in the presence of a non-degen bilinear form (say, an inner product)
a bilinear form is the same thing as an operator

what's the connection?
esp.:
is sylvester's theorem related to the (real?) spectral theorem?
(yes: they're both same data: a set of axes, each with a weight;
 either diag(a1,...,an) on the axes,
 or a1 x_1^2 + ... + an x_n^2 as form)
(yes; think about how the -proofs- are related)
(NB: given any bilinear form, operators yield bilinear forms)
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uppertriangulizability
  understand esp. this one on commuting
a commuting family of operators stabilizes a common flag
  (i.e., is conjugate into the Borel group)
pf: just need to show that they have a common eigenvector,
  then it's formal (just as "stabilizes a flag" follows from
  "has an eigenvector")
[Cor: spectral theorem: T and T* stabilize a common flag
 so T also stabilizes the perp flag, hence ortho-diagonalizable;

 (take the intersections to get basis)
 concretely,
 they stabilize V_0 < V_1 < ... < V_n,
 so also stabilize (V_n)* < ... (V_0)*
   (V_k has dim k; (V_k)* has -codim- k)
 (V_k) + (V_k)* is an ortho direct sum (by def'n of perp)
 V_k \cap (V_{k+1})* has dim 1: let e_k be a unit norm vector;
 the e_i, e_j are ortho b/c if i > j, then e_i in V_i
 and e_j in (V_{j+1})* < (V_i)*, so <e_i,e_j>=0]
[even more simply:
 Lemma 1: T preserves V iff T^* preserves V^*; Pf: defn
 Lemma 2: On inner product space, T and T^* both preserve V
   iff T preserves direct sum V + V^perp (iff they both do, clearly)
 Pf of spectral thm: T,T^* commute, so they have a common eigenvector
 v; so they preserve v + v^perp. induct.
 (this skips flags and does the induction -here-, which is a bit simpler)]

Kolchin's thm:
  a nilpotent family is conjugate into the nilpotent matrices
two operators A,B s.t. [A,B] has rank 1 stabilize a common flag
  (Drinfeld's problem that cheered him up one year:
   proof is: [A,B] = C has trace 0 (b/c commutator),
   and rank one means C= w^* \otimes v;
   trace zero means w^*v = 0, so C^2=0, so nilpotent;
   thus this follows from the general case)
  (? what about a family whose derived algebra has rank 1?)
NB: the Borel group isn't commutative, so we should expect something
  like this; a commuting family is a very special algebra
  (presumably we can bound the rank; I'd guess by n? (if in GL_n+1)
   oh yeah -- this really should be the def'n of the rank or such)
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SL2 inverse
When is the map A -> A^{-1} given by conjugation?
Well:
- you better be abelian, as A -> A^{-1} is an -anti-homomorphism,
  and any conjugation is a homomorphism
  [oh! so this is an -outer- automorphism:
   point is that every abelian group has a natural Z/2 action
   given by A -> A^{-1}, which is an outer automorphism,
   so get an extension A semi Z/2:
   question is to find these in linear land]
- you better have determinant +/- 1, as conjugation doesn't
  change your determinant and you need det A^{-1}=(det A)^{-1},
  so (det A)^2 = 1 so det A = +/- 1: so SLn (or SLn^\pm)
- you better be dimension 2, as crammer's rule says A -> A^{-1}
  is degree (n-1) (b/c of minors) on SLn

here's 2 EGs -- are they all?
(l 0)
(0 l^-1)
conjugate by
(0 1)
(1 0)
[many conjugate such groups: expand on one, contract on other]
(cos th -sin th)
(sin th  cos th)
(conj by -any- reflection works)
-weird-:
conj by
(0 1)
(1 0) works on all of these individually, but can't work on products
  b/c non-abelian
duh: gABg' = gAg'gBg' = A'B' != B'A' = (AB)'

(cosh th sinh th)
(sinh th cosh th)

- so you better be SL2 (or SL1^\pm, where A -> A^{-1} is trivial)
then (a b) -> (d -b)
     (c d)    (-c a)
[mneumonic: recall det = ad-bc (cross), and look at first column,
 which better be d,-b (for formula to work out) -- can check work
 by looking at other entries. (easy to get confused b/c you switch
 some and convert others to minus -- which is which?)]
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permutation matrices: transpose is inverse
pf 1: obvious from matrix
pf 2: a_ji=1 means you take row i to row j
  so obviously the transpose reverses this
pf 3: they're a standard basis, hence give orthogonal matrix
also fun for easy computations, esp. S_3;
also good way of demonstrating -sign-
[funny irony: some defns of det use sign on S_n to define it;
 conversely, can use det to define sign on S_n!
 they're really a very similar idea;
 really, one should say that S_n = GL_n F_1 or something,
 and then sign -is- det (well, sorta -- shit, this is the archimedean
 place).]
also O_n \cap GL_n Z = (1,-1)^n semidirect S_n as permutation matrices
[a sorta maximal torus and weyl group;
 I wanna say U_n \cap some complex thing = (S^1)^n semidirect S_n:
 the complex thing should be "matrices with entries of integer length"]