Proof of Jordan decomposition
   
We're working over an algebraically closed field;
$T$ is an operator on $V$.

Step 1: every operator has an eigenvector
[This is the fundamental step.]
Pf 1a: $\det(\lambda I - T)$ is a monic polynomial of degree $n$,
  hence has a root, say $\lambda_1$.
  So $\det(\lambda_1 I - T) = 0$,
  so it has a non-trivial kernel.
  Pick $v_1 \neq 0$ in that kernel; this is an eigenvector with
  eigenvalue $\lambda_1$.
Pf 1b: see Axler

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If the eigenvalues are all different, we win immediately: the distinct
eigenvectors form a basis.
The main problems are:
- showing that we still get a decomposition if some eigenvalues are
  duplicated
  [Abstractly, you can use the primary/generalized eigenspace
   decomposition.]
- studying an operator with a single, repeated eigenvalue
  [This is a pain, and I don't get it.]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Step 2: every operator preserves a flag
Look at $T$ on $V/v_1$ (well-defined b/c $v_1$ is an eigenvector),
and repeat. An eigenvector on $V/v_1$ pulls back to given a
2-dimensional space (containing $v_1$), and so forth, yielding a flag.
Dually, we could have found an eigenvector for $T^*$, whose perp is an
invariant hyperspace under $T$ -- then you'd work with subspaces, not
quotient spaces.

Step 3: pick a basis for the flag
This shows that $T$ can be represented by an upper triangular matrix.
Actually, if you pick a orthonormal basis for the flag (with respect
to a given hermitian product), then you only need to conjugate a
given matrix form of $T$ by unitary matrices, yielding the Schur
decomposition.

Step 4: rearrange eigenvalues
By conjugating by a permutation, you can group the same eigenvalues
together on the diagonal -- EZ.

Step 5: generalized eigenspaces
We can kill some off-diagonal entries easily, so it becomes blocks.
(This corresponds to the generalized eigenspaces.)
Pf:
If $M$ is upper triangular, and $m_{ij}=K$, where
$m_{ii}=\lambda, m_{jj}=\mu$, where $\lambda \neq \mu$
(picture:
 lambda     K
       ...
           mu)
then conjugation by $I+K/(\lambda-\mu) e_{ij}$ kills $K$
(and shouldn't cause too much other harm -- maybe need to work in some
 order.)

Step 6: elementary Jordan blocks
This step I don't know how to do:
if $T$ has only one eigenvalue (as it does on the restriction to
$W_i$), then it should be similar to a direct sum of elementary Jordan
blocks, but I don't see how to prove this off-hand.
You either need to do some algebra (d'apres Hoffman-Kunze) or some
linear algebra (d'apres Axler).

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Alternative proof (Axler, I think):
- Decompose $V = \bigoplus W_i$ into generalized eigenspaces:
  if $\lambda$ an eigenvalue, with eigenvector $v$,
  look at $\ker (T-\lambda)^N$ for $N$ large: eventually this
  stabilizes, and it is the ``generalized eigenspace''.
  $T$ also preserves the image of $(T-\lambda)^N$
  b/c this has also stabilized [check];
  thus get direct sum, and induct.
  (this requires proof).
- Study $T$ on $W_i$, and show it has a nice form.
  a: show that nilpotent operators have nice jordan form
  b: thus $T$ on $W_i$ has nice form in the same basis, duh.
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So why do nilpotent operators have a nice basis?
Here's what we'd like to say:
pick $v_1$, and look at $v,Tv,\dots,T^kv=0$, and the first $k$ are
linearly independent. Now pick a $v_2$ that's not in the span, and
repeat.

Here's the point: we have a flag
$0 < \ker T < \ker T^2 < \dots < \ker T^k = V$ etc.
  (all strict inclusions)
$T(\ker T^i) = \ker T^{i-1}$ -- like duh.

Oh!
Pick a basis for $\ker T$
Pull back to basis of $\ker T^2$ (however you want)
okay, this can be made to work...
pick compliment for $\ker T^{k-1}$ in $\ker T^k$,
push it forward (still linearly independent b/c...),
  fill it out to compliment here, repeat.
[more global form of below]

Oh!
this is actually -very- simple.
given a parabolic subalgebra (it preserves a flag, not necessarily
complete), there's a basis such that the whole subalgebra is in
(block) upper triangular form.
  (like duh: do this "basis for compliment" that I'm talking about)
$T$ is nilpotent iff it strictly stabilizes a flag (duh);
1: extend this flag to a complete flag, which it strictly stabilizes;
 then in -any- basis subordinate to this flag, it's -strictly- upper
 triangular
 (the Jordan block form comes from the above more detailed description)

FIXME: Kolchin (sp?)'s theorem is exactly that if a whole algebra is
nilpotent, then it strictly stabilizes a -common- flag
  (that's the hard part).

Hmmm...better: if $T^k = 0$, but no lower one does, then look at
some $v$ that's a witness.
Claim 1: $v,Tv,\dots,T^{k-1}v$ are linearly independent.
Pf: $T^iv \in \ker T^{k-i} \setminus \ker T^{k-i-1}$
  (they're subordinate to some flag, hence linearly independent)
Okay, so on their span, it's a Jordan block.
Claim 2: This has a compliment that's fixed.
Pf: take a compliment in $\ker T$ and pull it back to rest?
Claim 3: done
Pf: duh -- induct.

Better proof: nilpotent means you strictly stabilize a complete flag
(send each thing down one);
work with this (pick bases and such?)
