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\subsection{Properties of the dual map, double
dual}

Let $V, W$ be vector spaces over $F$, $\alpha \in
L(V, W)$.
\[ \mathcal{E} = (e_1, \dots, e_n) \]
basis of $V$
\[ \mathcal{F} = (f_1, \dots, f_n) \]
another basis of $V$. Let
\[ P = [\id]_{\mathcal{F}, \mathcal{E}} \]
(change of basis matrix from $\mathcal{F}$ to
$\mathcal{E}$)
\[ \mathcal{E}^* = (\eps_1, \dots, \eps_n) \]
\[ \mathcal{F}^* = (\eta_1, \dots, \eta_n) \]

\begin{flashcard}[change-of-basis-dual]
\begin{lemma*}
    Let $P$ be the change of basis matrix from
    $\mathcal{F}$ to $\mathcal{E}$. Then the change
    of basis matrix from
    $\mathcal{F}^*$to $\mathcal{E}^*$ is:
    \[ \cloze{(P^{-1})^\top} \]
\end{lemma*}

\begin{proof}
    \[ \cloze{[\id]_{\mathcal{F}^*, \mathcal{E}^*} =
    [\id]_{\mathcal{E}, \mathcal{F}}^\top =
    ([\id]^{-1}_{\mathcal{F}, \mathcal{E}})^\top =
    (P^{-1})^\top} \]
\end{proof}
\end{flashcard}

\subsubsection*{Properties of the dual map}

\begin{flashcard}[dual-map-properties]
\begin{lemma*}
    Let $V, W$ be vector spaces over $F$. Let
    $\alpha \in \mathcal{L}(V, W)$ and $\alpha^*
    \in \mathcal{L}(W^*, V^*)$ be the dual map.
    Then:
    \begin{enumerate}[(i)]
        \item \cloze{$N(\alpha^*) = (\Image \alpha)^0$
            (so $\alpha^*$ injective $\iff$
            $\alpha$ surjective)}
        \item \cloze{$\Image \alpha^* \le (N(\alpha))^0$
            with \emph{equality} if $V, W$ are
            finite dimensional (hence in this
            case, $\alpha^*$ surjective $\iff$
            $\alpha$ injective).}
    \end{enumerate}
\end{lemma*}

\fcscrap{
\noindent
Dual method: there are many problems
(controllability) where the understanding of
$\alpha^*$ is simpler than the understanding of
$\alpha$.
}

\begin{proof}
    \begin{enumerate}[(i)]
        \item \cloze{Let $\eps \in W^*$. Then:
            \begin{align*}
                \eps \in N(\alpha^*)
                &\iff \alpha^*(\eps) = 0 \\
                &\iff \alpha^*(\eps) =
                \eps(\alpha) = 0 \\
                &\iff ~\forall x \in V, ~
                \eps(\alpha)(x) = \eps(\alpha(x))
                = 0 \\
                &\iff \eps \in (\Image \alpha)
            \end{align*}}
        \item \cloze{Let us first show that:
            \[ \Image(\alpha^*) \le (N(\alpha))^0 \]
            Indeed, let $\eps \in
            \Image(\alpha^*)$
            \[ \implies \eps = \alpha^*(\varphi),
            ~\varphi \in W^* \]
            \[ \implies ~\forall u \in N(\alpha)
            \mid \eps(u) = \alpha^*(\varphi)(u) =
            \varphi \circ \alpha(u)
            \varphi(\alpha(u)) = 0 \]
            \[ \implies \eps \in (N(\alpha))^0 \]}
            \fcscrap{In finite dimension, we can compute
            the dimensions of $\Image(\alpha^*)$
            and $(N(\alpha))^0$. 
            \[ \dim(\Image(\alpha^*)) =
            r(\alpha^*) \]
            \[ r(\alpha^*) =
            r([\alpha^*]_{\mathcal{C}^*,
            \mathcal{B}^*}) =
            r([\alpha]_{\mathcal{B},
            \mathcal{C}}^\top) =
            r([\alpha]_{\mathcal{B}, \mathcal{C}})
            = r(\alpha) \]
            \[ \implies r(\alpha) = r(\alpha^*)
            \]}
    \end{enumerate}
    \fcscrap{
    so
    \begin{align*}
        \dim (\Image \alpha^*)
        &= r(\alpha^*) \\
        &+ r(\alpha) \\
        &= \dim V - \dim N(\alpha) \\
        &= \dim [(N(\alpha))^0]
    \end{align*}
    so $\Image(\alpha^*) \le (N(\alpha))^0$ and
    $\dim(\Image(\alpha^*)) = \dim[(N(\alpha))^0]$
    so $\Image(\alpha^*) = [N(\alpha)]^0$.}
\end{proof}
\end{flashcard}

\subsubsection*{Double dual}

\begin{itemize}
    \item $V$ vector space over $F$
    \item $V^* = \mathcal{L}(V, f)$ dual of $V$.
        We define the bidual:
        \[ V^{**} = (V^*)^* = \mathcal{L}(V^*, F) \]
\end{itemize}
Very important space in infinite dimension: in
general, there is no obvious connection between
$V$ and $V^*$ (unless Hilbertian structure).
However, there is a large class of function spaces
wuch that $V \simeq V^{**}$.

