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\subsection{Dual spaces and dual maps}

\begin{definition*}
    Let $V$ be a vector space. The we define
    \begin{align*}
        V^*
        &= \text{dual of $V$} \\
        &= L(V, F) \\
        &= \{\alpha : V \to F \text{ linear}\}
    \end{align*}
\end{definition*}

\begin{notation*}
    $\alpha : V \to F$ linear. Then $\alpha$ is a
    linear form.
\end{notation*}

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item \[ \Trace : \mathcal{M}_{n, n}(F) \to F \]
        \[ A = (a_{ij}) \mapsto \sum_{i = 1}^n
        a_{ii} \]
        \[ \implies \Trace \in \mathcal{M}_{n,
        n}^* (F) \]
    \item $f : [0, 1] \to \RR$
        \[ T_f : \mathcal{C}^\infty ([0, 1], \RR) \]
        \[ \varphi \mapsto \int_0^1 f(x)
        \varphi(x) \dd x \]
        then $T_f$ is a linear form on
        $\mathcal{C}^\infty ([0, 1], \RR)$ ($\RR$
        vector space). Quantum mechanics. A
        function defines a linear form.
\end{enumerate}

\begin{flashcard}[dual-basis]
\begin{lemma*}[Dual basis]
    Let $V$ be a vector space over $F$ with a
    finite basis
    \[ \mathcal{B} = \{e_1, \dots, e_n\} \]
    Then there exists a basis for $V^*$ given by
    \[ \mathcal{B}^* = \{\eps_1, \dots, \eps_n\} \]
    with
    \[ \cloze{\eps_j \left( \sum_{i = 1}^n a_i e_i
    \right) = a_j, \quad 1 \le j \le n} \]
    We call $\mathcal{B}^*$ the dual basis of
    $\mathcal{B}$.
\end{lemma*}
\end{flashcard}

\begin{remark*}
    Kronecker symbol
    \[ \delta_{ij} = \begin{cases}
        1 & \text{if $i = j$} \\
        0 & \text{otherwise}
    \end{cases} \]
    \[ \eps_j \left( \sum_{i = 1}^n a_i e_i
    \right) = a_j \iff \eps_j(a_i) = \delta_{ij} \]
\end{remark*}

\begin{proof}
    Let $\{\eps_1, \dots, \eps_n\}$ be defined as
    above.
    \begin{enumerate}[(1)]
        \item Check that it is free: indeed,
            $\sum_{j = 1}^n \lambda_j \eps_j = 0$
            \[ \implies \sum_{j = 1}^n \lambda_j
            \eps_j (e_i) = 0 \]
            \[ \implies \sum_{j = 1}^n \lambda_i =
            0 \quad \forall 1 \le i \le n \]
            \[ \implies \text{family is free} \]
        \item Check that it is generating: Pick
            $\alpha \in V^*$, then $x \in V$:
            \[ \alpha(x) = \alpha \left( \sum_{j =
            1}^n \lambda_j e_j \right) = \sum_{j =
            1}^n \lambda_j a(e_j) \]
            On the other hand, let the linear
            form:
            \[ \sum_{j = 1}^n
            \alpha(e_j) \eps_j \in V^* \]
            Then:
            \begin{align*}
                \sum_{j = 1}^n \alpha(e_j)
                \eps_j(x)
                &= \sum_{j = 1}^n \alpha(e_j)
                \eps_j \left( \sum_{k = 1}^n
                \lambda_k e_k \right) \\
                &= \sum_{j = 1}^n \alpha(e_j)
                \sum_{k = 1}^n \lambda_k
                \eps_k(e_k) \\
                &= \sum_{j = 1}^n \alpha(e_j)
                \lambda_j \\
                &= \alpha(x) \\
                \implies \alpha
                &= \sum_{j = 1}^n \alpha(e_j)
                \eps_j \qedhere
            \end{align*}
    \end{enumerate}
\end{proof}

\begin{corollary*}
    $V$ finite dimensional,
    \[ \implies \dim V^* = \dim V \]
\end{corollary*}

\begin{warning*}
    These results about $V^*$ are not relevant /
    very different when talking about infinite
    dimensional vector spaces instead.
\end{warning*}

