\subsection{Basic Notation}
Denote the pairing between a vector space and its dual by:
$$\left$$
where $w^*\in V^*, V \in V$.
The notation $w^*$ doesn't mean that there is a $w\in V$ of which $w^*$ is dual;
we just want to underline that it's an element of the dual: $w^* \in V^*$.
If $V$ has a bilinear form on it, then we write
$$\left$$
where both $v,w \in V$.
In matrix terms, for the standard dot product
$$\left = w^tv$$
For a general bilinear form, it's instead:
$$\left = w^tMv$$
throughout, assume inner product space
Given an operator $T$,
the adjoint $T^*$ is defined by:
$$(T^*w^*)(v) = w^*(Tv)$$
(it's the pullback of $w^*$), i.e.,
$$\left = \left$$
In Dirac bra-ket notation, this is more symmetrically written:
\begin{align*}
\left &= \left = w^*Tv\\
&= \left = (T^*w^*)(v)
\end{align*}
Given an operator, the norm/quadratic form it induces is:
$$\norm{v}_T^2 := \left$$
This corresponds to writing the matrix and considering it the matrix
for a form, and is different from pulling back the existing form
(which is of interest in other contexts):
$$\norm{Tv} = \left = \left$$
Note that $T^*T$ is always self-adjoint; one can also consider
$TT^*$, which has the dual interpretation.
\subsection{bra-ket notation}
The punnily named bra-ket notation, common in physics, writes vectors as
$\left|v\right>$ (called \Def{ket})
and covectors as
$\left &:= \left|Tv\right>\\
\left\left$.
This is similar to how the shape of a matrix tells you its type,
but done more abstractly.
\subsection{Annihilators and perps}
Annihilator/Perp
Perp is internal annihilator
$T$ preserves $W$ iff $T^*$ preserves $W^*$
$T$ preserves flag iff $T^*$ preserves dual flag
where $V_i^* := \Ann V_i$.
- adjointness
$T$ preserves $W < V$ iff $T^*$ preserves $W^* < V^*$.
\begin{def}[Dual flag]
Given a flag
$0 = V_0 < V_1 < \dots V_n = V$, we get a dual flag
$V^* = V_0^* > V_1^* > \dots V_n^* = 0$.
\end{def}
\begin{prop}
$T$ preserves a flag iff $T^*$ preserves the dual flag.
\end{prop}
\begin{proof}
Let $v \in V, w^* \in V^*$.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Proof that an operator on a complex finite dimensional vector space has a
non-trivial eigenvector:
\begin{proof}[Polynomial algebra]
$T$ has an eigenvalue, hence an eigenvector.
Take the characteristic polynomial:
it has a root $\lambda$ by the fundamental theorem of algebra,
and thus $T-\lambda I$ has a non-trivial kernel, which is the eigenspace
for $\lambda$.
\end{proof}
\begin{proof}[Topology]
Lefschetz fixed point theorem
If it's not invertible, it as a non-trivial kernel.
If it is invertible, it yield a self-homeomorphism of projective space $\CP^{n-1}$
to itself
and it's the Id on $H^2$
(which generates the cohomology as an algebra)
hence has at least $(n-1)+1=n$ fixed points.
\end{proof}
Note that taking the companion matrix of a polynomial and applying the topological proof of the existence of an eigenvector proves the fundamental theorem of algebra.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- 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