The Jordan Canonical Form

Existence, by strong induction on $\dim W$. If $W = \{0\}$ the statement holds vacuously. Suppose $\dim W = d \geq 1$ and the result holds for all nilpotent operators on spaces of dimension less than $d$. Let $p$ be minimal with $N^p = 0$ (if $N = 0$, take $p=1$ and every basis vector generates its own length-$1$ chain, done). Consider $U = \operatorname{im}(N) = N(W)$. Since $N$ is not invertible (it is nilpotent on a nonzero space, as $N^p=0$ forces $\det(N)^p = 0$), $U \subsetneq W$, so $\dim U < d$. Also $N(U) \subseteq U$, so $N|_U$ is a well-defined nilpotent operator on $U$ (nilpotent because $(N|_U)^p = N^p|_U = 0$). By the inductive hypothesis applied to $N|_U$, there exist $u_1, \dots, u_s \in U$ and lengths $\ell_1, \dots, \ell_s$ with

Since each $u_i \in U = N(W)$, choose $w_i \in W$ with $N w_i = u_i$. Then $N^{\ell_i} w_i = N^{\ell_i - 1} u_i \ne 0$ and $N^{\ell_i+1} w_i = N^{\ell_i} u_i = 0$, so $w_i$ generates a cyclic chain of length $\ell_i + 1$:

Claim 1: The vectors $\{N^j w_i : 1 \le i \le s,\ 0 \le j \le \ell_i\}$ are linearly independent, and $\sum_i C_i$ has dimension $\sum_i (\ell_i+1) = \dim U + s$. Indeed, note $\{N^j w_i\}_{j=1}^{\ell_i}$ span the same set as $\{N^{j-1}u_i\}_{j=1}^{\ell_i}$, which together over all $i$ are linearly independent because they form (a subfamily equal to) the basis of $U$ produced by the inductive hypothesis. Adding the vectors $w_1,\dots,w_s$ (for $j=0$) preserves independence: any linear dependence among $\{w_i\} \cup \{N^j w_i : j\ge1\}$, applying $N$, gives a dependence among the $N^j w_i$ ($j \geq 1$) alone (since $N(\text{dependence}) = 0$ eliminates the $w_i$ terms only if their coefficients vanish after applying $N$ — more precisely, suppose $\sum_i c_i w_i + \sum_{i,j\ge 1} c_{ij} N^j w_i = 0$. Applying $N$: $\sum_i c_i u_i + \sum_{i,j \ge 1} c_{ij} N^j u_i = 0$ in $U$, forcing (by independence of the basis of $U$) all $c_i = 0$ and all $c_{ij}=0$ for $j \geq 1$ with $N^j u_i$ nonzero terms — since the basis of $U$ is exactly $\{u_i, Nu_i,\dots\}$, this forces every coefficient in that relation to be $0$, in particular $c_i = 0$ for all $i$; then the original relation reduces to $\sum_{i,j\ge1} c_{ij}N^jw_i = 0$, i.e. $\sum_{i,j\ge1}c_{ij}N^{j-1}u_i = 0$, forcing $c_{ij}=0$ too). Hence Claim 1 holds. Claim 2: $\dim(C_1 + \cdots + C_s) = \dim U + s \geq \dim U$, and we may extend a basis to reach all of $W$. Let $S = C_1 \oplus \cdots \oplus C_s$ (direct because the spanning sets found in Claim 1 are independent). We have $\dim S = \dim U + s$. Now pick any complement: extend the linearly independent set $\{N^{\ell_i} w_i\}_{i=1}^s$ (a basis for $S \cap \ker N \cap$ ... — more directly) as follows. Since $\ker N \cap U$ has dimension $s$ (it is spanned by the "top" vectors $N^{\ell_i-1}u_i = N^{\ell_i}w_i$ of each chain in $U$, by the rank–nullity structure of the cyclic decomposition of $N|_U$), and $\ker N$ has dimension $\dim W - \dim U$ (rank–nullity for $N: W \to U$, since $N$ is surjective onto $U$ by definition of $U$), we may choose a complement $K$ of $\ker N \cap U$ inside $\ker N$, with

