What is the easiest proof you know for the Jordan Canonical Form I read numerous demonstration of the existence of the Jordan Canonical Form, but all of them involve more than 2 pages of demonstration with numerous lemmas in between. 
I'm writing some notes for some students, but the subject is only tangentially related to Jordan Normal Form and so I was wondering if anybody knew a simple 1-page demonstration of the existence of this form!
If the demonstration is original I will find a way to cite you :)
 A: $\newcommand{\span}{\operatorname{span}}$Are you assuming an algebraically closed closed ground field? If so then here is a proof. Let $W_{\lambda}=\{v :(T-\lambda)^kv=0, \text{for some $k$} \}$.
Its not hard to show that $V=\bigoplus W_{\lambda}$, ill skip this but it does take some time. So look at the operator $N=T-\lambda$ in the space $W=W_{\lambda}$. $N$ is nilpotent. Chose $v_1, \ldots , v_p$ basis for $\ker N$, which has to be non zero. Now consider those $w$ such that $Nw\in \span\{ v_1\}$ these $w$ are one dimensional, modulo the $\ker N$ : since if there were $w$ and $w^{\prime}$ ,linearly independent, with a choice of constant $N(w-bw^{\prime})=0$ and so  $w-bw^{\prime}\in  \ker N$. So now chose some vectors
such that $Nw_i \in  \span\{ v_i\}$ where possible say $w_1\ldots w_q$ for $q\leq p$. Next chose $x_i$ such that $Nx_i\in  \span\{ w_i\}$ and so on. Now your basis for $W$ is $v_1,w_1,x_1,y_1,\ldots  v_2,w_2,x_2, \ldots v_3,w_3,x_3,\ldots$ its easy to see that under this basis $N$ and thus $T$ has the desired form on $W$. Its not 2 pages but it is also quite terse.
A: $\newcommand{\kerA}{\ker(A_\lambda)}\newcommand{\rangeA}{\operatorname{range}(A_\lambda)}\newcommand{\im}{\operatorname{im}}\newcommand{\range}{\operatorname{range}}$Let $A$ be a $N\times N$ matrix acting in a vector space $V$. We know that there is always some $\lambda\in\mathbb C$ such that $Av=\lambda v$ for some $v\neq 0$. Defining $A_\lambda\equiv A-\lambda I$, this is equivalent to saying that $\kerA\neq\{0\}$. Let $\dim\kerA=\ell>0$.
Our goal is to decompose $V$ as direct sum of invariants of $A$ (equivalently, invariants of $A_\lambda$). This means that we are looking for subspaces $V_i$ such that $V=\bigoplus_i V_i$ and $AV_i\subseteq V_i$. Moreover, we want these invariants to be as small as possible (that is, we want them to be simple).
From the rank-nullity theorem, we know that, for any $A$, we have $\dim\ker(A)+\dim\range(A)=N$. It is also easy to see that, for any $A$, both $\ker(A)$ and $\range(A)$ are invariant subspaces. These two facts imply that, provided that $\ker(A)\cap\range(A)=\{0\}$, we have the decomposition $V=\ker(A)\oplus\range(A)$ of $V$ into invariants of $A$.
But what if $\ker(A)\cap\range(A)\neq\{0\}$? In this case the decomposition of $V$ clearly doesn't work. The problem is that there aren't enough elements in $\ker(A)$ and $\range(A)$ combined to make up a basis for $V$: even though their combined dimensionality is correct, they are "redundant", due to their have a non-zero intersection.
The fix is to think of the situation in a slightly different way. A generalisation of the statement that $\ker(A)$ is invariant is that the set of $v\in V$ such that $A^k v=0$ for some $k$ is invariant. But $\ker(A^\ell)$ is always a linear space and $\ker(A^\ell)\subseteq\ker(A^{\ell+1})$ for all $\ell$, so what if we try our decomposition using such powers of $A$ rather than $A$? Indeed, pick some $k$ high enough that $\ker(A^k)=\ker(A^{k+1})$ (this must happen at some point, being $V$ finite-dimensional).
Then $\ker(A^k)$ is the set of $v$ that are eventually sent to $0$ by repeated applications of $A$, while $\range(A^k)$ the set of those vectors that will never be sent to $0$, thus $\ker(A^k)\cap\range(A^k)=\{0\}$. This means that rank-nullity does work here, and gives us the decomposition that we are looking for:
$V = \ker(A^k)\oplus \range(A^k)$.
You can now apply the above reasoning replacing $A$ with $A_\lambda$ for all $\lambda$ in the spectrum of $A$. A subspace is invariant under $A$ if and only if it is invariant under $A_\lambda$ for any $\lambda\in\mathbb C$, and thus we can be build a decomposition of $V$ in terms of invariants of $A$ that are of the form $\ker(A_\lambda^{k_\lambda})$ for $\lambda\in\operatorname{spectrum}(A)$.
