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$\dim_\CC W < \infty$, $\dim_\CC V < \infty$,
$\varphi$ sesquilinear, $V \times W \to \CC$
\begin{itemize}
    \item linear first variable: $\varphi(\lambda
        u, v) = \lambda \varphi(u, v)$
    \item multilinear second variable $\varphi(u,
        \lambda v) = \ol{\lambda} \varphi(u, v)$
\end{itemize}

\begin{definition*}
    $\mathcal{B} = (v_1, \dots, v_m)$ basis of
    $V$, $\mathcal{C} = (w_1, \dots, w_n)$ basis
    of $W$.
    \[ [\varphi]_{\mathcal{B}, \mathcal{C}} =
    (\varphi(v_i, w_j)) \]
    $m \times n$ matrix.
\end{definition*}

\begin{lemma*}
    $\varphi(v, w) = [v]_{\mathcal{B}}^\top
    [\varphi]_{\mathcal{B}, \mathcal{C}}
    \ol{[w]_{\mathcal{B}}}$
\end{lemma*}

\begin{proof}
    Exercise.
\end{proof}

\begin{lemma*}
    $\mathcal{B}$, $\mathcal{B}'$ basis for $V$,
    $P = [\id]_{\mathcal{B}', \mathcal{B}}$,
    $\mathcal{C}, \mathcal{C}'$ basis for $W$, $Q
    = [\id]_{\mathcal{C}', \mathcal{C}}$. Then
    \[ [\varphi]_{\mathcal{B}', \mathcal{C}'} =
    P^\top [\varphi]_{\mathcal{B}, \mathcal{C}}
    \ol{Q} \]
\end{lemma*}

\begin{proof}
    Exercise.
\end{proof}

\subsection{Hermitian Forms / $\CC$, Skew
Symmetric forms / $\RR$}

\subsubsection*{Hermitian form}

$\dim_\CC V < \infty$, $\varphi : V \times V \to
\CC$ sesquilinear ($W = V$).

\begin{definition*}[Hermitian form]
    A sesquilinear form $\varphi : V \times V \to
    \CC$ is called \emph{Hermitian} if\
    \[ \forall (u, v) \in V \times V, \quad
    \varphi(u, v) = \ol{\varphi(v, u)} \]
\end{definition*}

\begin{remark*}
    $\varphi$ Hermitian
    \[ \implies \varphi(u, u) = \ol{\varphi(u, u)} \]
    \[ \implies \forall u \in V, \varphi(u, u) \in
    \RR \]
    Allows us to speak of positive / negative
    (semi) definite Hermitian form.
\end{remark*}

\begin{lemma*}
    A sesquilinear form $\varphi : V \times V \to
    \CC$ is Hermitian if and only if $\forall
    \mathcal{B}$ basis of $V$,
    \[ [\varphi]_{\mathcal{B}} =
    \ol{[\varphi]_{\mathcal{B}}^\top} \]
\end{lemma*}

\begin{proof}
    $A = [\varphi]_{\mathcal{B}} = (a_{ij})_{1 \le
    i, j \le n}$, $a_{ij} = \varphi(e_i, e_j)$.
    Then $a_{ij} = \varphi(e_i, e_j)$, $a_{ji} =
    \varphi(e_j, e_i) = \ol{\varphi(e_i, e_j)} =
    \ol{a_{ij}}$.
    \[ \implies [\varphi]_{\mathcal{B}}^\top =
    \ol{[\varphi]}_{\mathcal{B}} \]
    Conversely $[\varphi]_{\mathcal{B}} = A$, $A =
    \ol{A^\top}$
    \[ u = \sum_{i = 1}^n \lambda_i e_i \]
    \[ v = \sum_{i = 1}^n \mu_i e_i \]
    $\mathcal{B} = (e_1, \dots, e_n)$
    \begin{align*}
        \varphi(u, v)
        &= \varphi \left( \sum_{i = 1}^n \lambda_i
        e_i, \sum_{i = 1}^n \mu_i e_i \right) \\
        &= \sum_{i, j = 1}^n \lambda_i \ol{\mu_j}
        \varphi(e_i, e_j) \\
        &= \sum_{i, j = 1}^n \lambda_i \ol{\mu_j}
        a_{ij}
    \end{align*}
    Then
    \begin{align*}
        \ol{\varphi(v, u)}
        &= \ol{\varphi \left( \sum_{i = 1}^n \mu_i
        e_i, \sum_{i = 1}^n \lambda_i e_i
        \right)} \\
        &= \ol{\sum_{i = 1}^n \mu_i \ol{\lambda_j}
        \varphi(e_i, e_j)} \\
        &= \sum_{i, j = 1}^n \ol{\mu_i} \lambda_j
        \ol{a_{ij}} \\
        &= \sum_{i, j = 1}^n \lambda_i \ol{\mu_j}
        \ol{a_{ji}} \\
        &= \sum_{i, j = 1}^n \lambda_i \ol{\mu_j}
        a_{ij} \\
        &= \varphi(u, v) \qedhere
    \end{align*}
\end{proof}