Choose a basis $w_{s+1}, \dots, w_r$ of $K$ (so $r - s = \dim K$), each satisfying $N w_i = 0$, i.e. each generates a chain of length $1$: $C_i = \operatorname{span}\{w_i\}$. Now verify $W = S \oplus K$: dimensions add correctly,

so it suffices to show $S \cap K = \{0\}$ and $S + K$ is direct with the rest — but we show directly $W = \bigoplus_{i=1}^r C_i$ by showing the total spanning set is independent and has size $\dim W$. Independence of $\{N^jw_i : 1\le i\le s, j\le \ell_i\} \cup \{w_i : s<i\le r\}$: any dependence, after applying $N$, kills all the $w_i$ ($i>s$) terms (since $Nw_i=0$ for $i > s$) and all $j\ge 1$ terms shift down by one power; arguing as in Claim 1 this forces the coefficients of $N^jw_i$ for $j \ge 1$ to vanish, and separately the coefficients of $w_i$ ($i \le s$) to vanish by Claim 1's argument, leaving a dependence purely among $\{w_i\}_{i>s}$, which are independent by construction (basis of $K$) and independent from $S$ because $K$ was chosen as a complement to $\ker N \cap U$ inside $\ker N$, hence $K \cap U = \{0\}$ and in particular $K \cap S = \{0\}$ (as $S \subseteq U + \operatorname{span}(w_i's)$... since each generator $w_i$, $i\le s$, was chosen with $Nw_i = u_i \in U$, so $S = \bigoplus C_i \subseteq \operatorname{span}(w_1,\ldots,w_s) + U$, and $K$ is disjoint from that span by the complement construction). Counting total dimension: $\sum_{i=1}^s(\ell_i+1) + (r-s) = (\dim U + s) + (\dim W - \dim U - s) = \dim W$. A linearly independent spanning-sized set is a basis, so $W = \bigoplus_{i=1}^r C_i$, completing the inductive step with chain lengths $k_i = \ell_i + 1$ for $i \le s$ and $k_i = 1$ for $i > s$.

Uniqueness of the multiset $\{k_i\}$. For $j \geq 0$, $\operatorname{rank}(N^j)$ restricted to a single chain summand $C_i = \operatorname{span}\{w_i, Nw_i,\dots,N^{k_i-1}w_i\}$ (with $N^{k_i}w_i = 0$) contributes exactly $\max(k_i - j, 0)$ to the rank of $N^j$ on $W$ (since $N^j$ maps the basis of $C_i$ to $\{N^jw_i,\dots,N^{k_i-1}w_i, 0,\dots,0\}$, of which $\max(k_i-j,0)$ are nonzero and, being a sub-list of an independent chain basis mapped into disjoint summands, contribute independently to the rank of $N^j$ on the whole space — direct sums of $N$-invariant subspaces make $N^j$ block diagonal, and rank adds over blocks). Hence

The right side is exactly the "partition rank function" of the partition $\{k_i\}$ of $\dim W$; a classical identity for partitions gives

which depends only on $N$ (not on any choice of chains), proving the multiset $\{k_1,\dots,k_r\}$ is an invariant of $N$ alone. $\blacksquare$

Proof of the main theorem, using the Lemma. Existence. By the Generalized Eigenspace Decomposition, since $\mathbb{F}$ is algebraically closed, the characteristic polynomial of $T$ splits into linear factors, and $V$ decomposes as a direct sum of generalized eigenspaces

where $\lambda_1,\dots,\lambda_m$ are the distinct eigenvalues of $T$ and $n = \dim V$. Each $G(\lambda_i,T)$ is $T$-invariant, and on $G(\lambda_i,T)$ the operator $N_i := (T - \lambda_i I)\big|_{G(\lambda_i,T)}$ is nilpotent by definition of generalized eigenspace (every vector is annihilated by some power of $T - \lambda_iI$, and finite-dimensionality gives a uniform bound, namely $N_i^{\dim G(\lambda_i,T)} = 0$). Apply the Lemma to $N_i$ on $W = G(\lambda_i, T)$: we obtain a basis of $G(\lambda_i,T)$ consisting of chains $\{w, N_iw,\dots,N_i^{k-1}w\}$ with $N_i^k w = 0$. On such a chain, writing $v_j = N_i^{k-j}w$ for $j=1,\dots,k$ (so $v_1 = N_i^{k-1}w$ is the "bottom," annihilated by $N_i$), we get

