An intuitive approach to the Jordan Normal form
Let me sketch a proof of existence of the Jordan canonical form which, I believe, makes it somewhat natural.
Let us say that a linear endomorphism $f:V\to V$ of a nonzero finite dimensional vector space is decomposable if there exist proper subspaces $U_1$, $U_2$ of $V$ such that $V=U_1\oplus U_2$, $f(U_1)\subseteq U_1$ and $f(U_2)\subseteq U_2$, and let us say that $f$ is indecomposable if it is not decomposable. In terms of bases and matrices, it is easy to see that the map $f$ is decomposable iff there exists a basis of $V$ such that the matrix of $f$ with respect to which has a non-trivial diagonal block decomposition (that it, it is block diagonal two blocks)
Now it is not hard to prove the following:
Lemma 1. If $f:V\to V$ is an endomorphism of a nonzero finite dimensional vector space, then there exist $n\geq1$ and nonzero subspaces $U_1$, $\dots$, $U_n$ of $V$ such that $V=\bigoplus_{i=1}^nU_i$, $f(U_i)\subseteq U_i$ for all $i\in\{1,\dots,n\}$ and for each such $i$ the restriction $f|_{U_i}:U_i\to U_i$ is indecomposable.
Indeed, you can more or less imitate the usual argument that shows that every natural number larger than one is a product of prime numbers.
This lemma allows us to reduce the study of linear maps to the study of indecomposable linear maps. So we should start by trying to see how an indecomposable endomorphism looks like.
There is a general fact that comes useful at times:
Lemma. If $h:V\to V$ is an endomorphism of a finite dimensional vector space, then there exists an $m\geq1$ such that $V=\ker h^m\oplus\def\im{\operatorname{im}}\im h^m$.
I'll leave its proof as a pleasant exercise.
So let us fix an indecomposable endomorphism $f:V\to V$ of a nonzero finite dimensional vector space. As $k$ is algebraically closed, there is a nonzero $v\in V$ and a scalar $\lambda\in k$ such that $f(v)=\lambda v$. Consider the map $h=f-\lambda\mathrm{Id}:V\to V$: we can apply the lemma to $h$, and we conclude that $V=\ker h^m\oplus\def\im{\operatorname{im}}\im h^m$ for some $m\geq1$. moreover, it is very easy to check that $f(\ker h^m)\subseteq\ker h^m$ and that $f(\im h^m)\subseteq\im h^m$. Since we are supposing that $f$ is indecomposable, one of $\ker h^m$ or $\im h^m$ must be the whole of $V$. As $v$ is in the kernel of $h$, so it is also in the kernel of $h^m$, so it is not in $\im h^m$, and we see that $\ker h^m=V$.
This means, precisely, that $h^m:V\to V$ is the zero map, and we see that $h$ is nilpotent. Suppose its nilpotency index is $k\geq1$, and let $w\in V$ be a vector such that $h^{k-1}(w)\neq0=h^k(w)$.
Lemma. The set $\mathcal B=\{w,h(w),h^2(w),\dots,h^{k-1}(w)\}$ is a basis of $V$.
This is again a nice exercise.
Now you should be able to check easily that the matrix of $f$ with respect to the basis $\mathcal B$ of $V$ is a Jordan block.
In this way we conclude that every indecomposable endomorphism of a nonzero finite dimensional vector space has, in an appropriate basis, a Jordan block as a matrix. According to Lemma 1, then, every endomorphism of a nonzero finite dimensional vector space has, in an appropriate basis, a block diagonal matrix with Jordan blocks.
The true meaning of the Jordan canonical form is explained in the context of representation theory, namely, of finite dimensional representations of the algebra $k[t]$ (where $k$ is your algebraically closed ground field):
- Uniqueness of the normal form is the Krull-Schmidt theorem, and
- existence is the description of the indecomposable modules of $k[t]$.
