Spectral theorem

Wow. This is a -really- short proof of the spectral theorem!
hey, I figured out this proof myself, and it's the same one Axler uses!
(Axler: pp.~132--137)

Complex spectral theorem:
An operator $T$ on a complex inner product space has an orthonormal
basis of eigenvectors iff it's normal ($[T,T^*]=0$).

Pf (Matrices):
``only if'' is trivial:
  diagonal matrices commute with their adjoint.
``if'' (normal -> orthonormal basis):
- Schur decomposition:
  every matrix is conjugate to upper triangular -by orthogonal- ones
  [geometrically: you preserve a flag, and you can pick an orthonormal
   basis for this]
- normalcy is preserved under -unitary-/real-orthogonal change of basis
- the only upper triangular matrices that are normal are diagonal

Real spectral theorem:
An operator $T$ on a real inner product space has an orthonormal
basis of eigenvectors iff it's self-adjoint (aka, symmetric):
$T=T^*$.

Pf:
(slightly trickier in real case: need to prove that have real
 eigenvalues)
As above, but conjugate pair of complex eigenvalues yields
(if you only conjugate by real orthogonal matrices) a
non-symmetric 2x2 square on the diagonal (which is all you get if
you just assume normal), so you can't have any.

[you should think of rotation by an angle other than $k\pi$;
 this is real and normal but not self-adjoint,
 and this is basically the only thing that happens (up to scaling):
 a -b
 b  a]

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More deeply:
Why is normal the -right- condition?
It's actually -obvious- if you understand:
- adjoint operators
- commuting operators
[and apparently Drinfeld really thinks you should understand deeper
 linear algebra: the structure of two simultaneous operators
 (by looking at the filtrations they preserve and how they relate)]

Here's why:
Lemmata
- adjointness
  $T$ preserves $W < V$ iff $T^*$ preserves $W^* < V^*$.

- dual flag
  Given a flag $0 = V_0 < V_1 < \dots V_n = V$, we get a dual flag
  $V^* = V_0^* > V_1^* > \dots V_n^* = 0$.
  $T$ preserves a flag iff $T^*$ preserves the dual flag.

- perp flag
  Given a non-degenerate form, annihilator becomes an internal perp:
  $V = V_0^\perp > V_1^\perp > \dots V_n^\perp = 0$
  If the form is an inner product, then there's a unique orthogonal
  set of axes
  (an orthogonal direct sum decomposition into lines: $V=\bigoplus L_i$)
  that generates a flag and its perp (in the opposite orders).
  
  Concretely, $L_i := V_i \cap V_{i-1}^\perp$ is one-dimensional
  and $L_i \perp L_j$ if $i\neq j$: WLOG, $i<j$, so $L_i < V_i$,
  $L_j < V_{j-1}^\perp$, so $L_i \perp L_j$.
  [You can think of this as ``the compliment of $V_{i-1}$ in $V_i$'':
   building up the axes inductively]
  [Incidentally, $V_k = \oplus_{i=1}^k L_i$, and
   $V_k^\perp = \oplus_{i=k+1}^n L_i$; see below for a vector
   interpretation]

Pf (abstract):
- given two commuting operators, they preserve a common flag
  [this is the most subtle part]
  So if $T$ is normal, $T,T^*$ preserve a flag -- but since $T^*$ preserves
  the flag, $T$ preserves the perp flag also, so $T$ preserves the orthogonal
  set of axes, i.e., $T$ is orthogonally diagonalizable.

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Reinterpreting matrices and transposes
Given an inner product space,
If $T$ preserves a flag, with orthogonal basis $v_1,\dots,v_n$,
then $T^*$ preserves the perp flag, with orthogonal basis
$v_n,\dots,v_1$;

This is what it means (linear-algebraically) that the transpose of an
upper triangular matrix is lower triangular.

- If a single operator $T$ preserves both a flag and the perp flag,
  then it preserves an orthogonal set of axes;
  This is what it means that the intersection of upper and lower
  triangular matrices is diagonal matrices.
  [dude, we're getting serious insights]

- commuting matrices are simultaneously upper triangulizable,
  so if $T$ is normal (so $T,T^*$ commute), then there's a basis in
  which they're both upper triangular. But if $T^*$ is upper
  triangular, then $T$ is lower triangular -- and since it's both, it
  must be diagonal.
  [More abstractly: there's some flag that both $T,T^*$ preserve --
   but then $T^{**}=T$ preserves the perp flag, and we're done.]

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Commuting operators

Okay, so why are commuting operators sim. upper tri.?

Pf 1: "Topological Generator"
"For closed statements,
 true for 1 implies true for commuting family"
Given a commuting family of operators,
take a topological generator.
It stabilizes a flag, so the whole family does.

Pf 2: topologically clear?
I think this is topologically clear:
it's a torus acting on a flag manifold, and (for some reason) has a
fixed point.
This requires justification, but that can presumably be provided.

Simpler would be to show that they have a sim. eigenvector.
Oh: they preserve each other's (generalized) eigenspaces,
so these interlace. Duh.
Not -beautiful- as a proof, but I'm isolating the ick to a tiny area.

Pf 3: algebra
There's algebraic proofs of this;
I don't know or understand any.

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Applications:
- get principle curvatures on Riemannian manifolds
- conjugate by diagonal matrices with all 1s except one entry,
  which yields multiplication by $\alpha^2$ in that entry
  ...and get Sylvester's theorem
