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% 18/11/2022 11AM

\subsection{Sylvester's law / Sesquilinear forms}

Recall:

\begin{theorem*}
    $\dim_F V < \infty$, $\varphi : V \times V \to
    F$ is a \emph{symmetric} bilinear form
    $\implies$ there exists $\mathcal{B}$ basis of
    $V$ in which $[\varphi]_{\mathcal{B}}$ is
    diagonal.
\end{theorem*}

\begin{remark*}
    We take $F = \RR$ or $\CC$ in this subsection.
\end{remark*}

\begin{corollary*}
    $F = \CC$, $\dim_\CC V < \infty$, $\varphi$
    symmetric bilinear form on $V$. Then there
    exists basis of $V$ such that
    \[ [\varphi]_{\mathcal{B}} = \left(
    \begin{tabular}{c|c}
        $I_r$ & $0$ \\
        \hline
        $0$ & $0$
    \end{tabular} \right),
    \qquad r = \operatorname{rank}(\varphi)
    \]
\end{corollary*}

\begin{proof}
    Pick a basis $\mathcal{E} = (e_1, \dots, e_n)$
    such that
    \[ [\varphi]_{\mathcal{E}} =
    \begin{pmatrix}
        a_1 & 0 & \cdots & 0 \\
        0 & a_2 & \cdots & 0 \\
        \vdots & \vdots & \ddots & \vdots \\
        0 & 0 & \cdots & a_n
    \end{pmatrix}
    \]
    Reorder $e_i$ such that $a_i \neq 0$ for $1
    \le i \le r$, $a_i = 0$ for $i > r$. For $i
    \le r$, I let $\sqrt{a_i}$ be \emph{a} choice
    of complex root of $a_i$, we define:
    \[ v_i = \begin{cases}
        \frac{e_i}{\sqrt{a_i}} & \text{for $1 \le
        i \le r$} \\
        e_i & \text{for $i < r$}
    \end{cases} \]
    $\mathcal{B} = (v_1, \dots, v_r, e_{r + 1},
    \dots, e_n)$, $\mathcal{B}$ basis of $V$
    \[ \implies [\varphi]_{\mathcal{B}} = \left(
    \begin{tabular}{c|c}
        $I_r$ & $0$ \\
        \hline
        $0$ & $0$
    \end{tabular} \right) \qedhere \]
\end{proof}

\begin{corollary*}
    Every \emph{symmetric} matrix of
    $\mathcal{M}_n(\CC)$ is \emph{congruent} to a
    \emph{UNIQUE} matrix of the form:
    \[ \left(
    \begin{tabular}{c|c}
        $I_r$ & $0$ \\
        \hline \\
        $0$ & $0$
    \end{tabular} \right) \]
\end{corollary*}

\noindent
We want to address the same problem with $F =
\RR$. $\to$ we cannot take complex roots this time.

\begin{corollary*}
    $F = \RR$, $\dim_\RR V < \infty$, $\varphi$
    \emph{symmetric} bilinear form on $V$. Then
    there exists $\mathcal{B} = (v_1, \dots, v_n)$
    basis of $V$ such that
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/b47439966a9e11ed.png}
    \end{center}
    $p, q \ge 0$, $p + q = r(\varphi)$.
\end{corollary*}

\begin{proof}
    $\mathcal{E} = (e_1, \dots, e_n)$ basis of $V$
    such that
    \[ [\varphi]_{\mathcal{E}} =
    \begin{pmatrix}
        a_1 & 0 & \cdots & 0 \\
        0 & a_2 & \cdots & 0 \\
        \vdots & \vdots & \ddots & \vdots \\
        0 & 0 & \cdots & a_n
    \end{pmatrix}
    \qquad a_i \in \RR \]
    Reorder $a_i$ so that:
    \begin{itemize}
        \item $a_i > 0$ for $1 \le i \le p$
        \item $a_i < 0$ for $p + 1 \le i \le q$
        \item $a_i = 0$ for $i \ge q + 1$
    \end{itemize}
    We define:
    \[ v_i = \begin{cases}
        \frac{e_i}{\sqrt{a_i}} & 1 \le i \le p \\
        \frac{e_i}{\sqrt{|a_i|}} & p + 1 \le i \le
        q \\
        e_i & i \ge q + 1
    \end{cases} \]
    Then let $\mathcal{B} = (v_1, \dots, v_n)$ and
    then $[\varphi]_{\mathcal{B}}$ has the
    announced form.
\end{proof}

\begin{definition*}[signature]
    We define (under the assumptions above)
    \[ s(\varphi) = p - q \equiv \text{signature
    of $\varphi$} \]
    (we also speak of the signature of the
    associated quadratic form $Q(u) = \varphi(u,
    u)$)
\end{definition*}