\begin{example*}
    $p > 2$,
    \[ L^p(\RR) = \left\{f : \RR \to \RR \left| \int_\RR
    |f(x)|^p \dd x < \infty \right. \right\} \]
    Is a reflexive space.
\end{example*}

\noindent
In general, there is a canonical embedding of $V$
into $V^{**}$. Indeed, pick $v \in V$, we define:
\[ \hat{v} : V^* \to F \]
\[ \eps \mapsto \eps(v) \]
linear:
\begin{itemize}
    \item $\eps \in V^*$ implies $\eps(v)  \in F$.
    \item linearity: $\lambda_1, \lambda_2 \in F,
        \eps_1, \eps_2 \in V^*$
        \begin{align*}
            \hat{v} = (\lambda_1 \eps_1 +
            \lambda_2 \eps_2)
            &= (\lambda_1 \eps_1 + \lambda_2
            \eps_2)(v) \\
            &= \lambda_1 \eps_1(v) + \lambda_2
            \eps_2(v) \\
            &= \lambda_1 \hat{v}(\eps_1) +
            \lambda_2 \hat{v}(\eps_2)
        \end{align*}
        so $\hat{v} \in \mathcal{L}(V^*, F)$.
\end{itemize}

\begin{flashcard}[double-dual-injection]
\begin{theorem*}
    If $V$ is a \emph{finite dimensional} vector
    space over $F$, then:
    \cloze{
    \[ \hat{} : V \to V^{**} \]
    \[ v \mapsto \hat{v} \]
    }
    \prompt{\cloze{where $\hat{v} : V^* \to F$,
    $\eps \mapsto \eps(v)$}}
    is an isomorphism.
\end{theorem*}

\fcscrap{
\noindent
(in infinite dimension we can show under canonical
assumptions (Banach space) that: $\hat{}$ is
injective)
}

\begin{proof}
    \begin{itemize}
        \item $V$ finite dimensional. Given $v \in
            V$, $\hat{v} \in V^{**} \in
            \mathcal{L}(V^*, F)$.
        \item $\hat{}$ linear: \prompt{(check)}
            \fcscrap{let $v_1, v_2 \in
            V$, $\lambda_1, \lambda_2 F$, $\eps
            \in V^*$:
            \begin{align*}
                \widehat{(\lambda_1 v_1 + \lambda_2
                v_2)}(\eps)
                &= \eps(\lambda_1 v_1 + \lambda_2
                v_2) \\
                &= \lambda_1 \eps(v_1) + \lambda_2
                \eps(v_2) \\
                &= \lambda_1 \hat{v}_1(\eps) +
                \lambda_2 \hat{v}_2(\eps) \\
                \implies \widehat{(\lambda_1 v_1 +
                \lambda_2 v_2)}
                &= \lambda_1
                \hat{v}_1 + \lambda_2 \hat{v}_2
            \end{align*}}
        \item $\hat{}$ injective: \cloze{indeed, let $e
            \in V \setminus \{0\}$. I extend $(e,
            e_2, \dots, e_n)$ basis of $V$. Let
            $(\eps, \eps_2, \dots, \eps_n)$ the
            dual basis of (of $V^*$), then
            \[ \hat{e}(\eps) = \eps(e) = 1 \]
            \[ \implies \hat{e} \neq \{0\} \]
            \[ \implies N(\hat{}) = \{0\} \]
            so $\hat{}$ is injective.}
        \item \cloze{$\hat{}$ isomorphism. We can compute
            dimensions}\fcscrap{:
            \[ \dim V = \dim V^* = \dim[(V^*)^*] =
            \dim( V^{**}) \]
            As a conclusion: $\hat{} : V \to V^{**}$ is
            injective, $\dim V = \dim V^{**}$, so $\hat{}$
            is surjective, so $\hat{}$ is an
            isomorphism.}
    \end{itemize}
\end{proof}
\end{flashcard}

\begin{flashcard}[Uhat-le-U00]
\begin{lemma*}
    Let $V$ be a finite dimensional vector space
    over $F$, let $U \le V$. Then
    \[ \cloze{\hat{U} = U^{00}} \]
    so after identification of $V$ and $V^{**}$,
    we have
    \[ \cloze{U \simeq U^{00}} \]
\end{lemma*}

\begin{proof}
    Let us show that: \cloze{$U \le U^{00}$.}
    \begin{itemize}
        \item \cloze{Indeed, let $u \in U$:
            \[ \forall \eps \in U^0, \eps(u) = 0 \]
            \[ \implies \forall \eps \in U^0,
            \eps(u) = \hat{u}(\eps) = 0 \]
            \[ \implies \hat{u} \in U^{00} \]
            \[ \implies \hat{U} \subset U^{00} \]}
        \item \cloze{Commute dimensions
            \[ \dim U^{00} = \dim V - \dim U^0 =
            \dim U \]}
    \end{itemize}
\end{proof}
\end{flashcard}

\begin{remark*}
    $T \le V^*$
    \[ T^0 = \{v \in V \mid \theta(v) = 0, \forall
    \theta \in T\} \]
\end{remark*}