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\ihead{A note on bilinear forms}
\ohead{\today}
\begin{document}

\title{A note on bilinear forms}
\author{Darij Grinberg}
\date{\today}
\maketitle
\tableofcontents

\section{Introduction}

In this note, we shall prove some elementary linear-algebraic properties of
bilinear forms. These properties generalize some of the standard facts about
nondegenerate bilinear forms on finite-dimensional vector spaces (e.g., the
fact that $\left(  V^{\perp}\right)  ^{\perp}=V$ for a vector subspace $V$ of
a vector space $A$ equipped with a nondegenerate symmetric bilinear form) to
bilinear forms which may be degenerate. They are unlikely to be new, but I
have not found them explored anywhere, whence this note.

\section{Bilinear maps and forms}

We fix a field $\mathbf{k}$. This field will be fixed for the rest of this
note. The notion of a \textquotedblleft vector space\textquotedblright\ will
always be understood to mean \textquotedblleft$\mathbf{k}$-vector
space\textquotedblright. The word \textquotedblleft subspace\textquotedblright%
\ will always mean \textquotedblleft$\mathbf{k}$-vector
subspace\textquotedblright. The word \textquotedblleft
linear\textquotedblright\ will always mean \textquotedblleft$\mathbf{k}%
$-linear\textquotedblright. If $V$ and $W$ are two $\mathbf{k}$-vector spaces,
then $\operatorname*{Hom}\left(  V,W\right)  $ denotes the vector space of all
linear maps from $V$ to $W$. If $S$ is a subset of a vector space $V$, then
$\operatorname*{span}S$ will mean the span of $S$ (that is, the subspace of
$V$ spanned by $S$).

We recall the definition of a bilinear map:

\begin{definition}
\label{def.bilmap}Let $V$, $W$ and $U$ be three vector spaces. Let $f:V\times
W\rightarrow U$ be any map.

\textbf{(a)} We say that the map $f$ is \textit{linear in its first argument}
if, for every $w\in W$, the map $V\rightarrow U,\ v\mapsto f\left(
v,w\right)  $ is linear.

\textbf{(b)} We say that the map $f$ is \textit{linear in its second argument}
if, for every $v\in V$, the map $W\rightarrow U,\ w\mapsto f\left(
v,w\right)  $ is linear.

\textbf{(c)} We say that the map $f$ is \textit{bilinear} if it is both linear
in its first argument and linear in its second argument.
\end{definition}

More explicitly, this definition can be rewritten as follows: A map $f:V\times
W\rightarrow U$ is linear in its first argument if and only if every
$\lambda_{1},\lambda_{2}\in\mathbf{k}$ and $v_{1},v_{2}\in V$ and $w\in W$
satisfy%
\begin{equation}
f\left(  \lambda_{1}v_{1}+\lambda_{2}v_{2},w\right)  =\lambda_{1}f\left(
v_{1},w\right)  +\lambda_{2}f\left(  v_{2},w\right)  .\label{eq.def.bilmap.1}%
\end{equation}
A map $f:V\times W\rightarrow U$ is linear in its second argument if and only
if every $\lambda_{1},\lambda_{2}\in\mathbf{k}$ and $v\in V$ and $w_{1}%
,w_{2}\in W$ satisfy%
\begin{equation}
f\left(  v,\lambda_{1}w_{1}+\lambda_{2}w_{2}\right)  =\lambda_{1}f\left(
v,w_{1}\right)  +\lambda_{2}f\left(  v,w_{2}\right)  .\label{eq.def.bilmap.2}%
\end{equation}
A map $f:V\times W\rightarrow U$ is bilinear if and only if it satisfies both
(\ref{eq.def.bilmap.1}) and (\ref{eq.def.bilmap.2}).

Bilinear maps are also known as \textquotedblleft$\mathbf{k}$-bilinear
maps\textquotedblright\ (actually, this latter notion is used when
$\mathbf{k}$ is not clear from the context).

Here are some examples:

\begin{example}
\label{exam.bilmap.0}Let $V$, $W$ and $U$ be three vector spaces. Let
$\mathbf{0}:V\times W\rightarrow U$ be the map which sends every $p\in V\times
W$ to $0$. Then, the map $\mathbf{0}$ is bilinear.
\end{example}

\begin{example}
\label{exam.bilmap.prods}Consider $\mathbf{k}^{n}$ as the vector space of all
row vectors of size $n$ (with entries in $\mathbf{k}$).

\begin{itemize}
\item The map $\mathbf{k}^{n}\times\mathbf{k}^{n}\rightarrow\mathbf{k}$
sending every $\left(  \left(  x_{1},x_{2},\ldots,x_{n}\right)  ,\left(
y_{1},y_{2},\ldots,y_{n}\right)  \right)  \in\mathbf{k}^{n}\times
\mathbf{k}^{n}$ to $x_{1}y_{1}+x_{2}y_{2}+\cdots+x_{n}y_{n}\in\mathbf{k}$ is a
bilinear map. (This map is called the \textit{dot product map}, and the image
of a pair $\left(  v,w\right)  \in\mathbf{k}^{n}\times\mathbf{k}^{n}$ under
this map is called the \textit{dot product} of $v$ and $w$.)

\item The map $\mathbf{k}^{n}\times\mathbf{k}^{n}\rightarrow\mathbf{k}^{n}$
sending every $\left(  \left(  x_{1},x_{2},\ldots,x_{n}\right)  ,\left(
y_{1},y_{2},\ldots,y_{n}\right)  \right)  \in\mathbf{k}^{n}\times
\mathbf{k}^{n}$ to $\left(  x_{1}y_{1},x_{2}y_{2},\ldots,x_{n}y_{n}\right)
\in\mathbf{k}^{n}$ is a bilinear map.

\item The map $\mathbf{k}^{n}\times\mathbf{k}^{n}\rightarrow\mathbf{k}%
^{n\times n}$ (where $\mathbf{k}^{n\times n}$ denotes the ring of $n\times
n$-matrices over $\mathbf{k}$) sending every $\left(  \left(  x_{1}%
,x_{2},\ldots,x_{n}\right)  ,\left(  y_{1},y_{2},\ldots,y_{n}\right)  \right)
\in\mathbf{k}^{n}\times\mathbf{k}^{n}$ to the matrix%
\[
\left(
\begin{array}
[c]{cccc}%
x_{1}y_{1} & x_{1}y_{2} & \cdots & x_{1}y_{n}\\
x_{2}y_{1} & x_{2}y_{2} & \cdots & x_{2}y_{n}\\
\vdots & \vdots & \ddots & \vdots\\
x_{n}y_{1} & x_{n}y_{2} & \cdots & x_{n}y_{n}%
\end{array}
\right)  \in\mathbf{k}^{n\times n}%
\]
is a bilinear map.

\item The map $\mathbf{k}^{n}\times\mathbf{k}^{n}\rightarrow\mathbf{k}$
sending every $\left(  \left(  x_{1},x_{2},\ldots,x_{n}\right)  ,\left(
y_{1},y_{2},\ldots,y_{n}\right)  \right)  \in\mathbf{k}^{n}\times
\mathbf{k}^{n}$ to $x_{1}+y_{1}+x_{2}+y_{2}+\cdots+x_{n}+y_{n}\in\mathbf{k}$
is not bilinear (unless $n=0$). It is neither linear in its first argument nor
linear in its second argument.

\item The map $\mathbf{k}^{n}\times\mathbf{k}^{n}\rightarrow\mathbf{k}$
sending every $\left(  \left(  x_{1},x_{2},\ldots,x_{n}\right)  ,\left(
y_{1},y_{2},\ldots,y_{n}\right)  \right)  \in\mathbf{k}^{n}\times
\mathbf{k}^{n}$ to $\left(  x_{1}+x_{2}+\cdots+x_{n}\right)  y_{1}y_{2}\cdots
y_{n}\in\mathbf{k}$ is linear in its first argument, but not bilinear (unless
$n\leq1$).
\end{itemize}
\end{example}

\begin{example}
\label{exam.bilmap.ev}Let $V$ and $W$ be two vector spaces. The map
$\operatorname*{Hom}\left(  V,W\right)  \times V\rightarrow W,\ \left(
f,v\right)  \mapsto f\left(  v\right)  $ is bilinear.
\end{example}

Let us see two ways to construct new bilinear maps out of old:

\begin{definition}
\label{def.op}Let $V$, $W$ and $U$ be three sets. Let $f:V\times W\rightarrow
U$ be any map. Then, $f^{\operatorname*{op}}$ will be denote the map%
\[
W\times V\rightarrow U,\ \ \ \ \ \ \ \ \ \ \left(  w,v\right)  \mapsto
f\left(  v,w\right)  .
\]
Thus, $f^{\operatorname*{op}}\left(  w,v\right)  =f\left(  v,w\right)  $ for
every $\left(  w,v\right)  \in W\times V$.
\end{definition}

The following proposition is obvious:

\begin{proposition}
\label{prop.op}Let $V$, $W$ and $U$ be three sets. Let $f:V\times W\rightarrow
U$ be any map.

\textbf{(a)} We have $\left(  f^{\operatorname*{op}}\right)
^{\operatorname*{op}}=f$.

\textbf{(b)} Assume that the sets $V$, $W$ and $U$ are vector spaces. Assume
that the map $f$ is bilinear. Then, the map $f^{\operatorname*{op}}:W\times
V\rightarrow U$ is bilinear.
\end{proposition}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.op}.]\textbf{(a)} The map
$f^{\operatorname*{op}}$ is a map $W\times V\rightarrow U$ (by its
definition). Thus, the map $\left(  f^{\operatorname*{op}}\right)
^{\operatorname*{op}}$ is a map $V\times W\rightarrow U$ (by its definition).

For every $\left(  v,w\right)  \in V\times W$, we have%
\begin{align*}
\left(  f^{\operatorname*{op}}\right)  ^{\operatorname*{op}}\left(
v,w\right)   &  =\left(  f^{\operatorname*{op}}\right)  \left(  w,v\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\left(
f^{\operatorname*{op}}\right)  ^{\operatorname*{op}}\right) \\
&  =f\left(  v,w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition
of }f^{\operatorname*{op}}\right)  .
\end{align*}
In other words, $\left(  f^{\operatorname*{op}}\right)  ^{\operatorname*{op}%
}=f$ (since both $\left(  f^{\operatorname*{op}}\right)  ^{\operatorname*{op}%
}$ and $f$ are maps $V\times W\rightarrow U$). This proves Proposition
\ref{prop.op} \textbf{(a)}.

\textbf{(b)} The map $f$ is bilinear. In other words, the map $f$ is both
linear in its first argument and linear in its second argument (by the
definition of \textquotedblleft bilinear\textquotedblright).

The map $f$ is linear in its first argument. In other words,
\begin{equation}
\text{for every }w\in W\text{, the map }V\rightarrow U,\ v\mapsto f\left(
v,w\right)  \text{ is linear} \label{pf.prop.op.b.1}%
\end{equation}
(by the definition of \textquotedblleft linear in its first
argument\textquotedblright).

The map $f$ is linear in its second argument. In other words,
\begin{equation}
\text{for every }v\in V\text{, the map }W\rightarrow U,\ w\mapsto f\left(
v,w\right)  \text{ is linear} \label{pf.prop.op.b.2}%
\end{equation}
(by the definition of \textquotedblleft linear in its second
argument\textquotedblright).

For every $v\in V$, the map $W\rightarrow U,\ w\mapsto f^{\operatorname*{op}%
}\left(  w,v\right)  $ is linear\footnote{\textit{Proof.} Let $v\in V$. For
every $w\in W$, we have $f^{\operatorname*{op}}\left(  w,v\right)  =f\left(
v,w\right)  $ (by the definition of $f^{\operatorname*{op}}$). But
(\ref{pf.prop.op.b.2}) shows that the map $W\rightarrow U,\ w\mapsto f\left(
v,w\right)  $ is linear. In other words, the map $W\rightarrow U,\ w\mapsto
f^{\operatorname*{op}}\left(  w,v\right)  $ is linear (since
$f^{\operatorname*{op}}\left(  w,v\right)  =f\left(  v,w\right)  $ for every
$w\in W$). Qed.}. Renaming the indices $v$ and $w$ as $w$ and $v$ in this
statement, we obtain the following: For every $w\in V$, the map $W\rightarrow
U,\ v\mapsto f^{\operatorname*{op}}\left(  v,w\right)  $ is linear. In other
words, the map $f^{\operatorname*{op}}$ is linear in its first argument (by
the definition of \textquotedblleft linear in its first
argument\textquotedblright).

For every $w\in W$, the map $V\rightarrow U,\ v\mapsto f^{\operatorname*{op}%
}\left(  w,v\right)  $ is linear\footnote{\textit{Proof.} Let $w\in W$. For
every $v\in V$, we have $f^{\operatorname*{op}}\left(  w,v\right)  =f\left(
v,w\right)  $ (by the definition of $f^{\operatorname*{op}}$). But
(\ref{pf.prop.op.b.1}) shows that the map $V\rightarrow U,\ v\mapsto f\left(
v,w\right)  $ is linear. In other words, the map $V\rightarrow U,\ v\mapsto
f^{\operatorname*{op}}\left(  w,v\right)  $ is linear (since
$f^{\operatorname*{op}}\left(  w,v\right)  =f\left(  v,w\right)  $ for every
$v\in V$). Qed.}. Renaming the indices $v$ and $w$ as $w$ and $v$ in this
statement, we obtain the following: For every $v\in W$, the map $V\rightarrow
U,\ w\mapsto f^{\operatorname*{op}}\left(  v,w\right)  $ is linear. In other
words, the map $f^{\operatorname*{op}}$ is linear in its second argument (by
the definition of \textquotedblleft linear in its second
argument\textquotedblright).

Thus, we have shown that the map $f^{\operatorname*{op}}$ is both linear in
its first argument and linear in its second argument. In other words, the map
$f^{\operatorname*{op}}$ is bilinear (by the definition of \textquotedblleft
bilinear\textquotedblright). This proves Proposition \ref{prop.op}
\textbf{(b)}.
\end{proof}
\end{verlong}

The following proposition is equally trivial:

\begin{proposition}
\label{prop.restrict}Let $V$, $W$ and $U$ be three vector spaces. Let $A$ be a
subspace of $V$, and let $B$ be a subspace of $W$. Let $f:V\times W\rightarrow
U$ be a bilinear map. Then, the restriction $f\mid_{A\times B}$ is a bilinear
map $A\times B\rightarrow U$.
\end{proposition}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.restrict}.]The map $f$ is bilinear. In other
words, the map $f$ is both linear in its first argument and linear in its
second argument (by the definition of \textquotedblleft
bilinear\textquotedblright).

The map $f$ is linear in its first argument. In other words,
\begin{equation}
\text{for every }w\in W\text{, the map }V\rightarrow U,\ v\mapsto f\left(
v,w\right)  \text{ is linear} \label{pf.prop.restrict.1}%
\end{equation}
(by the definition of \textquotedblleft linear in its first
argument\textquotedblright).

The map $f$ is linear in its second argument. In other words,
\begin{equation}
\text{for every }v\in V\text{, the map }W\rightarrow U,\ w\mapsto f\left(
v,w\right)  \text{ is linear} \label{pf.prop.restrict.2}%
\end{equation}
(by the definition of \textquotedblleft linear in its second
argument\textquotedblright).

For every $w\in B$, the map $A\rightarrow U,\ v\mapsto\left(  f\mid_{A\times
B}\right)  \left(  v,w\right)  $ is linear\footnote{\textit{Proof.} Let $w\in
B$. Then, $w\in B\subseteq W$. Hence, (\ref{pf.prop.restrict.1}) shows that
the map $V\rightarrow U,\ v\mapsto f\left(  v,w\right)  $ is linear. Hence,
the restriction of this map $V\rightarrow U,\ v\mapsto f\left(  v,w\right)  $
to the subspace $A$ is also linear (since the restriction of a linear map to a
subspace of its domain is always linear). In other words, the map
$A\rightarrow U,\ v\mapsto f\left(  v,w\right)  $ is linear (since the
restriction of the map $V\rightarrow U,\ v\mapsto f\left(  v,w\right)  $ to
the subspace $A$ is the map $A\rightarrow U,\ v\mapsto f\left(  v,w\right)
$). In other words, the map $A\rightarrow U,\ v\mapsto\left(  f\mid_{A\times
B}\right)  \left(  v,w\right)  $ is linear (since every $v\in A$ satisfies
$\left(  f\mid_{A\times B}\right)  \left(  v,w\right)  =f\left(  v,w\right)
$). Qed.}. In other words, the map $f\mid_{A\times B}$ is linear in its first
argument (by the definition of \textquotedblleft linear in its first
argument\textquotedblright).

For every $v\in A$, the map $B\rightarrow U,\ w\mapsto\left(  f\mid_{A\times
B}\right)  \left(  v,w\right)  $ is linear\footnote{\textit{Proof.} Let $v\in
A$. Then, $v\in A\subseteq V$. Hence, (\ref{pf.prop.restrict.2}) shows that
the map $W\rightarrow U,\ w\mapsto f\left(  v,w\right)  $ is linear. Hence,
the restriction of this map $W\rightarrow U,\ w\mapsto f\left(  v,w\right)  $
to the subspace $B$ is also linear (since the restriction of a linear map to a
subspace of its domain is always linear). In other words, the map
$B\rightarrow U,\ w\mapsto f\left(  v,w\right)  $ is linear (since the
restriction of the map $W\rightarrow U,\ w\mapsto f\left(  v,w\right)  $ to
the subspace $B$ is the map $B\rightarrow U,\ w\mapsto f\left(  v,w\right)
$). In other words, the map $B\rightarrow U,\ w\mapsto\left(  f\mid_{A\times
B}\right)  \left(  v,w\right)  $ is linear (since every $w\in B$ satisfies
$\left(  f\mid_{A\times B}\right)  \left(  v,w\right)  =f\left(  v,w\right)
$). Qed.}. In other words, the map $f\mid_{A\times B}$ is linear in its second
argument (by the definition of \textquotedblleft linear in its second
argument\textquotedblright).

Thus, we have shown that the map $f\mid_{A\times B}$ is both linear in its
first argument and linear in its second argument. In other words, the map
$f\mid_{A\times B}$ is bilinear (by the definition of \textquotedblleft
bilinear\textquotedblright). This proves Proposition \ref{prop.restrict}.
\end{proof}
\end{verlong}

Let us introduce some more terminology, for convenience:

\begin{definition}
\label{def.f(-)}Let $V$, $W$ and $U$ be three sets. Let $f:V\times
W\rightarrow U$ be a bilinear map.

The map $f:V\times W\rightarrow U$ is bilinear. In other words, the map $f$ is
both linear in its first argument and linear in its second argument (by the
definition of \textquotedblleft bilinear\textquotedblright).

\textbf{(a)} The map $f$ is linear in its first argument. In other words, for
every $w\in W$, the map $V\rightarrow U,\ v\mapsto f\left(  v,w\right)  $ is
linear (by the definition of \textquotedblleft linear in its first
argument\textquotedblright). This latter map will be denoted by $f\left(
-,w\right)  $. This map $f\left(  -,w\right)  $ is thus a linear map
$V\rightarrow U$; in other words, $f\left(  -,w\right)  \in\operatorname*{Hom}%
\left(  V,U\right)  $.

\textbf{(b)} The map $f$ is linear in its second argument. In other words, for
every $v\in V$, the map $W\rightarrow U,\ w\mapsto f\left(  v,w\right)  $ is
linear (by the definition of \textquotedblleft linear in its second
argument\textquotedblright). This latter map will be denoted by $f\left(
v,-\right)  $. This map $f\left(  v,-\right)  $ is thus a linear map
$W\rightarrow U$; in other words, $f\left(  v,-\right)  \in\operatorname*{Hom}%
\left(  W,U\right)  $.
\end{definition}

The notations $f\left(  v,-\right)  $ and $f\left(  -,w\right)  $ introduced
in Definition \ref{def.f(-)} should not be confused with the notation
$f\left(  v,w\right)  $ for the image of a pair $\left(  v,w\right)  \in
V\times W$ under $f$. (Fortunately, they can easily be distinguished from the
notation $f\left(  v,w\right)  $, because a dash cannot be mistaken for an
element of $V$ or an element of $W$.) Of course, the dash in the notations
$f\left(  v,-\right)  $ and $f\left(  -,w\right)  $ indicates
\textquotedblleft insert element here\textquotedblright.

\textquotedblleft Bilinear form\textquotedblright\ is just an alternative
terminology for a bilinear map whose target is $\mathbf{k}$:

\begin{definition}
Let $V$ and $W$ be two vector spaces. A \textit{bilinear form} on $V\times W$
means a bilinear map $V\times W\rightarrow\mathbf{k}$.
\end{definition}

Using the notion of a bilinear form, we can restate Proposition \ref{prop.op}
\textbf{(b)} and Proposition \ref{prop.restrict} in the particular case of
$U=\mathbf{k}$ as follows:

\begin{proposition}
\label{prop.bilform.op-and-restrict}Let $V$ and $W$ be two vector spaces. Let
$f:V\times W\rightarrow\mathbf{k}$ be a bilinear form.

\textbf{(a)} The map $f^{\operatorname*{op}}:W\times V\rightarrow\mathbf{k}$
is a bilinear form.

\textbf{(b)} Let $A$ be a subspace of $V$, and let $B$ be a subspace of $W$.
Then, the restriction $f\mid_{A\times B}$ is a bilinear form $A\times
B\rightarrow\mathbf{k}$.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.bilform.op-and-restrict}.]Proposition
\ref{prop.bilform.op-and-restrict} \textbf{(a)} follows immediately from
Proposition \ref{prop.op} \textbf{(b)} (applied to $U=\mathbf{k}$).
Proposition \ref{prop.bilform.op-and-restrict} \textbf{(b)} follows
immediately from Proposition \ref{prop.restrict} (applied to $U=\mathbf{k}$).
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.bilform.op-and-restrict}.]The map $f:V\times
W\rightarrow\mathbf{k}$ is a bilinear form. In other words, the map $f:V\times
W\rightarrow\mathbf{k}$ is a bilinear map (according to the definition of a
\textquotedblleft bilinear form\textquotedblright).

\textbf{(a)} Proposition \ref{prop.op} \textbf{(b)} (applied to $U=\mathbf{k}%
$) shows that the map $f^{\operatorname*{op}}:W\times V\rightarrow\mathbf{k}$
is bilinear. In other words, $f^{\operatorname*{op}}$ is a bilinear map
$W\times V\rightarrow\mathbf{k}$. In other words, $f^{\operatorname*{op}}$ is
a bilinear form (by the definition of a \textquotedblleft bilinear
form\textquotedblright). This proves Proposition
\ref{prop.bilform.op-and-restrict} \textbf{(a)}.

\textbf{(b)} Proposition \ref{prop.restrict} (applied to $U=\mathbf{k}$) shows
that the restriction $f\mid_{A\times B}$ is a bilinear map $A\times
B\rightarrow\mathbf{k}$. In other words, the restriction $f\mid_{A\times B}$
is a bilinear form $A\times B\rightarrow\mathbf{k}$ (by the definition of a
\textquotedblleft bilinear form\textquotedblright). This proves Proposition
\ref{prop.bilform.op-and-restrict} \textbf{(b)}.
\end{proof}
\end{verlong}

\section{Left and right orthogonal spaces}

We shall now introduce the notions of left and right orthogonal spaces with
respect to a bilinear map:

\begin{definition}
\label{def.LR}Let $V$, $W$ and $U$ be three vector spaces. Let $f:V\times
W\rightarrow U$ be a map.

\textbf{(a)} If $B$ is a subset of $W$, then $\mathcal{L}_{f}\left(  B\right)
$ denotes the subset $\left\{  v\in V\ \mid\ f\left(  v,b\right)  =0\text{ for
all }b\in B\right\}  $ of $V$. We call $\mathcal{L}_{f}\left(  B\right)  $ the
\textit{left orthogonal space of }$B$.

\textbf{(b)} If $A$ is a subset of $V$, then $\mathcal{R}_{f}\left(  A\right)
$ denotes the subset $\left\{  w\in W\ \mid\ f\left(  a,w\right)  =0\text{ for
all }a\in A\right\}  $ of $W$. We call $\mathcal{R}_{f}\left(  A\right)  $ the
\textit{right orthogonal space of }$B$.
\end{definition}

We will mostly use the notations $\mathcal{L}_{f}\left(  B\right)  $ and
$\mathcal{R}_{f}\left(  A\right)  $ in the case when $f$ is a
\textbf{bilinear} map. However, we have chosen to introduce them in full
generality, since others occasionally use them in more general settings (e.g.,
for sesquilinear maps).

