Theorem: Assume that the characteristic polynomial $x_f$ splits into linear factors. Then there exists a Jordan normal form for f. The Jordan normal form is unique up to the order of the Jordan blocks.
Definition: A string in V is an l-tuple of vectors $S = (v_1, v_2, ..., v_l)$ (where l is allowed to be countably infinite) subject to the following requirements:
(i) $v_1$ is an eigenvector with eigenvalue $\lambda$, i.e. $f_\lambda (v_1) = 0$ and $w_1 \neq 0$
(ii) $f_\lambda (v_{k+1}) = v_k $ for $l \leq k\leq l-1$
We call $v_1$ the leading vector,$ l$ the length and $\lambda$ the eigenvalue of $S$
Proof: We prove the existence of a Jordan basis by induction on n. For n = 0 the empty set is a Jordan basis. Now let $n \geq 1$ and assume that the result is proved for all vector spaces of dimension $< n$.
The assumption that $x_f$ splits into linear factors aim plies that f has at least one eigenvalue. Let $\lambda$ be any eigenvalue and let $W = Im(f_\lambda)$ be the image of $f_\lambda$. Then W is an f-invariant subspace of dimension r = rank $f_\lambda$. Not that r = n - nullity $f_\lambda$ < n, because $f_\lambda$ is not injective. Let $g = f|W: W \to W$ be the restriction of f to W. The characteristic polynomial of g divides that of f, and therefore $x_g$ splits into linear factors. Hence, by the induction hypothesis, there exists a Jordan basis C for g. We can write C as a disjoint union $T_1\cup T_2 \cup ... \cup T_q$ ors strings for g. We are going to extend C to a Jordan basis for f. This involves 2 steps: 1. some of the strings in C have to be extended by one element in V; 2. we have to add a few strings of length 1 to C.
Step 1: Consider the $\lambda$-eigenspace of g, $W_\lambda =$ ker $g_\lambda$. Let m be its dimension. Since elements of $W_\lambda$ are just eigenvectors of f that happen to lie in W, we have $W_\lambda = W \cap V_\lambda$. Let us find a basis of $W_\lambda$. A basis of $W_\lambda$ is given by the leading vectors of those strings in C which have eigenvalue equal to $\lambda$. Because m = dim $W_\lambda$, there have to be m such strings. After renumbering the strings in C if necessary, we may assume that the first m strings $T_1, T_2, ... T_m$ have eigenvalue $\lambda$. Let $w_1, w_2, ..., w_m$ be the last vectors in these strings. As these vectors are in W = Im $f_\lambda$, there exists $x_1, x_2, ..., x_m \in V$ such that $f_\lambda(x_k) = w_k$ for $1 \leq k \leq m$. For $1 \leq k \leq m$, define $S_k = (T_k, x_k)$, i,e. append $x_k$ to the string $T_k$ to create a string which is longer by one element.
Step 2. Consider the $\lambda$-eigenspace of f, $V_\lambda =$ ker $f_\lambda$. Its dimension is n-r and it contains the m-dimensional space $W_\lambda$. Let $v_1, v_2, ..., v_m$ be the leading vectors of the strings T_1, T_2, ..., T_m. As we saw under step1, {$v_1, v_2, ..., v_m$} is a basis of $W_\lambda$. Extend this to a basis {$v_1, v_2, ..., v_m, u_1, u_2, ..., u_{n-r-m}$} of V_\lambda. For $1 \leq k \leq n-r-m$ define $R_k = {u_k}$.
We assert that
$$B = R_1 \cup R_2 \cup ... \cup R_{n-r-m} \cup S_1 \cup S_2 \cup ... \cup S_m \cup T_{m+1} \cup ... \cup T_q$$
is a Jordan basis for f. Indeed, the R's, S's, and T's are strings and their leading vectors are by construction independent. Therefore B is independent. Since B is obtained from C by adding $n-r-m+m = n-r$ vectors, the cardinality of B is $|B| = n -r + |C| = n - r + r = n$, so B is a basis of V. We have now found a Jordan basis for f and thereby completed the inductive step.
The proof of uniqueness is omitted here.
There are a few questions that I do not understand about this proof:
Why C is a disjoint union of q strings?
Why some of the strings in C have to be extended by one element in V? And why is $x_k$ chosen to be that extra element?
3.