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\begin{definition*}[Rank and nullity]
    \begin{itemize}
        \item $r(\alpha) = \dim \Image \alpha$
            (rank)
        \item $n(\alpha) = \dim \Ker \alpha$
            (nullity)
    \end{itemize}
\end{definition*}

\begin{theorem*}[Rank nullity theorem]
    \begin{itemize}
        \item Let $U, V$ be vector spaces over
            $F$, $\dim_F U < +\infty$.
        \item Let $\alpha : U \to V$ be a linea
            map, then
            \[ \dim U = r(\alpha) + n(\alpha \]
    \end{itemize}
\end{theorem*}

\begin{proof}
    We have proved that $U / \Ker \alpha \simeq
    \Image \alpha$. So $\dim (U / \Ker \alpha) =
    \dim \Image \alpha$. But $\dim (U / \Ker
    \alpha) = \dim U - \dim \Ker \alpha$. So $\dim
    U = \dim \Ker \alpha + \dim \Image \alpha =
    r(\alpha) + n(\alpha)$.
\end{proof}

\begin{lemma*}
    Let $V$, $W$ be vector spaces over $F$ of
    \emph{equal} finite dimension. Let $\alpha : V
    \to W$ be a linear map. Then the following are
    equivalent:
    \begin{itemize}
        \item $\alpha$ is injective
        \item $\alpha$ is surjective
        \item $\alpha$ is an isomorphism
    \end{itemize}
\end{lemma*}

\begin{proof}
    Follows immediately from the rank-nullity
    theorem. (Exercise)
\end{proof}

\begin{example*}
    Let $V = \{(x, y, z) \in \RR^3 : x + y + z =
    0\}$. Then consider $\alpha : \RR^3 \to \RR$
    defined by $(x, y, z) \mapsto x + y + z$. Then
    $\Ker \alpha = V$ and $\Image \alpha = \RR$,
    hence by rank nullity $3 = n(\alpha) + 1$
    hence $\dim V = 2$.
\end{example*}

\subsection{Linear maps from $V$ to $W$ and matrices}

The space of linear maps from $V$ to $W$. Let $V$,
$W$ be vector spaces over $F$.
\[ L(V, W) = \{\alpha : V \to W \text{ linear}\} \]

\begin{proposition*}
    $L(V, W)$ is a vector space over $F$ with:
    \[ (\alpha_1 + \alpha_2)(v) = \alpha_1(v) +
    \alpha_2(v) \]
    \[ (\lambda \alpha)(v) = \lambda \alpha (v) \]
    Moreover if $V$ and $W$ are finite dimensional
    over $F$, then so is $L(V, W)$ and:
    \[ \dim_F L(V, W) = (\dim_F V) (\dim_F W) \]
\end{proposition*}

\begin{proof}
    Proof that it is a vector space is an
    exercise. \\
    We will prove the statement about dimensions
    soon.
\end{proof}

\subsubsection*{Matrices and linear maps}

\begin{definition*}[Matrix]
    A $m \times n$ matrix over $F$ is an array
    with $m$ rows and $n$ columns with entries in
    $F$.
\end{definition*}

\begin{notation*}
    $\mathcal{M}_{m, n}(F)$ is the set of $m
    \times n$ matrices over $F$.
\end{notation*}

\begin{proposition*}
    $\mathcal{M}_{m, n}(F)$ is an $F$ vector space
    under operations:
    \begin{itemize}
        \item $(a_{ij}) + (b_{ij}) = (a_{ij} +
            b_{ij})$
        \item $\lambda (a_{ij}) = (\lambda
            a_{ij})$
    \end{itemize}
\end{proposition*}

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

\begin{proposition*}
    $\dim_F \mathcal{M}_{m, n}(F) = m \times n$.
\end{proposition*}

\begin{proof}
    We exhibit a basis using \emph{elementary}
    matrices. Pick $1 \le i \le m$, $1 \le j \le
    n$. Then we define $E_{ij}$ to be the matrix
    which is 0 everywhere, except it is 1 in the
    entry that is in the $i$-th row and $j$-th
    column. Then $(E_{ij})$ is a basis of
    $\mathcal{M}_{m, n}(F)$. Clearly spans
    $\mathcal{M}_{m, n}(F)$. Family is free is an
    exercise.
\end{proof}

