Eigenvectors

Oh!
Suddenly -I understand- why:
- any operator
- on a finite dimensional vector space
- over an algebraically closed field
...has an eigenvector.

The idea was to think about the relation/opposition between
eigenvectors and cyclic vectors,
and the realization that you can get an eigenvector -starting from any
vector-!

It's (very roughly) an averaging trick.

Pf:
[I think that I've seen this proof before,
 but didn't really grok it at the time?]
[BTW, remember that ``characteristic classes'' of a vector bundle
 -are- the coefficients of the characteristic polynomial,
 in some sense: one definition -is- this construction.]
Take $v \in V, v\neq 0$.
Look at $\sp{v,Tv,\dots,T^k v}$ until this set is linearly dependent,
so $T^d v = a_{d-1}T^{d-1}v + \dots + a_0 v$,
i.e., $p(T)v=0$.
[B/c we take the minimal $k$, the top one is involved;
 b/c this is a field, we can make it monic]
$p$ is a polynomial over the field, so it has a root $\lambda$,
so $p(T)=(T-\lambda)q(T)$, and $q(T)=T^{d-1}+\dots+b_0$.
Let $w=q(T)v$.
- Then $p(T)v=0$, so $(T-\lambda)q(T)v = 0$, so $(T-\lambda)w=0$,
  so $Tw=\lambda w$, so $w$ is an eigenvector for $T$!
- $w\neq 0$ b/c $v,\dots,T^{d-1}v$ is linearly independent
  ($d$ is minimal for dependence), and we're taking a non-trivial
  linear combination (b/c the leading term $T^{d-1}$ has non-zero
  coefficient, in fact $1$).

Very concretely:
- start with any vector
- repeat until you cycle
- take the step -right before- the last one,
  and then add -previous- steps to it to make this an eigenvector
  (the eigenvector we're looking for lies on the affine space
   $T^{d-1}v + \sp{v,\dots,T^{d-2}v}$)
- the equation you're trying to solve is:
  $T(T^{d-1}v + ... + b_0 v)=\lambda(T^{d-1}v+\dots+b_0v)$
  In other words, (moving stuff to the left), you're trying to find a
  -factor- of $T-\lambda$

To see concretely why you need algebraic closure:
Take the typical ``rotation by $\pi/2$'' in $\bR^2$
You get $T^2v = -v$,
so you look at $Tv+bv$.
Then $T(Tv+bv) = T^2v+Tbv = bTv-v=b(Tv-(1/b)v)$,
so $b=-1/b$, so $b^2=-1$.

Concrete EG of this:
take a shear
[1 1
   1]
Start with $v=(0,1)$
$Tv=(1,1)$; still independent.
$T^2v=(2,1)$: dependent!
$T^2v=2Tv-v$, so $(T^2-2T+I)v=0$
$T-I$ factors this (and yields $T-I$; forgive confusion)
so $Tv-v=(1,1)-(0,1)=(0,1)$ is an eigenvector with eigenvalue 1,
as it should be.

More abstractly:
shows that $T-\lambda$ has a kernel.

It's completely elementary and obvious and doesn't require
determinants at all.

Look, here's the real story:
A vector space with an operator $T$ on it is a module over
$K[x]$.
A finite dim'l vect space yields a f.g. module,
and there's no free part, so it's $K[x]/q_1 \oplus K[x]/q_k$
(either dividing or rel prime).
We're looking for an $K[x]/(x-\lambda)$ -submodule-.
Just factor $q_1=(x-\lambda)r_1$. Duh.
Then $r_1 v_1$ is such a submodule (where $v_1$ is a generator for the
first summand), which just says
[Think: in $\bZ/6$, $3$ generates a $\bZ/2$ submodule.]

Lots of other stuff with ``cyclic vectors'' is -just- Chinese
Remainder Theorem.

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Oh, and this shows immediately how we can construct an operator with
no eigenvectors:
- abstractly,
  take a free $K[x]$ module
- concretely,
  take an infinite-dimensional space ($\bR^\infty := \bigoplus_\bN
  \bR$: cofinitely many coordinates vanish),
  and let $T$ act as ``right shift''
Pf that there's no eigenvectors:
  if $(a_0,\dotsc) = \lambda(0,a_0,\dotsc)$
  then $a_0=\lambda 0=0$, and by induction
    $a_{i+1}=\lambda a_i=\lambda 0 = 0$.
Pf 2: This is injective, but every coordinate is eventually zero

These are -exactly the same thing-:
$x$ is the ``shift'' (that's how it acts, given a cyclic vector)

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This is a generalization of the averaging trick:
if $T^n v=v$, then $v+Tv+\dots+T^{n-1}v$
is a fixed point.
(but may be zero)

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BTW, what's wrong with the usual determinant proof?
It's indirect:
you compute a value s.t. the determinant is not invertible,
and conclude that $T-\lambda$ has a kernel.

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BTW, why does a minimal poly have degree at most $n$?
(Sans determinant)
[Obviously $I,T,\dots,T^{n^2}$ has a relation
 b/c $\dim \End V = n^2$, but it doesn't take that far]

Start with any $v$, and go 'til it cycles.
You now have a $d$-dimensional subspace $V_1$ on which $T$ satisfies a
degree $d$ poly, $q_1$
-Quotient- out by this (you can do this b/c $T$ preserves the
subspace).
Repeat.
Take the product.

Concretely, $q_k(T)=0$ means $q_k(T)v$ lies inside $V_{k-1}$,
so if you apply $q_{k-1}$ to -that-, it'll lie inside $V_{k-2}$ etc.
...and by commutativity, we could've start the product at any stage,
so we didn't mess anything up by taking the product
($p=q_1\dots q_k = q_{i+1}\dots q_k q_1\dots q_i$,
 so this kills the whole space).