\begin{remark*}
    It is sometimes convenient to think of $V^*$
    as the space of \emph{row} vectors of length
    $n$ over $F$, i.e. let $(e_1, \dots, e_n)$ be
    a basis of $V$, $x = \sum_{i = 1}^n x_i e_i
    \in V$, and let $(\eps_1, \dots, \eps_n)$ be a
    basis of $V$ with $\alpha = \sum_{i = 1}^n
    \alpha_i \eps_i \in V^*$. Then
    \begin{align*}
        \alpha(x)
        &= \sum_{i = 1}^n \alpha \eps_i \left(
        \sum_{j = 1}^n x_j e_j \right) \\
        &= \sum_{i = 1}^n a_i \sum_{j = 1}^n x_j
        \eps_i(e_j) \\
        &= \sum_{i = 1}^n a_i x_i \\
        &= (\alpha_1, \dots, \alpha_n)
    \end{align*}
    (scalar product structure)
\end{remark*}

\begin{flashcard}[annihilator-of-subspace]
\begin{definition*}
    If $U \le V$ (vector subspace), we define the
    \emph{annihilator} of $U$ by:
    \[ \cloze{U^0 = \{\alpha \in V^* : \forall u \in U,
    \alpha(u) = 0\}} \]
\end{definition*}
\end{flashcard}

\begin{flashcard}[annihilator-is-subspace-with-dimensional-formula]
\begin{lemma*}
    \begin{enumerate}[(i)]
        \item $U^0 \le V^*$ (vector subspace)
        \item \cloze{If $U \le V$ and $\dim V < \infty$
            then $\dim V = \dim U + \dim U^0$.}
    \end{enumerate}
\end{lemma*}

\begin{proof}
    \begin{enumerate}[(i)]
        \item \cloze{$0 \in U^0$. If $\alpha, \alpha' \in
            U^0$, then, for all $u \in U$,
            \[ (\alpha + \alpha')(u) = \alpha(u) +
            \alpha'(u) = 0 \]
            and for all $\lambda \in F$,
            \[ (\lambda \alpha)(u) = \lambda
            \alpha(u) = 0 \]
            so $U^0 \le V^*$.}
        \item \cloze{Let $U \le V$, $\dim V = n$. Let
            $(e_1, \dots, e_k)$ be a basis of $U$,
            complete it to a basis
            \[ \mathcal{B} = (e_1, \dots, e_k,
            e_{k + 1}, \dots, e_n) \]
            of $V$. Let $(\eps_1, \dots, \eps_n)$
            be the dual basis of $\mathcal{B}$. We
            claim that $U^0 = \langle \eps_{k +
            1}, \dots, \eps_n \rangle$.
            \begin{itemize}
                \item Pick $i > k$, then:
                    \[ \eps_i(e_k) = \delta_{ik} =
                    0 \]
                    so $\eps_i \in U^0$, since $U
                    = \langle e_1, \dots, e_k
                    \rangle$. So
                    \[ \langle \eps_{k + 1},
                    \dots, \eps_n \rangle \le U^0 \]
                \item Let $\alpha \in U^0$, then
                    let $\alpha \in V^*$, with
                    \[ \alpha = \sum_{i = 1}^n
                    \alpha_i \eps_i \]
                    Now for $i \le k$:
                    \[ \alpha \in U^0 \implies
                    \alpha(e_i) = 0 ~\forall 1 \le
                    i \le k \]
                    \[ \implies \sum_{j = 1}^n
                    \alpha_j \eps_j(e_i) = 0 \]
                    \[ \implies \alpha_i = 0 \quad
                    \forall 1 \le i \le k \]
                    \[ \implies \alpha = \sum_{i =
                    1}^n \alpha_i \eps_i = \sum_{i
                    = k + 1}^n \alpha_i \eps_i \]
                    \[ \implies \alpha \in \langle
                    \eps_{k + 1}, \dots, \eps_n
                    \rangle \]
                    \[ \implies U^0 \le \langle
                    \eps_{k + 1}, \dots, \eps_n
                    \rangle \]
            \end{itemize}
            }
    \end{enumerate}
\end{proof}
\end{flashcard}

