Upper triangularizability
(-not- triangulizability; need the `-ar')

Ooooooopppppsssss:
okay, the correct, ultimate statement is:
Heydar's thm:
A semigroup of linear transforms is triangularizable iff
the trace is permutable (meaning $\tr ABC = \tr BAC$ on the group)

ohohoh: nonono, that's the answer for -semigroups-
I want the answer for -algebras-.

[simpler ones: Engel's thm that a Lie Algebra of nilpotent operators
 is triangularizable]

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For lots of info:
"Simultaneous Triangularization"
by H. Radjavi and P. Rosenthal
Springer-Verlag New York, Inc., 2000

"A survey of Heydar Radjavi"
Linear Algebra and its Applications 383 (2004) 1--15
-------------------------------------------

The correct statement should be: (or -not-!!!)
given a Lie subalgebra $\fa < \gl$, it is conjugate into the Borel
algebra $\fb$ (of upper-triangular matrices) iff the derived algebra
$\fa' := [\fa,\fa]$ is nilpotent.
Pf:
Easy direction:
$\fb' = \fn$, the strictly upper triangular matrices, which are a
nilpotent algebra, and nilpotence is preserved under conjugation.
Hard direction:
???

cor: every operator is upper-triangularizable
pf: the algebra it generates is abelian, hence the derived algebra
   is trivial

cor: a commuting family of operators is simultaneously
  upper-triangularizable
pf: the algebra they generate is abelian

cor: (kolchin's thm) a nilpotent algebra is conjugate into $\fn$
(group POV: a unipotent semigroup is triangularizable)
pf: it's conjugate into $\fb$, and the nilpotent elements of $\fb$ are
  exactly $\fn$.

cor: (rank 1) If $[A,B]$ has rank 1, then they're simultaneously
  upper-triangulable
pf: a rank 1 commutator is nilpotent (see below)

[NB: Drinfeld told me to think about this problem.
 It's confusing as stated, b/c it's true for a very general reason.
 I don't know if he wanted a direct proof, or if he wanted me to think
 about it this way. It's quite a good problem, as there's several
 steps of progress to be made:
 - understand rank 1 operators
 - understand commutators ($\sl$, i.e., trace 0)
 - conjugation into borel]

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Rank 1 operators

Prop: If $A \in \End V$ is rank one, then $A$ acts as multiplication
by $\tr A$ on $\im A$.
Cor: $A$ is nilpotent (necessarily of degree $2$) iff $\tr A = 0$

Pf 1 (tensor: slickest):
  $A = v^* \otimes v$,
  so $A^2 = v^*(v) (v^* \otimes v) = v^*(v) A$
  and $v^*(v) = \tr A$.
  [Understand rank, composition, and trace in terms of tensor products.]
  [this really is a very geometric proof:
   $A$ sends everything to some multiple of $v$; that multiple is some
   linear functional, $v^*$.]
Pf 2 (ker, image: most direct):
  if $\im A < \ker A$, then $A^2=0$, so $A$ is nilpotent, hence has
  trace $0$, and acts as that on $\im A$.
  If $\im A \not < \ker A$, then $V=\im A \oplus \ker A$,
  and $A = k \oplus 0$, where $k = A\restrict_{\im A}$,
  and $\tr A = k$.
[Note that this is a case analysis, while the tensor POV is uniform.]
Pf 3 (matrix): Look at the matrix: upper triangulize it, and look at
  putative non-zero diagonal entries.

Cor: A rank 1 commutator is nilpotent (of rank $2$)
Pf: $\gl'=\sl$, the traceless operators. Thus if $A$ is a rank 1
commutator, it acts as $0$ on its image, so $A^2 = 0$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Commuting operators

There's a much simpler Pf for -commuting- operators over the complex
(or real) numbers:
recall the mantra:
any -closed- statement that's true of a single matrix
is true of a commuting family ("torus")
(via: "take a topological generator")

Since a single operator stabilizes a flag/is upper triangularizable
  (Schur's thm),
a commuting algebra does likewise.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Orthogonal and Borel

Prop 1: $\fo \cap \fb = \fd$
  $O(n) \cap B = O(1)^n$
  $U(n) \cap B = U(1)^n$
an orthogonal operator that's also upper-triangular
is diagonal with unit norm entries
Pf: obvious; Pf 1: take orthogonal axes for flag; must fix 'em
  Pf 2 (matrix): upper left entry has $\abs{z}=1$;
   conclude that $a_{2,1}=0$ (by orthogonality) and work your way
   forward

Prop 2:
Every real-orthogonal/unitary operator is orthogonally diagonalizable
Pf 1: Schur decomposition, and above observation
Pf 2: $A^*=A^T=A^{-1}$, so $A$ is normal; apply spectral theorem.

Prop 3:
A commuting family of orthogonals is conjugate into diagonals
Pf: conjugate into borel, and then they're diagonal
This is a concrete case of ``conjugate into maximal torus''.