We shall first prove some simple facts:

\begin{proposition}
\label{prop.LR.LR}Let $V$, $W$ and $U$ be three vector spaces. Let $f:V\times
W\rightarrow U$ be a map.

\textbf{(a)} If $B$ is a subset of $W$, then $\mathcal{L}_{f}\left(  B\right)
=\mathcal{R}_{f^{\operatorname*{op}}}\left(  B\right)  $.

\textbf{(b)} If $A$ is a subset of $V$, then $\mathcal{R}_{f}\left(  A\right)
=\mathcal{L}_{f^{\operatorname*{op}}}\left(  A\right)  $.
\end{proposition}

Notice that $A$ and $B$ are not required to be subspaces, only subsets, in
Proposition \ref{prop.LR.LR}; furthermore, the map $f$ is not required to be bilinear.

\begin{proof}
[Proof of Proposition \ref{prop.LR.LR}.]\textbf{(b)} Let $A$ be a subset of
$V$. Then, the definition of $\mathcal{L}_{f^{\operatorname*{op}}}\left(
A\right)  $ yields%
\begin{align*}
\mathcal{L}_{f^{\operatorname*{op}}}\left(  A\right)   &  =\left\{  v\in
W\ \mid\ \underbrace{f^{\operatorname*{op}}\left(  v,b\right)  }%
_{\substack{=f\left(  b,v\right)  \\\text{(by the definition of }%
f^{\operatorname*{op}}\text{)}}}=0\text{ for all }b\in A\right\} \\
&  =\left\{  v\in W\ \mid\ f\left(  b,v\right)  =0\text{ for all }b\in
A\right\} \\
&  =\left\{  v\in W\ \mid\ f\left(  a,v\right)  =0\text{ for all }a\in
A\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the index }b\text{
as }a\right) \\
&  =\left\{  w\in W\ \mid\ f\left(  a,w\right)  =0\text{ for all }a\in
A\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the index }v\text{
as }w\right) \\
&  =\mathcal{R}_{f}\left(  A\right)
\end{align*}
(since $\mathcal{R}_{f}\left(  A\right)  =\left\{  w\in W\ \mid\ f\left(
a,w\right)  =0\text{ for all }a\in A\right\}  $ (by the definition of
$\mathcal{R}_{f}\left(  A\right)  $)). This proves Proposition
\ref{prop.LR.LR} \textbf{(b)}.

\textbf{(a)} Let $B$ be a subset of $W$. Proposition \ref{prop.op}
\textbf{(a)} yields $\left(  f^{\operatorname*{op}}\right)
^{\operatorname*{op}}=f$. But Proposition \ref{prop.LR.LR} \textbf{(b)}
(applied to $W$, $V$, $f^{\operatorname*{op}}$ and $B$ instead of $V$, $W$,
$f$ and $A$) shows that $\mathcal{R}_{f^{\operatorname*{op}}}\left(  B\right)
=\mathcal{L}_{\left(  f^{\operatorname*{op}}\right)  ^{\operatorname*{op}}%
}\left(  B\right)  $. Since $\left(  f^{\operatorname*{op}}\right)
^{\operatorname*{op}}=f$, this rewrites as $\mathcal{R}_{f^{\operatorname*{op}%
}}\left(  B\right)  =\mathcal{L}_{f}\left(  B\right)  $. This proves
Proposition \ref{prop.LR.LR} \textbf{(a)}.
\end{proof}

\begin{proposition}
\label{prop.LR.basics}Let $V$, $W$ and $U$ be three vector spaces. Let
$f:V\times W\rightarrow U$ be a bilinear map.

\textbf{(a)} If $B$ is a subset of $W$, then $\mathcal{L}_{f}\left(  B\right)
$ is a subspace of $V$.

\textbf{(b)} If $A$ is a subset of $V$, then $\mathcal{R}_{f}\left(  A\right)
$ is a subspace of $W$.
\end{proposition}

Notice that $A$ and $B$ are not required to be subspaces, only subsets, in
Proposition \ref{prop.LR.basics}.

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.LR.basics}.]\textbf{(a)} Let $B$ be a subset
of $W$.

Recall that a linear map $f\left(  -,w\right)  \in\operatorname*{Hom}\left(
V,U\right)  $ is defined for every $w\in W$ (according to Definition
\ref{def.f(-)} \textbf{(a)}). Thus, $\operatorname*{Ker}\left(  f\left(
-,w\right)  \right)  $ is a subspace of $V$ for every $w\in W$ (because the
kernel of a linear map is always a subspace of its domain). Consequently,
$\bigcap_{w\in B}\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  $ is
an intersection of subspaces of $V$, and therefore itself a subspace of $V$.

We claim that%
\begin{equation}
\mathcal{L}_{f}\left(  B\right)  =\bigcap_{w\in B}\operatorname*{Ker}\left(
f\left(  -,w\right)  \right)  \label{pf.prop.LR.basics.short.a.mainclaim}%
\end{equation}
\footnote{Here, the intersection $\bigcap_{w\in B}\operatorname*{Ker}\left(
f\left(  -,w\right)  \right)  $ is taken in the ambient set $V$. Thus, if
$B=\varnothing$, then $\bigcap_{w\in B}\operatorname*{Ker}\left(  f\left(
-,w\right)  \right)  $ is understood to be $V$.}.

\textit{Proof of (\ref{pf.prop.LR.basics.short.a.mainclaim}):} Let
$x\in\mathcal{L}_{f}\left(  B\right)  $. Thus,%
\[
x\in\mathcal{L}_{f}\left(  B\right)  =\left\{  v\in V\ \mid\ f\left(
v,b\right)  =0\text{ for all }b\in B\right\}
\]
(by the definition of $\mathcal{L}_{f}\left(  B\right)  $). In other words,
$x$ is an element of $V$ and satisfies
\begin{equation}
f\left(  x,b\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for all }b\in B.
\label{pf.prop.LR.basics.short.a.mainclaim.pf.0}%
\end{equation}


Now, we have $x\in\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  $
for every $w\in B$\ \ \ \ \footnote{\textit{Proof.} Let $w\in B$. Applying
(\ref{pf.prop.LR.basics.short.a.mainclaim.pf.0}) to $b=w$, we obtain $f\left(
x,w\right)  =0$.
\par
But the definition of $f\left(  -,w\right)  $ yields $\left(  f\left(
-,w\right)  \right)  \left(  x\right)  =f\left(  x,w\right)  =0$. In other
words, $x\in\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  $, qed.}.
In other words, $x\in\bigcap_{w\in B}\operatorname*{Ker}\left(  f\left(
-,w\right)  \right)  $.

Let us now forget that we fixed $x$. We thus have shown that $x\in
\bigcap_{w\in B}\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  $ for
every $x\in\mathcal{L}_{f}\left(  B\right)  $. In other words,%
\begin{equation}
\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  \subseteq
\bigcap_{w\in B}\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  .
\label{pf.prop.LR.basics.short.a.mainclaim.pf.1}%
\end{equation}


On the other hand, let $y\in\bigcap_{w\in B}\operatorname*{Ker}\left(
f\left(  -,w\right)  \right)  $. Thus, $y\in\bigcap_{w\in B}%
\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  \subseteq V$.
Moreover, we have $f\left(  y,b\right)  =0$ for all $b\in B$%
\ \ \ \ \footnote{\textit{Proof.} Let $b\in B$. We have $y\in\bigcap_{w\in
B}\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  \subseteq
\operatorname*{Ker}\left(  f\left(  -,b\right)  \right)  $ (since $b\in B$).
Thus, $\left(  f\left(  -,b\right)  \right)  \left(  y\right)  =0$. But the
definition of $f\left(  -,b\right)  $ yields $\left(  f\left(  -,b\right)
\right)  \left(  y\right)  =f\left(  y,b\right)  $. Hence, $f\left(
y,b\right)  =\left(  f\left(  -,b\right)  \right)  \left(  y\right)  =0$,
qed.}. Thus, $y$ is an element of $V$ and satisfies $f\left(  y,b\right)  =0$
for all $b\in B$. In other words, $y\in\left\{  v\in V\ \mid\ f\left(
v,b\right)  =0\text{ for all }b\in B\right\}  $.

Since $\mathcal{L}_{f}\left(  B\right)  =\left\{  v\in V\ \mid\ f\left(
v,b\right)  =0\text{ for all }b\in B\right\}  $, this rewrites as
$y\in\mathcal{L}_{f}\left(  B\right)  $.

Let us now forget that we fixed $y$. We thus have shown that $y\in
\mathcal{L}_{f}\left(  B\right)  $ for every $y\in\bigcap_{w\in B}%
\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  $. In other words,%
\[
\mathcal{L}_{f}\left(  B\right)  \subseteq\bigcap_{w\in B}\operatorname*{Ker}%
\left(  f\left(  -,w\right)  \right)  .
\]
Combining this with (\ref{pf.prop.LR.basics.short.a.mainclaim.pf.1}), we
obtain $\mathcal{L}_{f}\left(  B\right)  =\bigcap_{w\in B}\operatorname*{Ker}%
\left(  f\left(  -,w\right)  \right)  $. This proves
(\ref{pf.prop.LR.basics.short.a.mainclaim}).

Now, recall that the subset $\bigcap_{w\in B}\operatorname*{Ker}\left(
f\left(  -,w\right)  \right)  $ of $V$ is a subspace of $V$. In view of
(\ref{pf.prop.LR.basics.short.a.mainclaim}), this rewrites as follows: The
subset $\mathcal{L}_{f}\left(  B\right)  $ of $V$ is a subspace of $V$. This
proves Proposition \ref{prop.LR.basics} \textbf{(a)}.

\textbf{(b)} Let $A$ be a subset of $V$. Proposition \ref{prop.LR.LR}
\textbf{(b)} yields $\mathcal{R}_{f}\left(  A\right)  =\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  A\right)  $. But the map
$f^{\operatorname*{op}}$ is bilinear (by Proposition \ref{prop.op}
\textbf{(b)}). Hence, Proposition \ref{prop.LR.basics} \textbf{(a)} (applied
to $W$, $V$, $f^{\operatorname*{op}}$ and $A$ instead of $V$, $W$, $f$ and
$B$) shows that $\mathcal{L}_{f^{\operatorname*{op}}}\left(  A\right)  $ is a
subspace of $W$. Since $\mathcal{R}_{f}\left(  A\right)  =\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  A\right)  $, this yields that $\mathcal{R}%
_{f}\left(  A\right)  $ is a subspace of $W$. This proves Proposition
\ref{prop.LR.basics} \textbf{(b)}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.LR.basics}.]\textbf{(a)} Let $B$ be a subset
of $W$.

Recall that a linear map $f\left(  -,w\right)  \in\operatorname*{Hom}\left(
V,U\right)  $ is defined for every $w\in W$ (according to Definition
\ref{def.f(-)} \textbf{(a)}). Thus, $\operatorname*{Ker}\left(  f\left(
-,w\right)  \right)  $ is a subspace of $V$ for every $w\in W$ (because the
kernel of a linear map is always a subspace of its domain). Consequently,
$\bigcap_{w\in B}\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  $ is
an intersection of subspaces of $V$, and therefore itself a subspace of $V$.

We claim that%
\begin{equation}
\mathcal{L}_{f}\left(  B\right)  =\bigcap_{w\in B}\operatorname*{Ker}\left(
f\left(  -,w\right)  \right)  \label{pf.prop.LR.basics.a.mainclaim}%
\end{equation}
\footnote{Here, the intersection $\bigcap_{w\in B}\operatorname*{Ker}\left(
f\left(  -,w\right)  \right)  $ is taken in the ambient set $V$. Thus, if
$B=\varnothing$, then $\bigcap_{w\in B}\operatorname*{Ker}\left(  f\left(
-,w\right)  \right)  $ is understood to be $V$.}.

\textit{Proof of (\ref{pf.prop.LR.basics.a.mainclaim}):} Let $x\in
\mathcal{L}_{f}\left(  B\right)  $. Thus,%
\[
x\in\mathcal{L}_{f}\left(  B\right)  =\left\{  v\in V\ \mid\ f\left(
v,b\right)  =0\text{ for all }b\in B\right\}
\]
(by the definition of $\mathcal{L}_{f}\left(  B\right)  $). In other words,
$x$ is an element $v$ of $V$ satisfying $f\left(  v,b\right)  =0$ for all
$b\in B$. In other words, $x$ is an element of $V$ and satisfies $f\left(
x,b\right)  =0$ for all $b\in B$.

Now, we have $x\in\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  $
for every $w\in B$\ \ \ \ \footnote{\textit{Proof.} Let $w\in B$. Recall that
$f\left(  x,b\right)  =0$ for all $b\in B$. Applying this to $b=w$, we obtain
$f\left(  x,w\right)  =0$.
\par
But the definition of $f\left(  -,w\right)  $ yields $\left(  f\left(
-,w\right)  \right)  \left(  x\right)  =f\left(  x,w\right)  =0$. In other
words, $x\in\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  $, qed.}.
In other words, $x\in\bigcap_{w\in B}\operatorname*{Ker}\left(  f\left(
-,w\right)  \right)  $.

Let us now forget that we fixed $x$. We thus have shown that $x\in
\bigcap_{w\in B}\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  $ for
every $x\in\mathcal{L}_{f}\left(  B\right)  $. In other words,%
\begin{equation}
\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  \subseteq
\bigcap_{w\in B}\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  .
\label{pf.prop.LR.basics.a.mainclaim.pf.1}%
\end{equation}


On the other hand, let $y\in\bigcap_{w\in B}\operatorname*{Ker}\left(
f\left(  -,w\right)  \right)  $. Thus, $y\in\bigcap_{w\in B}%
\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  \subseteq V$.
Moreover, we have $f\left(  y,b\right)  =0$ for all $b\in B$%
\ \ \ \ \footnote{\textit{Proof.} Let $b\in B$. We have $y\in\bigcap_{w\in
B}\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  \subseteq
\operatorname*{Ker}\left(  f\left(  -,b\right)  \right)  $ (since $b\in B$).
Thus, $\left(  f\left(  -,b\right)  \right)  \left(  y\right)  =0$. But the
definition of $f\left(  -,b\right)  $ yields $\left(  f\left(  -,b\right)
\right)  \left(  y\right)  =f\left(  y,b\right)  $. Hence, $f\left(
y,b\right)  =\left(  f\left(  -,b\right)  \right)  \left(  y\right)  =0$,
qed.}. Thus, $y$ is an element of $V$ and satisfies $f\left(  y,b\right)  =0$
for all $b\in B$. In other words, $y$ is an element $v$ of $V$ satisfying
$f\left(  v,b\right)  =0$ for all $b\in B$. In other words, $y\in\left\{  v\in
V\ \mid\ f\left(  v,b\right)  =0\text{ for all }b\in B\right\}  $. Since
$\mathcal{L}_{f}\left(  B\right)  =\left\{  v\in V\ \mid\ f\left(  v,b\right)
=0\text{ for all }b\in B\right\}  $ (by the definition of $\mathcal{L}%
_{f}\left(  B\right)  $), this rewrites as $y\in\mathcal{L}_{f}\left(
B\right)  $.

Let us now forget that we fixed $y$. We thus have shown that $y\in
\mathcal{L}_{f}\left(  B\right)  $ for every $y\in\bigcap_{w\in B}%
\operatorname*{Ker}\left(  f\left(  -,w\right)  \right)  $. In other words,%
\[
\mathcal{L}_{f}\left(  B\right)  \subseteq\bigcap_{w\in B}\operatorname*{Ker}%
\left(  f\left(  -,w\right)  \right)  .
\]
Combining this with (\ref{pf.prop.LR.basics.a.mainclaim.pf.1}), we obtain
$\mathcal{L}_{f}\left(  B\right)  =\bigcap_{w\in B}\operatorname*{Ker}\left(
f\left(  -,w\right)  \right)  $. This proves
(\ref{pf.prop.LR.basics.a.mainclaim}).

Now, recall that the subset $\bigcap_{w\in B}\operatorname*{Ker}\left(
f\left(  -,w\right)  \right)  $ of $V$ is a subspace of $V$. In view of
(\ref{pf.prop.LR.basics.a.mainclaim}), this rewrites as follows: The subset
$\mathcal{L}_{f}\left(  B\right)  $ of $V$ is a subspace of $V$. This proves
Proposition \ref{prop.LR.basics} \textbf{(a)}.

\textbf{(b)} Let $A$ be a subset of $V$. Proposition \ref{prop.LR.LR}
\textbf{(b)} yields $\mathcal{R}_{f}\left(  A\right)  =\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  A\right)  $. But the map
$f^{\operatorname*{op}}$ is bilinear (by Proposition \ref{prop.op}
\textbf{(b)}). Hence, Proposition \ref{prop.LR.basics} \textbf{(a)} (applied
to $W$, $V$, $f^{\operatorname*{op}}$ and $A$ instead of $V$, $W$, $f$ and
$B$) shows that $\mathcal{L}_{f^{\operatorname*{op}}}\left(  A\right)  $ is a
subspace of $W$. Since $\mathcal{R}_{f}\left(  A\right)  =\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  A\right)  $, this yields that $\mathcal{R}%
_{f}\left(  A\right)  $ is a subspace of $W$. This proves Proposition
\ref{prop.LR.basics} \textbf{(b)}.
\end{proof}
\end{verlong}

Of course, Proposition \ref{prop.LR.basics} \textbf{(b)} is an analogue of
Proposition \ref{prop.LR.basics} \textbf{(a)}, and could be proven in the same
way as we proved Proposition \ref{prop.LR.basics} \textbf{(a)}. However, we
have chosen to derive it from Proposition \ref{prop.LR.basics} \textbf{(a)}
instead (using Proposition \ref{prop.LR.LR} \textbf{(b)}), since this way is shorter.

\begin{proposition}
\label{prop.LR.contravari}Let $V$, $W$ and $U$ be three vector spaces. Let
$f:V\times W\rightarrow U$ be a map.

\textbf{(a)} If $B$ and $B^{\prime}$ are two subsets of $W$ satisfying
$B\subseteq B^{\prime}$, then $\mathcal{L}_{f}\left(  B\right)  \supseteq
\mathcal{L}_{f}\left(  B^{\prime}\right)  $.

\textbf{(b)} If $A$ and $A^{\prime}$ are two subsets of $V$ satisfying
$A\subseteq A^{\prime}$, then $\mathcal{R}_{f}\left(  A\right)  \supseteq
\mathcal{R}_{f}\left(  A^{\prime}\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.LR.contravari}.]\textbf{(a)} Let $B$ and
$B^{\prime}$ be two subsets of $W$ satisfying $B\subseteq B^{\prime}$. The
definition of $\mathcal{L}_{f}\left(  B\right)  $ yields
\begin{equation}
\mathcal{L}_{f}\left(  B\right)  =\left\{  v\in V\ \mid\ f\left(  v,b\right)
=0\text{ for all }b\in B\right\}  . \label{pf.prop.LR.contravari.a.0}%
\end{equation}
The definition of $\mathcal{L}_{f}\left(  B^{\prime}\right)  $ yields%
\[
\mathcal{L}_{f}\left(  B^{\prime}\right)  =\left\{  v\in V\ \mid\ f\left(
v,b\right)  =0\text{ for all }b\in B^{\prime}\right\}  .
\]


Now, let $x\in\mathcal{L}_{f}\left(  B^{\prime}\right)  $. Thus,
$x\in\mathcal{L}_{f}\left(  B^{\prime}\right)  =\left\{  v\in V\ \mid
\ f\left(  v,b\right)  =0\text{ for all }b\in B^{\prime}\right\}  $. In other
words, $x$ is an element $v$ of $V$ satisfying $f\left(  v,b\right)  =0$ for
all $b\in B^{\prime}$. In other words, $x$ is an element of $V$ and satisfies
\begin{equation}
f\left(  x,b\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for all }b\in B^{\prime}.
\label{pf.prop.LR.contravari.a.1}%
\end{equation}
Therefore, $f\left(  x,b\right)  =0$ for all $b\in B$%
\ \ \ \ \footnote{\textit{Proof.} Let $b\in B$. Then, $b\in B\subseteq
B^{\prime}$. Hence, $f\left(  x,b\right)  =0$ (by
(\ref{pf.prop.LR.contravari.a.1})). Qed.}. Hence, $x$ is an element of $V$ and
satisfies $f\left(  x,b\right)  =0$ for all $b\in B$. In other words, $x$ is
an element $v$ of $V$ satisfying $f\left(  v,b\right)  =0$ for all $b\in B$.
In other words, $x\in\left\{  v\in V\ \mid\ f\left(  v,b\right)  =0\text{ for
all }b\in B\right\}  $. In light of (\ref{pf.prop.LR.contravari.a.0}), this
rewrites as $x\in\mathcal{L}_{f}\left(  B\right)  $.

Now, let us forget that we fixed $x$. We thus have proven that $x\in
\mathcal{L}_{f}\left(  B\right)  $ for every $x\in\mathcal{L}_{f}\left(
B^{\prime}\right)  $. In other words, $\mathcal{L}_{f}\left(  B\right)
\supseteq\mathcal{L}_{f}\left(  B^{\prime}\right)  $. This proves Proposition
\ref{prop.LR.contravari} \textbf{(a)}.

\textbf{(b)} Let $A$ and $A^{\prime}$ be two subsets of $V$ satisfying
$A\subseteq A^{\prime}$. Proposition \ref{prop.LR.LR} \textbf{(b)} yields
$\mathcal{R}_{f}\left(  A\right)  =\mathcal{L}_{f^{\operatorname*{op}}}\left(
A\right)  $. Proposition \ref{prop.LR.LR} \textbf{(b)} (applied to $A^{\prime
}$ instead of $A$) yields $\mathcal{R}_{f}\left(  A^{\prime}\right)
=\mathcal{L}_{f^{\operatorname*{op}}}\left(  A^{\prime}\right)  $. But
Proposition \ref{prop.LR.contravari} \textbf{(a)} (applied to $W$, $V$,
$f^{\operatorname*{op}}$, $A$ and $A^{\prime}$ instead of $V$, $W$, $f$, $B$
and $B^{\prime}$) shows that $\mathcal{L}_{f^{\operatorname*{op}}}\left(
A\right)  \supseteq\mathcal{L}_{f^{\operatorname*{op}}}\left(  A^{\prime
}\right)  $. Thus, $\mathcal{R}_{f}\left(  A\right)  =\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  A\right)  \supseteq\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  A^{\prime}\right)  =\mathcal{R}_{f}\left(
A^{\prime}\right)  $. This proves Proposition \ref{prop.LR.contravari}
\textbf{(b)}.
\end{proof}

\begin{proposition}
\label{prop.LR.revert}Let $V$, $W$ and $U$ be three vector spaces. Let
$f:V\times W\rightarrow U$ be a map. Let $A$ be a subset of $V$. Let $B$ be a
subset of $W$. Then, $A\subseteq\mathcal{L}_{f}\left(  B\right)  $ holds if
and only if $B\subseteq\mathcal{R}_{f}\left(  A\right)  $.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.LR.revert}.]We have the following chain of
logical equivalences:%
\begin{align}
&  \ \left(  A\subseteq\mathcal{L}_{f}\left(  B\right)  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{every }x\in A\text{ satisfies }%
x\in\underbrace{\mathcal{L}_{f}\left(  B\right)  }_{\substack{=\left\{  v\in
V\ \mid\ f\left(  v,b\right)  =0\text{ for all }b\in B\right\}  \\\text{(by
the definition of }\mathcal{L}_{f}\left(  B\right)  \text{)}}}\right)
\nonumber\\
&  \Longleftrightarrow\ \left(  \text{every }x\in A\text{ satisfies }%
x\in\left\{  v\in V\ \mid\ f\left(  v,b\right)  =0\text{ for all }b\in
B\right\}  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{every }x\in A\text{ satisfies }f\left(
x,b\right)  =0\text{ for all }b\in B\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }x\in A\text{ satisfies }x\in
V\text{ already (because }A\subseteq V\text{)}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{every }x\in A\text{ and }b\in B\text{
satisfy }f\left(  x,b\right)  =0\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{every }a\in A\text{ and }b\in B\text{
satisfy }f\left(  a,b\right)  =0\right) \label{pf.prop.LR.revert.short.1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }x\text{ as
}a\right)  .\nonumber
\end{align}
Also, we have the following chain of logical equivalences:%
\begin{align*}
&  \ \left(  B\subseteq\mathcal{R}_{f}\left(  A\right)  \right) \\
&  \Longleftrightarrow\ \left(  \text{every }y\in B\text{ satisfies }%
y\in\underbrace{\mathcal{R}_{f}\left(  A\right)  }_{\substack{=\left\{  w\in
W\ \mid\ f\left(  a,w\right)  =0\text{ for all }a\in A\right\}  \\\text{(by
the definition of }\mathcal{R}_{f}\left(  A\right)  \text{)}}}\right) \\
&  \Longleftrightarrow\ \left(  \text{every }y\in B\text{ satisfies }%
y\in\left\{  w\in W\ \mid\ f\left(  a,w\right)  =0\text{ for all }a\in
A\right\}  \right) \\
&  \Longleftrightarrow\ \left(  \text{every }y\in B\text{ satisfies }f\left(
a,y\right)  =0\text{ for all }a\in A\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }y\in B\text{ satisfies }y\in
W\text{ already (because }B\subseteq W\text{)}\right) \\
&  \Longleftrightarrow\ \left(  \text{every }a\in A\text{ and }y\in B\text{
satisfy }f\left(  a,y\right)  =0\right) \\
&  \Longleftrightarrow\ \left(  \text{every }a\in A\text{ and }b\in B\text{
satisfy }f\left(  a,b\right)  =0\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }y\text{ as
}b\right)  .
\end{align*}
Comparing this with (\ref{pf.prop.LR.revert.short.1}), we obtain $\left(
A\subseteq\mathcal{L}_{f}\left(  B\right)  \right)  \ \Longleftrightarrow
\ \left(  B\subseteq\mathcal{R}_{f}\left(  A\right)  \right)  $. This proves
Proposition \ref{prop.LR.revert}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.LR.revert}.]Recall that $\mathcal{L}%
_{f}\left(  B\right)  =\left\{  v\in V\ \mid\ f\left(  v,b\right)  =0\text{
for all }b\in B\right\}  $ (by the definition of $\mathcal{L}_{f}\left(
B\right)  $). Also, $\mathcal{R}_{f}\left(  A\right)  =\left\{  w\in
W\ \mid\ f\left(  a,w\right)  =0\text{ for all }a\in A\right\}  $ (by the
definition of $\mathcal{R}_{f}\left(  A\right)  $).

