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% 02/11/2022 11AM

\begin{corollary*}
    $\sigma \in S_n$, $d$ volume form, then:
    \[ d(v_{\sigma(1)}, \dots, v_{\sigma(n)}) =
    \eps(\sigma) d(v_1, \dots, v_n) \]
\end{corollary*}

\begin{proof}
    $\sigma = \prod_{i = 1}^{n\sigma} \tau_i$.
\end{proof}

\begin{theorem*}
    Let $d$ be a volume form on $F^n$. Let $A =
    (A^{(1)}| \cdots | A^{(n)})$. Then
    \[ d(A^{(1)}| \cdots |A^{(n)}) d(e_1, \dots,
    e_n) \det A \]
\end{theorem*}

\noindent
Up to a constant, $\det$ is the \emph{only} volume
form on $F^n$.

\begin{proof}
    \begin{align*}
        d(A^{(1)}, \dots, A^{(n)})
        &= d \left( \sum_{i = 1}^n a_{i1} e_i
        , \dots, A^{(n)} \right) \\
        &= \sum_{i = 1}^n a_{i1} d(e_i, A^{(2)},
        \dots, A^{(n)}) \\
        &= \sum_{i = 1}^n a_{i1} d \left( e_i,
        \sum_{j = 1}^n a_{j2} e_j, \dots,
        A^{(n)} \right) \\
        &= \sum_{i = 1}^n \sum_{j = 1}^n a_{i1}
        a_{j2} d(e_i, e_j, \dots, A^{(n)}) \\
        &= \sum_{\substack{1 \le i_1 \le n\\1 \le
        i_2 \le n\\ \vdots\\1 \le i_n \le n}}
        \left( \prod_{k = 1}^n a_{i_k k}
        d(e_{i_1}, e_{i_2}, \dots, e_{i_n}) \right)
    \end{align*}
    The last $d$ term is nonzero only if all the
    $i_k$ are different, so we can write the $i_k$
    as a permutation. This means we can
    continue and get
    \begin{align*}
        d(A^{(1)}, \dots, A^{(n)})
        &= \sum_{\sigma \in S_n} \prod_{k = 1}^n
        a_{\sigma(k)k} d(e_{\sigma(1)}, \dots,
        e_{\sigma(n)}) \\
        &= \sum_{\sigma \in S_n} \left[ \prod_{k =
        1}^n a_{\sigma(k)k} \right] \eps(\sigma)
        d(e_1, \dots, e_n) \\
        &= d(e_1, \dots, e_n) \left[ \sum_{\sigma
        \in S_n} \eps(\sigma) \prod_{k = 1}^n
        a_{\sigma(k)k} \right] \\
        &= d(e_1, \dots, e_n) \det A \qedhere
    \end{align*}
\end{proof}

\begin{corollary*}
    $\det$ is the \emph{only} volume form such
    that
    \[ d(e_1, \dots, e_n) = 1 \]
\end{corollary*}

\subsection{Some properties of determinants}

\begin{flashcard}[det-is-multiplicative]
\begin{lemma*}
    $A, B \in \mathcal{M}_n(F)$, then:
    \[ \det(AB) = (\det A)(\det B) \]
\end{lemma*}

\begin{proof}
    \cloze{Indeed, pick $A$. Consider the map:
    \[ d_A : \ub{F^n \times \cdots \times F^n}_n
    \to F \]
    defined by
    \[ (v_1, \dots, v_n) \mapsto \det(A v_1,
    \dots, A v_n) \]
    Then:
    \begin{itemize}
        \item $d_A$ is multilinear: $v_i \mapsto A
            v_i$ is linear.
        \item $d_A$ is alternate: if $v_i = v_j$
            then $A v_i = A v_j$.
    \end{itemize}
    so $d_A$ is a volume form.\prompt{ Hence we
    know $d_A = C\det$, then compute $C = \det A$.}}
    \fcscrap{In particular,
    \[ d_A(v_1, \dots, v_n) = C \det(v_1, \dots,
    v_n) \]
    Now we compute $C$. $A e_i = (A)$ so
    \[ d_A(e_1, \dots, e_n) = \det(A e_1, \dots, A
    e_n) = \det(a_1, \dots, A_n) = \det A \]
    So
    \[ C = \det A \]
    We have proved:
    \begin{align*}
        d_A(v_1, \dots, v_n)
        &= d(A v_1, \dots, A v_n) \\
        &= (\det A) \det(v_1, \dots, v_n)
    \end{align*}
    Now observe:
    \[ AB = ((AB)_1, \dots, (AB)_n) \]
    \[ (AB)_i = AB_i \]
    so
    \begin{align*}
        \det(AB)
        &= \det(AB_1, \dots, AB_n) \\
        &= \det(A) \det(B_1, \dots, B_n) \\
        &= \det(A) \det(B)
    \end{align*}
}
\end{proof}
\end{flashcard}

