\documentclass{amsart}\newtheorem*{defn}{Defintion}
\newtheorem*{prop}{Proposition}
\newtheorem*{thm}{Theorem}
\newtheorem*{cor}{Corollary}
\theoremstyle{definition}
\newtheorem*{defin}{Definition}
\newtheorem*{example}{Example}

\newcommand{\F}{\mathbb F}
\newcommand{\bilin}[1]{\langle #1\rangle}

\title{Lecture : Bilinear Forms}

\begin{document}
\maketitle

Today we will focus on extra structure we can put on a vector space that render certain constructions more natural.

\begin{defin}
A \underline{bilinear form} $\langle-,-\rangle$ on $V$ is a function
\[
\langle-,-\rangle\colon V\times V\to \F
\]
that is \underline{bilinear}:
\[
\langle a\vec{v}+b\vec{u},\vec{w}\rangle=a\bilin{\vec{v},\vec{w}}+b\bilin{\vec{u},\vec{w}}\text{ and}
\]
\[
\bilin{\vec{v},a\vec{u}+b\vec{w}}=a\bilin{\vec{v},\vec{u}}+b\bilin{\vec{v},\vec{w}}.
\]
\end{defin}

It is important to note that $\bilin{-,-}$ is a function on $V\times V$ as a set and {\emph{not}} a linear map on the direct sum $V\oplus V$. This makes this definition a little unsettling: we consider structure on vector spaces that does not stay in the category of vector spaces.

We have a few types of special bilinear forms.
\begin{defin}
\begin{enumerate}
\item $\bilin{-,-}$ is symmetric if $\bilin{\vec{v},\vec{w}}=\bilin{\vec{w},\vec{v}}$ for all $\vec{v},\vec{w}\in V$.
\item $\bilin{-,-}$ is skew-symmetric if $\bilin{\vec{v},\vec{w}}=-\bilin{\vec{w},\vec{v}}$ for all $\vec{v},\vec{w}\in V$.
\end{enumerate}
\end{defin}

We should think of $\bilin{-,-}$ as a generalization of the dot product on $\mathbb R^n$. This provides a fantastic example.

\begin{example}
\begin{enumerate}
\item The dot product on $\mathbb R^n$ is a bilinear form.
\item The generalization of the dot product on $\F^n$ is a bilinear form.
\item If $a_1,\dots,a_n$ are elements of $\mathbb R$ (or $\F$), then $\bilin{\vec{v},\vec{w}}=a_1v_1w_1+\dots+a_nv_nw_n$ is a bilinear form.
\end{enumerate}
\end{example}

The last part is very helpful, and it helps us tie bilinear forms to matrices.

\begin{prop}
Choose a basis $\mathcal B$ $\vec{v}_1,\dots,\vec{v}_n$ for $V$, and let $[\vec{v}]_{\mathcal B}\in \mathbb F^n$ denote the column vector that represents $\vec{v}$ with respect to $\mathcal B$. Let $B$ denote the matrix with 
\[
B_{i,j}=\bilin{\vec{v}_i,\vec{v}_j}.
\]
Then
\[
[\vec{v}]_{\mathcal B}^t B[\vec{w}]_{\mathcal B}=\bilin{\vec{v},\vec{w}}.
\]
\end{prop}
\begin{proof}
This is an exercise in bilinearity. Both sides are clearly bilinear (the left one is so since multiplication distributes over addition). By linearity in the first factor, to know $\bilin{\vec{v},\vec{w}}$ (or the column vector form), it suffices to know $\bilin{\vec{v}_i,\vec{w}}$ for all $i$. Similarly, by linearity in the second factor, to know $\bilin{\vec{v},\vec{w}}$, it suffices to know $\bilin{\vec{v},\vec{v}_j}$ for all $j$. Thus we are reduced to showing
\[
[\vec{v}_i]_{\mathcal B}^t B[\vec{v}_j]_{\mathcal B}=\bilin{\vec{v}_i,\vec{v}_j}.
\]
However, by definition, $[\vec{v}_i]_{\mathcal B}=\vec{e}_i$. Matrix multiplication tells us that
\[
[\vec{v}_i]_{\mathcal B}^t B[\vec{v}_j]_{\mathcal B}=B_{i,j}=\bilin{\vec{v}_i,\vec{v}_j},
\]
as required.
\end{proof}
Thus when we have a basis $\mathcal B$, we have a connection between matrices and bilinear forms.
\begin{prop}
A choice of basis $\mathcal B$ of $V$ provides a $1-1$ correspondence between bilinear forms and the associated matrices.
\end{prop}