We shall first prove the following logical implication:%
\begin{equation}
\left(  A\subseteq\mathcal{L}_{f}\left(  B\right)  \right)  \ \Longrightarrow
\ \left(  f\left(  a,b\right)  =0\text{ for every }\left(  a,b\right)  \in
A\times B\right)  . \label{pf.prop.LR.revert.eq1}%
\end{equation}


\textit{Proof of (\ref{pf.prop.LR.revert.eq1}):} Assume that $A\subseteq
\mathcal{L}_{f}\left(  B\right)  $. We need to show that \newline$\left(
f\left(  a,b\right)  =0\text{ for every }\left(  a,b\right)  \in A\times
B\right)  $.

Let $a\in A$. Thus, $a\in A\subseteq\mathcal{L}_{f}\left(  B\right)  =\left\{
v\in V\ \mid\ f\left(  v,b\right)  =0\text{ for all }b\in B\right\}  $. In
other words, $a$ is an element $v$ of $V$ satisfying $f\left(  v,b\right)  =0$
for all $b\in B$. In other words, $a$ is an element of $V$ and satisfies
$f\left(  a,b\right)  =0$ for all $b\in B$.

Now, let us forget that we fixed $a$. We thus have shown that every $a\in A$
satisfies%
\begin{equation}
f\left(  a,b\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for all }b\in B.
\label{pf.prop.LR.revert.eq1.pf.1}%
\end{equation}


Now, fix $\left(  a,b\right)  \in A\times B$. Thus, $a\in A$ and $b\in B$.
Hence, $f\left(  a,b\right)  =0$ (by (\ref{pf.prop.LR.revert.eq1.pf.1})).

Now, let us forget that we fixed $\left(  a,b\right)  $. We thus have shown
that \newline$\left(  f\left(  a,b\right)  =0\text{ for every }\left(
a,b\right)  \in A\times B\right)  $. This proves the implication
(\ref{pf.prop.LR.revert.eq1}).

We shall next prove the following logical implication:%
\begin{equation}
\left(  f\left(  a,b\right)  =0\text{ for every }\left(  a,b\right)  \in
A\times B\right)  \Longrightarrow\left(  A\subseteq\mathcal{L}_{f}\left(
B\right)  \right)  . \label{pf.prop.LR.revert.eq2}%
\end{equation}


\textit{Proof of (\ref{pf.prop.LR.revert.eq2}):} Assume that $\left(  f\left(
a,b\right)  =0\text{ for every }\left(  a,b\right)  \in A\times B\right)  $.
We need to show that $A\subseteq\mathcal{L}_{f}\left(  B\right)  $.

We have assumed that%
\begin{equation}
f\left(  a,b\right)  =0\text{ for every }\left(  a,b\right)  \in A\times B.
\label{pf.prop.LR.revert.eq2.pf.1}%
\end{equation}


Now, let $a\in A$. Thus, $a\in A\subseteq V$. Also, every $b\in B$ satisfies
$\left(  a,b\right)  \in A\times B$ (since $a\in A$ and $b\in B$) and
therefore $f\left(  a,b\right)  =0$ (by (\ref{pf.prop.LR.revert.eq2.pf.1})).
In other words, $f\left(  a,b\right)  =0$ for every $b\in B$. Thus, $a$ is an
element of $V$ and satisfies $f\left(  a,b\right)  =0$ for all $b\in B$. In
other words, $a$ is an element $v$ of $V$ satisfying $f\left(  v,b\right)  =0$
for all $b\in B$. In other words, $a\in\left\{  v\in V\ \mid\ f\left(
v,b\right)  =0\text{ for all }b\in B\right\}  $. Since $\mathcal{L}_{f}\left(
B\right)  =\left\{  v\in V\ \mid\ f\left(  v,b\right)  =0\text{ for all }b\in
B\right\}  $, this rewrites as $a\in\mathcal{L}_{f}\left(  B\right)  $.

Now, let us forget that we fixed $a$. We thus have shown that every $a\in A$
satisfies $a\in\mathcal{L}_{f}\left(  B\right)  $. In other words,
$A\subseteq\mathcal{L}_{f}\left(  B\right)  $. This proves the implication
(\ref{pf.prop.LR.revert.eq2}).

Combining the two implications (\ref{pf.prop.LR.revert.eq1}) and
(\ref{pf.prop.LR.revert.eq2}), we obtain the equivalence%
\begin{equation}
\left(  A\subseteq\mathcal{L}_{f}\left(  B\right)  \right)
\ \Longleftrightarrow\ \left(  f\left(  a,b\right)  =0\text{ for every
}\left(  a,b\right)  \in A\times B\right)  . \label{pf.prop.LR.revert.eq12}%
\end{equation}


Next, we shall prove the following logical implication:%
\begin{equation}
\left(  B\subseteq\mathcal{R}_{f}\left(  A\right)  \right)  \ \Longrightarrow
\ \left(  f\left(  a,b\right)  =0\text{ for every }\left(  a,b\right)  \in
A\times B\right)  . \label{pf.prop.LR.revert.eq3}%
\end{equation}


\textit{Proof of (\ref{pf.prop.LR.revert.eq3}):} Assume that $B\subseteq
\mathcal{R}_{f}\left(  A\right)  $. We need to show that \newline$\left(
f\left(  a,b\right)  =0\text{ for every }\left(  a,b\right)  \in A\times
B\right)  $.

Let $b\in B$. Thus, $b\in B\subseteq\mathcal{R}_{f}\left(  A\right)  =\left\{
w\in W\ \mid\ f\left(  a,w\right)  =0\text{ for all }a\in A\right\}  $. In
other words, $b$ is an element $w$ of $W$ satisfying $f\left(  a,w\right)  =0$
for all $a\in A$. In other words, $b$ is an element of $W$ and satisfies
$f\left(  a,b\right)  =0$ for all $a\in A$.

Now, let us forget that we fixed $b$. We thus have shown that every $b\in B$
satisfies%
\begin{equation}
f\left(  a,b\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for all }a\in A.
\label{pf.prop.LR.revert.eq3.pf.1}%
\end{equation}


Now, fix $\left(  a,b\right)  \in A\times B$. Thus, $a\in A$ and $b\in B$.
Hence, $f\left(  a,b\right)  =0$ (by (\ref{pf.prop.LR.revert.eq3.pf.1})).

Now, let us forget that we fixed $\left(  a,b\right)  $. We thus have shown
that \newline$\left(  f\left(  a,b\right)  =0\text{ for every }\left(
a,b\right)  \in A\times B\right)  $. This proves the implication
(\ref{pf.prop.LR.revert.eq3}).

We shall next prove the following logical implication:%
\begin{equation}
\left(  f\left(  a,b\right)  =0\text{ for every }\left(  a,b\right)  \in
A\times B\right)  \Longrightarrow\left(  B\subseteq\mathcal{R}_{f}\left(
A\right)  \right)  . \label{pf.prop.LR.revert.eq4}%
\end{equation}


\textit{Proof of (\ref{pf.prop.LR.revert.eq4}):} Assume that $\left(  f\left(
a,b\right)  =0\text{ for every }\left(  a,b\right)  \in A\times B\right)  $.
We need to show that $B\subseteq\mathcal{R}_{f}\left(  A\right)  $.

We have assumed that%
\begin{equation}
f\left(  a,b\right)  =0\text{ for every }\left(  a,b\right)  \in A\times B.
\label{pf.prop.LR.revert.eq4.pf.1}%
\end{equation}


Now, let $b\in B$. Thus, $b\in B\subseteq W$. Also, every $a\in A$ satisfies
$\left(  a,b\right)  \in A\times B$ (since $a\in A$ and $b\in B$) and
therefore $f\left(  a,b\right)  =0$ (by (\ref{pf.prop.LR.revert.eq4.pf.1})).
In other words, $f\left(  a,b\right)  =0$ for every $a\in A$. Thus, $b$ is an
element of $W$ and satisfies $f\left(  a,b\right)  =0$ for all $a\in A$. In
other words, $b$ is an element $w$ of $W$ satisfying $f\left(  a,w\right)  =0$
for all $a\in A$. In other words, $b\in\left\{  w\in W\ \mid\ f\left(
a,w\right)  =0\text{ for all }a\in A\right\}  $. Since $\mathcal{R}_{f}\left(
A\right)  =\left\{  w\in W\ \mid\ f\left(  a,w\right)  =0\text{ for all }a\in
A\right\}  $, this rewrites as $b\in\mathcal{R}_{f}\left(  A\right)  $.

Now, let us forget that we fixed $b$. We thus have shown that every $b\in B$
satisfies $b\in\mathcal{R}_{f}\left(  A\right)  $. In other words,
$B\subseteq\mathcal{R}_{f}\left(  A\right)  $. This proves the implication
(\ref{pf.prop.LR.revert.eq4}).

Combining the two implications (\ref{pf.prop.LR.revert.eq3}) and
(\ref{pf.prop.LR.revert.eq4}), we obtain the equivalence%
\begin{equation}
\left(  B\subseteq\mathcal{R}_{f}\left(  A\right)  \right)
\ \Longleftrightarrow\ \left(  f\left(  a,b\right)  =0\text{ for every
}\left(  a,b\right)  \in A\times B\right)  . \label{pf.prop.LR.revert.eq34}%
\end{equation}


Now, we have the following chain of equivalences:%
\begin{align*}
\left(  A\subseteq\mathcal{L}_{f}\left(  B\right)  \right)  \  &
\Longleftrightarrow\ \left(  f\left(  a,b\right)  =0\text{ for every }\left(
a,b\right)  \in A\times B\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.LR.revert.eq12})}\right) \\
&  \Longleftrightarrow\ \left(  B\subseteq\mathcal{R}_{f}\left(  A\right)
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.LR.revert.eq34}%
)}\right)  .
\end{align*}
This proves Proposition \ref{prop.LR.revert}.
\end{proof}
\end{verlong}

\begin{corollary}
\label{cor.LR.revert.one-el}Let $V$, $W$ and $U$ be three vector spaces. Let
$f:V\times W\rightarrow U$ be a map.

\textbf{(a)} If $B$ is a subset of $W$, then $\mathcal{L}_{f}\left(  B\right)
=\left\{  v\in V\ \mid\ B\subseteq\mathcal{R}_{f}\left(  \left\{  v\right\}
\right)  \right\}  $.

\textbf{(b)} If $A$ is a subset of $V$, then $\mathcal{R}_{f}\left(  A\right)
=\left\{  w\in W\ \mid\ A\subseteq\mathcal{L}_{f}\left(  \left\{  w\right\}
\right)  \right\}  $.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.LR.revert.one-el}.]\textbf{(a)} Let $B$ be a
subset of $W$. For every $v\in V$, we have the following logical equivalence:%
\begin{equation}
\left(  B\subseteq\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)  \right)
\ \Longleftrightarrow\ \left(  v\in\mathcal{L}_{f}\left(  B\right)  \right)
\label{pf.cor.LR.revert.one-el.a.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.cor.LR.revert.one-el.a.1}):} Let $v\in V$.
Proposition \ref{prop.LR.revert} (applied to $A=\left\{  v\right\}  $) yields
that $\left\{  v\right\}  \subseteq\mathcal{L}_{f}\left(  B\right)  $ holds if
and only if $B\subseteq\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)  $.
In other words, we have the following logical equivalence $\left(  \left\{
v\right\}  \subseteq\mathcal{L}_{f}\left(  B\right)  \right)
\ \Longleftrightarrow\ \left(  B\subseteq\mathcal{R}_{f}\left(  \left\{
v\right\}  \right)  \right)  $. But clearly, we have the logical equivalence
$\left(  \left\{  v\right\}  \subseteq\mathcal{L}_{f}\left(  B\right)
\right)  \ \Longleftrightarrow\ \left(  v\in\mathcal{L}_{f}\left(  B\right)
\right)  $. Thus, we have the chain of logical equivalences%
\[
\left(  v\in\mathcal{L}_{f}\left(  B\right)  \right)  \ \Longleftrightarrow
\ \left(  \left\{  v\right\}  \subseteq\mathcal{L}_{f}\left(  B\right)
\right)  \ \Longleftrightarrow\ \left(  B\subseteq\mathcal{R}_{f}\left(
\left\{  v\right\}  \right)  \right)  .
\]
This proves (\ref{pf.cor.LR.revert.one-el.a.1}).}. Now,
\begin{align*}
&  \left\{  v\in V\ \mid\ \underbrace{B\subseteq\mathcal{R}_{f}\left(
\left\{  v\right\}  \right)  }_{\substack{\text{this is equivalent to }\left(
v\in\mathcal{L}_{f}\left(  B\right)  \right)  \\\text{(by
(\ref{pf.cor.LR.revert.one-el.a.1}))}}}\right\} \\
&  =\left\{  v\in V\ \mid\ v\in\mathcal{L}_{f}\left(  B\right)  \right\}
=\mathcal{L}_{f}\left(  B\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mathcal{L}_{f}\left(  B\right)  \subseteq V\right)  .
\end{align*}
This proves Corollary \ref{cor.LR.revert.one-el} \textbf{(a)}.

\textbf{(b)} Let $A$ be a subset of $V$. Then, Corollary
\ref{cor.LR.revert.one-el} \textbf{(a)} (applied to $W$, $V$,
$f^{\operatorname*{op}}$ and $A$ instead of $V$, $W$, $f$ and $B$) shows that
\begin{equation}
\mathcal{L}_{f^{\operatorname*{op}}}\left(  A\right)  =\left\{  v\in
W\ \mid\ A\subseteq\mathcal{R}_{f^{\operatorname*{op}}}\left(  \left\{
v\right\}  \right)  \right\}  =\left\{  w\in W\ \mid\ A\subseteq
\mathcal{R}_{f^{\operatorname*{op}}}\left(  \left\{  w\right\}  \right)
\right\}  \label{pf.cor.LR.revert.one-el.b.1}%
\end{equation}
(here, we renamed the index $v$ as $w$).

But Proposition \ref{prop.LR.LR} \textbf{(b)} shows that $\mathcal{R}%
_{f}\left(  A\right)  =\mathcal{L}_{f^{\operatorname*{op}}}\left(  A\right)
$. Furthermore, every $w\in W$ satisfies $\mathcal{L}_{f}\left(  \left\{
w\right\}  \right)  =\mathcal{R}_{f^{\operatorname*{op}}}\left(  \left\{
w\right\}  \right)  $ (by Proposition \ref{prop.LR.LR} \textbf{(a)}, applied
to $B=\left\{  w\right\}  $). Hence,%
\begin{align*}
\left\{  w\in W\ \mid\ A\subseteq\underbrace{\mathcal{L}_{f}\left(  \left\{
w\right\}  \right)  }_{=\mathcal{R}_{f^{\operatorname*{op}}}\left(  \left\{
w\right\}  \right)  }\right\}   &  =\left\{  w\in W\ \mid\ A\subseteq
\mathcal{R}_{f^{\operatorname*{op}}}\left(  \left\{  w\right\}  \right)
\right\} \\
&  =\mathcal{L}_{f^{\operatorname*{op}}}\left(  A\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.cor.LR.revert.one-el.b.1}%
)}\right) \\
&  =\mathcal{R}_{f}\left(  A\right)  .
\end{align*}
This proves Corollary \ref{cor.LR.revert.one-el} \textbf{(b)}.
\end{proof}

\begin{corollary}
\label{cor.LR.revert.double}Let $V$, $W$ and $U$ be three vector spaces. Let
$f:V\times W\rightarrow U$ be a map.

\textbf{(a)} If $B$ is a subset of $W$, then $B\subseteq\mathcal{R}_{f}\left(
\mathcal{L}_{f}\left(  B\right)  \right)  $.

\textbf{(b)} If $A$ is a subset of $V$, then $A\subseteq\mathcal{L}_{f}\left(
\mathcal{R}_{f}\left(  A\right)  \right)  $.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.LR.revert.double}.]\textbf{(a)} Let $B$ be a
subset of $W$. Proposition \ref{prop.LR.revert} (applied to $A=\mathcal{L}%
_{f}\left(  B\right)  $) shows that $\mathcal{L}_{f}\left(  B\right)
\subseteq\mathcal{L}_{f}\left(  B\right)  $ holds if and only if
$B\subseteq\mathcal{R}_{f}\left(  \mathcal{L}_{f}\left(  B\right)  \right)  $.
Hence, $B\subseteq\mathcal{R}_{f}\left(  \mathcal{L}_{f}\left(  B\right)
\right)  $ (since $\mathcal{L}_{f}\left(  B\right)  \subseteq\mathcal{L}%
_{f}\left(  B\right)  $). This proves Corollary \ref{cor.LR.revert.double}
\textbf{(a)}.

\textbf{(b)} Let $A$ be a subset of $V$. Proposition \ref{prop.LR.revert}
(applied to $B=\mathcal{R}_{f}\left(  A\right)  $) shows that $A\subseteq
\mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(  A\right)  \right)  $ holds if
and only if $\mathcal{R}_{f}\left(  A\right)  \subseteq\mathcal{R}_{f}\left(
A\right)  $. Hence, $A\subseteq\mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(
A\right)  \right)  $ (since $\mathcal{R}_{f}\left(  A\right)  \subseteq
\mathcal{R}_{f}\left(  A\right)  $). This proves Corollary
\ref{cor.LR.revert.double} \textbf{(b)}.
\end{proof}

We have so far not focussed on computing the subspaces $\mathcal{L}_{f}\left(
B\right)  $ and $\mathcal{R}_{f}\left(  A\right)  $ in Definition
\ref{def.LR}. This is straightforward to do when $B$ (resp., $A$) is a finite
set and $V$, $W$ and $U$ are finite-dimensional vector spaces (because in this
case, the statement that \textquotedblleft$f\left(  v,b\right)  =0$ for all
$b\in B$\textquotedblright\ is a conjunction of finitely many linear
equations, and thus easy to solve using linear algebra). However, when $B$
(resp., $A$) is infinite, this becomes harder (as one would need to solve an
infinite system of linear equations). However, when $B$ (resp., $A$) is a
finite-dimensional subspace of $W$ (resp. $V$), then it is still easy, due to
the following fact:

\begin{proposition}
\label{prop.LR.span}Let $V$, $W$ and $U$ be three vector spaces. Let
$f:V\times W\rightarrow U$ be a bilinear map.

\textbf{(a)} If $B$ is a subset of $W$, then $\mathcal{L}_{f}\left(  B\right)
=\mathcal{L}_{f}\left(  \operatorname*{span}B\right)  $.

\textbf{(b)} If $A$ is a subset of $V$, then $\mathcal{R}_{f}\left(  A\right)
=\mathcal{R}_{f}\left(  \operatorname*{span}A\right)  $.
\end{proposition}

Using Proposition \ref{prop.LR.span} \textbf{(a)}, it is easy to compute
$\mathcal{L}_{f}\left(  B\right)  $ whenever $V$, $W$ and $U$ are three
finite-dimensional vector spaces and $B$ is a subspace of $W$%
\ \ \ \ \footnote{\textit{Proof.} Namely, let us assume that $V$, $W$ and $U$
are three finite-dimensional vector spaces, and that $B$ is a subspace of $W$.
The vector space $B$ is a subspace of the finite-dimensional vector space $W$,
and thus itself finite-dimensional. Hence, $B$ has a finite basis $B^{\prime}%
$. Consider this $B^{\prime}$.
\par
Proposition \ref{prop.LR.span} \textbf{(a)} (applied to $B^{\prime}$ instead
of $B$) shows that $\mathcal{L}_{f}\left(  B^{\prime}\right)  =\mathcal{L}%
_{f}\left(  \underbrace{\operatorname*{span}\left(  B^{\prime}\right)
}_{\substack{=B\\\text{(since }B^{\prime}\text{ is a basis of }B\text{)}%
}}\right)  =\mathcal{L}_{f}\left(  B\right)  $. But since $B^{\prime}$ is
finite, it is easy to compute $\mathcal{L}_{f}\left(  B^{\prime}\right)  $
using linear algebra. In other words, it is easy to compute $\mathcal{L}%
_{f}\left(  B\right)  $ using linear algebra (since $\mathcal{L}_{f}\left(
B^{\prime}\right)  =\mathcal{L}_{f}\left(  B\right)  $). Qed.}. Similarly,
using Proposition \ref{prop.LR.span} \textbf{(b)}, it is easy to compute
$\mathcal{R}_{f}\left(  A\right)  $ whenever $V$, $W$ and $U$ are three
finite-dimensional vector spaces and $A$ is a subspace of $V$.

\begin{proof}
[Proof of Proposition \ref{prop.LR.span}.]\textbf{(a)} Let $B$ be a subset of
$W$. Recall that $\operatorname*{span}B$ is the smallest subspace of $W$
containing $B$ as a subset. Thus, if $G$ is any subspace of $W$ containing $B$
as a subset, then%
\begin{equation}
\operatorname*{span}B\subseteq G. \label{pf.prop.LR.span.a.span}%
\end{equation}


Clearly, $B\subseteq\operatorname*{span}B$. Thus, Proposition
\ref{prop.LR.contravari} \textbf{(a)} (applied to $B^{\prime}%
=\operatorname*{span}B$) shows that $\mathcal{L}_{f}\left(  B\right)
\supseteq\mathcal{L}_{f}\left(  \operatorname*{span}B\right)  $.

