$\DeclareMathOperator{\Ker}{Ker}\DeclareMathOperator{\id}{Id}\DeclareMathOperator{\N}{\mathcal{N}}$ I make reference to this paper, which I have noticed, as I've been taking notes from it, struggles with cohesive and structured proofs - I always have to read between the lines, and jump back and forth to piece their argument together. Nevertheless, it has served me well in understanding the existence of the Jordan decomposition, except for this one, last final step. Everything I discuss is on the final two pages. A few definitions first:
The paper defines a "relative basis" of a vector space $V$ w.r.t $U$ as a collection of linearly independent vectors $e_1,e_2,e_3,\cdots$ such that: $\forall v\in V,\exists u\in U:v=c_1e_1+c_2e_2+c_3e_3+\cdots+u$. Vectors $e_1,e_2,e_3,\cdots$ are "relatively linearly independent" to a vector space $U$ if the only linear combination of them that exists in $U$ is the trivial one: $c_1e_1+c_2e_2+c_3e_3+\cdots\in U\iff c_1,c_2,c_3,\cdots=0$.
The goal of the proof is to show:
- For any linear operator $A:V\to V$, $V$ finite dimensional, and for $\lambda_i,1\le i\le k$ the distinct eigenvalues of $A$ in $V$, there exist naturals $n_i$ such that: $V=\oplus_{i=1}^k\Ker(A-\lambda_i\id)^{n_i}$.
- Denote $\Ker(A-\lambda_i\id)^{n_i}$ as $V_i$. Then each $V_i$ has a basis $e_1^{(i)},e_2^{(i)},e_3^{(i)},\cdots,e_{n_i}^{(i)}:(A-\lambda_i\id)e_1^{(i)}=0$ and $(A-\lambda_i\id)e_j^{(i)}=e_{j-1}^{(i)}$
Note: in each kernel, $n_i$ is the number such that $\Ker(A-\lambda_i\id)^{n_i}=\Ker(A-\lambda_i\id)^{n_i+r},\forall r\ge1$.
They have proven $1)$ satisfactorily. I recognise the basis of $2)$ as the Jordan Chain, and I recognise that proving its existence indeed proves the Jordan Form's existence. However, in their proof that the Jordan basis (as the paper calls it) exists, I have to read between the lines of a lot of alphabet soup, with an unmotivated line of argument.
They first want to show that any nilpotent operator $B:V\to V$ has a Jordan basis in its $V_i$. They start by declaring that there exists: $e_1,e_2,e_3,\cdots,e_s$ which are a relative basis of $V=\Ker(B^k)$ w.r.t $\Ker(B^{k-1})$. For convenience, I will let $\N_i$ denote $\Ker(B^i)$ as the paper does. They then show that $e_1,e_2,e_3,\cdots,e_s,B(e_1),B(e_2),B(e_3),\cdots,B(e_s)$ are a relative basis of $V_i$ relative to $\N_{k-2}$. And in fact they extend this all the way down until they have a huge basis for $V_i$ that ends up being relative to a zero space only - i.e. a standard basis for $V_i$. The problem for me is, there are far more than $\dim V_i=k$ vectors in this basis, so it is not in fact a basis. They airily say (I paraphrase):
Renumber these vectors to form a Jordan basis. Let $v_1=B^{k-1}(e_1),\,v_2=B^{k-2}(e_1),\cdots,v_k=e_1$. This is a Jordan basis for $V_i$.
It is not clear to me at all what they expect the reader to do with the numerous other vectors in their basis; one could make Jordan bases out of $e_2,e_3,\cdots,$ and their $f_1,f_2,\cdots,g_1,g_2,\cdots$ basis vectors are simply standing pretty taking space on the page but not seeming to have any purpose whatsoever - their Jordan basis based on $e_1$ has not been shown to be unique, or indeed even a basis for $V_i$. I am quite confused by their argument; I recall when finding the Jordan Form of a matrix that only one Jordan chain is necessary (per eigenvalue), not their soup of vectors... I cannot even tell if by their use of "$V$" later in the article they mean the entire space such that $A:V\to V$ or the subspace $V_i=\Ker(A-\lambda_i\id)^{n_i}$.
Many thanks to anyone who can decipher their argument and explain how the Jordan chain based on $e_1$ is a basis for $V_i$ on its own, and how to casually ignore the other many vectors in their basis on Page $5$, Lemma $6$.
Additional, not so important side-question: they assume $A-\lambda\id$ is nilpotent, if $\lambda$ is an eigenvalue of $A$. I am not so sure that this is the case (but haven't made any tests yet), and certainly it goes unproven in the article. A quick thought: a linear operator is nilpotent if its eigenvalues are all zero. Well, the eigenvalues of $(A-\lambda\id)$ are the $\mu$ such that $|(A-\lambda\id)-\mu\id|=0$. $\mu$ here of course can be zero, but it could also surely be the difference such that $\lambda+\mu$ is another eigenvalue of $A$ - i.e. non-zero. I am baffled!
EDIT:
With great thanks to @levap, it must be remembered that multiple Jordan blocks for the same eigenvalue is in fact very much a thing... and all the bases in the article are valid. And the point is that $A-\lambda\id$ is not necessarily nilpotent in general, but it is nilpotent in the space $\Ker(A-\lambda\id)^k$