\noindent
This definition makes sense:

\begin{theorem*}[Sylvester's law of inertia]
    $F = \RR$, $\dim_\RR V < \infty$, $\varphi$
    symmetric bilinear form on $V$. If $\varphi$
    is represented by:
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/a7fe5f066a9f11ed.png}
    \end{center}
    with $\mathcal{B}, \mathcal{B}'$ bases of $V$.
    Then $p = p'$ and $q = q'$.
\end{theorem*}

\begin{definition*}
    $\varphi$ be a symmetric bilinear form on a
    real valued vector space ($F = \RR$). We say
    that:
    \begin{enumerate}[(i)]
        \item $\varphi$ is \emph{positive definite}
            \[ \iff \forall u \in V \setminus
            \{0\}, \quad \varphi(u, u) > 0 \]
        \item $\varphi$ is \emph{positive semi
            definite}
            \[ \iff \forall u \in V, \quad
            \varphi(u, u) \ge 0 \]
        \item $\varphi$ is \emph{negative definite}
            \[ \iff \forall u \in V \setminus
            \{0\}, \quad \varphi(u, u) < 0 \]
        \item $\varphi$ is negative semi definite
            \[ \iff \forall u \in V, \quad
            \varphi(u, u) \le 0 \]
    \end{enumerate}
\end{definition*}

\begin{example*}
    \[ \left(
    \begin{tabular}{c|c}
        $I_p$ & $0$ \\
        \hline
        $0$ & $0$
    \end{tabular} \right) \]
    \begin{itemize}
        \item positive definite for $p = n$
        \item positive semi definite for $p \le
            n$.
    \end{itemize}
\end{example*}

\begin{hiddenflashcard}[proof-of-sylvesters-law-of-inertia]
Proof of Sylvester's law of inertia? \\
\cloze{
Prove $p$ is the largest possible dimension of a
subspace on which $\varphi$ is positive definite
(call this number $p'$, so want to prove $p = p'$
always).
Let $X$ be the span of the first $p$ columns of
$[\varphi]$ and let $Y$ be the span of the rest.
\\
Then $\varphi$ is positive definite on $X$, so $p
\ge p'$. \\
Suppose $\varphi$ pos def on $X'$. Show $X' \cap Y
= \{0\}$, hence $\dim(X' + Y) = \dim(X') +
\dim(Y)$, so $\dim(X') \le \dim(V) - \dim(Y) =
\dim(X) = p$. So $p' \le p$. \\
So $p = p'$.
}
\end{hiddenflashcard}

\begin{proof}
    (Of Sylvester's law of inertia) \\
    In order to prove that $p$ is independent of
    the choice of the basis, we show that $p$ has
    a geometric interpretation: \\
    Claim: $p$ is the largest dimension of
    subspace on which $\varphi$ is positive
    definite. \\
    Proof: \\
    Say $\mathcal{B} = (v_1, \dots, v_n)$ in
    which:
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/557c868a6aa011ed.png}
    \end{center}
    \begin{enumerate}[(1)]
        \item Let $X = \langle v_1, \dots, v_p
            \rangle$. Then $\varphi$ is positive
            definite on $X$. Indeed, $u = \sum_{i
            = 1}^p \lambda_i v_i$,
            \begin{align*}
                Q(u)
                &= \varphi(u, u) \\
                &= \varphi \left( \sum_{i = 1}^p
                \lambda_i v_i, \sum_{i = 1}^p
                \lambda_i v_i \right) \\
                &= \sum_{i, j = 1}^n \lambda_i
                \lambda_j \varphi(v_i, v_j) \\
                &= \sum_{i = 1}^p \lambda_i^2
                &> 0
                &&\text{for $u \neq 0$}
            \end{align*}
            $\dim X = p$, $\varphi|_{X \times X}$
            is positive definite.
        \item Suppose that $\varphi$ is definite
            positive when restricted to another
            subspace $X'$. Let $X = \langle v_1,
            \dots, v_p \rangle$, $Y = \langle v_{p
            + 1}, \dots, v_n \rangle$,
            $\mathcal{B} = (v_1, \dots, v_n)$.
            Then
            \begin{center}
                \includegraphics[width=0.6\linewidth]
                {images/f11937826aa011ed.png}
            \end{center}
            $\implies$ We know that $\varphi$ is
            negative semi definite on $Y$. So $Y
            \cap X' = \{0\}$. Indeed, if $u \in Y
            \cap X'$ and $u \neq 0$, then $u \in Y$ so
            $\varphi(u, u) \le 0$, but $u \in X'$
            so $\varphi(u, u) > 0$. So $Y \cap X'
            = \{0\}$. So $Y + X' = Y \oplus X'$,
            so $\dim Y + \dim X' \le n$, and $\dim
            Y = n - p$ so $\dim X' \le p$.
    \end{enumerate}
    So now we know that $p$ has a geometric
    interpretation / is unique. Then by
    considering $-\varphi$, we find that $q$ is
    unique too.
\end{proof}