Corollary \ref{cor.LR.revert.one-el} \textbf{(a)} yields $\mathcal{L}%
_{f}\left(  B\right)  =\left\{  v\in V\ \mid\ B\subseteq\mathcal{R}_{f}\left(
\left\{  v\right\}  \right)  \right\}  $. Also, Corollary
\ref{cor.LR.revert.one-el} \textbf{(a)} (applied to $\operatorname*{span}B$
instead of $B$) yields%
\begin{equation}
\mathcal{L}_{f}\left(  \operatorname*{span}B\right)  =\left\{  v\in
V\ \mid\ \operatorname*{span}B\subseteq\mathcal{R}_{f}\left(  \left\{
v\right\}  \right)  \right\}  . \label{pf.prop.LR.span.a.3}%
\end{equation}


But for any $v\in V$, we have the following logical equivalence:%
\begin{equation}
\left(  B\subseteq\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)  \right)
\ \Longleftrightarrow\ \left(  \operatorname*{span}B\subseteq\mathcal{R}%
_{f}\left(  \left\{  v\right\}  \right)  \right)  \label{pf.prop.LR.span.a.eq}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.LR.span.a.eq}):} Let $v\in V$.
\par
Assume that $B\subseteq\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)  $.
We shall show that $\operatorname*{span}B\subseteq\mathcal{R}_{f}\left(
\left\{  v\right\}  \right)  $.
\par
We know that $\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)  $ is a
subspace of $W$ (by Proposition \ref{prop.LR.basics} \textbf{(b)}, applied to
$A=\left\{  v\right\}  $). Hence, $\mathcal{R}_{f}\left(  \left\{  v\right\}
\right)  $ is a subspace of $W$ containing $B$ as a subset (since
$B\subseteq\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)  $). Thus,
$\operatorname*{span}B\subseteq\mathcal{R}_{f}\left(  \left\{  v\right\}
\right)  $ (by (\ref{pf.prop.LR.span.a.span}), applied to $G=\mathcal{R}%
_{f}\left(  \left\{  v\right\}  \right)  $).
\par
Now, let us forget that we assumed that $B\subseteq\mathcal{R}_{f}\left(
\left\{  v\right\}  \right)  $. We thus have proven $\operatorname*{span}%
B\subseteq\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)  $ under the
assumption that $B\subseteq\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)
$. In other words, we have proven the implication $\left(  B\subseteq
\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)  \right)  \ \Longrightarrow
\ \left(  \operatorname*{span}B\subseteq\mathcal{R}_{f}\left(  \left\{
v\right\}  \right)  \right)  $.
\par
On the other hand, if $\operatorname*{span}B\subseteq\mathcal{R}_{f}\left(
\left\{  v\right\}  \right)  $, then $B\subseteq\operatorname*{span}%
B\subseteq\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)  $. Hence, we
have proven the implication $\left(  \operatorname*{span}B\subseteq
\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)  \right)  \Longrightarrow
\left(  B\subseteq\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)  \right)
$. Combining this implication with the (already proven) implication $\left(
B\subseteq\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)  \right)
\ \Longrightarrow\ \left(  \operatorname*{span}B\subseteq\mathcal{R}%
_{f}\left(  \left\{  v\right\}  \right)  \right)  $, we obtain the equivalence
$\left(  B\subseteq\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)
\right)  \ \Longleftrightarrow\ \left(  \operatorname*{span}B\subseteq
\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)  \right)  $. This proves
(\ref{pf.prop.LR.span.a.eq}).}. Now,%
\begin{align*}
\mathcal{L}_{f}\left(  B\right)   &  =\left\{  v\in V\ \mid
\ \underbrace{B\subseteq\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)
}_{\substack{\text{this is equivalent to }\left(  \operatorname*{span}%
B\subseteq\mathcal{R}_{f}\left(  \left\{  v\right\}  \right)  \right)
\\\text{(by (\ref{pf.prop.LR.span.a.eq}))}}}\right\} \\
&  =\left\{  v\in V\ \mid\ \operatorname*{span}B\subseteq\mathcal{R}%
_{f}\left(  \left\{  v\right\}  \right)  \right\}  =\mathcal{L}_{f}\left(
\operatorname*{span}B\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.LR.span.a.3})}\right)  .
\end{align*}
This proves Proposition \ref{prop.LR.span} \textbf{(a)}.

\textbf{(b)} Let $A$ be a subset of $W$. Then, Proposition \ref{prop.LR.LR}
\textbf{(b)} yields $\mathcal{R}_{f}\left(  A\right)  =\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  A\right)  $. Also, Proposition
\ref{prop.LR.LR} \textbf{(b)} (applied to $\operatorname*{span}A$ instead of
$A$) yields $\mathcal{R}_{f}\left(  \operatorname*{span}A\right)
=\mathcal{L}_{f^{\operatorname*{op}}}\left(  \operatorname*{span}A\right)  $.

But the map $f^{\operatorname*{op}}$ is bilinear (by Proposition \ref{prop.op}
\textbf{(b)}). Hence, Proposition \ref{prop.LR.span} \textbf{(a)} (applied to
$W$, $V$, $f^{\operatorname*{op}}$ and $A$ instead of $V$, $W$, $f$ and $B$)
shows that $\mathcal{L}_{f^{\operatorname*{op}}}\left(  A\right)
=\mathcal{L}_{f^{\operatorname*{op}}}\left(  \operatorname*{span}A\right)  $.
Thus, $\mathcal{R}_{f}\left(  A\right)  =\mathcal{L}_{f^{\operatorname*{op}}%
}\left(  A\right)  =\mathcal{L}_{f^{\operatorname*{op}}}\left(
\operatorname*{span}A\right)  =\mathcal{R}_{f}\left(  \operatorname*{span}%
A\right)  $. This proves Proposition \ref{prop.LR.span} \textbf{(b)}.
\end{proof}

Next, we explore the orthogonal spaces of unions:

\begin{proposition}
\label{prop.LR.union}Let $V$, $W$ and $U$ be three vector spaces. Let
$f:V\times W\rightarrow U$ be a map.

\textbf{(a)} If $B_{1}$ and $B_{2}$ are two subsets of $W$, then
$\mathcal{L}_{f}\left(  B_{1}\cup B_{2}\right)  =\mathcal{L}_{f}\left(
B_{1}\right)  \cap\mathcal{L}_{f}\left(  B_{2}\right)  $.

\textbf{(b)} If $A_{1}$ and $A_{2}$ are two subsets of $V$, then
$\mathcal{R}_{f}\left(  A_{1}\cup A_{2}\right)  =\mathcal{R}_{f}\left(
A_{1}\right)  \cap\mathcal{R}_{f}\left(  A_{2}\right)  $.
\end{proposition}

Of course, Proposition \ref{prop.LR.union} does not hold (in general) when the
$\cup$ and $\cap$ signs are switched.

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.LR.union}.]\textbf{(a)} Let $B_{1}$ and
$B_{2}$ be two subsets of $W$. The definition of $\mathcal{L}_{f}\left(
B_{1}\cup B_{2}\right)  $ yields%
\begin{align*}
&  \mathcal{L}_{f}\left(  B_{1}\cup B_{2}\right) \\
&  =\left\{  v\in V\ \mid\ \underbrace{f\left(  v,b\right)  =0\text{ for all
}b\in B_{1}\cup B_{2}}_{\substack{\text{this is equivalent to}\\\left(
f\left(  v,b\right)  =0\text{ for all }b\in B_{1}\right)  \wedge\left(
f\left(  v,b\right)  =0\text{ for all }b\in B_{2}\right)  }}\right\} \\
&  =\left\{  v\in V\ \mid\ \left(  f\left(  v,b\right)  =0\text{ for all }b\in
B_{1}\right)  \wedge\left(  f\left(  v,b\right)  =0\text{ for all }b\in
B_{2}\right)  \right\} \\
&  =\underbrace{\left\{  v\in V\ \mid\ f\left(  v,b\right)  =0\text{ for all
}b\in B_{1}\right\}  }_{\substack{=\mathcal{L}_{f}\left(  B_{1}\right)
\\\text{(by the definition of }\mathcal{L}_{f}\left(  B_{1}\right)  \text{)}%
}}\cap\underbrace{\left\{  v\in V\ \mid\ f\left(  v,b\right)  =0\text{ for all
}b\in B_{2}\right\}  }_{\substack{=\mathcal{L}_{f}\left(  B_{2}\right)
\\\text{(by the definition of }\mathcal{L}_{f}\left(  B_{2}\right)  \text{)}%
}}\\
&  =\mathcal{L}_{f}\left(  B_{1}\right)  \cap\mathcal{L}_{f}\left(
B_{2}\right)  .
\end{align*}
This proves Proposition \ref{prop.LR.union} \textbf{(a)}.

\textbf{(b)} Proposition \ref{prop.LR.union} \textbf{(b)} can be derived from
Proposition \ref{prop.LR.union} \textbf{(a)} using the same tactic that we
used (for example) to derive Proposition \ref{prop.LR.contravari} \textbf{(b)}
from Proposition \ref{prop.LR.contravari} \textbf{(a)}. (Alternatively,
Proposition \ref{prop.LR.union} \textbf{(b)} can be proven analogously to
Proposition \ref{prop.LR.union} \textbf{(a)}.)
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.LR.union}.]\textbf{(a)} Let $B_{1}$ and
$B_{2}$ be two subsets of $W$. Set $B^{\prime}=B_{1}\cup B_{2}$.

We have $B_{1}\subseteq B_{1}\cup B_{2}=B^{\prime}$. Hence, Proposition
\ref{prop.LR.contravari} \textbf{(a)} (applied to $B=B_{1}$) yields
$\mathcal{L}_{f}\left(  B_{1}\right)  \supseteq\mathcal{L}_{f}\left(
B^{\prime}\right)  $. In other words, $\mathcal{L}_{f}\left(  B^{\prime
}\right)  \subseteq\mathcal{L}_{f}\left(  B_{1}\right)  $.

We have $B_{2}\subseteq B_{1}\cup B_{2}=B^{\prime}$. Hence, Proposition
\ref{prop.LR.contravari} \textbf{(a)} (applied to $B=B_{2}$) yields
$\mathcal{L}_{f}\left(  B_{2}\right)  \supseteq\mathcal{L}_{f}\left(
B^{\prime}\right)  $. In other words, $\mathcal{L}_{f}\left(  B^{\prime
}\right)  \subseteq\mathcal{L}_{f}\left(  B_{2}\right)  $.

Combining $\mathcal{L}_{f}\left(  B^{\prime}\right)  \subseteq\mathcal{L}%
_{f}\left(  B_{1}\right)  $ with $\mathcal{L}_{f}\left(  B^{\prime}\right)
\subseteq\mathcal{L}_{f}\left(  B_{2}\right)  $, we obtain%
\begin{equation}
\mathcal{L}_{f}\left(  B^{\prime}\right)  \subseteq\mathcal{L}_{f}\left(
B_{1}\right)  \cap\mathcal{L}_{f}\left(  B_{2}\right)  .
\label{pf.prop.LR.union.a.easydir}%
\end{equation}


Now, let $x\in\mathcal{L}_{f}\left(  B_{1}\right)  \cap\mathcal{L}_{f}\left(
B_{2}\right)  $ be arbitrary. We shall show that $x\in\mathcal{L}_{f}\left(
B^{\prime}\right)  $.

Indeed, we have%
\[
x\in\mathcal{L}_{f}\left(  B_{1}\right)  \cap\mathcal{L}_{f}\left(
B_{2}\right)  \subseteq\mathcal{L}_{f}\left(  B_{1}\right)  =\left\{  v\in
V\ \mid\ f\left(  v,b\right)  =0\text{ for all }b\in B_{1}\right\}
\]
(by the definition of $\mathcal{L}_{f}\left(  B_{1}\right)  $). In other
words, $x$ is an element $v$ of $V$ satisfying $f\left(  v,b\right)  =0$ for
all $b\in B_{1}$. In other words, $x$ is an element of $V$ and satisfies%
\begin{equation}
f\left(  x,b\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for all }b\in B_{1}.
\label{pf.prop.LR.union.a.1}%
\end{equation}


Also,%
\[
x\in\mathcal{L}_{f}\left(  B_{1}\right)  \cap\mathcal{L}_{f}\left(
B_{2}\right)  \subseteq\mathcal{L}_{f}\left(  B_{2}\right)  =\left\{  v\in
V\ \mid\ f\left(  v,b\right)  =0\text{ for all }b\in B_{2}\right\}
\]
(by the definition of $\mathcal{L}_{f}\left(  B_{2}\right)  $). In other
words, $x$ is an element $v$ of $V$ satisfying $f\left(  v,b\right)  =0$ for
all $b\in B_{2}$. In other words, $x$ is an element of $V$ and satisfies%
\begin{equation}
f\left(  x,b\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for all }b\in B_{2}.
\label{pf.prop.LR.union.a.2}%
\end{equation}


Now, let $b\in B^{\prime}$. We shall show that $f\left(  x,b\right)  =0$.

Indeed, we have $b\in B^{\prime}=B_{1}\cup B_{2}$. In other words, either
$b\in B_{1}$ or $b\in B_{2}$. Thus, we are in one of the following two cases:

\textit{Case 1:} We have $b\in B_{1}$.

\textit{Case 2:} We have $b\in B_{2}$.

Let us first consider Case 1. In this case, we have $b\in B_{1}$. Hence,
$f\left(  x,b\right)  =0$ (by (\ref{pf.prop.LR.union.a.1})). Thus, $f\left(
x,b\right)  =0$ is proven in Case 1.

Now, let us consider Case 2. In this case, we have $b\in B_{2}$. Hence,
$f\left(  x,b\right)  =0$ (by (\ref{pf.prop.LR.union.a.2})). Thus, $f\left(
x,b\right)  =0$ is proven in Case 2.

We have now proven $f\left(  x,b\right)  =0$ in both Cases 1 and 2. Since
these two Cases cover all possibilities, this shows that $f\left(  x,b\right)
=0$ always holds.

Now, let us forget that we fixed $b\in B^{\prime}$. We thus have shown that
$f\left(  x,b\right)  =0$ for all $b\in B^{\prime}$. Hence, $x$ is an element
of $V$ and satisfies $f\left(  x,b\right)  =0$ for all $b\in B^{\prime}$. In
other words, $x$ is an element $v$ of $V$ satisfying $f\left(  v,b\right)  =0$
for all $b\in B^{\prime}$. In other words, $x\in\left\{  v\in V\ \mid
\ f\left(  v,b\right)  =0\text{ for all }b\in B^{\prime}\right\}  $. Since
$\mathcal{L}_{f}\left(  B^{\prime}\right)  =\left\{  v\in V\ \mid\ f\left(
v,b\right)  =0\text{ for all }b\in B^{\prime}\right\}  $ (by the definition of
$\mathcal{L}_{f}\left(  B^{\prime}\right)  $), this rewrites as $x\in
\mathcal{L}_{f}\left(  B^{\prime}\right)  $.

Now, let us forget that we fixed $x$. We thus have proven that $x\in
\mathcal{L}_{f}\left(  B^{\prime}\right)  $ for every $x\in\mathcal{L}%
_{f}\left(  B_{1}\right)  \cap\mathcal{L}_{f}\left(  B_{2}\right)  $. In other
words, $\mathcal{L}_{f}\left(  B_{1}\right)  \cap\mathcal{L}_{f}\left(
B_{2}\right)  \subseteq\mathcal{L}_{f}\left(  B^{\prime}\right)  $. Combining
this inclusion with (\ref{pf.prop.LR.union.a.easydir}), we obtain
$\mathcal{L}_{f}\left(  B_{1}\right)  \cap\mathcal{L}_{f}\left(  B_{2}\right)
=\mathcal{L}_{f}\left(  B^{\prime}\right)  $. Since $B^{\prime}=B_{1}\cup
B_{2}$, this rewrites as $\mathcal{L}_{f}\left(  B_{1}\right)  \cap
\mathcal{L}_{f}\left(  B_{2}\right)  =\mathcal{L}_{f}\left(  B_{1}\cup
B_{2}\right)  $. This proves Proposition \ref{prop.LR.union} \textbf{(a)}.

\textbf{(b)} Let $A_{1}$ and $A_{2}$ be two subsets of $V$. Proposition
\ref{prop.LR.union} \textbf{(a)} (applied to $W$, $V$, $f^{\operatorname*{op}%
}$, $A_{1}$ and $A_{2}$ instead of $V$, $W$, $f$, $B_{1}$ and $B_{2}$) shows
that $\mathcal{L}_{f^{\operatorname*{op}}}\left(  A_{1}\cup A_{2}\right)
=\mathcal{L}_{f^{\operatorname*{op}}}\left(  A_{1}\right)  \cap\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  A_{2}\right)  $.

But Proposition \ref{prop.LR.LR} \textbf{(b)} (applied to $A=A_{1}$) yields
$\mathcal{R}_{f}\left(  A_{1}\right)  =\mathcal{L}_{f^{\operatorname*{op}}%
}\left(  A_{1}\right)  $. Also, Proposition \ref{prop.LR.LR} \textbf{(b)}
(applied to $A=A_{2}$) yields $\mathcal{R}_{f}\left(  A_{2}\right)
=\mathcal{L}_{f^{\operatorname*{op}}}\left(  A_{2}\right)  $. Also,
Proposition \ref{prop.LR.LR} \textbf{(b)} (applied to $A=A_{1}\cup A_{2}$)
yields $\mathcal{R}_{f}\left(  A_{1}\cup A_{2}\right)  =\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  A_{1}\cup A_{2}\right)  $. Thus,%
\[
\mathcal{R}_{f}\left(  A_{1}\cup A_{2}\right)  =\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  A_{1}\cup A_{2}\right)
=\underbrace{\mathcal{L}_{f^{\operatorname*{op}}}\left(  A_{1}\right)
}_{=\mathcal{R}_{f}\left(  A_{1}\right)  }\cap\underbrace{\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  A_{2}\right)  }_{=\mathcal{R}_{f}\left(
A_{2}\right)  }=\mathcal{R}_{f}\left(  A_{1}\cup A_{2}\right)  =\mathcal{R}%
_{f}\left(  A_{1}\right)  \cap\mathcal{R}_{f}\left(  A_{2}\right)  .
\]
This proves Proposition \ref{prop.LR.union} \textbf{(b)}.
\end{proof}
\end{verlong}

A similar result holds for orthogonal spaces of sums of subspaces when $f$ is bilinear:

\begin{proposition}
\label{prop.LR.sum}Let $V$, $W$ and $U$ be three vector spaces. Let $f:V\times
W\rightarrow U$ be a bilinear map.

\textbf{(a)} If $B_{1}$ and $B_{2}$ are two subspaces of $W$, then
$\mathcal{L}_{f}\left(  B_{1}+B_{2}\right)  =\mathcal{L}_{f}\left(
B_{1}\right)  \cap\mathcal{L}_{f}\left(  B_{2}\right)  $.

\textbf{(b)} If $A_{1}$ and $A_{2}$ are two subspaces of $V$, then
$\mathcal{R}_{f}\left(  A_{1}+A_{2}\right)  =\mathcal{R}_{f}\left(
A_{1}\right)  \cap\mathcal{R}_{f}\left(  A_{2}\right)  $.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.LR.sum}.]\textbf{(a)} Let $B_{1}$ and $B_{2}$
be two subspaces of $W$. Then, we have $\operatorname*{span}\left(  B_{1}\cup
B_{2}\right)  =B_{1}+B_{2}$\ \ \ \ \footnote{\textit{Proof.} Let $B=B_{1}\cup
B_{2}$. A well-known fact from linear algebra tells us that
$\operatorname*{span}B$ is the smallest subspace of $W$ containing $B$ as a
subset. Thus, if $G$ is any subspace of $W$ containing $B$ as a subset, then%
\begin{equation}
\operatorname*{span}B\subseteq G. \label{pf.prop.LR.sum.short.a.span.pf.1}%
\end{equation}
\par
Now, $B_{1}$ and $B_{2}$ are two subspaces of $W$. Thus, $B_{1}+B_{2}$ is
again a subspace of $W$. Also, $B=B_{1}\cup B_{2}\subseteq B_{1}+B_{2}$ (since
$B_{1}\subseteq B_{1}+B_{2}$ and $B_{2}\subseteq B_{1}+B_{2}$). Thus,
$B_{1}+B_{2}$ is a subspace of $W$ containing $B_{1}\cup B_{2}$ as a subset.
Therefore, (\ref{pf.prop.LR.sum.short.a.span.pf.1}) (applied to $G=B_{1}%
+B_{2}$) shows that $\operatorname*{span}B\subseteq B_{1}+B_{2}$.
\par
On the other hand, combining $B_{1}\subseteq B_{1}\cup B_{2}=B\subseteq
\operatorname*{span}B$ and $B_{2}\subseteq B_{1}\cup B_{2}=B\subseteq
\operatorname*{span}B$, we obtain $B_{1}+B_{2}\subseteq\operatorname*{span}B$
(since $\operatorname*{span}B$ is a vector subspace of $W$). Combined with
$\operatorname*{span}B\subseteq B_{1}+B_{2}$, this yields $B_{1}%
+B_{2}=\operatorname*{span}B$. Since $B=B_{1}\cup B_{2}$, this rewrites as
$B_{1}+B_{2}=\operatorname*{span}\left(  B_{1}\cup B_{2}\right)  $. Qed.}.
Now, Proposition \ref{prop.LR.span} \textbf{(a)} (applied to $B=B_{1}\cup
B_{2}$) shows that $\mathcal{L}_{f}\left(  B_{1}\cup B_{2}\right)
=\mathcal{L}_{f}\left(  \underbrace{\operatorname*{span}\left(  B_{1}\cup
B_{2}\right)  }_{=B_{1}+B_{2}}\right)  =\mathcal{L}_{f}\left(  B_{1}%
+B_{2}\right)  $. Hence,%
\[
\mathcal{L}_{f}\left(  B_{1}+B_{2}\right)  =\mathcal{L}_{f}\left(  B_{1}\cup
B_{2}\right)  =\mathcal{L}_{f}\left(  B_{1}\right)  \cap\mathcal{L}_{f}\left(
B_{2}\right)
\]
(by Proposition \ref{prop.LR.union} \textbf{(a)}). This proves Proposition
\ref{prop.LR.sum} \textbf{(a)}.

\textbf{(b)} The map $f^{\operatorname*{op}}$ is bilinear (by Proposition
\ref{prop.op} \textbf{(b)}). Hence, Proposition \ref{prop.LR.sum} \textbf{(b)}
can be derived from Proposition \ref{prop.LR.sum} \textbf{(a)} using the same
tactic that we used (for example) to derive Proposition
\ref{prop.LR.contravari} \textbf{(b)} from Proposition
\ref{prop.LR.contravari} \textbf{(a)}. (Alternatively, Proposition
\ref{prop.LR.sum} \textbf{(b)} can be proven analogously to Proposition
\ref{prop.LR.sum} \textbf{(a)}.)
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.LR.sum}.]\textbf{(a)} Let $B_{1}$ and $B_{2}$
be two subspaces of $W$. Then, $\operatorname*{span}\left(  B_{1}\cup
B_{2}\right)  =B_{1}+B_{2}$\ \ \ \ \footnote{\textit{Proof.} Let $B=B_{1}\cup
B_{2}$. A well-known fact from linear algebra tells us that
$\operatorname*{span}B$ is the smallest subspace of $W$ containing $B$ as a
subset. Thus, if $G$ is any subspace of $W$ containing $B$ as a subset, then%
\begin{equation}
\operatorname*{span}B\subseteq G. \label{pf.prop.LR.sum.a.span.pf.1}%
\end{equation}
\par
Now, $B_{1}$ and $B_{2}$ are two subspaces of $W$. Thus, $B_{1}+B_{2}$ is
again a subspace of $W$. Also, combining $B_{1}\subseteq B_{1}+B_{2}$ and
$B_{2}\subseteq B_{1}+B_{2}$, we obtain $B_{1}\cup B_{2}\subseteq B_{1}+B_{2}%
$. Hence, $B=B_{1}\cup B_{2}\subseteq B_{1}+B_{2}$. Thus, $B_{1}+B_{2}$ is a
subspace of $W$ containing $B_{1}\cup B_{2}$ as a subset. Therefore,
(\ref{pf.prop.LR.sum.a.span.pf.1}) (applied to $G=B_{1}+B_{2}$) shows that
$\operatorname*{span}B\subseteq B_{1}+B_{2}$.
\par
On the other hand, $B_{1}\subseteq B_{1}\cup B_{2}=B\subseteq
\operatorname*{span}B$ and $B_{2}\subseteq B_{1}\cup B_{2}=B\subseteq
\operatorname*{span}B$.
\par
Now, recall the following simple fact from linear algebra: If $X$, $Y$ and $Z$
are three subspaces of $W$ satisfying $X\subseteq Z$ and $Y\subseteq Z$, then
$X+Y\subseteq Z$. This fact (applied to $X=B_{1}$, $Y=B_{2}$ and
$Z=\operatorname*{span}B$) shows that $B_{1}+B_{2}\subseteq
\operatorname*{span}B$ (since $B_{1}\subseteq\operatorname*{span}B$ and
$B_{2}\subseteq\operatorname*{span}B$). Combining this with
$\operatorname*{span}B\subseteq B_{1}+B_{2}$, we obtain $B_{1}+B_{2}%
=\operatorname*{span}B$. Since $B=B_{1}\cup B_{2}$, this rewrites as
$B_{1}+B_{2}=\operatorname*{span}\left(  B_{1}\cup B_{2}\right)  $. Qed.}.
Now, Proposition \ref{prop.LR.span} \textbf{(a)} (applied to $B=B_{1}\cup
B_{2}$) shows that $\mathcal{L}_{f}\left(  B_{1}\cup B_{2}\right)
=\mathcal{L}_{f}\left(  \underbrace{\operatorname*{span}\left(  B_{1}\cup
B_{2}\right)  }_{=B_{1}+B_{2}}\right)  =\mathcal{L}_{f}\left(  B_{1}%
+B_{2}\right)  $. Hence,%
\[
\mathcal{L}_{f}\left(  B_{1}+B_{2}\right)  =\mathcal{L}_{f}\left(  B_{1}\cup
B_{2}\right)  =\mathcal{L}_{f}\left(  B_{1}\right)  \cap\mathcal{L}_{f}\left(
B_{2}\right)
\]
(by Proposition \ref{prop.LR.union} \textbf{(a)}). This proves Proposition
\ref{prop.LR.sum} \textbf{(a)}.

