An intuitive approach to the Jordan Normal form

I want to understand the meaning behind the Jordan Normal form, as I think this is crucial for a mathematician.

As far as I understand this, the idea is to get the closest representation of an arbitrary endomorphism towards the diagonal form. As diagonalization is only possible if there are sufficient eigenvectors, we try to get a representation of the endomorphism with respect to its generalized eigenspaces, as their sum always gives us the whole space. Therefore bringing an endomorphism to its Jordan normal form is always possible.

How often an eigenvalue appears on the diagonal in the JNF is determined by its algebraic multiplicity. The number of blocks is determined by its geometric multiplicity. Here I am not sure whether I’ve got the idea right. I mean, I have trouble interpreting this statement.

What is the meaning behind a Jordan normal block and why is the number of these blocks equal to the number of linearly independent eigenvectors?

I do not want to see a rigorous proof, but maybe someone could answer for me the following sub-questions.

(a) Why do we have to start a new block for each new linearly independent eigenvector that we can find?

(b) Why do we not have one block for each generalized eigenspace?

(c) What is the intuition behind the fact that the Jordan blocks that contain at least $$k+1k+1$$ entries of the eigenvalue $$λ\lambda$$ are determined by the following? $$dim(ker(A−λI)k+1)−dim(ker(A−λI)k)\dim(\ker(A-\lambda I)^{k+1}) - \dim(\ker(A-\lambda I)^k)$$

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.