I've been working through the details of the Jordan decomposition theorem. Up to now, I was able to show that if $V$ is a finite dimensional vector space and $T: V\rightarrow V$ is a linear operator with distinct eigenvalues $\lambda_1, \ldots, \lambda_k$, then
$$V=K_{\lambda_1}\oplus \ldots \oplus K_{\lambda_k}$$
where
$$K_{\lambda_j}=\textrm{ker}[(T-\lambda_j I)^{\mu_a(\lambda_j)}]$$
for each $j\in \{1, \ldots, k\}$. Above, $\mu_a(\lambda_j)$ is the algebraic multiplicity of $\lambda_j$.
Now, as far as I understood, the idea is to show each $K_{\lambda_j}$ decomposes as a direct sum of some cyclic invariant subspaces. I think the next holds:
Conjecture. Let $V$ be a finite dimensional vector space and $T: V\rightarrow V$ a linear operator having $\lambda$ as eigenvalue. Then there are vectors $x_1, \ldots, x_k\in V$ such that
$$K_\lambda=Z_\lambda(x_1)\oplus \ldots \oplus Z_{\lambda}(x_k)$$
where
$$Z(x_j)=\textrm{Span}\{(T-\lambda_j I)^{p_j-1}(x_j), \ldots, (T-\lambda_j I)(x_j), x_j\}$$
with $p_j\in \mathbb N$ the least integer such that $(T-\lambda_j I)^{p_j}(x_j)=0$.
So I have a few questions about it:
First, is the above conjecture correct?
How is the number $k$ determined?
How are the vectors $x_1, \ldots, x_k$ determined?
Thanks.
First off I tend to go a bit all out on my answers so sorry in advance if this might be a bit of a long answer. Also excuse my somewhat janky english since I am learning all this in german and have to try to formulate my understanding in english which is also why I use odd vocabulary sometimes:
Something to note is that it's important for the Jordan Decomposition and the Jordan Normal Form to exist that the characteristical polynomial decomposes into it's linear factors, so to speak, that all roots of the polynomial are in the field that the vector space corresponds to (if the Field is $\mathbb{C}$ for example this would be the case since $\mathbb{C}$ is algebraically closed).
First I find it to be easier to think about the Jordan decompisition and the conjectures leading up to the theorem if one has the goal or rather the Jordan-Normal-Form in mind. So what we want is a basis so that the matrix of our linear operator regarding that basis is a block matrix consisting of Jordan Blocks. These Blocks have the form:
$$ \begin{pmatrix} \lambda_j & 1 & \ldots & \\ 0 & \lambda_j & \ldots \\ \vdots & \vdots & \ddots & 1\\ & & & \lambda_j \end{pmatrix} $$ The size of the Block can be between 1 and $j_i$ where $j_i$ is the algebraic multiplicity (the exponent of the linear factor corresponding to the eigenvalue in the characteristical polynomial in product form) of the Eigenvalue $\lambda_i$.
There are different formulations of the Jordan decomposition theorem but the theory behind constructing a basis for the Jordan Normal Form and the Jordan Decomposition itself is very similar.
Now you have already correclty stated that the (finite!) vector space decomposes into the generalized eigenspaces (if the field of the vector space is algebraically closed or the characteristical polynomial decomposes into its linear factors as stated above).
Now you want to show that the generalized Eigenspace $K_{\lambda_j}$ decomposes as a direct sum of cyclic subspaces which is correct aswell. The thing to note here is that these cyclic subspaces correspond to said Jordan-Blocks above.
Now to your questions:
1. The idea of the conjecture is correct it might not be formulated perfectly. As already mentioned above we now want to construct our cyclic subspaces for the Jordan Decomposition/Form. In order to do that we look at the generalized Eigenspace of the Eigenvalue $\lambda$. There are different things that have to be shown first (which you might know already) which are useful in the following.
You have used a different notation but I think this one is easy to understand aswell. What you have denoted above as $K_\lambda$ (the generalized Eigenspace to the eigenvalue $\lambda$ can also be defined as follows):
$$ K_\lambda= \bigcup _{n\geq 0} \ker((T-\lambda I)^n) $$ I think this is what you meant with your notation above however since I am not sure I'm going this one. Here it's clear that the generalized Eigenspace is just the union of the nullspaces when applying the linear operator $T-\lambda I$ multiple times.
Now we want to take a closer look at these Nullspaces. For one it holds that $$ \ker((T-\lambda I)^j) \subseteq \ker((T-\lambda I)^{j+1}) $$ (Which should make sense). Furthermore there exists an $n \in \mathbb{N}$ with $$ \ker((T-\lambda I)^n) = K_\lambda $$ This is due to the fact above and the finite dimension of our vector space. Also now for $i < n$ it holds that $$ \ker((T-\lambda I)^{i-1}) \subseteq \ker((T-\lambda I)^{i}) \wedge \ker((T-\lambda I)^{i-1}) \neq \ker((T-\lambda I)^{i}) $$ One can now show that the dimension of these Nullspaces increases with the increasing of the exponents up to said bound n where the Nullspace is the generalized Eigenspace.