\begin{definition*}
    $A \in \mathcal{M}_n(F)$, we say that:
    \begin{enumerate}[(i)]
        \item $A$ is \emph{singular} if $\det A =
            0$
        \item $A$ is \emph{non singular} if $\det
            A \neq 0$.
    \end{enumerate}
\end{definition*}

\begin{lemma*}
    $A$ is invertible implies $A$ is non singular.
\end{lemma*}

\begin{proof}
    $A$ is invertible.
    \[ \implies \exists A^{-1}, AA^{-1} = A^{-1}A
    = I_n \]
    \[ \implies \det(AA^{-1}) = \det(A^{-1}A) =
    \det I_n = 1 \]
    \[ \implies (\det A)(\det A^{-1}) = 1 \]
    \[ \implies \det A \neq 0 \]
\end{proof}

\begin{remark*}
    We have proved that
    \[ \det(A^{-1}) = \frac{1}{\det A} \]
\end{remark*}

\begin{theorem*}
    Let $A \in \mathcal{M}_n(F)$. Then the
    following are equivalent:
    \begin{enumerate}[(i)]
        \item $A$ is invertible
        \item $A$ is non singular
        \item $r(A) = n$
    \end{enumerate}
\end{theorem*}

\begin{proof}
    (i) $\iff$ (iii) done (rank nullity Theorem).
    (i) $\implies$ (iii) is lemma above. We need
    to show (ii) $\implies$ (iii). Indeed, assume
    $r(A) < n$. Then
    \[ \iff \dim \Span\{c_1, \dots, c_n\} < n \]
    \[ \implies \exists (\lambda_1, \dots,
    \lambda_n) \neq (0, \dots, 0) \]
    \[ \sum_{i = 1}^n \lambda_i c_i = 0 \]
    I pick $j$ such that $\lambda_j \neq 0$
    \[ \implies c_j = -\frac{1}{\lambda_j} \sum_{i
    \neq j} \lambda_i c_i \]
    \begin{align*}
        \implies \det A
        &= \det (c_1, \dots, c_j,
        \dots, c_n) \\
        &= \det \left( c_1, \dots,
        -\frac{1}{\lambda_j} \sum_{i \neq j}
        \lambda_i c_i, \dots, c_n \right) \\
        &= \sum_{i \neq j} -\frac{1}{\lambda_j}
        \det(c_1, \dots, c_i, \dots, c_n) \\
        &= 0
    \end{align*}
\end{proof}

\begin{remark*}
    This gives us the sharp criterion for
    invertibility of a linear system of $n$
    equations with $n$ unknowns:
    \[ Y \in F^n \]
    \[ A \in \mathcal{M}_n(F) \]
    \[ AX = Y, X \in F^n \]
    exists a unique solution if and only if $A$ is
    invertible, which happens if and only if
    $\det A \neq 0$.
\end{remark*}

\subsubsection*{Determinant of linear maps}

\begin{lemma*}
    Conjugate matrices have the same determinant.
\end{lemma*}

\begin{proof}
    \begin{align*}
        \det(P^{-1} AP)
        &= \det(P^{-1}) \det A \det P \\
        &= \frac{1}{\det P} \det A \det P \\
        &= \det A
    \end{align*}
    ($P$ invertible implies $\det P \neq 0$).
\end{proof}

\begin{definition*}
    $\alpha : V \to V$ linear (endomorphism). We
    define
    \[ \det \alpha = \det([\alpha]_{\mathcal{B}}) \]
    $\mathcal{B}$ is any basis of $V$. This number
    does not depend on the choice of the basis!
\end{definition*}

