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% 28/10/2022 11AM

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

\begin{flashcard}[annihilator-of-sum-and-intersection]
\begin{lemma*}
    Let $V$ be a \fcemph{finite dimensional} vector space
    over $F$. Let $U_1, U_2 \le V$. Then
    \begin{enumerate}[(i)]
        \item $(U_1 \cloze{+} U_2)^0 = U_1^0
            \cloze{\cap} U_2^0$
        \item $(U_1 \cloze{\cap} U_2)^0 = U_1^0
            \cloze{+} U_2^0$
    \end{enumerate}
\end{lemma*}

\begin{proof}
    \begin{enumerate}[(i)]
        \item \cloze{Exercise.}
        \item \cloze{Take $^0$ of (i) and use $U^{00} =
            U$.}
    \end{enumerate}
\end{proof}
\end{flashcard}

\subsection{Bilinear Forms}

$\implies$ Quadratic algebra.

\begin{flashcard}[bilinear-form-defn]
\begin{definition*}
    $U, V$ vector spaces over $F$. Then:
    \[ \varphi : U \times V \to F \]
    is a \emph{bilinear form} if \cloze{it ``linear in
    both components'':
    \begin{itemize}
        \item $\varphi(u, \bullet) : V \to F$ is
            linear for all $u \in U$ ($v \mapsto
            \varphi(u, v)$).
        \item $\varphi(\bullet, v) : U \to F$ is
            linear for all $v \in V$ ($u \mapsto
            \varphi(u, v)$)
    \end{itemize}}
\end{definition*}
\end{flashcard}

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item $V \times V^* \to F$
        \[ (v, \theta) \mapsto \theta(v) \]
    \item Scalar product / dot product on $U = V = \RR^n$
    \item $U = V = \mathcal{C}([0, 1], \RR)$
        \[ \varphi(f, g) = \int_0^1 f(t) g(t) \dd
        t\]
        (``infinite dimensional scalar product)
\end{enumerate}

\begin{flashcard}[bilinear-form-matrix]
\begin{definition*}[matrix of a bilinear form in a
    basis]
    $\mathcal{B} = (e_1, \dots, e_m)$ basis of
    $U$, $\mathcal{C} = (f_1, \dots, f_n)$ basis
    of $V$. $\varphi : U \times V \to F$ bilinear
    form. \\
    The matrix of $\varphi$ with respect to
    $\mathcal{B}$ and $\mathcal{C}$ is:
    \[ \cloze{[\varphi]_{\mathcal{B}, \mathcal{C}} =
    (\ub{\varphi(e_i, f_j)}_{\in F})_{\substack{1
    \le i \le m \\ 1 \le j \le n}}} \]
\end{definition*}
\end{flashcard}

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

\noindent
Link between the bilinear form and its matrix in
given basis.

\begin{proof}
    $u = \sum_{i = 1}^m \lambda_i e_i$, $v =
    \sum_{j = 1}^n \mu_j f_j$. Then by linearity:
    \begin{align*}
        \varphi(u, v)
        &= \varphi \left( \sum_{i = 1}^m \lambda_i
        e_i, \sum_{j = 1}^n \mu_j e_j \right) \\
        &= \sum_{i = 1}^m \sum_{j = 1}^n \lambda_i
        \mu_j \varphi(e_i, e_j) \\
        &= [u]_{\mathcal{B}}^\top
        [\varphi]_{\mathcal{B}, \mathcal{C}}
        [v]_{\mathcal{C}}
    \end{align*}
\end{proof}

\begin{remark*}
    $[\varphi]_{\mathcal{B}, \mathcal{C}}$ is the
    only matrix such that ($*$) holds.
\end{remark*}

\begin{notation*}
    $\varphi : U \times V \to F$ bilinear form,
    then it determines two linear maps:
    \[ \varphi_L : U \to V^* \]
    \[ \varphi_L(u) : V \to F \]
    \[ v \mapsto \varphi(u, v) \]
    define $\varphi_R$ similarly.
\end{notation*}

\begin{lemma*}
    $\mathcal{B} = (e_1, \dots, e_m)$ basis of
    $U$, $B^* = (\eps_1, \dots, \eps_m)$ dual
    basis of $U^*$. $\mathcal{C} = (f_1, \dots,
    f_n)$ basis of $V$, $\mathcal{C}^* = (\eta_1,
    \dots, \eta_n)$ dual basis of $V^*$. Let $A =
    [\varphi]_{\mathcal{B}, \mathcal{C}}$ then:
    \[ [\varphi_R]_{\mathcal{C}, \mathcal{B}^*} =
    A \]
    \[ [\varphi]_{\mathcal{B}, \mathcal{C}^*} =
    A^\top \]
\end{lemma*}

\begin{proof}
    \begin{align*}
        \varphi_L(e_i)(f_j)
        &= \varphi(e_i, f_j) = A_{ij} \\
        \implies \varphi_L(e_i)
        &= \sum A_{ij} \eta_j
    \end{align*}
    similarly for $\varphi_R$.
\end{proof}