\textbf{(b)} Let $A_{1}$ and $A_{2}$ be two subspaces of $V$. Then, the map
$f^{\operatorname*{op}}$ is bilinear (by Proposition \ref{prop.op}
\textbf{(b)}). Hence, Proposition \ref{prop.LR.sum} \textbf{(a)} (applied to
$W$, $V$, $f^{\operatorname*{op}}$, $A_{1}$ and $A_{2}$ instead of $V$, $W$,
$f$, $B_{1}$ and $B_{2}$) shows that $\mathcal{L}_{f^{\operatorname*{op}}%
}\left(  A_{1}+A_{2}\right)  =\mathcal{L}_{f^{\operatorname*{op}}}\left(
A_{1}\right)  \cap\mathcal{L}_{f^{\operatorname*{op}}}\left(  A_{2}\right)  $.

But Proposition \ref{prop.LR.LR} \textbf{(b)} (applied to $A=A_{1}$) yields
$\mathcal{R}_{f}\left(  A_{1}\right)  =\mathcal{L}_{f^{\operatorname*{op}}%
}\left(  A_{1}\right)  $. Also, Proposition \ref{prop.LR.LR} \textbf{(b)}
(applied to $A=A_{2}$) yields $\mathcal{R}_{f}\left(  A_{2}\right)
=\mathcal{L}_{f^{\operatorname*{op}}}\left(  A_{2}\right)  $. Also,
Proposition \ref{prop.LR.LR} \textbf{(b)} (applied to $A=A_{1}+A_{2}$) yields
$\mathcal{R}_{f}\left(  A_{1}+A_{2}\right)  =\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  A_{1}+A_{2}\right)  $. Thus,%
\[
\mathcal{R}_{f}\left(  A_{1}+A_{2}\right)  =\mathcal{L}_{f^{\operatorname*{op}%
}}\left(  A_{1}+A_{2}\right)  =\underbrace{\mathcal{L}_{f^{\operatorname*{op}%
}}\left(  A_{1}\right)  }_{=\mathcal{R}_{f}\left(  A_{1}\right)  }%
\cap\underbrace{\mathcal{L}_{f^{\operatorname*{op}}}\left(  A_{2}\right)
}_{=\mathcal{R}_{f}\left(  A_{2}\right)  }=\mathcal{R}_{f}\left(  A_{1}\cup
A_{2}\right)  =\mathcal{R}_{f}\left(  A_{1}\right)  \cap\mathcal{R}_{f}\left(
A_{2}\right)  .
\]
This proves Proposition \ref{prop.LR.sum} \textbf{(b)}.
\end{proof}
\end{verlong}

\section{Interlude: Symmetric and antisymmetric bilinear forms}

We make a digression to define the notions of \textit{symmetric} and
\textit{antisymmetric} bilinear forms and maps:

\begin{definition}
\label{def.symmetric}Let $V$ and $U$ be two vector spaces. Let $f:V\times
V\rightarrow U$ be a map.

\textbf{(a)} The map $f$ is said to be \textit{symmetric} if and only if it
satisfies%
\[
\left(  f\left(  v,w\right)  =f\left(  w,v\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  v,w\right)  \in V\times W\right)
.
\]


\textbf{(b)} The map $f$ is said to be \textit{antisymmetric} if and only if
it satisfies%
\[
\left(  f\left(  v,w\right)  =-f\left(  w,v\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  v,w\right)  \in V\times W\right)
.
\]


Antisymmetric maps are also called \textit{skew-symmetric} maps.
\end{definition}

\begin{proposition}
\label{prop.symmetric.L=R}Let $V$ and $U$ be two vector spaces. Let $f:V\times
V\rightarrow U$ be a map. Let $A$ be a subset of $V$.

\textbf{(a)} If $f$ is symmetric, then $\mathcal{L}_{f}\left(  A\right)
=\mathcal{R}_{f}\left(  A\right)  $.

\textbf{(b)} If $f$ is antisymmetric, then $\mathcal{L}_{f}\left(  A\right)
=\mathcal{R}_{f}\left(  A\right)  $.
\end{proposition}

Proposition \ref{prop.symmetric.L=R} allows for the following definition
(which is actually standard):

\begin{definition}
Let $V$ and $U$ be two vector spaces. Let $f:V\times V\rightarrow U$ be a map.
Let $A$ be a subset of $V$. Assume that $f$ is symmetric or antisymmetric.
Then, Proposition \ref{prop.symmetric.L=R} shows that $\mathcal{L}_{f}\left(
A\right)  =\mathcal{R}_{f}\left(  A\right)  $. The subset $\mathcal{L}%
_{f}\left(  A\right)  =\mathcal{R}_{f}\left(  A\right)  $ of $V$ is denoted by
$A^{\perp}$, at least when $f$ is clear from the context.
\end{definition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.symmetric.L=R}.]Straightforward and left to
the reader.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.symmetric.L=R}.]\textbf{(a)} Assume that $f$
is symmetric. In other words, $f$ satisfies%
\begin{equation}
\left(  f\left(  v,w\right)  =f\left(  w,v\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  v,w\right)  \in V\times W\right)
\label{pf.prop.symmetric.L=R.a.symm}%
\end{equation}
(by the definition of \textquotedblleft symmetric\textquotedblright).

Now, the definition of $\mathcal{R}_{f}\left(  A\right)  $ yields
$\mathcal{R}_{f}\left(  A\right)  =\left\{  w\in V\ \mid\ f\left(  a,w\right)
=0\text{ for all }a\in A\right\}  $. However, the definition of $\mathcal{L}%
_{f}\left(  A\right)  $ yields%
\begin{align*}
\mathcal{L}_{f}\left(  A\right)   &  =\left\{  v\in V\ \mid\ f\left(
v,b\right)  =0\text{ for all }b\in A\right\} \\
&  =\left\{  v\in V\ \mid\ \underbrace{f\left(  v,a\right)  }%
_{\substack{=f\left(  a,v\right)  \\\text{(by
(\ref{pf.prop.symmetric.L=R.a.symm}), applied to }w=a\text{)}}}=0\text{ for
all }a\in A\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the index }b\text{
as }a\right) \\
&  =\left\{  v\in V\ \mid\ f\left(  a,v\right)  =0\text{ for all }a\in
A\right\} \\
&  =\left\{  w\in V\ \mid\ f\left(  a,w\right)  =0\text{ for all }a\in
A\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the index }v\text{
as }w\right) \\
&  =\mathcal{R}_{f}\left(  A\right)  .
\end{align*}
This proves Proposition \ref{prop.symmetric.L=R} \textbf{(a)}.

\textbf{(b)} Assume that $f$ is antisymmetric. In other words, $f$ satisfies%
\begin{equation}
\left(  f\left(  v,w\right)  =-f\left(  w,v\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  v,w\right)  \in V\times W\right)
\label{pf.prop.symmetric.L=R.b.anti}%
\end{equation}
(by the definition of \textquotedblleft antisymmetric\textquotedblright).

Now, the definition of $\mathcal{R}_{f}\left(  A\right)  $ yields
$\mathcal{R}_{f}\left(  A\right)  =\left\{  w\in V\ \mid\ f\left(  a,w\right)
=0\text{ for all }a\in A\right\}  $. However, the definition of $\mathcal{L}%
_{f}\left(  A\right)  $ yields%
\begin{align*}
\mathcal{L}_{f}\left(  A\right)   &  =\left\{  v\in V\ \mid\ f\left(
v,b\right)  =0\text{ for all }b\in A\right\} \\
&  =\left\{  v\in V\ \mid\ \underbrace{f\left(  v,a\right)  }%
_{\substack{=-f\left(  a,v\right)  \\\text{(by
(\ref{pf.prop.symmetric.L=R.b.anti}), applied to }w=a\text{)}}}=0\text{ for
all }a\in A\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the index }b\text{
as }a\right) \\
&  =\left\{  v\in V\ \mid\ \underbrace{-f\left(  a,v\right)  =0}%
_{\substack{\text{this is equivalent}\\\text{to }\left(  f\left(  a,v\right)
=0\right)  }}\text{ for all }a\in A\right\} \\
&  =\left\{  v\in V\ \mid\ f\left(  a,v\right)  =0\text{ for all }a\in
A\right\} \\
&  =\left\{  w\in V\ \mid\ f\left(  a,w\right)  =0\text{ for all }a\in
A\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the index }v\text{
as }w\right) \\
&  =\mathcal{R}_{f}\left(  A\right)  .
\end{align*}
This proves Proposition \ref{prop.symmetric.L=R} \textbf{(b)}.
\end{proof}
\end{verlong}

\section{The morphism of quotients I}

We shall now construct a certain morphism between quotient spaces induced by
any bilinear map. First, we recall the universal property of quotient spaces:

\begin{proposition}
\label{prop.quotient.uniprop}Let $V$ be a vector space. Let $A$ be a subspace
of $V$. Let $\pi_{V,A}$ be the canonical projection $V\rightarrow V/A$.

Let $W$ be a further vector space. Let $g:V\rightarrow W$ be a linear map such
that $g\left(  A\right)  =0$. Then, there exists a unique linear map
$g^{\prime}:V/A\rightarrow W$ such that $g=g^{\prime}\circ\pi_{V,A}$.
\end{proposition}

Let us fix a notation for projections onto quotient spaces:

\begin{definition}
\label{def.quotient.mod}Let $V$ be a vector space. Let $A$ be a subspace of
$V$. Let $v\in V$. Then, the residue class of $v$ modulo $A$ (that is, the
image of $v$ under the canonical projection $V\rightarrow V/A$) will be
denoted by $\left[  v\right]  _{A}$. (Other widespread notations for this
residue class are $v\operatorname{mod}A$, $v+A$ and $\overline{v}_{A}$.)
\end{definition}

We now state the main theorem of this section:

\begin{theorem}
\label{thm.quotient.1}Let $V$, $W$ and $U$ be three vector spaces. Let
$f:V\times W\rightarrow U$ be a bilinear map.

\textbf{(a)} Then, there exists a unique linear map $\alpha:V/\mathcal{L}%
_{f}\left(  W\right)  \rightarrow\operatorname*{Hom}\left(  W/\mathcal{R}%
_{f}\left(  V\right)  ,U\right)  $ satisfying%
\begin{equation}
\left(  \left(  \alpha\left(  \left[  v\right]  _{\mathcal{L}_{f}\left(
W\right)  }\right)  \right)  \left(  \left[  w\right]  _{\mathcal{R}%
_{f}\left(  V\right)  }\right)  =f\left(  v,w\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  v,w\right)  \in V\times W\right)  .
\label{eq.thm.quotient.1.eq}%
\end{equation}


\textbf{(b)} This map $\alpha$ is injective.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.quotient.1}.]First, we notice that $\mathcal{L}%
_{f}\left(  W\right)  $ is a subspace of $V$ (by Proposition
\ref{prop.LR.basics} \textbf{(a)}, applied to $B=W$). Hence, the quotient
space $V/\mathcal{L}_{f}\left(  W\right)  $ is well-defined. Furthermore,
$\mathcal{R}_{f}\left(  V\right)  $ is a subspace of $W$ (by Proposition
\ref{prop.LR.basics} \textbf{(b)}, applied to $A=V$). Hence, the quotient
space $W/\mathcal{R}_{f}\left(  V\right)  $ is well-defined.

\textbf{(a)} It is easy to see that there exists \textbf{at most one} linear
map $\alpha:V/\mathcal{L}_{f}\left(  W\right)  \rightarrow\operatorname*{Hom}%
\left(  W/\mathcal{R}_{f}\left(  V\right)  ,U\right)  $ satisfying
(\ref{eq.thm.quotient.1.eq})\footnote{\textit{Proof.} Let $\alpha_{1}$ and
$\alpha_{2}$ be two linear maps $\alpha:V/\mathcal{L}_{f}\left(  W\right)
\rightarrow\operatorname*{Hom}\left(  W/\mathcal{R}_{f}\left(  V\right)
,U\right)  $ satisfying (\ref{eq.thm.quotient.1.eq}). We shall show that
$\alpha_{1}=\alpha_{2}$.
\par
We know that $\alpha_{1}$ is a linear map $\alpha:V/\mathcal{L}_{f}\left(
W\right)  \rightarrow\operatorname*{Hom}\left(  W/\mathcal{R}_{f}\left(
V\right)  ,U\right)  $ satisfying (\ref{eq.thm.quotient.1.eq}). In other
words, $\alpha_{1}$ is a linear map $V/\mathcal{L}_{f}\left(  W\right)
\rightarrow\operatorname*{Hom}\left(  W/\mathcal{R}_{f}\left(  V\right)
,U\right)  $ and satisfies%
\begin{equation}
\left(  \left(  \alpha_{1}\left(  \left[  v\right]  _{\mathcal{L}_{f}\left(
W\right)  }\right)  \right)  \left(  \left[  w\right]  _{\mathcal{R}%
_{f}\left(  V\right)  }\right)  =f\left(  v,w\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  v,w\right)  \in V\times W\right)  .
\label{pf.thm.quotient.1.uni.pf.1}%
\end{equation}
\par
Now, let $x\in V/\mathcal{L}_{f}\left(  W\right)  $. Let $y\in W/\mathcal{R}%
_{f}\left(  V\right)  $.
\par
We have $x\in V/\mathcal{L}_{f}\left(  W\right)  $. Hence, we can write $x$ in
the form $\left[  v\right]  _{\mathcal{L}_{f}\left(  W\right)  }$ for some
$v\in V$. Consider this $v$. Thus, $x=\left[  v\right]  _{\mathcal{L}%
_{f}\left(  W\right)  }$.
\par
We have $y\in W/\mathcal{R}_{f}\left(  V\right)  $. Hence, we can write $y$ in
the form $\left[  w\right]  _{\mathcal{R}_{f}\left(  V\right)  }$ for some
$w\in W$. Consider this $w$. Thus, $y=\left[  w\right]  _{\mathcal{R}%
_{f}\left(  V\right)  }$.
\par
Now,%
\[
\left(  \alpha_{1}\left(  \underbrace{x}_{=\left[  v\right]  _{\mathcal{L}%
_{f}\left(  W\right)  }}\right)  \right)  \left(  \underbrace{y}_{=\left[
w\right]  _{\mathcal{R}_{f}\left(  V\right)  }}\right)  =\left(  \alpha
_{1}\left(  \left[  v\right]  _{\mathcal{L}_{f}\left(  W\right)  }\right)
\right)  \left(  \left[  w\right]  _{\mathcal{R}_{f}\left(  V\right)
}\right)  =f\left(  v,w\right)
\]
(by (\ref{pf.thm.quotient.1.uni.pf.1})). The same reasoning (but applied to
$\alpha_{2}$ instead of $\alpha_{1}$) shows that $\left(  \alpha_{2}\left(
x\right)  \right)  \left(  y\right)  =f\left(  v,w\right)  $. Thus, $\left(
\alpha_{1}\left(  x\right)  \right)  \left(  y\right)  =f\left(  v,w\right)
=\left(  \alpha_{2}\left(  x\right)  \right)  \left(  y\right)  $.
\par
Let us now forget that we fixed $y$. We thus have shown that $\left(
\alpha_{1}\left(  x\right)  \right)  \left(  y\right)  =\left(  \alpha
_{2}\left(  x\right)  \right)  \left(  y\right)  $ for every $y\in
W/\mathcal{R}_{f}\left(  V\right)  $. In other words, $\alpha_{1}\left(
x\right)  =\alpha_{2}\left(  x\right)  $.
\par
Let us now forget that we fixed $x$. We thus have shown that $\alpha
_{1}\left(  x\right)  =\alpha_{2}\left(  x\right)  $ for every $x\in
V/\mathcal{L}_{f}\left(  W\right)  $. In other words, $\alpha_{1}=\alpha_{2}$.
\par
Let us now forget that we fixed $\alpha_{1}$ and $\alpha_{2}$. We thus have
shown that $\alpha_{1}=\alpha_{2}$ whenever $\alpha_{1}$ and $\alpha_{2}$ are
two linear maps $\alpha:V/\mathcal{L}_{f}\left(  W\right)  \rightarrow
\operatorname*{Hom}\left(  W/\mathcal{R}_{f}\left(  V\right)  ,U\right)  $
satisfying (\ref{eq.thm.quotient.1.eq}). In other words, there exists
\textbf{at most one} linear map $\alpha:V/\mathcal{L}_{f}\left(  W\right)
\rightarrow\operatorname*{Hom}\left(  W/\mathcal{R}_{f}\left(  V\right)
,U\right)  $ satisfying (\ref{eq.thm.quotient.1.eq}). Qed.}.

We shall now construct such a map.

Let $\pi_{V,\mathcal{L}_{f}\left(  W\right)  }$ be the canonical projection
$V\rightarrow V/\mathcal{L}_{f}\left(  W\right)  $. Thus,%
\begin{equation}
\pi_{V,\mathcal{L}_{f}\left(  W\right)  }\left(  v\right)  =\left[  v\right]
_{\mathcal{L}_{f}\left(  W\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }v\in
V. \label{pf.thm.quotient.1.a.piV}%
\end{equation}


Let $\pi_{W,\mathcal{R}_{f}\left(  V\right)  }$ be the canonical projection
$W\rightarrow W/\mathcal{R}_{f}\left(  V\right)  $. Thus,%
\begin{equation}
\pi_{W,\mathcal{R}_{f}\left(  V\right)  }\left(  w\right)  =\left[  w\right]
_{\mathcal{R}_{f}\left(  V\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }w\in
W. \label{pf.thm.quotient.1.a.piW}%
\end{equation}


Let $v\in V$. The map $f$ is bilinear.

Recall that a linear map $f\left(  v,-\right)  \in\operatorname*{Hom}\left(
W,U\right)  $ is defined (according to Definition \ref{def.f(-)}
\textbf{(b)}). We have $\left(  f\left(  v,-\right)  \right)  \left(
\mathcal{R}_{f}\left(  V\right)  \right)  =0$\ \ \ \ \footnote{\textit{Proof.}
Let $y\in\mathcal{R}_{f}\left(  V\right)  $. We shall prove that $\left(
f\left(  v,-\right)  \right)  \left(  y\right)  =0$.
\par
We have $y\in\mathcal{R}_{f}\left(  V\right)  =\left\{  w\in W\ \mid\ f\left(
a,w\right)  =0\text{ for all }a\in V\right\}  $ (by the definition of
$\mathcal{R}_{f}\left(  V\right)  $). In other words, $y$ is an element $w$ of
$W$ satisfying $f\left(  a,w\right)  =0$ for all $a\in V$. In other words, $y$
is an element of $W$ and satisfies $f\left(  a,y\right)  =0$ for all $a\in V$.
\par
We know that $f\left(  a,y\right)  =0$ for all $a\in V$. Applying this to
$a=v$, we obtain $f\left(  v,y\right)  =0$. But the definition of $f\left(
v,-\right)  $ yields $\left(  f\left(  v,-\right)  \right)  \left(  y\right)
=f\left(  v,y\right)  =0$.
\par
Now, let us forget that we fixed $y$. We thus have shown that $\left(
f\left(  v,-\right)  \right)  \left(  y\right)  =0$ for every $y\in
\mathcal{R}_{f}\left(  V\right)  $. In other words, $\left(  f\left(
v,-\right)  \right)  \left(  \mathcal{R}_{f}\left(  V\right)  \right)  =0$.
Qed.}. Hence, Proposition \ref{prop.quotient.uniprop} (applied to $W$,
$\mathcal{R}_{f}\left(  V\right)  $, $\pi_{W,\mathcal{R}_{f}\left(  V\right)
}$, $U$ and $f\left(  v,-\right)  $ instead of $V$, $A$, $\pi_{V,A}$, $W$ and
$g$) shows that there exists a unique linear map $g^{\prime}:W/\mathcal{R}%
_{f}\left(  V\right)  \rightarrow U$ such that $f\left(  v,-\right)
=g^{\prime}\circ\pi_{W,\mathcal{R}_{f}\left(  V\right)  }$. Let us denote this
$g^{\prime}$ by $g_{v}$. Thus, $g_{v}$ is a linear map $W/\mathcal{R}%
_{f}\left(  V\right)  \rightarrow U$ and satisfies $f\left(  v,-\right)
=g_{v}\circ\pi_{W,\mathcal{R}_{f}\left(  V\right)  }$.

We have%
\begin{equation}
g_{v}\left(  \left[  w\right]  _{\mathcal{R}_{f}\left(  V\right)  }\right)
=f\left(  v,w\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }w\in W
\label{pf.thm.quotient.1.a.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.quotient.1.a.1}):} Let $w\in W$. From
(\ref{pf.thm.quotient.1.a.piW}), we obtain $\left[  w\right]  _{\mathcal{R}%
_{f}\left(  V\right)  }=\pi_{W,\mathcal{R}_{f}\left(  V\right)  }\left(
w\right)  $. Applying the map $g_{v}$ to both sides of this equality, we
obtain%
\begin{align*}
g_{v}\left(  \left[  w\right]  _{\mathcal{R}_{f}\left(  V\right)  }\right)
&  =g_{v}\left(  \pi_{W,\mathcal{R}_{f}\left(  V\right)  }\left(  w\right)
\right)  =\underbrace{\left(  g_{v}\circ\pi_{W,\mathcal{R}_{f}\left(
V\right)  }\right)  }_{=f\left(  v,-\right)  }\left(  w\right) \\
&  =\left(  f\left(  v,-\right)  \right)  \left(  w\right)  =f\left(
v,w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }f\left(
v,-\right)  \right)  .
\end{align*}
This proves (\ref{pf.thm.quotient.1.a.1}).}.

Now, let us forget that we fixed $v$. We thus have constructed a linear map
$g_{v}:W/\mathcal{R}_{f}\left(  V\right)  \rightarrow U$ for every $v\in V$.
We have furthermore shown that this map satisfies (\ref{pf.thm.quotient.1.a.1}%
) for every $v\in V$. For every $v\in V$, the map $g_{v}$ is a linear map
$W/\mathcal{R}_{f}\left(  V\right)  \rightarrow U$, and thus belongs to
$\operatorname*{Hom}\left(  W/\mathcal{R}_{f}\left(  V\right)  ,U\right)  $.

Now, we define a map $g:V\rightarrow\operatorname*{Hom}\left(  W/\mathcal{R}%
_{f}\left(  V\right)  ,U\right)  $ by%
\[
\left(  g\left(  v\right)  =g_{v}\ \ \ \ \ \ \ \ \ \ \text{for every }v\in
V\right)  .
\]
(This is well-defined, because for every $v\in V$, the map $g_{v}$ belongs to
\newline$\operatorname*{Hom}\left(  W/\mathcal{R}_{f}\left(  V\right)
,U\right)  $.) Thus, for every $v\in V$ and $w\in W$, we have%
\begin{equation}
\underbrace{\left(  g\left(  v\right)  \right)  }_{=g_{v}}\left(  \left[
w\right]  _{\mathcal{R}_{f}\left(  V\right)  }\right)  =g_{v}\left(  \left[
w\right]  _{\mathcal{R}_{f}\left(  V\right)  }\right)  =f\left(  v,w\right)
\label{pf.thm.quotient.1.a.2}%
\end{equation}
(by (\ref{pf.thm.quotient.1.a.1})).