\subsubsection*{Polarization identity}

A Hermitian form $\varphi$ on a complex vector
space $V$ is \emph{entirely determined} by:
$Q : V \to \RR$, $u \mapsto \varphi(u, u)$ via the
formula:
\begin{flashcard}[PolarizFormulaHermitian]
    \begin{align*}
        \varphi(u, v)
        &= \cloze{\frac{1}{4}
        [Q(u + v) - Q(u - v) + iQ(u + iv) -i Q(u - iv)]
        } \\
        &= \text{polarization identity for
        Hermitian forms}
    \end{align*}
\end{flashcard}

\begin{proof}
    Exercise (just check).
\end{proof}

\begin{theorem*}[Sylvester's law of inertia for
    Hermitian forms]
    $\dim_\CC V < \infty$, $\varphi : V \times V
    \to \CC$ a Hermitian form on $V$. Then
    $\exists \mathcal{B} = (v_1, \dots, v_n)$
    basis of $V$:
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/b541e748698f11ed.png}
    \end{center}
    where $P$ and $q$ depend only on $\varphi$.
\end{theorem*}

\begin{proof}
    (Sketch: nearly identical to the real case of
    symmetric forms).
    \begin{itemize}
        \item Existence: $\varphi \equiv 0$, done.
            Assume $\varphi \neq 0$, then the
            polarization identity ensures that
            there exists $e_1 \neq 0$ such that
            \[ \varphi(e_1, e_1) \neq 0 \]
            Rescale:
            \[ v_1 =
            \frac{e_1}{\sqrt{|\varphi(e_1, e_1)|}} \]
            $\implies \varphi(v_1, v_1) = \pm 1$.
            Then we consider the orthogonal:
            \[ W = \{w \in V \mid \varphi(v_1, w)
            = 0\} \]
            and we check (verbatim like in the
            real case)
            \[ V = \langle v_1 \rangle \oplus W \]
            ($\dim W = n - 1$). Now argue by
            induction on the dimension on $V$ by
            considering $\varphi \mid_W$ which is
            Hermitian on $W \times W$.
        \item Uniqueness of $p$: As in the real
            case,
            \[ p \equiv \text{maximal dimension of
            a subspace on which $\varphi$ is
            definite positive ($\varphi(u, u) \in
            \RR$)} \]
            Similarly for $q$.
    \end{itemize}
\end{proof}

\subsubsection*{Skew Symmetric Real Valued Forms}

$F = \RR$, $V$ vector space over $\RR$.

\begin{definition*}[skew symmetric]
    A bilinear form $\varphi : V \times V \to \RR$
    is \emph{skew symmetric} if:
    \[ \varphi(u, v) = -\varphi(v, u) \quad
    \forall (u, v) \in V \times V \]
\end{definition*}

\noindent
This is also often called antisymmetric.