\begin{flashcard}[dual-map]
\begin{lemma*}
    Let $V, W$ be vector spaces over $F$. Let
    $\alpha \in \mathcal{L}(V, W)$. Then \cloze{the map:
    \[ \alpha^* : W^* \to V^* \]
    \[ \eps \mapsto \eps \circ \alpha \]
    is an element of $\mathcal{L}(W^*, V^*)$.} It
    is called the \emph{dual} map of $\alpha$.
\end{lemma*}
\end{flashcard}

\begin{proof}
    \begin{itemize}
        \item $\eps \circ \alpha : V \to F$ linear
            follows by linearity of $\eps$ and
            $\alpha$, so $\eps \circ \alpha \in
            V^*$.
        \item $\alpha^*$ linear: let $\theta_1,
            \theta_2 \in W^*$, then
            \begin{align*}
                \alpha^*(\theta_1 + \theta_2)
                &= (\theta_1 + \theta_2)(\alpha)
                \\
                &= \theta_1 \circ \alpha +
                \theta_2 \circ \alpha \\
                &= \alpha^*(\theta_1) +
                \alpha^*(\theta_2)
            \end{align*}
            and similarly for all $\lambda \in F$,
            \[ \alpha^*(\lambda \theta) = \lambda
            \alpha^*(\theta) \]
            so $\alpha^*$ is linear, i.e.
            $\alpha^* \in \mathcal{L}(V^*, W^*)$.
    \end{itemize}
\end{proof}

\begin{flashcard}[dual-map-matrix]
\begin{proposition*}[Dual map matrix]
    Let $V, W$ be finite dimensional spaces over
    $F$ with basis respectively $\mathcal{B}$ and
    $\mathcal{C}$. Let $\mathcal{B}^*$,
    $\mathcal{C}^*$ be the dual basis of
    $\mathcal{B}$ and $\mathcal{C}$. Then:
    \[ \cloze{[\alpha^*]_{\mathcal{C}^*, \mathcal{B}^*} =
    [\alpha]_{\mathcal{B}, \mathcal{C}}^\top} \]
\end{proposition*}
\end{flashcard}

\begin{proof}
    $\mathcal{B} = (b_1, \dots, b_n)$,
    $\mathcal{C} = (c_1, \dots, c_m)$,
    $\mathcal{B}^* = (\beta_1, \dots, \beta_n)$,
    $\mathcal{C}^* = (\gamma_1, \dots, \gamma_m)$.
    Say
    \[ [\alpha]_{\mathcal{B}, \mathcal{C}} = A =
    (a_{ij})_{1 \le i \le m, 1 \le j \le n} \]
    Recall: $\alpha^* : W^* \to V^*$. Let us
    compute:
    \begin{align*}
        \alpha^* (\gamma_r)(b_s)
        &= \gamma_r \circ \alpha(b_s) \\
        &= \gamma_r (\alpha(b_s)) \\
        &= \gamma_r \left( \sum_t a_{ts} c_t
        \right) \\
        &= \sum_t a_{ts} \gamma_r (c_t) \\
        &= a_{rs}
    \end{align*}
    Say
    \begin{align*}
        [\alpha^*]_{\mathcal{C}^*, \mathcal{B}^*}
        &= (\alpha^*(\gamma_1), \dots,
        \alpha^*(\gamma_m))
        \begin{pmatrix}
            \beta_1 \\
            \vdots \\
            \beta_n
        \end{pmatrix}
        \\
        &= (m_{ij})_{1 \le i \le n, 1 \le j \le m}
        \\
        \implies \alpha^*(\gamma_r)
        &= \sum_{i = 1}^n m_{ir} \beta_i \\
        \implies \alpha^*(\gamma_r)(b_s)
        &= \sum_{i = 1}^n m_{ir} \beta_i (b_s) \\
        &= m_{sr}
    \end{align*}
    Conclusion $\alpha^*(\gamma_r)(b_s) = a_{rs} =
    m_{sr}$ so $[\alpha^*]_{\mathcal{C}^*,
    \mathcal{B}^*} = [\alpha]_{\mathcal{B},
    \mathcal{C}}^\top$.
\end{proof}