The map $g$ is linear\footnote{\textit{Proof.} Let $v_{1}$ and $v_{2}$ be two
elements of $V$. Let $\lambda_{1}$ and $\lambda_{2}$ be two elements of
$\mathbf{k}$. We shall show that $g\left(  \lambda_{1}v_{1}+\lambda_{2}%
v_{2}\right)  =\lambda_{1}g\left(  v_{1}\right)  +\lambda_{2}g\left(
v_{2}\right)  $.
\par
Let $y\in W/\mathcal{R}_{f}\left(  V\right)  $. Thus, $y$ can be written in
the form $\left[  w\right]  _{\mathcal{R}_{f}\left(  V\right)  }$ for some
$w\in W$. Consider this $w$. Thus, $y=\left[  w\right]  _{\mathcal{R}%
_{f}\left(  V\right)  }$. Hence, for every $v\in V$, we have%
\begin{equation}
\left(  g\left(  v\right)  \right)  \left(  \underbrace{y}_{=\left[  w\right]
_{\mathcal{R}_{f}\left(  V\right)  }}\right)  =\left(  g\left(  v\right)
\right)  \left(  \left[  w\right]  _{\mathcal{R}_{f}\left(  V\right)
}\right)  =f\left(  v,w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.quotient.1.a.2})}\right)  . \label{pf.thm.quotient.1.a.3}%
\end{equation}
\par
Recall that a linear map $f\left(  -,w\right)  \in\operatorname*{Hom}\left(
V,U\right)  $ is defined (according to Definition \ref{def.f(-)}
\textbf{(a)}). This map $f\left(  -,w\right)  $ is linear, and thus we have
$\left(  f\left(  -,w\right)  \right)  \left(  \lambda_{1}v_{1}+\lambda
_{2}v_{2}\right)  =\lambda_{1}\left(  f\left(  -,w\right)  \right)  \left(
v_{1}\right)  +\lambda_{2}\left(  f\left(  -,w\right)  \right)  \left(
v_{2}\right)  $. Comparing this with $\left(  f\left(  -,w\right)  \right)
\left(  \lambda_{1}v_{1}+\lambda_{2}v_{2}\right)  =f\left(  \lambda_{1}%
v_{1}+\lambda_{2}v_{2},w\right)  $ (by the definition of $f\left(  -,w\right)
$), we obtain%
\begin{align*}
f\left(  \lambda_{1}v_{1}+\lambda_{2}v_{2},w\right)   &  =\lambda
_{1}\underbrace{\left(  f\left(  -,w\right)  \right)  \left(  v_{1}\right)
}_{\substack{=f\left(  v_{1},w\right)  \\\text{(by the definition of }f\left(
-,w\right)  \text{)}}}+\lambda_{2}\underbrace{\left(  f\left(  -,w\right)
\right)  \left(  v_{2}\right)  }_{\substack{=f\left(  v_{2},w\right)
\\\text{(by the definition of }f\left(  -,w\right)  \text{)}}}\\
&  =\lambda_{1}f\left(  v_{1},w\right)  +\lambda_{2}f\left(  v_{2},w\right)  .
\end{align*}
\par
Now, (\ref{pf.thm.quotient.1.a.3}) (applied to $v=v_{1}$) yields $\left(
g\left(  v_{1}\right)  \right)  \left(  y\right)  =f\left(  v_{1},w\right)  $.
Also, (\ref{pf.thm.quotient.1.a.3}) (applied to $v=v_{2}$) yields $\left(
g\left(  v_{2}\right)  \right)  \left(  y\right)  =f\left(  v_{2},w\right)  $.
But (\ref{pf.thm.quotient.1.a.3}) (applied to $v=\lambda_{1}v_{1}+\lambda
_{2}v_{2}$) yields%
\begin{align*}
\left(  g\left(  \lambda_{1}v_{1}+\lambda_{2}v_{2}\right)  \right)  \left(
y\right)   &  =f\left(  \lambda_{1}v_{1}+\lambda_{2}v_{2},w\right)
=\lambda_{1}\underbrace{f\left(  v_{1},w\right)  }_{=\left(  g\left(
v_{1}\right)  \right)  \left(  y\right)  }+\lambda_{2}\underbrace{f\left(
v_{2},w\right)  }_{=\left(  g\left(  v_{2}\right)  \right)  \left(  y\right)
}\\
&  =\lambda_{1}\left(  g\left(  v_{1}\right)  \right)  \left(  y\right)
+\lambda_{2}\left(  g\left(  v_{2}\right)  \right)  \left(  y\right)  =\left(
\lambda_{1}g\left(  v_{1}\right)  +\lambda_{2}g\left(  v_{2}\right)  \right)
\left(  y\right)  .
\end{align*}
\par
Let us now forget that we fixed $y$. We thus have shown that $\left(  g\left(
\lambda_{1}v_{1}+\lambda_{2}v_{2}\right)  \right)  \left(  y\right)  =\left(
\lambda_{1}g\left(  v_{1}\right)  +\lambda_{2}g\left(  v_{2}\right)  \right)
\left(  y\right)  $ for every $y\in W/\mathcal{R}_{f}\left(  V\right)  $. In
other words, $g\left(  \lambda_{1}v_{1}+\lambda_{2}v_{2}\right)  =\lambda
_{1}g\left(  v_{1}\right)  +\lambda_{2}g\left(  v_{2}\right)  $.
\par
Let us now forget that we fixed $v_{1}$, $v_{2}$, $\lambda_{1}$ and
$\lambda_{2}$. Thus, we have proven that $g\left(  \lambda_{1}v_{1}%
+\lambda_{2}v_{2}\right)  =\lambda_{1}g\left(  v_{1}\right)  +\lambda
_{2}g\left(  v_{2}\right)  $ for every $v_{1}\in V$, $v_{2}\in V$,
$\lambda_{1}\in\mathbf{k}$ and $\lambda_{2}\in\mathbf{k}$. In other words, the
map $g$ is linear, qed.}. Moreover, $g\left(  \mathcal{L}_{f}\left(  W\right)
\right)  =0$\ \ \ \ \footnote{\textit{Proof.} Let $p\in\mathcal{L}_{f}\left(
W\right)  $. Thus, $p\in\mathcal{L}_{f}\left(  W\right)  =\left\{  v\in
V\ \mid\ f\left(  v,b\right)  =0\text{ for all }b\in W\right\}  $ (by the
definition of $\mathcal{L}_{f}\left(  W\right)  $). In other words, $p$ is an
element $v$ of $V$ satisfying $f\left(  v,b\right)  =0$ for all $b\in W$. In
other words, $p$ is an element of $V$ and satisfies $f\left(  p,b\right)  =0$
for all $b\in W$.
\par
Let $y\in W/\mathcal{R}_{f}\left(  V\right)  $. Thus, $y$ can be written in
the form $\left[  w\right]  _{\mathcal{R}_{f}\left(  V\right)  }$ for some
$w\in W$. Consider this $w$. Thus, $y=\left[  w\right]  _{\mathcal{R}%
_{f}\left(  V\right)  }$.
\par
But we have $f\left(  p,b\right)  =0$ for all $b\in W$. Applying this to
$b=w$, we obtain $f\left(  p,w\right)  =0$. But (\ref{pf.thm.quotient.1.a.2})
(applied to $v=p$) yields $\left(  g\left(  p\right)  \right)  \left(  \left[
w\right]  _{\mathcal{R}_{f}\left(  V\right)  }\right)  =f\left(  p,w\right)
=0$. Hence, $\left(  g\left(  p\right)  \right)  \left(  \underbrace{y}%
_{=\left[  w\right]  _{\mathcal{R}_{f}\left(  V\right)  }}\right)  =\left(
g\left(  p\right)  \right)  \left(  \left[  w\right]  _{\mathcal{R}_{f}\left(
V\right)  }\right)  =0$.
\par
Now, let us forget that we fixed $y$. We thus have shown that $\left(
g\left(  p\right)  \right)  \left(  y\right)  =0$ for every $y\in
W/\mathcal{R}_{f}\left(  V\right)  $. In other words, $g\left(  p\right)  =0$.
\par
Now, let us forget that we fixed $p$. We thus have shown that $g\left(
p\right)  =0$ for every $p\in\mathcal{L}_{f}\left(  W\right)  $. In other
words, $g\left(  \mathcal{L}_{f}\left(  W\right)  \right)  =0$, qed.}. Hence,
Proposition \ref{prop.quotient.uniprop} (applied to $\mathcal{L}_{f}\left(
W\right)  $, $\pi_{V,\mathcal{L}_{f}\left(  W\right)  }$ and
$\operatorname*{Hom}\left(  W/\mathcal{R}_{f}\left(  V\right)  ,U\right)  $
instead of $A$, $\pi_{V,A}$ and $g$) shows that there exists a unique linear
map $g^{\prime}:V/\mathcal{L}_{f}\left(  W\right)  \rightarrow
\operatorname*{Hom}\left(  W/\mathcal{R}_{f}\left(  V\right)  ,U\right)  $
such that $g=g^{\prime}\circ\pi_{V,\mathcal{L}_{f}\left(  W\right)  }$. Let us
denote this $g^{\prime}$ by $\beta$. Thus, $\beta$ is a linear map
$V/\mathcal{L}_{f}\left(  W\right)  \rightarrow\operatorname*{Hom}\left(
W/\mathcal{R}_{f}\left(  V\right)  ,U\right)  $ and satisfies $g=\beta\circ
\pi_{V,\mathcal{L}_{f}\left(  W\right)  }$.

Now, for all $\left(  v,w\right)  \in V\times W$, we have%
\begin{align*}
&  \left(  \beta\left(  \underbrace{\left[  v\right]  _{\mathcal{L}_{f}\left(
W\right)  }}_{\substack{=\pi_{V,\mathcal{L}_{f}\left(  W\right)  }\left(
v\right)  \\\text{(by (\ref{pf.thm.quotient.1.a.piV}))}}}\right)  \right)
\left(  \left[  w\right]  _{\mathcal{R}_{f}\left(  V\right)  }\right) \\
&  =\left(  \underbrace{\beta\left(  \pi_{V,\mathcal{L}_{f}\left(  W\right)
}\left(  v\right)  \right)  }_{=\left(  \beta\circ\pi_{V,\mathcal{L}%
_{f}\left(  W\right)  }\right)  \left(  v\right)  }\right)  \left(  \left[
w\right]  _{\mathcal{R}_{f}\left(  V\right)  }\right)  =\left(
\underbrace{\left(  \beta\circ\pi_{V,\mathcal{L}_{f}\left(  W\right)
}\right)  }_{=g}\left(  v\right)  \right)  \left(  \left[  w\right]
_{\mathcal{R}_{f}\left(  V\right)  }\right) \\
&  =\left(  g\left(  v\right)  \right)  \left(  \left[  w\right]
_{\mathcal{R}_{f}\left(  V\right)  }\right)  =f\left(  v,w\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.quotient.1.a.2})}\right)  .
\end{align*}
In other words, the linear map $\beta$ satisfies
\[
\left(  \left(  \beta\left(  \left[  v\right]  _{\mathcal{L}_{f}\left(
W\right)  }\right)  \right)  \left(  \left[  w\right]  _{\mathcal{R}%
_{f}\left(  V\right)  }\right)  =f\left(  v,w\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  v,w\right)  \in V\times W\right)
.
\]
Hence, the map $\beta$ is a linear map $\alpha:V/\mathcal{L}_{f}\left(
W\right)  \rightarrow\operatorname*{Hom}\left(  W/\mathcal{R}_{f}\left(
V\right)  ,U\right)  $ satisfying (\ref{eq.thm.quotient.1.eq}). Therefore,
there exists at least one linear map $\alpha:V/\mathcal{L}_{f}\left(
W\right)  \rightarrow\operatorname*{Hom}\left(  W/\mathcal{R}_{f}\left(
V\right)  ,U\right)  $ satisfying (\ref{eq.thm.quotient.1.eq}) (namely,
$\beta$). Consequently, there exists a \textbf{unique} linear map
$\alpha:V/\mathcal{L}_{f}\left(  W\right)  \rightarrow\operatorname*{Hom}%
\left(  W/\mathcal{R}_{f}\left(  V\right)  ,U\right)  $ satisfying
(\ref{eq.thm.quotient.1.eq}) (since we already know that there exists
\textbf{at most one} linear map $\alpha:V/\mathcal{L}_{f}\left(  W\right)
\rightarrow\operatorname*{Hom}\left(  W/\mathcal{R}_{f}\left(  V\right)
,U\right)  $ satisfying (\ref{eq.thm.quotient.1.eq})). This proves Theorem
\ref{thm.quotient.1} \textbf{(a)}.

\textbf{(b)} Consider the unique linear map $\alpha:V/\mathcal{L}_{f}\left(
W\right)  \rightarrow\operatorname*{Hom}\left(  W/\mathcal{R}_{f}\left(
V\right)  ,U\right)  $ satisfying (\ref{eq.thm.quotient.1.eq}).

We have
\begin{align}
\mathcal{L}_{f}\left(  W\right)   &  =\left\{  v\in V\ \mid\ f\left(
v,b\right)  =0\text{ for all }b\in W\right\}  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\mathcal{L}_{f}\left(  W\right)  \right)
\nonumber\\
&  =\left\{  p\in V\ \mid\ f\left(  p,b\right)  =0\text{ for all }b\in
W\right\} \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the index }v\text{
as }p\right) \nonumber\\
&  =\left\{  p\in V\ \mid\ f\left(  p,w\right)  =0\text{ for all }w\in
W\right\} \label{pf.thm.quotient.1.b.1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the index }b\text{
as }w\right)  .\nonumber
\end{align}


Let $x\in\operatorname*{Ker}\alpha$. We have $x\in\operatorname*{Ker}%
\alpha\subseteq V/\mathcal{L}_{f}\left(  W\right)  $. Hence, we can write $x$
in the form $\left[  v\right]  _{\mathcal{L}_{f}\left(  W\right)  }$ for some
$v\in V$. Consider this $v$. Thus, $x=\left[  v\right]  _{\mathcal{L}%
_{f}\left(  W\right)  }$.

We have $x\in\operatorname*{Ker}\alpha$, so that $\alpha\left(  x\right)  =0$.
Now, let $w\in W$. Then, $\left(  v,w\right)  \in V\times W$ (since $v\in V$
and $w\in W$). Thus, (\ref{eq.thm.quotient.1.eq}) shows that $\left(
\alpha\left(  \left[  v\right]  _{\mathcal{L}_{f}\left(  W\right)  }\right)
\right)  \left(  \left[  w\right]  _{\mathcal{R}_{f}\left(  V\right)
}\right)  =f\left(  v,w\right)  $ (since we know that $\alpha$ satisfies
(\ref{eq.thm.quotient.1.eq})). Hence,%
\[
f\left(  v,w\right)  =\left(  \alpha\left(  \underbrace{\left[  v\right]
_{\mathcal{L}_{f}\left(  W\right)  }}_{=x}\right)  \right)  \left(  \left[
w\right]  _{\mathcal{R}_{f}\left(  V\right)  }\right)  =\underbrace{\left(
\alpha\left(  x\right)  \right)  }_{=0}\left(  \left[  w\right]
_{\mathcal{R}_{f}\left(  V\right)  }\right)  =0\left(  \left[  w\right]
_{\mathcal{R}_{f}\left(  V\right)  }\right)  =0.
\]


Let us now forget that we fixed $w$. We thus have shown that $f\left(
v,w\right)  =0$ for all $w\in W$. Thus, $v$ is an element of $V$ and satisfies
$f\left(  v,w\right)  =0$ for all $w\in W$. In other words, $v$ is an element
$p$ of $V$ satisfying $f\left(  p,w\right)  =0$ for all $w\in W$. In other
words,%
\[
v\in\left\{  p\in V\ \mid\ f\left(  p,w\right)  =0\text{ for all }w\in
W\right\}  .
\]
In view of (\ref{pf.thm.quotient.1.b.1}), this rewrites as $v\in
\mathcal{L}_{f}\left(  W\right)  $. Hence, $\left[  v\right]  _{\mathcal{L}%
_{f}\left(  W\right)  }=0$. Thus, $x=\left[  v\right]  _{\mathcal{L}%
_{f}\left(  W\right)  }=0$.

Let us now forget that we fixed $x$. We thus have shown that $x=0$ for every
$x\in\operatorname*{Ker}\alpha$. In other words, $\operatorname*{Ker}\alpha
=0$. Thus, the linear map $\alpha$ is injective. This proves Theorem
\ref{thm.quotient.1} \textbf{(b)}.
\end{proof}

\section{The morphism of quotients II}

So far, we have not used the assumption that $\mathbf{k}$ is a field; we could
have just as well let $\mathbf{k}$ be any commutative ring. (Of course, we
would have to talk about $\mathbf{k}$-modules instead of vector spaces, and
similarly; but apart from this, the results and proofs would have been the
same.) We shall now prove some slightly deeper result where we will actually
need this assumption (that $\mathbf{k}$ is a field). We will also make some
assumptions on finite-dimensionality.

\begin{definition}
\label{def.dualspace}Let $V$ be a vector space. Then, $V^{\ast}$ will denote
the vector space $\operatorname*{Hom}\left(  V,\mathbf{k}\right)  $. This
space $V^{\ast}$ is called the \textit{dual space} of $V$. It is well-known
that $\dim\left(  V^{\ast}\right)  =\dim V$ when $V$ is finite-dimensional.
Moreover, there is a canonical injective linear map $V\rightarrow V^{\ast\ast
}$ (which sends every $v\in V$ to the linear map $\widetilde{v}\in V^{\ast
\ast}$ which sends every $f\in V^{\ast}$ to $f\left(  v\right)  \in\mathbf{k}%
$). When $V$ is finite-dimensional, this linear map $V\rightarrow V^{\ast\ast
}$ is a vector space isomorphism.
\end{definition}

We shall first state some well-known properties of finite-dimensional vector
spaces. The first of these properties is the fact that an injective linear map
between two finite-dimensional vector spaces of equal dimension must be an isomorphism:

\begin{proposition}
\label{prop.findim.inj-iso}Let $A$ and $B$ be two finite-dimensional vector
spaces. Let $i:A\rightarrow B$ be an injective linear map. Assume that $\dim
A=\dim B$. Then, $i$ is a vector space isomorphism.
\end{proposition}

The next property is the particular case of Proposition
\ref{prop.findim.inj-iso} when the map $i$ is an inclusion:

\begin{proposition}
\label{prop.findim.include}Let $B$ be a finite-dimensional vector space. Let
$A$ be a subspace of $B$ such that $\dim A=\dim B$. Then, $A=B$.
\end{proposition}

Furthermore, the domain of an injective linear map always has at most the same
dimension as its target:

\begin{proposition}
\label{prop.findim.inj-dim}Let $A$ and $B$ be two finite-dimensional vector
spaces. Let $i:A\rightarrow B$ be an injective linear map. Then, $\dim
A\leq\dim B$.
\end{proposition}

Now, we state the next crucial theorem:

\begin{theorem}
\label{thm.quotient.2}Let $V$ and $W$ be two finite-dimensional vector spaces.
Let $f:V\times W\rightarrow\mathbf{k}$ be a bilinear form.

\textbf{(a)} Then, there exists a unique linear map $\alpha:V/\mathcal{L}%
_{f}\left(  W\right)  \rightarrow\left(  W/\mathcal{R}_{f}\left(  V\right)
\right)  ^{\ast}$ satisfying (\ref{eq.thm.quotient.1.eq}).

\textbf{(b)} This map $\alpha$ is a vector space isomorphism.

\textbf{(c)} We have $\dim\left(  V/\mathcal{L}_{f}\left(  W\right)  \right)
=\dim\left(  W/\mathcal{R}_{f}\left(  V\right)  \right)  $.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.quotient.2}.]The map $f:V\times W\rightarrow
\mathbf{k}$ is a bilinear form. In other words, the map $f:V\times
W\rightarrow\mathbf{k}$ is a bilinear map (according to the definition of a
\textquotedblleft bilinear form\textquotedblright).

\textbf{(a)} Theorem \ref{thm.quotient.1} \textbf{(a)} (applied to
$U=\mathbf{k}$) yields that there exists a unique linear map $\alpha
:V/\mathcal{L}_{f}\left(  W\right)  \rightarrow\operatorname*{Hom}\left(
W/\mathcal{R}_{f}\left(  V\right)  ,\mathbf{k}\right)  $ satisfying
(\ref{eq.thm.quotient.1.eq}). In other words, there exists a unique linear map
$\alpha:V/\mathcal{L}_{f}\left(  W\right)  \rightarrow\left(  W/\mathcal{R}%
_{f}\left(  V\right)  \right)  ^{\ast}$ satisfying (\ref{eq.thm.quotient.1.eq}%
) (since $\left(  W/\mathcal{R}_{f}\left(  V\right)  \right)  ^{\ast
}=\operatorname*{Hom}\left(  W/\mathcal{R}_{f}\left(  V\right)  ,\mathbf{k}%
\right)  $). This proves Theorem \ref{thm.quotient.2} \textbf{(a)}.

Theorem \ref{thm.quotient.1} \textbf{(b)} (applied to $U=\mathbf{k}$) yields
that the unique linear map $\alpha:V/\mathcal{L}_{f}\left(  W\right)
\rightarrow\operatorname*{Hom}\left(  W/\mathcal{R}_{f}\left(  V\right)
,\mathbf{k}\right)  $ satisfying (\ref{eq.thm.quotient.1.eq}) is injective. In
other words, the unique linear map $\alpha:V/\mathcal{L}_{f}\left(  W\right)
\rightarrow\left(  W/\mathcal{R}_{f}\left(  V\right)  \right)  ^{\ast}$
satisfying (\ref{eq.thm.quotient.1.eq}) is injective (since $\left(
W/\mathcal{R}_{f}\left(  V\right)  \right)  ^{\ast}=\operatorname*{Hom}\left(
W/\mathcal{R}_{f}\left(  V\right)  ,\mathbf{k}\right)  $). Consider this map
$\alpha$.

The vector spaces $V/\mathcal{L}_{f}\left(  W\right)  $ and $W/\mathcal{R}%
_{f}\left(  V\right)  $ are finite-dimensional (since $V$ and $W$ are
finite-dimensional). The vector space $\left(  W/\mathcal{R}_{f}\left(
V\right)  \right)  ^{\ast}$ is finite-dimensional (since $W/\mathcal{R}%
_{f}\left(  V\right)  $ is finite-dimensional). We know that the map $\alpha$
is injective. Thus, Proposition \ref{prop.findim.inj-dim} (applied to
$A=V/\mathcal{L}_{f}\left(  W\right)  $, $B=\left(  W/\mathcal{R}_{f}\left(
V\right)  \right)  ^{\ast}$ and $i=\alpha$) shows that $\dim\left(
V/\mathcal{L}_{f}\left(  W\right)  \right)  \leq\dim\left(  \left(
W/\mathcal{R}_{f}\left(  V\right)  \right)  ^{\ast}\right)  $.

But every finite-dimensional vector space $G$ satisfies $\dim\left(  G^{\ast
}\right)  =\dim G$. Applying this to $G=W/\mathcal{R}_{f}\left(  V\right)  $,
we obtain $\dim\left(  \left(  W/\mathcal{R}_{f}\left(  V\right)  \right)
^{\ast}\right)  =\dim\left(  W/\mathcal{R}_{f}\left(  V\right)  \right)  $.
Hence,%
\begin{equation}
\dim\left(  V/\mathcal{L}_{f}\left(  W\right)  \right)  \leq\dim\left(
\left(  W/\mathcal{R}_{f}\left(  V\right)  \right)  ^{\ast}\right)
=\dim\left(  W/\mathcal{R}_{f}\left(  V\right)  \right)  .
\label{pf.thm.quotient.2.b.1}%
\end{equation}


We thus have proven the inequality (\ref{pf.thm.quotient.2.b.1}) for every two
finite-dimensional vector spaces $V$ and $W$ and every bilinear form
$f:V\times W\rightarrow\mathbf{k}$.