Moreover, the description of indecomposable modules follows more or less easily (in a strong sense: if you did not know about the Jordan canonical form, you could guess it by looking at the following:) the simple modules are very easy to describe (this is where algebraically closedness comes in) and the extensions between them (in the sense of homological algebra) are also easy to describe (because $k[t]$ is an hereditary ring) Putting these things together (plus the Jordan-Hölder theorem) one gets existence.
There is no real meaning behind the Jordan normal form; this form is just as good as it gets in general (and then only over a field where the characteristic polynomial splits). That is, as good as it gets in our attempts to understand the action of a linear operator$~\phi$ on a finite dimensional vector space by decomposing the space as a direct sum of $\phi$-stable subspaces, so that we can study the action of$~\phi$ on each of the components separately, and reconstruct the whole action from the action on the components. (This is not the only possible approach to understanding$~\phi$, but one may say that whenever such a decomposition is possible, it does simplify our understanding.) Direct sum decompositions into $\phi$-stable subspaces correspond to reducing the matrix to a block diagonal form (the $\phi$-stability means that the images of basis vectors in each summand only involve basis vectors in the same summand, whence the diagonal blocks), and the finer the decomposition is, the smaller the diagonal blocks. If one can decompose into a sum of $1$-dimensional $\phi$-stable subspaces then one obtains a diagonal matrix, but this is not always possible. Jordan block correspond to $\phi$-stable subspaces that cannot be decomposed in any way as a direct sum of smaller such subspaces, so they are the end of the line of our decompositions.
Your concrete questions are easier to answer. Since (subspaces corresponding to) Jordan blocks for$~\lambda$ are obtained from a (non-unique) direct sum decomposition of the generalised eigenspace for $\lambda$, one can study the generalised eigenspace along that decomposition; in particular the (true) eigenspace is the direct sum of the eigenspaces for each Jordan block, and each of them is of dimension$~1$, whence the dimension of the eigenspace for$~\lambda$ equals the number of Jordan blocks for$~\lambda$. See this answer.
This also answers question (a), although I should note that one does not start with eigenvectors to find a decomposition into Jordan blocks. It is the other ways around: each Jordan block one can decompose into comes with (up to a scalar) a single eigenvector, and (since the decomposition is a direct sum) these vectors for different blocks are linearly independent. One cannot in general just take any basis of the eigenspace for$~\lambda$ and construct a Jordan block around each basis vector. To see why, consider the situation where the Jordan blocks are to be of sizes $2$ and $1$. Then the eigenvector coming from the larger Jordan block must be not only in the kernel, but also in the image of $\phi-\lambda I$, and not all eigenvectors for$~\lambda$ have that property; therefore only bases where one basis vector is such a special eigenvector can correspond to a decomposition into Jordan blocks. (Actually giving an algorithm for decomposing into Jordan blocks is not easy, although the possibility to do so is an important theoretic fact.)
The answer to question (b) is implied by this: since a Jordan block by nature only contributes $1$ to the geometric multiplicity of$~\lambda$, one must have multiple Jordan blocks inside the generalised eigenspace whenever the geometric multiplicity of$~\lambda$ is more than one. Just think of the simple case of a diagonalisable matrix with a (generalised) eigenspace of dimension $d>1$: a diagonal matrix with $d$ diagonal entries$~\lambda$ is not a Jordan block. and this can only be seen as $d$ Jordan blocks of size $1$ strung together. In fact one should not wish that there were only one Jordan block: this finer decomposition is actually much better (when it is possible). Note that in the diagonalisable case any decomposition of the eigenspace into $1$-dimensional subspaces will do, exemplifying the highly non-unique nature of decompositions.
Finally for question (c) note that inside a single Jordan block, the dimensions of the kernels of the powers of $A-\lambda I$ in your formula increase with the exponent by unit steps until reaching the size of the Jordan block (after which they remain constant), so that the Jordan block contributes at most$~1$ to the difference of dimensions, and it does so if and only if its size is at least $k+1$. Again by the nice nature of direct sums, you can just add up these contributions from each of the Jordan blocks, so the difference of dimensions is equal to the number of Jordan block of size at least $k+1$. (And this is a way to see that this number cannot depend on the choices involved in decomposing the space into Jordan blocks.)