What happens then if we change basis? Let $\mathcal C=\{\vec{w}_1,\dots,\vec{w}_n\}$ be another basis, and let $P={}_\mathcal C P_{\mathcal B}$ be the change-of-basis matrix. Then we know
\[
[\vec{v}]_{\mathcal C}={}_{\mathcal C}P_{\mathcal B}[\vec{v}]_{\mathcal B}.
\]
Let $C$ be the matrix associated to $\bilin{-,-}$ in the $\mathcal C$-basis, and similarly for $B$. Then we have
\[
[\vec{v}]_{\mathcal C}^t C[\vec{w}]_{\mathcal C}=\bilin{\vec{v},\vec{w}}=[\vec{v}]_{\mathcal B}^t B [\vec{w}]_{\mathcal B}.
\]
Combining these two equations, we see that for all $\vec{v}$ and $\vec{w}$, we have
\[
[\vec{v}]_{\mathcal C}^t C[\vec{w}]_{\mathcal C}=\big(P[\vec{v}]_\mathcal B\big)^t C \big(P[\vec{w}]_\mathcal B\big)=[\vec{v}]_{\mathcal B}^t\big(P^tCP\big)[\vec{w}]_{\mathcal B}=[\vec{v}]_{\mathcal B} B[\vec{w}_{\mathcal B}.
\]
We've therefore shown the following.
\begin{prop}
If $P$ is the change-of-basis matrix from $\mathcal B$ to $\mathcal C$, and if $B$ (resp $C$) represents $\bilin{-,-}$ in the $\mathcal B$-basis (resp $\mathcal C$-basis), then
\[
P^tCP=B.
\]
\end{prop}
Thus we'll say that two matrices are transpose equivalent if there is an invertible matrix $P$ such that $P^tCP=B$. This is not standard.

Row vectors showed up in another context: dual spaces. This is one of the most powerful aspects of bilinear forms: they provide a {\emph{natural}} way to connect a vector space and its dual.

\begin{thm}
Let $\vec{v}\in V$, and let $R_{\vec{v}}$ denote $\bilin{\vec{v},-}$. Then
\begin{enumerate}
\item For all $\vec{v}$, $R_{\vec{v}}$ is a linear transformation $V\to\F$.
\item The assignment $R\colon V\to V^\ast$ given by $\vec{v}\mapsto R_{\vec{v}}$ is linear.
\end{enumerate}
\end{thm}

\begin{proof}
Both of these are given by linearity in their respective factors. The bilinear form $\bilin{-,-}$ is linear in the second factor, so for all $\vec{v}$, $\bilin{\vec{v},-}$ is a linear map. Thus we have the first part, and we learn that $R$ is a function from $V$ to $V^\ast$. 

For the second, we need to know that for all $\vec{u}\in V$,
\[
R_{a\vec{v}+b\vec{w}}(\vec{u})=aR_{\vec{v}}(\vec{u})+bR_{\vec{w}}(\vec{u}).
\]
This will show that $V\to V^\ast$ is a linear map. So we check:
\[
R_{a\vec{v}+b\vec{w}}(\vec{u})=\bilin{a\vec{v}+b\vec{w},\vec{u}}=a\bilin{\vec{v},\vec{u}}+b\bilin{\vec{w},\vec{u}}=aR_{\vec{v}}(\vec{u})+bR_{\vec{w}}(\vec{u}),
\]
since $\bilin{-,-}$ is linear in the first factor.
\end{proof}

To fully understand the map $R$, we have to determine $\ker(R)$ and the image of $R$.

\begin{defin}
The \underline{radical} of $V$, $Rad(V)$ is defined by
\[
Rad(V)=\{\vec{v}\in V | \forall\vec{w}\in V, \quad \bilin{\vec{v},\vec{w}}=0\}.
\]
\end{defin}
\begin{prop}
We have
\[
\ker(R)=Rad(V).
\]
\end{prop}
\begin{proof}
This is immediate. The zero functional is the one that assigns the value $0$ to all $\vec{w}\in V$. So $R(\vec{v})=0$ iff for all $\vec{w}\in V$, $\bilin{\vec{v},\vec{w}}=0$. This is the same thing as $\vec{v}\in Rad(V)$.
\end{proof}

To make this more useful, we tether this to the matrix form. If $\vec{v}\in V$ is in the radical, then we have
\[
[\vec{v}]_\mathcal B^t B[\vec{w}]_\mathcal B=0
\] 
for all $\vec{w}\in V$. This is the same condition as $[\vec{v}]_\mathcal B^t B=\vec{0}^t$, or $[\vec{v}]_{\mathcal B}$ is in the null-space of $B^t$.

\begin{defin}
A bilinear form $\bilin{-,-}$ is \underline{singular} if $B$ is singular.

A bilinear form $\bilin{-,-}$ is \underline{non-singular} if $B$ is invertible.
\end{defin}

\begin{cor}[Reisz Representation Theorem]
If $\bilin{-,-}$  is non-singular, then
\[
R\colon V\to V^\ast
\]
is an injection. If $V$ is finite dimensional, then $R$ is an isomorphism.
\end{cor}

Thus in the finite dimensional case, for any $f\in V^\ast$, we have a vector $\vec{v}_f\in V$ such that $f(\vec{w})=\bilin{\vec{v}_f,\vec{w}}$ for all $\vec{w}\in V$. This is the Reisz vector for $f$.

Now if $S$ is a subspace of $V$, then $S$ inherits a bilinear form by restriction. It can be the case that $S$ is singular even if $V$ is not. If $V$ is non-singular and $S$ is a singular subspace, then the Reisz vector for $f\in S^\ast$ need not be a vector in $S$. In fact, it could be just a vector in $V$. $S$ is non-singular exactly when the Reisz vector is back in $S$.

\end{document}