with $\nu_i^{c_i} = 0$). If we find a Jordan basis for each $\nu_i$ (a basis in which $\nu_i$ is a direct sum of Jordan blocks $J_m(0)$), then in the same basis $\alpha|_{V_i}$ is a direct sum of blocks $J_m(\lambda_i)$, and the union of all such bases gives a Jordan form for $\alpha$. **Step 2: Jordan basis for a nilpotent endomorphism.** Let $\nu: W \to W$ be nilpotent with $\nu^r = 0$ and $\nu^{r-1} \neq 0$. We construct a Jordan basis by induction on $\dim W$. **Base case:** $\dim W = 1$. Then $\nu = 0$ and the Jordan form is $J_1(0) = (0)$. **Inductive step:** Let $W_j = \ker(\nu^j)$ for $j \geq 0$. We have $\{0\} = W_0 \subsetneq W_1 \subsetneq \cdots \subsetneq W_r = W$ (strict inclusions since $\nu^{r-1} \neq 0$). Choose $w \in W$ with $\nu^{r-1}(w) \neq 0$. The vectors $w, \nu(w), \dots, \nu^{r-1}(w)$ are linearly independent (if $\sum a_j \nu^j(w) = 0$, apply $\nu^{r-1-j}$ starting from the highest surviving term to get a contradiction). They span a $\nu$-invariant subspace $U = \langle w, \nu(w), \dots, \nu^{r-1}(w) \rangle$ on which $\nu$ has the matrix $J_r(0)$ with respect to the basis $(\nu^{r-1}(w), \nu^{r-2}(w), \dots, w)$. One can show (by careful construction using the filtration $W_0 \subset W_1 \subset \cdots$) that $U$ has a $\nu$-invariant complement $U'$ in $W$. By the inductive hypothesis, $\nu|_{U'}$ has a Jordan basis. Combining with the basis for $U$ gives a Jordan basis for $\nu$. **Step 3: Uniqueness.** The Jordan form is unique up to permutation of blocks by [Uniqueness of Jordan Form via Nullities](/theorems/413): the Jordan block structure is completely determined by the [sequence](/page/Sequence) of nullities $\dim\ker(\alpha - \lambda\,\mathrm{id})^r$ for each eigenvalue $\lambda$ and $r \geq 1$, which are basis-independent quantities. $\blacksquare$