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\subsection{Bilinear Forms}

Bilinear form: $\varphi : V \times V \to F$.
\begin{itemize}
    \item $\dim_F V < \infty$, $\mathcal{B}$ basis
        of $V$.
    \item $[\varphi]_{\mathcal{B}} =
        [\varphi]_{\mathcal{B}, \mathcal{B}} =
        (\varphi(e_i, e_j))_{1 \le i, j \le n}$.
        $\mathcal{B} = (e_i)_{1 \le i \le n}$.
\end{itemize}

\begin{lemma*}
    $\varphi : V \times V \to F$ bilinear,
    $\mathcal{B}, \mathcal{B}'$ two basis of $V$,
    $P = [\id]_{\mathcal{B}', \mathcal{B}}$ then
    \[ [\varphi]_{\mathcal{B}'} = P^\top
    [\varphi]_{\mathcal{B}} P \]
\end{lemma*}

\begin{proof}
    Special case of the general formula $\to$
    Lecture 10.
\end{proof}

\begin{flashcard}[congruent-matrices]
\begin{definition*}[Congruent matrices]
    \cloze{$A, B \in \mathcal{M}_n(F)$, we say that
    $A$ and $B$ are \emph{congruent} if and only
    if there exists $P$ invertible such that
    \[ A = P^\top B P \]}
\end{definition*}
\end{flashcard}

\begin{remark*}
    This defines an equivalence relation.
\end{remark*}

\begin{definition*}[Symmetric]
    A bilinear form $\varphi$ on $V$ is
    \emph{symmetric} if:
    \[ \varphi(u, v) = \varphi(v, u) \quad \forall
    (u, v) \subset V \times V \]
\end{definition*}

\begin{remark*}
    \begin{itemize}
        \item $A \in \mathcal{M}_n(F)$, we say
            that $A$ is symmetric if and only if
            $A = A^\top$
            \[ \iff A = (a_{ij})_{1 \le i, j \le
            n}, a_{ij} = a_{ji} \]
        \item $\varphi$ symmetric $\iff$
            $[\varphi]_{\mathcal{B}}$ is symmetric
            in \emph{any} basis $\mathcal{B}$ of
            $V$.
    \end{itemize}
\end{remark*}

\begin{remark*}
    To be able to represent $\varphi$ by a
    diagonal matrix, then $\varphi$ must be
    symmetric
    \[ P^\top AP = D \implies D^\top P^\top A^\top
    P \]
    \[ \implies A = A^\top \]
\end{remark*}

\begin{flashcard}[quadratic-form]
\begin{definition*}[Quadratic form]
    \cloze{A map $Q : V \to F$ is a \emph{quadratic form}
    if and only if there exists a bilinear form
    $\varphi : V \times V \to F$ such that
    $\forall u \in V$,
    \[ Q(u) = \varphi(u, u) \]}
\end{definition*}
\end{flashcard}

\begin{remark*}
    $\mathcal{B} = (e_i)_{1 \le i \le n}$, $A =
    [\varphi]_{\mathcal{B}} = (\ub{\varphi(e_i,
    e_j)}_{a_{ij}})_{1 \le i, j \le n}$. Then
    \[ u = \sum_{i = 1}^n x_i e_i, x = (x_1,
    \dots, x_n)^\top \]
    Then
    \begin{align*}
        Q(u)
        &= \varphi(u, u) \\
        &= \varphi \left( \sum_{i = 1}^n x_i e_i,
        \sum_{i = 1}^n x_i e_i \right) \\
        &= \sum_{i = 1}^n \sum_{j = 1}^n x_i x_j
        \ub{\varphi(e_i, e_j)}_{a_{ij}} \\
        &= \sum_{i = 1}^n \sum_{j = 1}^n x_i x_j
        a_{ij} \\
        &= x^\top A x
    \end{align*}
    so
    \[ \boxed{Q(u) = x^\top A x} \]
\end{remark*}

\noindent
Observation:
\begin{align*}
    x^\top A x
    &= \sum_{i, j = 1}^n a_{ij} x_i x_j \\
    &= \sum_{i, j = 1}^n a_{ji} x_i x_j \\
    &= \half \sum_{i, j = 1}^n (a_{ij} + a_{ji})
    x_i x_j \\
    &= \half x^\top (A + A^\top) x
\end{align*}
and $\half(A + A^\top)$ is symmetric.

