Eigen-stuff

Here's a much, much better POV, that I just realized.

Recall the classic intro definition of eigenvector:
-Given- an operator $T$,
an eigenvector is a vector $v$
s.t. $Tv=\lambda v$ for some scalar $\lambda \in K$

Ok, fine, so what's the point?
The justification is 3-fold:
[vector POV]
- the action of $T$ on eigenvectors is easily understood
- if you can break up a vector into eigenvectors,
  the action is similarly easily understood
- operators are generically semisimple,
  so you can almost always do this
- ...and you can still do pretty close otherwise
  (Jordan)

Here's a more advanced viewpoint:
[subspace POV]
- scalar operators are the easiest to understand
- if you can decompose a space into eigenspaces,
  you can understand the operator
  (this is a slight abuse of language:
   that is, it's true but confusing:
   people will think ``simple=scalar'' and ``semisimple=sum of
   scalars'',
   while actually simple = no subspaces)
- and you can do this, generically
- even when you can't, you can decompose it as generalized eigenspaces

More sophisticated:
[submodule POV]
- A vector space with an operator is a $K[x]$-module;
  understanding the action is -tantamount- to understanding $V$
  as a $K[x]$-module.
- [This is -not- quite true: difference between char and minimal poly
   and all that]
- We can classify $K[x]$-modules.

Recall some terminology:
A module is -simple- iff it has no (proper) -submodules-;
a module is -semisimple- iff it breaks up a sum of simples.
a module is -irreducible- iff it does not break up as a sum.
The difference between -simple- and -irreducible- is:
  for irreducible, it's okay to have submodules, so long as they don't
  have complements.
``Not semisimple'' -exactly- means ``at some point irreducible but not simple''
[EG: $\bZ$ is irreducible, but not simple: $k\bZ$ is a submodule for
 any $k$, but it doesn't break up as a direct sum.]
A -ring- is semisimple iff every module over it is semisimple;
this exactly means: ``whenever you have a submodule, you have a complement''
[EG: representations of finite volume groups are semisimple,
 b/c you have an invariant metric (via averaging),
 hence you can take perps.]

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Ok, here's the real point:
The notion of eigenvector for an operator generalizes to
representation theory, in 2 ways:
- groups
- algebras

So:
given a vector space $V$,
- ...and an operator $T$,
  we let $V_\lambda = $ all the $\lambda$-eigenvectors
- ...and a group representation $\rho$
  we let $V_\eta$
  [note that here it's a little ickier:
   you can't say ``the vectors on which $G$ acts like $\eta$'',
   b/c there's no natural $\eta$ action on $V$;
   you need to say ``isomorphic'']
************ this next is why I wrote this!!! ***********
- ...and an algebra representation $A \to \End V$
  an eigenvector (here) is a vector $v$ s.t. there is some
  -functional- $\lambda$
  s.t. $av=\lambda(a) v$

These are both generalizations:
- we can think of an operator as a representation of $\bN$
- or we can think of it as a generator of a 1-dimensional algebra

Ok, so the ``algebra representation POV'' is simply:
take linear algebra seriously: replace -objects- by -spaces-
and -scalars- by -functionals-

Here's the better way to say it:
- Since the pairing $(\End V) \times V \to V$ is bilinear,
  we can replace ``operator'' and ``vector'' by their span.
- having an eigenvector means a subspace on which the action
  of $A$ factors through scalars: $A \to^\lambda K \to \End V$

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Urk.
What I -should- say is:
There's -two parts- of representation theory.
There's understanding (Lie) -algebra- maps $A \to \End V$,
and understanding -group- maps $G \to \Aut V$.
Given a -single- operator $T$:
- you can extend this to an -algebra- map like duh
  (linearly): $kT$
- you can extend this to a map of -monoids- via -powers- of $T$
  ($T^n$) $\bN \to \End V$
- if $T$ is -invertible-, you can extend this to a map of -groups-
  $\bZ \to \Aut V$
- but you -cannot- generally extend it to a map of a -continuous-
  group.
  EG, if $\det T < 0$ and you're working over $\bR$
  [...and this leads naturally into ``logs''
   and the exponential map $\End V \to \Aut V$!!!]

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So -why- is this filed under ``commuting families''?
Ah!
B/c the key idea should be:
- given a Lie subalgebra $\fa$ (note: this is weaker than being a
  subalgebra, since we're only closed under bracket, not
  multiplication), it has an eigenvector iff the derived Lie algebra
  is nilpotent

Thus by induction on dimension, it stabilizes a flag,
-exactly as in the usual way for a single operator-.

So what's the usual proof that there's an eigenvector?
You take the characteristic polynomial, find a zero,
and use the eigen-value- to find an eigenvector.

BTW, the functional is exactly
``top left entry, when in upper-triangular form''

Here should be a key idea:
- In matrix terms,
  given an upper triangular matrix,
  adding a -strictly- upper triangular matrix to it doesn't change the
  characteristic poly.
  [They don't need to commute.
   OTOH, if you add two upper triangular matrices,
   their eigenvalues add pairwise.]
Another important idea is:
- commuting operators
  preserve each others eigenspaces and -generalized- eigenspaces.
  [1: they preserve each others -kernels-,
      hence also the kernels of $(A-\lambda)$, and $(A-\lambda)^n$]
  

[This seems related to Jordan-Chevalley.]

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Urk, ok.
- commuting operators have a common eigenvector
Pf 1: topology
  Take a topological generator, and take its eigenvector.