With these Nullspace one can now construct the $Z(x_j)$ or rather $Z_\lambda(x_j)$ you mentioned in your question.
The idea behind that is that if you take a vector $x_j$ from $\ker((T-\lambda I)^n)$ which is not in $\ker((T-\lambda I)^{n-1})$ and now look at $(T-\lambda I) x_j$ it follows that $(T-\lambda I) x_j \in \ker((T-\lambda I)^{n-1})$. Furthermore $x_j$ and $(T-\lambda I) x_j$ are linearly independent. This is then repeated up to $\ker((T-\lambda I)^{n-1} )x_j$. The span of these vectors is obviously cyclic and linearly independent because of the attributes above.
This concept is also referred to as Jordan-Chains because one takes a vector from the Nullspace $\ker((T-\lambda I)^n)$ which is not in $\ker((T-\lambda I)^{n-1})$ and builds up a chain of vectors up to one in $\ker((T-\lambda I)$ (which is the Eigenspace). Now the thing to note is that you don't how many of these chains you can construct which are linearly independent. Since for each chain you need a vector in $\ker((T-\lambda I)^n)$ which is not in $\ker((T-\lambda I)^{n-1})$ and linearly independent from the other chain starting vectors. So really it depends on the dimensions of the nullspace and if one can't start a new chain from $\ker((T-\lambda I)^n)$ one can try to start a 'shorter' chain from $\ker((T-\lambda I)^{n-1})$ and after that from $\ker((T-\lambda I)^{n-2})$ and so on. The point is that your decomposition should look as follows:
Conjecture: Let $V$ be a finite dimensional vector space (over a algebraically closed field) and $T: V \to V$ a linear opeator having $\lambda$ as eigenvalue. Then there are vectors $x_1, \ldots, x_k$ \in V such that: $$ K_\lambda = Z_\lambda (x_1) \oplus \ldots \oplus Z_\lambda(x_k) $$ where $$ Z_\lambda(x_j) = Span\{(T-\lambda_j I )^{p_{\lambda_j}} (x_j), \ldots , (T-\lambda_j I) (x_j), x_j \} $$ with $p_{\lambda_j} \leq n$ (with n being the least integer so that $K_\lambda = \ker(T-\lambda_j I)^n$)
So really the change is that you don't know whether the 'chain' begins at $\ker(T-\lambda_j I )^n$ or 'later' and as such the Conjecture above is technically not quite correct.
Now to the 2. question: Ne number k is the number of Jordan Blocks in the Jordan Normal Form as described above. It is the determined by the number of Jordan Chains that are needed. Another (easier way) to determine this number is by the Dimension of the Eigenspace. This is quite a usefull property when thinking about the Jordan Normal Form instead of the Jordan Decomposition in this context. As stated above the decomposition of the generalized Eigenspace gives us the these cyclic subspaces which we can construct with these Jordan 'Chains'. These basis of the Form: $$ (T-\lambda_j I )^{p_{\lambda_j}} (x_j), \ldots , (T-\lambda_j I) (x_j), x_j $$ Say we enumerate this basis as $b_1, \ldots, b_{p_{\lambda_j}+1}$. These Elements have the useful property that $T \cdot b_i = \lambda b_i + b_{i+1}$ (if $i \neq 1$ because $T \cdot b_1 = \lambda \cdot b_1$ since the first vector in our basis is in the eigenspace an as such an eigenvector). This is exactly why the matrix regarding the Jordan Basis has the nice form of Jordan Blocks as described above. Now we know that at the end of every chain has to a vector in the Eigenspace. And since we want the decomposition to be a direct sum the eigenvectors have to be linearly independent which results in us having as many chains as the dimension of the Eigenspace. So the answer is $k = \dim(\ker(T- \lambda I))$
3. question: I have already explained how the $x_1, \ldots ,x_k$ are constructed but I will go over it shortly once again:
The $x_1, \ldots ,x_k$ are the 'starts' of our Jordan 'Chains'. If one wants to determine these percisely one would have to calculate the Nullspaces which compose the generalized Eigenspace. Would then have to look at the n as defined above at vectors would have to construct chains of vectors from the 'highest exponent' Nullspace to the Eigenspace and start constructing shorter and shorter chains if there are no more linearly independent vectors left at the 'highest exponent' Nullspace, see for example Another Post about constructing a Jordan Basis
I hope this helped and feel free to ask any question or tell me about any mistakes.