\begin{remark*}
    Similarly, $q$ is the largest dimension of a
    subset on which $\varphi$ is negative
    definite.
\end{remark*}

\begin{definition*}
    $K = \text{kernel of a bilinear form
    $\varphi$} = \{v \in V \mid \forall u \in V,
    \varphi(u, v) = 0\}$.
\end{definition*}

\begin{remark*}
    $\dim K + r(\varphi) = n$
\end{remark*}

\begin{remark*}
    $F = \RR$. One notices that there is a
    subspace $T$ of dimension $n - (p + q) +
    \min\{p, q\}$ such that $\varphi |_T = 0$.
    Indeed: $\mathcal{B} = (v_1, \dots, v_n)$,
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/bb355a506aa111ed.png}
    \end{center}
    $T = \langle \ub{v_1 + v_{p + 1}, \dots, v_q +
    v_{p + q}}_{q}, \ub{v_{p + q + 1}, \dots,
    v_n}_{n - (p + q)} \rangle$ (if $p \ge q$).
    Check $\varphi|_T = 0$ ($\forall (u, v) \in T
    \times T, \varphi(u, v) = 0$). Moreover, one
    can show that this is the largest dimension of
    a subspace $T'$ on which $\varphi|_{T' \times
    T'} = 0$
\end{remark*}

\subsubsection*{Sesquilinear Forms}

\begin{itemize}
    \item $F = \CC$
    \item Standard \emph{inner product} on $\CC^n$
        is $\langle x, y \rangle = \sum_{i = 1}^n
        x_i \ol{y_i}$. In particular,
        \[ \|x\|^2 = \langle x, x \rangle = \ub{\sum_{i
        = 1}^n |x_i|^2}_{\in \RR^+} \]
\end{itemize}

\begin{warning*}
    $\CC^n \times \CC^n \to \CC$
    \[ (x, y) \mapsto \langle x, y \rangle =
    \sum_{i = 1}^n x_i \ol{y_i} \]
    is \emph{not} a bilinear form: $\lambda \in
    \CC$,
    \[ \langle \lambda x, y \rangle = \sum_{i =
    1}^n \lambda x_i \ol{y_i} = \lambda \langle x,
    y \rangle \]
    \[ \langle x, \lambda y \rangle = \sum_{i =
    1}^n x_i \ol{\lambda y_i} = \ol{\lambda}
    \langle x, y \rangle \]
    $\to$ antilinear with respect to the second
    coordinate.
\end{warning*}

\begin{definition*}
    $V, W$ $\CC$ vector spaces. A sesquilinear
    form $\varphi$ is a function $\varphi : V
    \times W \to \CC$ such that:
    \begin{enumerate}[(i)]
        \item $\varphi(\lambda_1 v_1 + \lambda_2
            v_2, w) = \lambda_1 \varphi(v_1, w) +
            \lambda_2 \varphi(v_2, w)$ (linear
            with respect to the first coordinate)
        \item $\varphi(v, \lambda_1 w_1 +
            \lambda_2 w_2) = \ol{\lambda_1}
            \varphi(v, w_1) + \ol{\lambda_2}
            \varphi(v, w_2)$
            (antilinear with respect to the second
            coordinate).
    \end{enumerate}
\end{definition*}