But $f^{\operatorname*{op}}:W\times V\rightarrow\mathbf{k}$ is a bilinear form
(by Proposition \ref{prop.bilform.op-and-restrict} \textbf{(a)}). Hence, we
can apply (\ref{pf.thm.quotient.2.b.1}) to $W$, $V$ and $f^{\operatorname*{op}%
}$ instead of $V$, $W$ and $f$. As a result, we obtain%
\[
\dim\left(  W/\mathcal{L}_{f^{\operatorname*{op}}}\left(  V\right)  \right)
\leq\dim\left(  V/\mathcal{R}_{f^{\operatorname*{op}}}\left(  W\right)
\right)  .
\]


But Proposition \ref{prop.LR.LR} \textbf{(a)} (applied to $U=\mathbf{k}$ and
$B=W$) yields $\mathcal{L}_{f}\left(  W\right)  =\mathcal{R}%
_{f^{\operatorname*{op}}}\left(  W\right)  $. Also, Proposition
\ref{prop.LR.LR} \textbf{(b)} (applied to $U=\mathbf{k}$ and $A=V$) yields
$\mathcal{R}_{f}\left(  V\right)  =\mathcal{L}_{f^{\operatorname*{op}}}\left(
V\right)  $. Now,%
\begin{align*}
\dim\left(  W/\underbrace{\mathcal{R}_{f}\left(  V\right)  }_{=\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  V\right)  }\right)   &  =\dim\left(
W/\mathcal{L}_{f^{\operatorname*{op}}}\left(  V\right)  \right) \\
&  \leq\dim\left(  V/\underbrace{\mathcal{R}_{f^{\operatorname*{op}}}\left(
W\right)  }_{=\mathcal{L}_{f}\left(  W\right)  }\right)  =\dim\left(
V/\mathcal{L}_{f}\left(  W\right)  \right)  .
\end{align*}
Combining this inequality with (\ref{pf.thm.quotient.2.b.1}), we obtain
$\dim\left(  V/\mathcal{L}_{f}\left(  W\right)  \right)  =\dim\left(
W/\mathcal{R}_{f}\left(  V\right)  \right)  $. This proves Theorem
\ref{thm.quotient.2} \textbf{(c)}.

It remains to prove Theorem \ref{thm.quotient.2} \textbf{(b)}. But this is now
easy: We have%
\[
\dim\left(  V/\mathcal{L}_{f}\left(  W\right)  \right)  =\dim\left(
W/\mathcal{R}_{f}\left(  V\right)  \right)  =\dim\left(  \left(
W/\mathcal{R}_{f}\left(  V\right)  \right)  ^{\ast}\right)  .
\]
Hence, Proposition \ref{prop.findim.inj-iso} (applied to $A=V/\mathcal{L}%
_{f}\left(  W\right)  $, $B=\left(  W/\mathcal{R}_{f}\left(  V\right)
\right)  ^{\ast}$ and $i=\alpha$) shows that $\alpha$ is a vector space
isomorphism (since $\alpha$ is injective). This proves Theorem
\ref{thm.quotient.2} \textbf{(b)}.
\end{proof}

Theorem \ref{thm.quotient.2} \textbf{(c)} leads to a particularly useful corollary:

\begin{corollary}
\label{cor.quotient.3}Let $V$ and $W$ be two finite-dimensional vector spaces.
Let $f:V\times W\rightarrow\mathbf{k}$ be a bilinear form. Let $A$ be a
subspace of $V$. Let $B$ be a subspace of $W$. Then,%
\[
\dim\left(  A/\left(  A\cap\mathcal{L}_{f}\left(  B\right)  \right)  \right)
=\dim\left(  B/\left(  B\cap\mathcal{R}_{f}\left(  A\right)  \right)  \right)
.
\]

\end{corollary}

We shall derive this corollary from Theorem \ref{thm.quotient.2} \textbf{(c)}
using the following simple fact:

\begin{proposition}
\label{prop.quotient.3.lem}Let $V$, $W$ and $U$ be three vector spaces. Let
$f:V\times W\rightarrow U$ be a map. Let $A$ be a subspace of $V$. Let $B$ be
a subspace of $W$.

\textbf{(a)} We have $A\cap\mathcal{L}_{f}\left(  B\right)  =\mathcal{L}%
_{f\mid_{A\times B}}\left(  B\right)  $.

\textbf{(b)} We have $B\cap\mathcal{R}_{f}\left(  A\right)  =\mathcal{R}%
_{f\mid_{A\times B}}\left(  A\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.quotient.3.lem}.]\textbf{(a)} The definition
of $\mathcal{L}_{f}\left(  B\right)  $ yields \newline$\mathcal{L}_{f}\left(
B\right)  =\left\{  v\in V\ \mid\ f\left(  v,b\right)  =0\text{ for all }b\in
B\right\}  $.

The map $f\mid_{A\times B}$ is a map $A\times B\rightarrow U$. Hence, the
definition of $\mathcal{L}_{f\mid_{A\times B}}\left(  B\right)  $ yields%
\begin{align*}
\mathcal{L}_{f\mid_{A\times B}}\left(  B\right)   &  =\left\{  v\in
A\ \mid\ \underbrace{\left(  f\mid_{A\times B}\right)  \left(  v,b\right)
}_{=f\left(  v,b\right)  }=0\text{ for all }b\in B\right\} \\
&  =\left\{  v\in A\ \mid\ f\left(  v,b\right)  =0\text{ for all }b\in
B\right\} \\
&  =\left\{  v\in V\ \mid\ v\in A\text{ and }f\left(  v,b\right)  =0\text{ for
all }b\in B\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }A\subseteq V\right) \\
&  =\underbrace{\left\{  v\in V\ \mid\ v\in A\right\}  }%
_{\substack{=A\\\text{(since }A\subseteq V\text{)}}}\cap\underbrace{\left\{
v\in V\ \mid\ f\left(  v,b\right)  =0\text{ for all }b\in B\right\}
}_{=\mathcal{L}_{f}\left(  B\right)  }\\
&  =A\cap\mathcal{L}_{f}\left(  B\right)  .
\end{align*}
This proves Proposition \ref{prop.quotient.3.lem} \textbf{(a)}.

\begin{vershort}
\textbf{(b)} This proof is similar to that of Proposition
\ref{prop.quotient.3.lem} \textbf{(a)}.
\end{vershort}

\begin{verlong}
\textbf{(b)} The definition of $\mathcal{R}_{f}\left(  A\right)  $ yields
$\mathcal{R}_{f}\left(  A\right)  =\left\{  w\in W\ \mid\ f\left(  a,w\right)
=0\text{ for all }a\in A\right\}  $.

The map $f\mid_{A\times B}$ is a map $A\times B\rightarrow U$. Hence, the
definition of $\mathcal{R}_{f\mid_{A\times B}}\left(  A\right)  $ yields%
\begin{align*}
\mathcal{R}_{f\mid_{A\times B}}\left(  A\right)   &  =\left\{  w\in
B\ \mid\ \underbrace{\left(  f\mid_{A\times B}\right)  \left(  a,w\right)
}_{=f\left(  a,w\right)  }=0\text{ for all }a\in A\right\} \\
&  =\left\{  w\in B\ \mid\ f\left(  a,w\right)  =0\text{ for all }a\in
A\right\} \\
&  =\left\{  w\in W\ \mid\ w\in B\text{ and }f\left(  a,w\right)  =0\text{ for
all }a\in A\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }B\subseteq W\right) \\
&  =\underbrace{\left\{  w\in W\ \mid\ w\in B\right\}  }%
_{\substack{=B\\\text{(since }B\subseteq W\text{)}}}\cap\underbrace{\left\{
w\in W\ \mid\ f\left(  a,w\right)  =0\text{ for all }a\in A\right\}
}_{=\mathcal{R}_{f}\left(  A\right)  }\\
&  =B\cap\mathcal{R}_{f}\left(  A\right)  .
\end{align*}
This proves Proposition \ref{prop.quotient.3.lem} \textbf{(b)}.
\end{verlong}
\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor.quotient.3}.]The vector spaces $V$ and $W$ are
finite-dimensional. Hence, their subspaces $A$ and $B$ also are
finite-dimensional. Proposition \ref{prop.bilform.op-and-restrict}
\textbf{(b)} shows that the restriction $f\mid_{A\times B}$ is a bilinear form
$A\times B\rightarrow\mathbf{k}$. Hence, Theorem \ref{thm.quotient.2}
\textbf{(c)} (applied to $A$, $B$ and $f\mid_{A\times B}$ instead of $V$, $W$
and $f$) shows that $\dim\left(  A/\mathcal{L}_{f\mid_{A\times B}}\left(
B\right)  \right)  =\dim\left(  B/\mathcal{R}_{f\mid_{A\times B}}\left(
A\right)  \right)  $.

But Proposition \ref{prop.quotient.3.lem} \textbf{(a)} (applied to
$U=\mathbf{k}$) shows that $A\cap\mathcal{L}_{f}\left(  B\right)
=\mathcal{L}_{f\mid_{A\times B}}\left(  B\right)  $. Also, Proposition
\ref{prop.quotient.3.lem} \textbf{(b)} (applied to $U=\mathbf{k}$) shows that
$B\cap\mathcal{R}_{f}\left(  A\right)  =\mathcal{R}_{f\mid_{A\times B}}\left(
A\right)  $. Hence,%
\begin{align*}
\dim\left(  A/\underbrace{\left(  A\cap\mathcal{L}_{f}\left(  B\right)
\right)  }_{=\mathcal{L}_{f\mid_{A\times B}}\left(  B\right)  }\right)   &
=\dim\left(  A/\mathcal{L}_{f\mid_{A\times B}}\left(  B\right)  \right)
=\dim\left(  B/\underbrace{\mathcal{R}_{f\mid_{A\times B}}\left(  A\right)
}_{=B\cap\mathcal{R}_{f}\left(  A\right)  }\right) \\
&  =\dim\left(  B/\left(  B\cap\mathcal{R}_{f}\left(  A\right)  \right)
\right)  .
\end{align*}
This proves Corollary \ref{cor.quotient.3}.
\end{proof}

\section{More on orthogonal spaces}

We are now in the position to prove less obvious results about orthogonal spaces.

First of all, using Corollary \ref{cor.quotient.3}, we can derive a formula
for the dimension of of an orthogonal space:

\begin{corollary}
\label{cor.quotient.4}Let $V$ and $W$ be two finite-dimensional vector spaces.
Let $f:V\times W\rightarrow\mathbf{k}$ be a bilinear form.

\textbf{(a)} If $A$ is a subspace of $V$, then%
\[
\dim\left(  \mathcal{R}_{f}\left(  A\right)  \right)  =\dim W-\dim
A+\dim\left(  A\cap\mathcal{L}_{f}\left(  W\right)  \right)  .
\]


\textbf{(b)} If $B$ is a subspace of $W$, then%
\[
\dim\left(  \mathcal{L}_{f}\left(  B\right)  \right)  =\dim V-\dim
B+\dim\left(  B\cap\mathcal{R}_{f}\left(  V\right)  \right)  .
\]

\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.quotient.4}.]\textbf{(a)} Let $A$ be a subspace
of $V$. Then, Corollary \ref{cor.quotient.3} (applied to $B=W$) yields
\begin{align*}
\dim\left(  A/\left(  A\cap\mathcal{L}_{f}\left(  W\right)  \right)  \right)
&  =\dim\left(  W/\underbrace{\left(  W\cap\mathcal{R}_{f}\left(  A\right)
\right)  }_{\substack{=\mathcal{R}_{f}\left(  A\right)  \\\text{(since
}\mathcal{R}_{f}\left(  A\right)  \subseteq W\text{)}}}\right) \\
&  =\dim\left(  W/\mathcal{R}_{f}\left(  A\right)  \right)  =\dim
W-\dim\left(  \mathcal{R}_{f}\left(  A\right)  \right)  ,
\end{align*}
so that%
\[
\dim W-\dim\left(  \mathcal{R}_{f}\left(  A\right)  \right)  =\dim\left(
A/\left(  A\cap\mathcal{L}_{f}\left(  W\right)  \right)  \right)  =\dim
A-\dim\left(  A\cap\mathcal{L}_{f}\left(  W\right)  \right)  .
\]
Solving this equation for $\dim\left(  \mathcal{R}_{f}\left(  A\right)
\right)  $, we obtain%
\begin{align*}
\dim\left(  \mathcal{R}_{f}\left(  A\right)  \right)   &  =\dim W-\left(  \dim
A-\dim\left(  A\cap\mathcal{L}_{f}\left(  W\right)  \right)  \right) \\
&  =\dim W-\dim A+\dim\left(  A\cap\mathcal{L}_{f}\left(  W\right)  \right)  .
\end{align*}
This proves Corollary \ref{cor.quotient.4} \textbf{(a)}.

\begin{vershort}
\textbf{(b)} The proof of Corollary \ref{cor.quotient.4} \textbf{(b)} is
analogous to that of Corollary \ref{cor.quotient.4} \textbf{(a)} (but now we
need to apply Corollary \ref{cor.quotient.3} to $A=V$).
\end{vershort}

\begin{verlong}
\textbf{(b)} Let $B$ be a subspace of $W$. Then, Corollary
\ref{cor.quotient.3} (applied to $A=V$) yields
\[
\dim\left(  V/\left(  V\cap\mathcal{L}_{f}\left(  B\right)  \right)  \right)
=\dim\left(  B/\left(  B\cap\mathcal{R}_{f}\left(  V\right)  \right)  \right)
=\dim B-\dim\left(  B\cap\mathcal{R}_{f}\left(  V\right)  \right)  ,
\]
so that%
\begin{align*}
\dim B-\dim\left(  B\cap\mathcal{R}_{f}\left(  V\right)  \right)   &
=\dim\left(  V/\underbrace{\left(  V\cap\mathcal{L}_{f}\left(  B\right)
\right)  }_{\substack{=\mathcal{L}_{f}\left(  B\right)  \\\text{(since
}\mathcal{L}_{f}\left(  B\right)  \subseteq V\text{)}}}\right)  =\dim\left(
V/\mathcal{L}_{f}\left(  B\right)  \right) \\
&  =\dim V-\dim\left(  \mathcal{L}_{f}\left(  B\right)  \right)  .
\end{align*}
Solving this equation for $\dim\left(  \mathcal{L}_{f}\left(  B\right)
\right)  $, we obtain%
\begin{align*}
\dim\left(  \mathcal{L}_{f}\left(  B\right)  \right)   &  =\dim V-\left(  \dim
B-\dim\left(  B\cap\mathcal{R}_{f}\left(  V\right)  \right)  \right) \\
&  =\dim V-\dim B+\dim\left(  B\cap\mathcal{R}_{f}\left(  V\right)  \right)  .
\end{align*}
This proves Corollary \ref{cor.quotient.4} \textbf{(b)}.
\end{verlong}
\end{proof}

Let us now state another fact from linear algebra:

\begin{proposition}
\label{prop.linalg.modlaw}Let $V$ be a vector space. Let $A$ and $B$ be two
subspaces of $V$.

\textbf{(a)} We have $\left(  A+B\right)  /B\cong A/\left(  A\cap B\right)  $
as vector spaces.

\textbf{(b)} Assume that $V$ is finite-dimensional. Then, $\dim\left(
A+B\right)  =\dim A+\dim B-\dim\left(  A\cap B\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.linalg.modlaw}.]\textbf{(a)} Let $\iota$ be
the canonical inclusion $A\rightarrow A+B$. (This is well-defined, since
$A\subseteq A+B$.) Let $\pi_{A,A\cap B}$ be the canonical projection
$A\rightarrow A/\left(  A\cap B\right)  $. Let $\pi_{A+B,B}$ be the canonical
projection $A+B\rightarrow\left(  A+B\right)  /B$.

The map $\pi_{A+B,B}\circ\iota:A\rightarrow\left(  A+B\right)  /B$ is linear
(since it is the composition of the two linear maps $\pi_{A+B,B}$ and $\iota
$). It furthermore satisfies $\left(  \pi_{A+B,B}\circ\iota\right)  \left(
A\cap B\right)  =0$\ \ \ \ \footnote{\textit{Proof.} Let $a\in A\cap B$. Thus,
$a\in A\cap B\subseteq A$; therefore, $\iota\left(  a\right)  $ is
well-defined. Now, $\iota$ is an inclusion map; therefore, $\iota\left(
a\right)  =a$. But $\pi_{A+B,B}$ is the canonical projection $A+B\rightarrow
\left(  A+B\right)  /B$. Hence, $\pi_{A+B,B}\left(  a\right)  =\left[
a\right]  _{B}=0$ (since $a\in A\cap B\subseteq B$). Therefore, $\left(
\pi_{A+B,B}\circ\iota\right)  \left(  a\right)  =\pi_{A+B,B}\left(
\underbrace{\iota\left(  a\right)  }_{=a}\right)  =\pi_{A+B,B}\left(
a\right)  =0$.
\par
Now, let us forget that we fixed $a$. We thus have shown that $\left(
\pi_{A+B,B}\circ\iota\right)  \left(  a\right)  =0$ for every $a\in A\cap B$.
In other words, $\left(  \pi_{A+B,B}\circ\iota\right)  \left(  A\cap B\right)
=0$, qed.}. Proposition \ref{prop.quotient.uniprop} (applied to $A$, $A\cap
B$, $\pi_{A,A\cap B}$, $\left(  A+B\right)  /B$ and $\pi_{A+B,B}\circ\iota$
instead of $V$, $A$, $\pi_{V,A}$, $W$ and $g$) thus shows that there exists a
unique linear map $g^{\prime}:A/\left(  A\cap B\right)  \rightarrow\left(
A+B\right)  /B$ such that $\pi_{A+B,B}\circ\iota=g^{\prime}\circ\pi_{A,A\cap
B}$. Consider this $g^{\prime}$.

We have%
\begin{equation}
g^{\prime}\left(  \left[  a\right]  _{A\cap B}\right)  =\left[  a\right]
_{B}\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A
\label{pf.prop.linalg.modlaw.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.linalg.modlaw.1}):} Let $a\in A$.
Recall that $\pi_{A,A\cap B}$ is the canonical projection $A\rightarrow
A/\left(  A\cap B\right)  $. Hence, $\pi_{A,A\cap B}\left(  a\right)  =\left[
a\right]  _{A\cap B}$. Thus, $\left[  a\right]  _{A\cap B}=\pi_{A,A\cap
B}\left(  a\right)  $. Applying the map $g^{\prime}$ to both sides of this
equality, we obtain
\begin{align*}
g^{\prime}\left(  \left[  a\right]  _{A\cap B}\right)   &  =g^{\prime}\left(
\pi_{A,A\cap B}\left(  a\right)  \right)  =\underbrace{\left(  g^{\prime}%
\circ\pi_{A,A\cap B}\right)  }_{=\pi_{A+B,B}\circ\iota}\left(  a\right)
=\left(  \pi_{A+B,B}\circ\iota\right)  \left(  a\right) \\
&  =\pi_{A+B,B}\left(  \underbrace{\iota\left(  a\right)  }%
_{\substack{=a\\\text{(since }\iota\text{ is an inclusion map)}}}\right)
=\pi_{A+B,B}\left(  a\right)  =\left[  a\right]  _{B}%
\end{align*}
(since $\pi_{A+B,B}$ is the canonical projection $A+B\rightarrow\left(
A+B\right)  /B$). This proves (\ref{pf.prop.linalg.modlaw.1}).}. Now, the map
$g^{\prime}$ is surjective\footnote{\textit{Proof.} Let $x\in\left(
A+B\right)  /B$. We shall show that $x\in g^{\prime}\left(  A/\left(  A\cap
B\right)  \right)  $.
\par
We have $x\in\left(  A+B\right)  /B$. Thus, $x$ can be written in the form
$\left[  u\right]  _{B}$ for some $u\in A+B$. Consider this $u$. Thus,
$x=\left[  u\right]  _{B}$.
\par
We have $u\in A+B$. Thus, $u$ can be written in the form $a+b$ for some $a\in
A$ and $b\in B$. Consider these $a$ and $b$. Thus, $u=a+b$. Now,
\begin{align*}
x  &  =\left[  \underbrace{u}_{=a+b}\right]  _{B}=\left[  a+b\right]
_{B}=\left[  a\right]  _{B}+\underbrace{\left[  b\right]  _{B}}%
_{\substack{=0\\\text{(since }b\in B\text{)}}}=\left[  a\right]
_{B}=g^{\prime}\left(  \underbrace{\left[  a\right]  _{A\cap B}}_{\in
A/\left(  A\cap B\right)  }\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.linalg.modlaw.1})}\right) \\
&  \in g^{\prime}\left(  A/\left(  A\cap B\right)  \right)  .
\end{align*}
\par
Now, let us forget that we fixed $x$. We thus have shown that every
$x\in\left(  A+B\right)  /B$ satisfies $x\in g^{\prime}\left(  A/\left(  A\cap
B\right)  \right)  $. In other words, $\left(  A+B\right)  /B\subseteq
g^{\prime}\left(  A/\left(  A\cap B\right)  \right)  $. In other words, the
map $g^{\prime}$ is surjective, qed.} and injective\footnote{\textit{Proof.}
Let $x\in\operatorname*{Ker}\left(  g^{\prime}\right)  $. We shall show that
$x=0$.
\par
We have $x\in\operatorname*{Ker}\left(  g^{\prime}\right)  \subseteq A/\left(
A\cap B\right)  $. Thus, $x$ can be written in the form $\left[  a\right]
_{A\cap B}$ for some $a\in A$. Consider this $a$. Thus, $x=\left[  a\right]
_{A\cap B}$.
\par
We have $x\in\operatorname*{Ker}\left(  g^{\prime}\right)  $ and thus
$g^{\prime}\left(  x\right)  =0$. Hence, $0=g^{\prime}\left(  \underbrace{x}%
_{=\left[  a\right]  _{A\cap B}}\right)  =g^{\prime}\left(  \left[  a\right]
_{A\cap B}\right)  =\left[  a\right]  _{B}$ (by (\ref{pf.prop.linalg.modlaw.1}%
)). Hence, $\left[  a\right]  _{B}=0$. In other words, $a\in B$. Combining
this with $a\in A$, we obtain $a\in A\cap B$. Hence, $\left[  a\right]
_{A\cap B}=0$. Thus, $x=\left[  a\right]  _{A\cap B}=0$.
\par
Let us now forget that we fixed $x$. We thus have proven that every
$x\in\operatorname*{Ker}\left(  g^{\prime}\right)  $ satisfies $x=0$. In other
words, $\operatorname*{Ker}\left(  g^{\prime}\right)  =0$. This shows that
$g^{\prime}$ is injective (since $g^{\prime}$ is a linear map), qed.}. Hence,
the map $g^{\prime}$ is bijective, and thus a vector space isomorphism (since
$g^{\prime}$ is linear). Thus, there exists a vector space isomorphism
$A/\left(  A\cap B\right)  \rightarrow\left(  A+B\right)  /B$ (namely,
$g^{\prime}$). In other words, $\left(  A+B\right)  /B\cong A/\left(  A\cap
B\right)  $ as vector spaces. Proposition \ref{prop.linalg.modlaw}
\textbf{(a)} is now proven.

\textbf{(b)} Proposition \ref{prop.linalg.modlaw} \textbf{(a)} yields $\left(
A+B\right)  /B\cong A/\left(  A\cap B\right)  $. Thus we conclude $\dim\left(
\left(  A+B\right)  /B\right)  =\dim\left(  A/\left(  A\cap B\right)  \right)
$ (since isomorphic vector spaces have equal dimensions). Thus, $\dim\left(
\left(  A+B\right)  /B\right)  =\dim\left(  A/\left(  A\cap B\right)  \right)
=\dim A-\dim\left(  A\cap B\right)  $. Comparing this with $\dim\left(
\left(  A+B\right)  /B\right)  =\dim\left(  A+B\right)  -\dim B$, we obtain
\[
\dim\left(  A+B\right)  -\dim B=\dim A-\dim\left(  A\cap B\right)  .
\]
Thus,%
\begin{align*}
\dim\left(  A+B\right)   &  =\dim B+\dim A-\dim\left(  A\cap B\right) \\
&  =\dim A+\dim B-\dim\left(  A\cap B\right)  .
\end{align*}
This proves Proposition \ref{prop.linalg.modlaw} \textbf{(b)}.
\end{proof}

We notice that Proposition \ref{prop.linalg.modlaw} \textbf{(a)} holds even if
$\mathbf{k}$ is not a field but just a commutative ring (and vector spaces are
replaced by $\mathbf{k}$-modules).