\begin{theorem*}
    $\det : L(V, V) \to F$ satisfies:
    \begin{enumerate}[(i)]
        \item $\det \id = 1$
        \item $\det (\alpha \circ \beta) =
            \det(\alpha) \det(\beta)$
        \item $\det(\alpha) \neq 0$ if and only if
            $\alpha$ is invertible and then
            \[ \det(\alpha^{-1}) =
            (\det\alpha)^{-1} \]
    \end{enumerate}
\end{theorem*}

\begin{proof}
    Pick a basis and express in terms of
    $[\alpha]_{\mathcal{B}}$ and
    $[\beta]_{\mathcal{B}}$.
\end{proof}

\subsubsection*{Determinant of block matrices}

\begin{lemma*}
    $A \in \mathcal{M}_k(F)$, $\mathcal{B} \in
    \mathcal{M}_l(F)$ and $C \in \mathcal{M}_{k,
    l}(F)$. Let
    \[ M =
    \begin{pmatrix}
        A & C \\
        0 & B
    \end{pmatrix}
    \in \mathcal{M}_n(F) \]
    ($n = k + l$) then
    \[ \det M = (\det A)(\det B) \]
\end{lemma*}

\begin{proof}
    \[ \det M = \sum_{\sigma \in S_n} \eps(\sigma)
    \prod_{i = 1}^n m_{\sigma(i) i} \tag{$*$} \]
    Observation:
    \[ m_{\sigma(i)i} = 0 \]
    if $i \le k$, $\sigma(i) > k$. So In ($*$),
    we need only sum over $\sigma \in S_n$ such
    that:
    \begin{enumerate}[(i)]
        \item $\forall j \in [1, k]$, $\sigma(j)
            \in [1, k]$
        \item and hence $\forall j \in [k + 1,
            n]$, $\sigma(j) \in [k + 1, n]$.
    \end{enumerate}
    In other words, we restrict to $\sigma$ of the
    form:
    \[ \sigma_1 : \{1, \dots, k\} \to \{1, \dots,
    k\} \]
    \[ \sigma_2 : \{k + 1, \dots, n\} \to \{k + 1,
    \dots, n\} \]
    \begin{enumerate}[(i)]
        \item $m_{\sigma(j)j}$ with $1 \le j \le
            k$, then $\sigma(j) \in \{1, \dots,
            k\}$, can be rewritten as
            \[ m_{\sigma(j)j} = a_{\sigma(j)j} =
            a_{\sigma_1(j)j} \]
        \item Similarly, for $k + 1 \le j \le n$,
            $k + 1 \le \sigma(j) \le n$,
            \[ m_{\sigma(j)j} = b_{\sigma(j)j} =
            b_{\sigma_1(j)j} \]
    \end{enumerate}
    Note that
    \[ \eps(\sigma) = \eps(\sigma_1)\eps(\sigma_2) \]
    so then
    \begin{align*}
        \det M
        &= \sum_{\sigma S_n} \eps(\sigma) \prod_{i
        = 1}^n m_{\sigma(i) i} \\
        &= \sum_{\substack{\sigma_1 \in
        S_k\\ \sigma_2 \in S_l}} \eps(\sigma_1
        \circ \sigma_2) \prod_{i = 1}^k
        a_{\sigma_1(i)i} \prod_{j = k + 1}^n
        b_{\sigma_2(j)j} \\
        &= \sum_{\substack{\sigma_1 \in S_k\\
        \sigma_2 \in S_l}} \eps(\sigma_1)
        \eps(\sigma_2) \prod_{i = 1}^k
        a_{\sigma_1(i)i} \prod_{j = k + 1}^n
        b_{\sigma_2(j)j} \\
        &= \left( \sum_{\sigma_1 \in S_k}
        \eps(\sigma_1) \prod_{i = 1}^k
        a_{\sigma_1(i)i} \right) \left( \sum_{\sigma_2
        \in S_l} \eps(\sigma_2) \prod_{j = k +
        1}^n a_{\sigma_2(j)j} \right) \\
        &= (\det A)(\det B) \qedhere
    \end{align*}
\end{proof}