\begin{proposition*}
    If $Q : V \to F$ is a quadratic form,
    then there exists a \emph{unique symmetric}
    bilinear form $\varphi : V \times V \to F$ such
    that:
    \[ \forall u \in V, Q(u) = \varphi(u, u) \]
\end{proposition*}

\begin{proof}
    Let $\psi$ bilinear form on $V$ such that
    \[ \forall u, Q(u) = \psi(u, u) \]
    Let
    \[ \varphi(u, v) = \half (\psi(u, v) +
    \psi(v, u)) \]
    \begin{itemize}
        \item $\varphi$ symmetric
        \item $\varphi(u, u) = \psi(u, u) = Q(u)$.
    \end{itemize}
    $\to$ existence of $\varphi$ symmetric. Now
    uniqueness. Let $\varphi$ be a symmetric
    bilinear form such that $\varphi(u, u) = Q(u)
    \forall u \in V$. Then
    \begin{align*}
        Q(u + v)
        &= \varphi(u + v, u + v) \\
        &= \varphi(u, u) + \varphi(v, u) +
        \varphi(v, u) + \varphi(v, v) \\
        &= Q(u) + 2 \varphi(u, v) + Q(v)
    \end{align*}
    so
    \[ \boxed{\varphi(u, v) = \half [Q(u + v) -
    Q(u) - Q(v)]} \]
    $\equiv$ POLARIZATION IDENTITY.
\end{proof}

\begin{theorem*}[Diagonalisation of symmetric
    bilinear forms]
    Let $\varphi : V \times V \to F$ be a
    symmetric bilinear form ($\dim_F V < \infty$).
    Then there exists a basis $\mathcal{B}$ of $V$
    such that:
    \[ [\varphi]_{\mathcal{B}} \text{ is diagonal} \]
\end{theorem*}

\begin{hiddenflashcard}[key-idea-diagonalisation-of-symmetric-bilinear-forms]
Key ideas in the algorithm for diagonalisation of
symmetric bilinear forms? \\
\cloze{
If $\varphi(u, u) = 0$ for all $u \in V$, then
$\varphi$ is identically zero, hence done. \\
Otherwise, $\exists u \in V \setminus \{0\}$ such
that $\varphi(u, u) \neq 0$. Define
\[ U = \ker\{\varphi(u, \bullet) : V \to F, v
\mapsto \varphi(u, v)\} \]
By rank nullity, $\dim U = \dim V - 1$. Now show
$U + \langle u \rangle = U \oplus \langle u
\rangle$ (check intersection is zero). Deduce $U +
\langle u \rangle = V$ by counting dimensions. \\
Induct.
}
\end{hiddenflashcard}

\noindent
$\implies$ extensions to infinite dimensional
cases.