We shall now prove the following fact:

\begin{proposition}
\label{prop.LRV.1}Let $V$ and $W$ be two finite-dimensional vector spaces. Let
$f:V\times W\rightarrow\mathbf{k}$ be a bilinear form.

\textbf{(a)} If $A$ is a subspace of $V$, then $\mathcal{L}_{f}\left(
\mathcal{R}_{f}\left(  A\right)  \right)  =A+\mathcal{L}_{f}\left(  W\right)
$.

\textbf{(b)} If $B$ is a subspace of $W$, then $\mathcal{R}_{f}\left(
\mathcal{L}_{f}\left(  B\right)  \right)  =B+\mathcal{R}_{f}\left(  V\right)
$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.LRV.1}.]The map $f:V\times W\rightarrow
\mathbf{k}$ is a bilinear form. In other words, the map $f:V\times
W\rightarrow\mathbf{k}$ is a bilinear map (according to the definition of a
\textquotedblleft bilinear form\textquotedblright).

\textbf{(a)} Let $A$ be a subspace of $V$.

\begin{vershort}
Straightforward applications of Proposition \ref{prop.LR.basics} \textbf{(a)}
show that both $\mathcal{L}_{f}\left(  W\right)  $ and $\mathcal{L}_{f}\left(
\mathcal{R}_{f}\left(  A\right)  \right)  $ are subspaces of $V$, and thus finite-dimensional.
\end{vershort}

\begin{verlong}
We know that $\mathcal{L}_{f}\left(  W\right)  $ is a subspace of $V$ (by
Proposition \ref{prop.LR.basics} \textbf{(a)}, applied to $U=\mathbf{k}$ and
$B=W$). Also, $\mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(  A\right)
\right)  $ is a subspace of $V$ (by Proposition \ref{prop.LR.basics}
\textbf{(a)}, applied to $U=\mathbf{k}$ and $B=W$). Notice that the vector
spaces $A+\mathcal{L}_{f}\left(  W\right)  $ and $\mathcal{L}_{f}\left(
\mathcal{R}_{f}\left(  A\right)  \right)  $ are subspaces of the
finite-dimensional vector space $V$, and thus themselves finite-dimensional.
\end{verlong}

Corollary \ref{cor.LR.revert.double} \textbf{(b)} yields $A\subseteq
\mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(  A\right)  \right)  $. Also,
$\mathcal{R}_{f}\left(  A\right)  \subseteq W$, and thus $\mathcal{L}%
_{f}\left(  \mathcal{R}_{f}\left(  A\right)  \right)  \supseteq\mathcal{L}%
_{f}\left(  W\right)  $ (by Proposition \ref{prop.LR.contravari} \textbf{(a)},
applied to $U=\mathbf{k}$, $B=\mathcal{R}_{f}\left(  A\right)  $ and
$B^{\prime}=W$). In other words, $\mathcal{L}_{f}\left(  W\right)
\subseteq\mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(  A\right)  \right)  $.
Combined with $A\subseteq\mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(
A\right)  \right)  $, this yields $A+\mathcal{L}_{f}\left(  W\right)
\subseteq\mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(  A\right)  \right)
$\ \ \ \ \footnote{\textit{Proof.} Recall the following simple fact from
linear algebra: If $X$, $Y$ and $Z$ are three subspaces of $V$ satisfying
$X\subseteq Z$ and $Y\subseteq Z$, then $X+Y\subseteq Z$. This fact (applied
to $X=A$, $Y=\mathcal{L}_{f}\left(  W\right)  $ and $Z=\mathcal{L}_{f}\left(
\mathcal{R}_{f}\left(  A\right)  \right)  $) shows that $A+\mathcal{L}%
_{f}\left(  W\right)  \subseteq\mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(
A\right)  \right)  $ (since $A\subseteq\mathcal{L}_{f}\left(  \mathcal{R}%
_{f}\left(  A\right)  \right)  $ and $\mathcal{L}_{f}\left(  W\right)
\subseteq\mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(  A\right)  \right)  $).
Qed.}.

We have $A\subseteq V$. Thus, Proposition \ref{prop.LR.contravari}
\textbf{(b)} (applied to $A^{\prime}=V$ and $U=\mathbf{k}$) yields
$\mathcal{R}_{f}\left(  A\right)  \supseteq\mathcal{R}_{f}\left(  V\right)  $.
Therefore, $\mathcal{R}_{f}\left(  A\right)  \cap\mathcal{R}_{f}\left(
V\right)  =\mathcal{R}_{f}\left(  V\right)  $.

Theorem \ref{thm.quotient.2} \textbf{(c)} shows that $\dim\left(
V/\mathcal{L}_{f}\left(  W\right)  \right)  =\dim\left(  W/\mathcal{R}%
_{f}\left(  V\right)  \right)  =\dim W-\dim\left(  \mathcal{R}_{f}\left(
V\right)  \right)  $. Hence,%
\begin{equation}
\dim W-\dim\left(  \mathcal{R}_{f}\left(  V\right)  \right)  =\dim\left(
V/\mathcal{L}_{f}\left(  W\right)  \right)  =\dim V-\dim\left(  \mathcal{L}%
_{f}\left(  W\right)  \right)  . \label{pf.prop.LRV.1.a.1}%
\end{equation}


Proposition \ref{prop.linalg.modlaw} (applied to $B=\mathcal{L}_{f}\left(
W\right)  $) shows that%
\begin{equation}
\dim\left(  A+\mathcal{L}_{f}\left(  W\right)  \right)  =\dim A+\dim\left(
\mathcal{L}_{f}\left(  W\right)  \right)  -\dim\left(  A\cap\mathcal{L}%
_{f}\left(  W\right)  \right)  . \label{pf.prop.LRV.1.a.2}%
\end{equation}


But $\mathcal{R}_{f}\left(  A\right)  $ is a subspace of $W$ (by Proposition
\ref{prop.LR.basics} \textbf{(a)}, applied to $U=\mathbf{k}$). Hence,
Corollary \ref{cor.quotient.4} \textbf{(b)} (applied to $B=\mathcal{R}%
_{f}\left(  A\right)  $) yields%
\begin{align}
&  \dim\left(  \mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(  A\right)
\right)  \right) \nonumber\\
&  =\dim V-\underbrace{\dim\left(  \mathcal{R}_{f}\left(  A\right)  \right)
}_{\substack{=\dim W-\dim A+\dim\left(  A\cap\mathcal{L}_{f}\left(  W\right)
\right)  \\\text{(by Corollary \ref{cor.quotient.4} \textbf{(a)})}}%
}+\dim\left(  \underbrace{\mathcal{R}_{f}\left(  A\right)  \cap\mathcal{R}%
_{f}\left(  V\right)  }_{=\mathcal{R}_{f}\left(  V\right)  }\right)
\nonumber\\
&  =\dim V-\left(  \dim W-\dim A+\dim\left(  A\cap\mathcal{L}_{f}\left(
W\right)  \right)  \right)  +\dim\left(  \mathcal{R}_{f}\left(  V\right)
\right) \nonumber\\
&  =\dim V-\dim W+\dim A-\dim\left(  A\cap\mathcal{L}_{f}\left(  W\right)
\right)  +\dim\left(  \mathcal{R}_{f}\left(  V\right)  \right) \nonumber\\
&  =\dim V-\underbrace{\left(  \dim W-\dim\left(  \mathcal{R}_{f}\left(
V\right)  \right)  \right)  }_{\substack{=\dim V-\dim\left(  \mathcal{L}%
_{f}\left(  W\right)  \right)  \\\text{(by (\ref{pf.prop.LRV.1.a.1}))}}}+\dim
A-\dim\left(  A\cap\mathcal{L}_{f}\left(  W\right)  \right) \nonumber\\
&  =\dim V-\left(  \dim V-\dim\left(  \mathcal{L}_{f}\left(  W\right)
\right)  \right)  +\dim A-\dim\left(  A\cap\mathcal{L}_{f}\left(  W\right)
\right) \nonumber\\
&  =\dim A+\dim\left(  \mathcal{L}_{f}\left(  W\right)  \right)  -\dim\left(
A\cap\mathcal{L}_{f}\left(  W\right)  \right) \nonumber\\
&  =\dim\left(  A+\mathcal{L}_{f}\left(  W\right)  \right)
\label{pf.prop.LRV.1.a.4}%
\end{align}
(by (\ref{pf.prop.LRV.1.a.2})).

Thus, we know that $\mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(  A\right)
\right)  $ is a finite-dimensional vector space; we know that $A+\mathcal{L}%
_{f}\left(  W\right)  $ is a subspace of $\mathcal{L}_{f}\left(
\mathcal{R}_{f}\left(  A\right)  \right)  $ (since $A+\mathcal{L}_{f}\left(
W\right)  \subseteq\mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(  A\right)
\right)  $); we furthermore know that $\dim\left(  A+\mathcal{L}_{f}\left(
W\right)  \right)  =\dim\left(  \mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(
A\right)  \right)  \right)  $ (by (\ref{pf.prop.LRV.1.a.4})). Thus,
Proposition \ref{prop.findim.include} (applied to $\mathcal{L}_{f}\left(
\mathcal{R}_{f}\left(  A\right)  \right)  $ and $A+\mathcal{L}_{f}\left(
W\right)  $ instead of $B$ and $A$) shows that $A+\mathcal{L}_{f}\left(
W\right)  =\mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(  A\right)  \right)  $.
This proves Proposition \ref{prop.LRV.1} \textbf{(a)}.

\textbf{(b)} Let $B$ be a subspace of $V$. Proposition
\ref{prop.bilform.op-and-restrict} \textbf{(a)} shows that the map
$f^{\operatorname*{op}}:W\times V\rightarrow\mathbf{k}$ is a bilinear form.
Hence, we can apply Proposition \ref{prop.LRV.1} \textbf{(a)} to $W$, $V$,
$f^{\operatorname*{op}}$ and $B$ instead of $V$, $W$, $f$ and $A$. As a
result, we obtain $\mathcal{L}_{f^{\operatorname*{op}}}\left(  \mathcal{R}%
_{f^{\operatorname*{op}}}\left(  B\right)  \right)  =B+\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  V\right)  $.

\begin{vershort}
But using straightforward (by now) applications of Proposition
\ref{prop.LR.LR}, we can rewrite this as $\mathcal{R}_{f}\left(
\mathcal{L}_{f}\left(  B\right)  \right)  =B+\mathcal{R}_{f}\left(  V\right)
$. This proves Proposition \ref{prop.LRV.1} \textbf{(b)}.
\end{vershort}

\begin{verlong}
But Proposition \ref{prop.LR.LR} \textbf{(a)} yields $\mathcal{L}_{f}\left(
B\right)  =\mathcal{R}_{f^{\operatorname*{op}}}\left(  B\right)  $. Also,
Proposition \ref{prop.LR.LR} \textbf{(b)} (applied to $A=V$) yields
$\mathcal{R}_{f}\left(  V\right)  =\mathcal{L}_{f^{\operatorname*{op}}}\left(
V\right)  $. Also, Proposition \ref{prop.LR.LR} \textbf{(b)} (applied to
$\mathcal{L}_{f}\left(  B\right)  $ instead of $A$) yields $\mathcal{R}%
_{f}\left(  \mathcal{L}_{f}\left(  B\right)  \right)  =\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  \mathcal{L}_{f}\left(  B\right)  \right)  $.
Thus,
\[
\mathcal{R}_{f}\left(  \mathcal{L}_{f}\left(  B\right)  \right)
=\mathcal{L}_{f^{\operatorname*{op}}}\left(  \underbrace{\mathcal{L}%
_{f}\left(  B\right)  }_{=\mathcal{R}_{f^{\operatorname*{op}}}\left(
B\right)  }\right)  =\mathcal{L}_{f^{\operatorname*{op}}}\left(
\mathcal{R}_{f^{\operatorname*{op}}}\left(  B\right)  \right)
=B+\underbrace{\mathcal{L}_{f^{\operatorname*{op}}}\left(  V\right)
}_{=\mathcal{R}_{f}\left(  V\right)  }=B+\mathcal{R}_{f}\left(  V\right)  .
\]
This proves Proposition \ref{prop.LRV.1} \textbf{(b)}.
\end{verlong}
\end{proof}

\begin{corollary}
\label{cor.LRV.2}Let $V$ and $W$ be two finite-dimensional vector spaces. Let
$f:V\times W\rightarrow\mathbf{k}$ be a bilinear form.

\textbf{(a)} If $A$ is a subspace of $V$ satisfying $\mathcal{L}_{f}\left(
W\right)  \subseteq A$, then $\mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(
A\right)  \right)  =A$.

\textbf{(b)} If $B$ is a subspace of $W$ satisfying $\mathcal{R}_{f}\left(
V\right)  \subseteq B$, then $\mathcal{R}_{f}\left(  \mathcal{L}_{f}\left(
B\right)  \right)  =B$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.LRV.2}.]\textbf{(a)} Let $A$ be a subspace of $V$
satisfying $\mathcal{L}_{f}\left(  W\right)  \subseteq A$. Proposition
\ref{prop.LRV.1} \textbf{(a)} shows that $\mathcal{L}_{f}\left(
\mathcal{R}_{f}\left(  A\right)  \right)  =A+\mathcal{L}_{f}\left(  W\right)
=A$ (since $\mathcal{L}_{f}\left(  W\right)  \subseteq A$). Corollary
\ref{cor.LRV.2} \textbf{(a)} is proven.

\textbf{(b)} Let $B$ be a subspace of $W$ satisfying $\mathcal{R}_{f}\left(
V\right)  \subseteq B$, then $\mathcal{R}_{f}\left(  \mathcal{L}_{f}\left(
B\right)  \right)  =B$. Proposition \ref{prop.LRV.1} \textbf{(b)} shows that
$\mathcal{R}_{f}\left(  \mathcal{L}_{f}\left(  B\right)  \right)
=B+\mathcal{R}_{f}\left(  V\right)  =B$ (since $\mathcal{R}_{f}\left(
V\right)  \subseteq B$). Corollary \ref{cor.LRV.2} \textbf{(b)} is proven.
\end{proof}

Our next claim is an analogue of Proposition \ref{prop.LR.sum} with the roles
of $\cap$ and $+$ interchanged (but also with stricter assumptions, since in
general it would not hold):

\begin{proposition}
\label{prop.LRV.3}Let $V$ and $W$ be two finite-dimensional vector spaces. Let
$f:V\times W\rightarrow\mathbf{k}$ be a bilinear form.

\textbf{(a)} If $B_{1}$ and $B_{2}$ are two subspaces of $W$ satisfying
$\mathcal{R}_{f}\left(  V\right)  \subseteq B_{1}$ and $\mathcal{R}_{f}\left(
V\right)  \subseteq B_{2}$, then $\mathcal{L}_{f}\left(  B_{1}\cap
B_{2}\right)  =\mathcal{L}_{f}\left(  B_{1}\right)  +\mathcal{L}_{f}\left(
B_{2}\right)  $.

\textbf{(b)} If $A_{1}$ and $A_{2}$ are two subspaces of $V$ satisfying
$\mathcal{L}_{f}\left(  W\right)  \subseteq A_{1}$ and $\mathcal{L}_{f}\left(
W\right)  \subseteq A_{2}$, then $\mathcal{R}_{f}\left(  A_{1}\cap
A_{2}\right)  =\mathcal{R}_{f}\left(  A_{1}\right)  +\mathcal{R}_{f}\left(
A_{2}\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.LRV.3}.]\textbf{(a)} Let $B_{1}$ and $B_{2}$
be two subspaces of $W$ satisfying $\mathcal{R}_{f}\left(  V\right)  \subseteq
B_{1}$ and $\mathcal{R}_{f}\left(  V\right)  \subseteq B_{2}$.

\begin{vershort}
Straightforward applications of Proposition \ref{prop.LR.basics} \textbf{(a)}
show that $\mathcal{L}_{f}\left(  B_{1}\right)  $ and $\mathcal{L}_{f}\left(
B_{2}\right)  $ are subspaces of $V$.
\end{vershort}

\begin{verlong}
We know that $\mathcal{L}_{f}\left(  B_{1}\right)  $ is a subspace of $V$ (by
Proposition \ref{prop.LR.basics} \textbf{(a)}, applied to $U=\mathbf{k}$ and
$B=B_{1}$). Furthermore, $\mathcal{L}_{f}\left(  B_{2}\right)  $ is a subspace
of $V$ (by Proposition \ref{prop.LR.basics} \textbf{(a)}, applied to
$U=\mathbf{k}$ and $B=B_{2}$). Hence, $\mathcal{L}_{f}\left(  B_{1}\right)
+\mathcal{L}_{f}\left(  B_{2}\right)  $ is a subspace of $V$ (since
$\mathcal{L}_{f}\left(  B_{1}\right)  $ and $\mathcal{L}_{f}\left(
B_{2}\right)  $ are subspaces of $V$).
\end{verlong}

From $B_{1}\subseteq W$, we obtain $\mathcal{L}_{f}\left(  B_{1}\right)
\supseteq\mathcal{L}_{f}\left(  W\right)  $ (by Proposition
\ref{prop.LR.contravari} \textbf{(a)}, applied to $U=\mathbf{k}$, $B=B_{1}$
and $B^{\prime}=W$). In other words, $\mathcal{L}_{f}\left(  W\right)
\subseteq\mathcal{L}_{f}\left(  B_{1}\right)  $.

Corollary \ref{cor.LRV.2} \textbf{(b)} (applied to $B=B_{1}$) yields
$\mathcal{R}_{f}\left(  \mathcal{L}_{f}\left(  B_{1}\right)  \right)  =B_{1}$
(since $\mathcal{R}_{f}\left(  V\right)  \subseteq B_{1}$). Corollary
\ref{cor.LRV.2} \textbf{(b)} (applied to $B=B_{2}$) yields $\mathcal{R}%
_{f}\left(  \mathcal{L}_{f}\left(  B_{2}\right)  \right)  =B_{2}$ (since
$\mathcal{R}_{f}\left(  V\right)  \subseteq B_{2}$).

Now, $\mathcal{L}_{f}\left(  W\right)  \subseteq\mathcal{L}_{f}\left(
B_{1}\right)  \subseteq\mathcal{L}_{f}\left(  B_{1}\right)  +\mathcal{L}%
_{f}\left(  B_{2}\right)  $. Hence, Corollary \ref{cor.LRV.2} \textbf{(a)}
(applied to $A=\mathcal{L}_{f}\left(  B_{1}\right)  +\mathcal{L}_{f}\left(
B_{2}\right)  $) yields%
\[
\mathcal{L}_{f}\left(  \mathcal{R}_{f}\left(  \mathcal{L}_{f}\left(
B_{1}\right)  +\mathcal{L}_{f}\left(  B_{2}\right)  \right)  \right)
=\mathcal{L}_{f}\left(  B_{1}\right)  +\mathcal{L}_{f}\left(  B_{2}\right)  .
\]
Hence,%
\begin{align*}
\mathcal{L}_{f}\left(  B_{1}\right)  +\mathcal{L}_{f}\left(  B_{2}\right)   &
=\mathcal{L}_{f}\left(  \underbrace{\mathcal{R}_{f}\left(  \mathcal{L}%
_{f}\left(  B_{1}\right)  +\mathcal{L}_{f}\left(  B_{2}\right)  \right)
}_{\substack{=\mathcal{R}_{f}\left(  \mathcal{L}_{f}\left(  B_{1}\right)
\right)  \cap\mathcal{R}_{f}\left(  \mathcal{L}_{f}\left(  B_{2}\right)
\right)  \\\text{(by Proposition \ref{prop.LR.sum} \textbf{(b)}, applied
to}\\A_{1}=\mathcal{L}_{f}\left(  B_{1}\right)  \text{ and }A_{2}%
=\mathcal{L}_{f}\left(  B_{2}\right)  \text{)}}}\right) \\
&  =\mathcal{L}_{f}\left(  \underbrace{\mathcal{R}_{f}\left(  \mathcal{L}%
_{f}\left(  B_{1}\right)  \right)  }_{=B_{1}}\cap\underbrace{\mathcal{R}%
_{f}\left(  \mathcal{L}_{f}\left(  B_{2}\right)  \right)  }_{=B_{2}}\right)
=\mathcal{L}_{f}\left(  B_{1}\cap B_{2}\right)  .
\end{align*}
This proves Proposition \ref{prop.LRV.3} \textbf{(a)}.

\begin{vershort}
\textbf{(b)} The map $f^{\operatorname*{op}}$ is a bilinear form (by
Proposition \ref{prop.bilform.op-and-restrict} \textbf{(a)}). Hence,
Proposition \ref{prop.LRV.3} \textbf{(b)} can be derived from Proposition
\ref{prop.LRV.3} \textbf{(a)} using the same tactic that we used (for example)
to derive Proposition \ref{prop.LR.contravari} \textbf{(b)} from Proposition
\ref{prop.LR.contravari} \textbf{(a)}.
\end{vershort}

\begin{verlong}
\textbf{(b)} Let $A_{1}$ and $A_{2}$ be two subspaces of $V$ satisfying
$\mathcal{L}_{f}\left(  W\right)  \subseteq A_{1}$ and $\mathcal{L}_{f}\left(
W\right)  \subseteq A_{2}$.

The map $f:V\times W\rightarrow\mathbf{k}$ is a bilinear form. In other words,
the map $f:V\times W\rightarrow\mathbf{k}$ is a bilinear map (according to the
definition of a \textquotedblleft bilinear form\textquotedblright).
Proposition \ref{prop.LR.LR} \textbf{(a)} (applied to $B=W$) yields
$\mathcal{L}_{f}\left(  W\right)  =\mathcal{R}_{f^{\operatorname*{op}}}\left(
W\right)  $. Hence, $\mathcal{R}_{f^{\operatorname*{op}}}\left(  W\right)
=\mathcal{L}_{f}\left(  W\right)  \subseteq A_{1}$ and $\mathcal{R}%
_{f^{\operatorname*{op}}}\left(  W\right)  =\mathcal{L}_{f}\left(  W\right)
\subseteq A_{2}$.

But the map $f^{\operatorname*{op}}:W\times V\rightarrow\mathbf{k}$ is a
bilinear form (by Proposition \ref{prop.bilform.op-and-restrict}
\textbf{(a)}). Hence, Proposition \ref{prop.LRV.3} \textbf{(a)} (applied to
$W$, $V$, $f^{\operatorname*{op}}$, $A_{1}$ and $A_{2}$ instead of $V$, $W$,
$f$, $B_{1}$ and $B_{2}$) shows that $\mathcal{L}_{f^{\operatorname*{op}}%
}\left(  A_{1}\cap A_{2}\right)  =\mathcal{L}_{f^{\operatorname*{op}}}\left(
A_{1}\right)  +\mathcal{L}_{f^{\operatorname*{op}}}\left(  A_{2}\right)  $.

But Proposition \ref{prop.LR.LR} \textbf{(b)} (applied to $A=A_{1}$) shows
that $\mathcal{R}_{f}\left(  A_{1}\right)  =\mathcal{L}_{f^{\operatorname*{op}%
}}\left(  A_{1}\right)  $. Also, Proposition \ref{prop.LR.LR} \textbf{(b)}
(applied to $A=A_{2}$) shows that $\mathcal{R}_{f}\left(  A_{2}\right)
=\mathcal{L}_{f^{\operatorname*{op}}}\left(  A_{2}\right)  $. Also,
Proposition \ref{prop.LR.LR} \textbf{(b)} (applied to $A=A_{1}\cap A_{2}$)
shows that
\begin{align*}
\mathcal{R}_{f}\left(  A_{1}\cap A_{2}\right)   &  =\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  A_{1}\cap A_{2}\right)
=\underbrace{\mathcal{L}_{f^{\operatorname*{op}}}\left(  A_{1}\right)
}_{=\mathcal{R}_{f}\left(  A_{1}\right)  }+\underbrace{\mathcal{L}%
_{f^{\operatorname*{op}}}\left(  A_{2}\right)  }_{=\mathcal{R}_{f}\left(
A_{2}\right)  }\\
&  =\mathcal{R}_{f}\left(  A_{1}\right)  +\mathcal{R}_{f}\left(  A_{2}\right)
.
\end{align*}
This proves Proposition \ref{prop.LRV.3} \textbf{(b)}.
\end{verlong}
\end{proof}


\end{document}