The key sources of matrices are:
- operators
- bilinear forms (on a single space $V$)
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(intrinsic) Types of operators

These are properties that make sense without choice of basis.

- invertible/singular

- simple
  scalar; in any basis, scalar
  (these are a subgroup; none of the rest are)
- semisimple
  diagonalizable; in any basis -subordinate to- direct sum
  decomposition, diagonal -- but not generally
- nilpotent
  some power is zero
  (I think this is a lie subalgebra)
- trace-zero
  obviously the characteristic poly is intrinsic,
  but this is hard to see geometrically
  (lie subalgebra)

It makes sense to add and compose operators
(there's an algebra structure on $\End V$),
and there's also the bracket.
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(intrinsic) Types of bilinear forms

Switching the two factors is an order two action on forms,
and interacts with the two natural actions x -> -x and z -> $z^*$
(complex conjugation) on scalars, yielding:
- symmetric
- skew-symmetric
- conjugate symmetric
- conjugate skew-symmetric

These are all vector subspaces,
symmetric and conjugate symmetric are lie subalgebras;
dunno about the skew ones

- positive/positive definite
- semipositive/positive semi-definite/non-negative
  (operator theorists call -these- "positive")
- indefinite
- seminegative/negative semi-definite/non-positive
- negative/negative definite
...accordingly with the quadratic form defined by the bilinear form.
(so is $Q(v) := B(v,v) > 0$ for $v\neq 0$, etc.?)
Positive/semi-positive, negative/semi-negative are all cones.
(indefinite is pretty junky: it's the rest)

It makes sense to add forms, but not to compose them
(there's a vector space structure on $(V^{\otimes 2})^*$).
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Types of matrices
MathWorld lists 113 types of matrices!

An operator defines a matrix only up to conjugation;
a form defines a matrix only up to (what do you call this?)
Operators transform by M -> $X^{-1}MX$,
forms changes by M -> $X^tMX$.

Given a basis, you can write down a matrix for
- operators
- and bilinear forms
...and then now correspond
  (as $(V^{\otimes 2})^* = V^* \otimes V^*
  \approx V^* \otimes V = \Hom(V,V) = \End V$;
  the middle approx is only preserved by orthogonal changes of basis)

When you write down a matrix for an operator, you have a basis, hence
an inner product -- hence the form of a matrix reflects the orthogonal
properties of the transform.

Conversely, given an operator and an inner product, you don't quite
get a matrix -- you get a matrix -up to- orthogonal conjugation.

Thinking of a matrix as representing an operator -or- a form
is -really weird- (and confusing), as they transform differently.
Orthogonal changes of basis (conjugating by X, where $X^t=X^{-1}$)
respects this.

Subgroups --------------
- D: diagonal
  [preserves lines spanned by basis vectors]
  topologically $K^n$, or $(K^*)^n$ if invertible
- U: upper triangular
  [preserves standard flag]
  topologically $K^n\oplus K^{\binom{n}{2}}$, or
    $(K^*)^n \oplus K^{\binom{n}{2}}$ if invertible
  (or L: lower triangular -- that's just reading the vectors backwards,
   correpsponding to conjugation by the permutation matrix:
   (  1)
   ( 1 )
   (1  ) )
  strictly upper (resp., lower) triangular
  (these are all nilpotent, and any group of nilpotent matrices is
   conjugate into this)

- block diagonal; block upper triangular
  [preserve some coarser direct sum decomposition/flag]

- O: orthogonal
  very interesting topology

Transposing -------------------
At the level of matrices, switching variables in a form corresponds
to transposing the matrix, and we get:
[these are all immediate to see from the matrix]
- symmetric
- skew-symmetric
- conjugate symmetric
- conjugate skew-symmetric

Note that (away from 2), each of these pairs give a direct sum
decomposition.

None of these are subgroups (which we'd expect: multiplying forms is
kinda silly); consider:
 (1 1)   (1 0)   (1 0)
 (1 1) * (0 0) = (1 0)
[ah, but they are Lie algebras or Jordan algebras!]

Positive etc. --------------------
[ok, this is b/c quad forms / Hermitian are formally real Jordan algebra]
Since several bilinear forms determine the same quadratic form,
(the natural splitting (away from 2) is the unique -symmetric-
 bilinear form, via polarization identities)
one often conventionally requires positive matrices to not just
define a positive form, but also be symmetric.

Geometrically, a matrix is (semi)positive iff in some orthogonal basis
it's just diagonal and (semi)positive on the diagonal:
i.e., really obvious behaviour.
The interest is that there are other characterizations in more
intrinsic terms.

A matrix is positive iff all upper-left minors are positive.

Positivity does make some sense for -operators-
(and here's where life gets confusing -- we originally defined it as a
 property of -forms-):
a (semi-)positive operator has a unique positive square root.

This all isn't so impressive -- if you look at a diagonal matrix, it's
obvious if it's positive or not (and you can look at the upper-left
minors, duh), and it obviously has a unique positive square root (or
$n$th root, or log); general case follows from spectral theorem.

Positive operators roughly correspond to positive reals
(sitting inside the complex plane, say):
they're exactly the operators of the form $T=S^*S$
(just as real numbers are exactly the complex numbers of the form
 $z^*z$; c.f. Polar Decomposition)

Here's some equivalent definitions of positive:
(p.145 of Axler)
a: positive (self-adjoint, positive form)
b: self-adjoint and non-negative eigenvalues
c: has a positive square root
d: has a self-adjoint square root
e: is of the form $S^*S$
a-d follow from the spectral theorem (which is serious);
d -> e is obvious (take $S$ to be a square root)
e -> a is obvious ($S^*S$ is self-adjoint,
  and $\left<S^*Sv,v\right> = \left<Sv,Sv\right> \geq 0$)

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Completely positive matrices
A real $n \times n$ matrix M is -completely positive- iff it can be
factored as $M=BB^t$, where $B$ is an $m\times n$ matrix with all
non-negative entries; the minimal $m$ is called the "factorization
index" or cp-rank of $M$.

Apparently combinatorialists and quadratic forms folk are interested
in these; they seem completely arbitrary to me.

The most basic questions (when is a matrix completely positive? how
can you compute the cp-rank?) are apparently still open.
-----------------------------------------------
Normal
$[T,T^*]=0$ (so $TT^*=T^*T$)
normal operators are kinda weird to me; I don't have a good sense for
them.
[okay, via the spectral theorem I know that they're just dumb:
 equivalently: in some orthogonal basis they're diagonal,
 but I don't have a good intuition for this]
[ok, now I do]

Anyways, here's some obvious notes:
- the following are obviously normal:
 + self-adjoint (over reals, this means symmetric)
   as $T^*=T$
 + unitary (over reals, this means orthogonal)
   as $T^*=T^{-1}$
[and b/c we preserve determinant, up to conjugation,
 we're not going to get $T^*$ to be any other power]
Notably, unitary/orthogonal operators have very obvious geometric
meaning, while self-adjoint/symmetric ones are more subtle.

Self-adjoint <-> normal, all real eigenvalues
-> obviously normal, so diagonalizable over C,
   so must have real eigenvalues
<- normal means diagonalizable over C, and
   all real eigenvalues (on that diagonal) means self-adjoint