\begin{proof}
    \begin{itemize}
        \item $\dim_F V < \infty$. 
        \item Induction on the dimension $n$.
        \item $n = 0, 1$ are clear.
        \item Suppose that the theorem holds
            for all dimensions $< n$.
            \begin{enumerate}[(1)]
                \item Let $\varphi : V \times V
                    \to F$ be a symmetric bilinear
                    form. If $\varphi(u, u) = 0$,
                    $\forall u \in V$, $\varphi$
                    is identically zero.
                    (polarization identity).
                    \[ \implies \exists u \in V
                    \setminus \{0\}  \mid
                    \varphi(u, u) \neq 0 \]
                    (because $\varphi \neq 0$).
                \item Let us call $u = e_1$.
                    ($e_1 \neq 0$, $\varpi(e_1,
                    e_1) \neq 0$). Let us define
                    \begin{align*}
                        U
                        &= (\langle e_1
                        \rangle)^\perp \\
                        &= \{v \in V \mid
                        \varphi(e_1, v) = 0\} \\
                        &= \Ker \{\varphi(e_1,
                        \bullet) : V \to F, v
                        \mapsto \varphi(e_1, v)\}
                    \end{align*}
                    (linear because $\varphi$ is
                    bilinear). Now rank nullity:
                    \[ \dim V = n = 1 + \dim U \]
                    ($r(\varphi(e_1, \bullet)) =
                    1$ $\varphi(e_1, e_1) \neq 0$)
                    So $\dim U = n - 1$.
                \item Claim $U + \langle e_1
                    \rangle = U \oplus \langle e_1
                    \rangle$. Indeed, $v \in
                    \langle e_1 \rangle \cap U$
                    then $v = \lambda e_1$,
                    $\lambda \in F$.
                    \[ \varphi(e_1, v) = 0 \quad
                    (v \in U) \]
                    so $0 = \varphi(e_1, \lambda
                    e_1) = \lambda \varphi(e_1,
                    e_1)$ so $\lambda = 0$, so $v
                    = 0$.
                \item Conclusion $V = \langle e_1
                    \rangle \oplus U$, by counting
                    dimensions.
                \item Complete $(e_2, \dots, e_n)$
                    basis of $U$. So $\mathcal{B}
                    = (e_1, e_2, \dots, e_n)$
                    basis of $V$. And:
                    \begin{center}
                        \includegraphics[width=0.6\linewidth]
                        {images/1fd617d665a411ed.png}
                    \end{center}
                    ($\varphi(e_j, e_1) =
                    \varphi(e_1, e_j) = 0$ for $2
                    \le j \le n$).
                    \[ A' = (\varphi(e_i, e_k))_{2
                    \le i, j \le n} \]
                    Then $(A')^\top = A'$ since
                    $\varphi$ symmetric.
                    \[ \implies \varphi |_U : U
                    \times U \to F \]
                    bilinear symmetric with matrix
                    $A'$. By the induction
                    hypotheses, I can find
                    $\mathcal{B}' = (e_2', \dots,
                    e_n')$ basis of $U$ in which
                    $[\varphi|_U]_{\mathcal{B}'}$
                    is diagonal. Then
                    \[ [\varphi]_{(e_1, e_2',
                    \dots, e_n')} \]
                    diagonal form. \qedhere
            \end{enumerate}
    \end{itemize}
\end{proof}

\begin{remark*}
    $\varphi(e_1, e_1) \neq 0$
    \[ \implies V = \langle e_1 \rangle \oplus U \]
    \[ U = \langle e_1 \rangle^\perp \]
\end{remark*}

\begin{example*}
    $V = \RR^3$
    \begin{itemize}
        \item $Q(x_1, x_2, x_3) = x_1^2 + x_2^2 +
            2x_3^2 + 2x_1 x_2 + 2x_1 x_3 - 2x_2
            x_3 = x^\top A x$ where
            \[ A =
            \begin{pmatrix}
                1 & 1 & 1 \\
                1 & 1 & -1 \\
                1 & -1 & 2
            \end{pmatrix}
            \]
        \item Diagonalise: Two ways.
            \begin{enumerate}[(1)]
                \item Follow the proof of
                    diagonalisation $\to$
                    algorithm.
                \item ``Complete the square''
                    \begin{align*}
                        Q(x_1, x_2, x_3)
                        &= x_1^2 + x_2^2 + 2x_3^2
                        + 2x_1 x_2 + 2x_1 x_3 -
                        2x_2 x_3 \\
                        &= (x_1 + x_2 + x_3)^2 +
                        x_3^2 - 4x_2 x_3 \\
                        &= \ub{(x_1 + x_2 +
                        x_3)}_{x_1'}^2 +
                        \ub{(x_3 - 2x_2)}_{x_2'}^2
                        - \ub{(2x_2)}_{x_3'}^2
                    \end{align*}
                    \begin{itemize}
                        \item $P$,
                            \[ P^\top AP =
                            \begin{pmatrix}
                                1 & 0 & 0 \\
                                0 & 1 & 0 \\
                                0 & 0 & -1
                            \end{pmatrix}
                        \]
                    \item To find $P$, remember:
                        \[
                        \begin{pmatrix}
                            x_1' \\
                            x_2' \\
                            x_3'
                        \end{pmatrix}
                        =
                        \begin{pmatrix}
                            1 & 1 & 1 \\
                            0 & -2 & 1 \\
                            0 & 2 & 0
                        \end{pmatrix}
                        = P^{-1} \]
                    \end{itemize}
            \end{enumerate}
    \end{itemize}
\end{example*}