\begin{flashcard}[bilinear-form-degenerate]
\begin{definition*}[Degenerate / non degenerate
    bilinear form]
    \cloze{$\Ker \varphi_L$: ``left kernel of
    $\varphi$'', $\Ker \varphi_R$: ``right kernel
    of $\varphi$''.} We say that $\varphi$ is
    non-degenerate if
    \[ \cloze{\Ker \varphi_L = \{0\} \quad \text{and}
    \quad \Ker \varphi_R = \{0\}} \]
    Otherwise, we say that $\varphi$ is
    degenerate.
\end{definition*}
\end{flashcard}

\begin{lemma*}
    $U$, $V$ finite dimensional. $\mathcal{B}$
    basis of $U$, $\mathcal{C}$ basis of $V$.
    $\varphi : U \times V \to F$ bilinear form, $A
    = [\varphi]_{\mathcal{B}, \mathcal{C}}$.
    Then $\varphi$ non degenerate $\iff$ $A$
    invertible.
\end{lemma*}

\begin{corollary*}
    $\varphi$ non degenerate
    \[ \implies \dim U = \dim V \]
\end{corollary*}

\begin{proof}
    \begin{align*}
        \varphi \text{ non degenerate}
        &\iff \Ker \varphi_L = \{0\} \text{ and }
        \Ker \varphi_R = \{0\} \\
        &\iff n(A^\top) = 0 \text{ and } n(A) = 0
        \\
        &\iff r(A^\top) = \dim U \text{ and } r(A)
        = \dim V \\
        &\iff \text{$A$ invertible and then: $\dim
        U = \dim V$}
    \end{align*}
\end{proof}

\begin{remark*}
    $\varphi : \RR^n \times \RR^n \to \RR$
    scalar product, then $\varphi$ is non
    degenerate, and if we take the usual bases,
    then $[\varphi]_{\mathcal{B}, \mathcal{B}} =
    I_n$.
\end{remark*}

\begin{corollary*}
    When $U$ and $V$ are finite dimensional, then
    choosing a \emph{non degenerate} bilinear form
    $\varphi : U \to V \to F$ is equivalent to
    choosing an isomorphism $\varphi_L : U \to
    V^*$.
\end{corollary*}

\begin{flashcard}[bilinear-form-orthogonal]
\begin{definition*}
    $T \subset U$, we define:
    \[ T^\perp = \cloze{\{v \in V \mid \varphi(t, v) = 0,
    \forall t \in T\}} \]
    Similarly define for $S \subset V$
    \[ {}^\perp S = \cloze{\{u \in U, \varphi(u, s) = 0,
    \forall s \in S\}} \]
\end{definition*}
\end{flashcard}

\subsubsection*{Change basis for bilinear forms}

\begin{flashcard}[bilinear-form-change-basis-formula]
\begin{proposition*}
    \begin{itemize}
        \item $\mathcal{B}$, $\mathcal{B}'$ basis of $U$, $P
            = [\id]_{\mathcal{B}', \mathcal{B}}$
        \item $\mathcal{C}, \mathcal{C}'$ basis of
            $V$, $Q = [\id]_{\mathcal{C}',
            \mathcal{C}}$.
    \end{itemize}
    Let $\varphi : U \times V \to F$ bilinear
    form, then
    \[ [\varphi]_{\mathcal{B}', \mathcal{C}'} =
    \cloze{P^\top [\varphi]_{\mathcal{B},
    \mathcal{C}} Q} \]
    change of basis formula for bilinear forms.
\end{proposition*}
\end{flashcard}

\begin{proof}
    \begin{align*}
        \varphi(u, v)
        &= [u]_{\mathcal{B}}^\top
        [\varphi]_{\mathcal{B},
        \mathcal{C}}[v]_{\mathcal{C}} \\
        &= (P[u]_{\mathcal{B}'})^\top
        [\varphi]_{\mathcal{B}, \mathcal{C}}
        (Q[v]_{\mathcal{C}'}) \\
        &= [u]_{\mathcal{B}'}^\top (P^\top
        [\varphi]_{\mathcal{B}, \mathcal{C}} Q)
        [v]_{\mathcal{C}'}
    \end{align*}
\end{proof}

\begin{definition*}
    The rank of $\varphi$ ($r(\varphi)$) is the rank of
    any matrix representing $\varphi$.
\end{definition*}

\noindent
Indeed, $r(P^\top AQ) = r(A)$ for any invertible
$P, Q$.

\begin{remark*}
    $r(\varphi) = r(\varphi_R) = r(\varphi_L)$.
    (we computed matrices in a basis and $r(A) =
    r(A^\top)$)
\end{remark*}

\noindent
More applications later: scalar product.

\subsection{Determinant and Traces}

\begin{definition*}
    $A \in \mathcal{M}_n(F) = \mathcal{M}_{n
    \times n}(F)$ We define the trace of $A$
    \[ \Trace A = \sum_{i = 1}^n A_{ii} \]
    \[ A = (A_{ij})_{1 \le i, j \le n} \]
\end{definition*}

\begin{remark*}
    $\mathcal{M}_n(F) \to F$ linear form ($A
    \mapsto \Trace A$).
\end{remark*}

\begin{lemma*}
    $\Trace(AB) = \Trace(BA)$.
\end{lemma*}

\begin{proof}
    \begin{align*}
        \Trace(AB) = \sum_{i = 1}^n \left( \sum_{j
        = 1}^n a_{ij} b_{ji} \right) \\
        &= \cdots \\
        &= \Trace(BA)
    \end{align*}
\end{proof}