i.e. $(T-\lambda_iI)v_1 = 0$ and $(T-\lambda_iI)v_j = v_{j-1}$, so $Tv_1 = \lambda_i v_1$ and $Tv_j = \lambda_i v_j + v_{j-1}$. In the ordered basis $(v_1,\dots,v_k)$ of this chain, the matrix of $T$ restricted to $\operatorname{span}(v_1,\ldots,v_k)$ is exactly $J_k(\lambda_i)$: column $j$ records $Tv_j = \lambda_i v_j + v_{j-1}$, giving $\lambda_i$ on the diagonal and a single $1$ in row $j-1$, i.e. the superdiagonal. Concatenating the chain-bases over all chains within $G(\lambda_i,T)$, and then over all $i = 1,\dots,m$, produces a basis $\beta$ of $V$ (a basis of the whole space, since the $G(\lambda_i,T)$ direct-sum to $V$ and each chain-basis is a basis of its summand) in which $[T]_\beta$ is block diagonal with each block a Jordan block $J_k(\lambda_i)$. This proves existence. Uniqueness. Suppose $\beta$ is any basis with $[T]_\beta$ a direct sum of Jordan blocks $J_{k_1}(\mu_1),\dots,J_{k_m}(\mu_m)$. First, the set of distinct $\mu_i$'s occurring must be exactly the set of eigenvalues of $T$: a block-diagonal matrix has characteristic polynomial equal to the product of the characteristic polynomials of its blocks, and $\det(xI - J_k(\lambda)) = (x-\lambda)^k$ since $J_k(\lambda)$ is upper triangular with $\lambda$ repeated on the diagonal; hence the multiset of all $\mu_i$ (with multiplicity $k_i$ each) is exactly the multiset of roots of the characteristic polynomial of $T$, which is similarity-invariant. Fix an eigenvalue $\lambda$ of $T$. Let $\beta_\lambda \subseteq \beta$ be the union of the basis vectors belonging to the Jordan blocks $J_{k_i}(\lambda)$ (those blocks among $[T]_\beta$ with eigenvalue exactly $\lambda$), and let $V_\lambda = \operatorname{span}(\beta_\lambda)$. Direct computation on a single block shows $(T - \lambda I)$ acts on the span of $J_k(\lambda)$'s basis exactly as the nilpotent shift, hence is nilpotent there, and for $\mu \ne \lambda$, $(T-\mu I)|_{V_\mu\text{-block}}$ has $J_k(\lambda) - \mu I = J_k(\lambda - \mu)$ which is invertible (upper triangular with nonzero diagonal entries $\lambda - \mu$). Since $[T]_\beta$ is block diagonal, $V_\lambda$ (the sum of blocks with that exact eigenvalue) is precisely

because on $V_\lambda$, $T - \lambda I$ is nilpotent (kernel eventually all of $V_\lambda$ under a high enough power), while on the complementary sum of other blocks, $T - \lambda I$ is invertible (each block $J_k(\mu)-\lambda I$ invertible for $\mu \neq \lambda$, so the whole block-diagonal restriction is invertible, being a direct sum of invertible maps), hence contributes nothing to the kernel of any power. But $\ker(T-\lambda I)^{\dim V} = G(\lambda, T)$ is exactly the generalized eigenspace, which is intrinsically defined from $T$ alone (no basis choice). Hence $V_\lambda = G(\lambda,T)$ is forced, independent of which valid Jordan basis $\beta$ we started with. Within $V_\lambda = G(\lambda,T)$, the restriction $N = (T-\lambda I)|_{V_\lambda}$ is a fixed nilpotent operator (determined by $T$ and $\lambda$ alone), and the basis $\beta_\lambda$ exhibits $N$ as a direct sum of shift-chains of lengths $k_i$ (the sizes of the blocks with eigenvalue $\lambda$) — precisely the cyclic decomposition data of $N$ from the Lemma. By the uniqueness clause of the Lemma, the multiset $\{k_i\}$ of chain lengths is determined by $N$ alone via

which does not depend on $\beta$ at all — it is a formula purely in terms of $T$ and $\lambda$. Therefore the number of Jordan blocks of size $k$ with eigenvalue $\lambda$ is the same for any valid Jordan basis $\beta$. Since $\lambda$ and $k$ were arbitrary, the entire multiset of Jordan blocks (eigenvalue and size for each) is an invariant of $T$, proving uniqueness up to reordering. Similarity criterion. If $A, B$ are similar, say $B = P^{-1}AP$, then a Jordan basis for $A$ pushed forward by $P$ is a Jordan basis for $B$ giving the identical block multiset, since similar operators represent the same abstract linear map on isomorphic spaces up to change of basis, so their Jordan forms (existing by the theorem) coincide up to block order. Conversely, if $A$ and $B$ have the same Jordan block multiset, then each is similar to the same block-diagonal matrix $J$ (via their respective Jordan bases), hence $A = P_1 J P_1^{-1}$ and $B = P_2 J P_2^{-1}$ for invertible $P_1, P_2$, giving $A = (P_1P_2^{-1}) B (P_1P_2^{-1})^{-1}$, so $A$ and $B$ are similar. $\blacksquare$