\begin{remark*}
    \begin{enumerate}[(i)]
        \item $\varphi(u, u) = -\varphi(u, u)$ so
            $\varphi(u, u) = 0$. $\forall u \in
            V$.
        \item $\forall \mathcal{B}$ basis of $V$,
            $[\varphi]_{\mathcal{B}} =
            -[\varphi]_{\mathcal{B}}^\top$.
        \item $A \in \mathcal{M}_n(\RR)$,
            \[ A = \half (A + A^\top) + \half (A -
            A^\top) \]
            i.e. decomposition into symmetric and
            antisymmetric / skew symmetric parts.
    \end{enumerate}
\end{remark*}

\begin{theorem*}[Sylvester for skew symmetric form]
    \begin{itemize}
        \item $V$ vector space over $\RR$,
            $\dim_\RR V < \infty$
        \item $\varphi : V \times V \to \RR$ skew
            symmetric bilinear form.
    \end{itemize}
    Then there exists $\mathcal{B}$ basis of $V$,\
    \[ \mathcal{B} = (v_1, w_1, v_2, w_2, \dots,
    v_m, w_m, v_{2m + 1}, v_{2m + 2}, \dots, v_n) \]
    such that
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/a105f8c6699111ed.png}
    \end{center}
\end{theorem*}

\begin{corollary*}
    Skew symmetric matrices have an even rank.
\end{corollary*}

\begin{proof}
    (Sketch). Induction on the dimension of $V$.
    \begin{itemize}
        \item $\varphi \equiv 0$ then done.
        \item $\varphi \neq 0 \implies \exists
            (v_1, v_w) \in V \times V$ such that
            $\varphi(v_1, w_1) \neq 0$.
        \item $v_1 \neq 0$, $w_1 \neq 0$, after
            scaling:
            \[ \varphi(v_1, w_1) = 1 \]
            \[ \implies \varphi(w_1, v_1) = -1 \]
            since skew symmetric.
        \item $(v_1, w_1)$ linearly independent.
            \[ \varphi(v_1, \lambda v_1) = \lambda
            \varphi(v_1, v_1) = 0 \]
            since skew symmetric.
        \item Define $U = \langle v_1, w_1
            \rangle$.
            \[ W = \{v \in V \mid \varphi(v_1, v)
            = \varphi(w_1, v) = 0\} \]
            Exercise: show that $V = U \oplus W$.
        \item Now apply the induction hypothesis
            to $\varphi |_{W \times W}$ skew
            symmetric. \qedhere
    \end{itemize}
\end{proof}

\subsubsection*{Inner Product Spaces}

\begin{itemize}
    \item definite positive bilinear forms:
        \begin{itemize}
            \item[$\to$] Scalar product
            \item[$\to$] Norm (distance)
        \end{itemize}
\end{itemize}
$\implies$ SPECTACULAR generalisation /
application to infinite dimensional spaces:

\subsubsection*{Hilbert Spaces}

$\to$ part II (linear analysis, analysis of
functions).

\begin{flashcard}[InnerProduct]
\begin{definition*}[Inner product]
    Let $V$ be a vector space over $\RR$
    (respectively $\CC$). An \emph{inner product}
    is a \cloze{positive definite symmetric (respectively
    Hermitian) bilinear form $\varphi$ on $V$.}
\end{definition*}
\end{flashcard}

\begin{notation*}
    $\varphi(u, v) = \langle u, v \rangle$.
\end{notation*}

\noindent
If such a bilinear form exists, $V$ is called a
real (respectively complex) inner product space.

\begin{example*}
    \begin{enumerate}[(i)]
        \item $\RR^n$, $x = (x_1, \dots,
            x_n)^\top$, $y = (y_1, \dots,
            y_n)^\top$,
            \[ \langle x, y \rangle = \sum_{i =
            1}^n x_i y_i \]
            $\to$ inner product.
        \item $\CC^n$, $\langle x, y \rangle =
            \sum_{i = 1}^n x_i \ol{y_i}$ $\to$
            inner product.
        \item $V = \mathcal{C}([0, 1], \CC)$
            \[ \langle f, g \rangle = \int_0^1
            f(t) \ol{g(t)} \dd t \]
            ``$L^2$ scalar product''
    \end{enumerate}
\end{example*}

\noindent
One can check that (i), (ii), (iii) are inner
products. 
\[ \langle u, u \rangle = 0 \implies u = 0 \]
$\to$ \emph{definite} positive assumption.