\subsubsection*{Representation of linear maps}

\begin{itemize}
    \item $V, W$ vector spaces over $F$, $\alpha :
        V \to W$ linear map.
    \item Basis $\mathcal{B}(v_1, \dots, v_n)$
        basis of $V$, $\mathcal{C} = (w_1, \dots,
        w_m)$ basis of $W$.
    \item Let $v \in V$, then we can write
        \[ v = \sum_{j = 1}^n \lambda_j v_j \]
        so we can consider the coordinates of $v$
        in the basis $\mathcal{B}$ $(\lambda_1,
        \dots, \lambda_n \in F^n$). We may write
        this as $[v]_{\mathcal{B}}$.
    \item Similarly for $w \in W$, we note
        $[w]_{\mathcal{C}}$ in a similar way.
\end{itemize}

\begin{definition*}[Matrix of $\alpha$ in
    $\mathcal{B}, \mathcal{C}$ basis]
    \[ [\alpha]_{\mathcal{B}, \mathcal{C}} \equiv
    \text{matrix of $\alpha$ with respect to
    $\mathcal{B}, \mathcal{C}$} \]
    We define it as:
    \[ [\alpha]_{\mathcal{B}, \mathcal{C}} =
    \begin{pmatrix}
        \vdots & \vdots & & \vdots \\
        [\alpha(v_1)]_{\mathcal{C}} & [\alpha(v_2)]_{\mathcal{C}} & \cdots & [\alpha(v_n)]_{\mathcal{C}} \\
        \vdots & \vdots & & \vdots
    \end{pmatrix}
    \]
\end{definition*}

\noindent
Observation:
\[ \alpha(v_j) = \sum_{i = 1}^m a_{ij} w_i \]

\begin{lemma*}
    For any $v \in V$,
    \[ [\alpha(v)]_{\mathcal{C}} =
    [\alpha]_{\mathcal{B}, \mathcal{C}} \cdot
    [v]_{\mathcal{B}} \]
    where
    \[ (Av)_i = \sum_{j = 1}^n a_{ij} \lambda_j \]
\end{lemma*}

\begin{proof}
    Let $v \in V$, with
    \[ v = \sum_{j = 1}^n \lambda_j v_j \]
    Then
    \begin{align*}
        \alpha(v)
        &= \alpha \left( \sum_{j = 1}^n \lambda_j
        v_j\right) \\
        &= \sum_{j = 1}^n \lambda_j \alpha(v_j) \\
        &= \sum_{j = 1}^n \lambda_j \sum_{i = 1}^m
        a_{ij} w_i \\
        &= \sum_{i = 1}^m \left( \sum_{j = 1}^n
        a_{ij} \lambda_j \right) w_i
    \end{align*}
\end{proof}

\begin{lemma*}
    Let $\beta : U \to V$, $\alpha : V \to W$
    linear, and hence $\alpha \circ \beta : U \to
    W$ linear. Let $\mathcal{A}$ be a basis of
    $U$, $\mathcal{B}$ be a basis of $V$, and
    $\mathcal{C}$ a basis of $W$. Then
    \[ [\alpha \circ \beta]_{\mathcal{A},
    \mathcal{C}} = [\alpha]_{\mathcal{B},
    \mathcal{C}}
    [\beta]_{\mathcal{A},\mathcal{B}} \]
\end{lemma*}

\begin{proof}
    $A = [\alpha]_{\mathcal{B}, \mathcal{C}}$, $B
    = [\beta]_{\mathcal{A}, \mathcal{B}}$. Pick
    $u_l \in \mathcal{A}$. Then
    \begin{align*}
        (\alpha \circ \beta)(u_l)
        &= \alpha(\beta(u_l)) \\
        &= \alpha \left(\sum_j b_{jl} v_j \right)
        \\
        &= \sum_j b_{jl} \alpha(v_j) \\
        &= \sum_j b_{jl} \sum_i a_{ij} w_i \\
        &= \sum_i \left( \sum_j a_{ij} b_{jl}
        \right) w_i \qedhere
    \end{align*}
\end{proof}

\begin{proposition*}
    If $V$ and $W $are vector spaces over $F$ and
    $\dim_F V = n$ and $\dim_F W = m$. Then $L(V,
    W) \simeq \mathcal{M}_{m, n}(F)$, and in
    particular, $\dim L(V, W) = m \times n$.
\end{proposition*}

\begin{proof}
    Fix $\mathcal{B}, \mathcal{C}$ basis of $V$,
    $W$.
    \begin{claim*}
        $\theta : L(V, W) \to \mathcal{M}_{m,
        n}(F)$ defined by $\alpha \mapsto
        [\alpha]_{\mathcal{B}, \mathcal{C}}$ is an
        isomorphism.
    \end{claim*}
    \begin{itemize}
        \item $\theta$ is linear:
            \[ [\lambda_1 \alpha_1 + \lambda_2
            \alpha_2]_{\mathcal{B}, \mathcal{C}} =
            \lambda_1 [\alpha_1]_{\mathcal{B},
            \mathcal{C}} + \lambda_2
            [\alpha_2]_{\mathcal{B}, \mathcal{C}} \]
        \item $\theta$ is surjective: let
            \[ A = (a_{ij}) \]
            Consider the map:
            \[ \alpha : v_j \mapsto \sum_{i = 1}^m
            a_{ij} w_i \]
            and extend by linearity. Then
            $[\alpha]_{\mathcal{B}, \mathcal{C}} =
            A$.
        \item $\theta$ is injective because
            \[ [\alpha]_{\mathcal{B}, \mathcal{C}}
            = 0 \implies \alpha \equiv 0 \]
    \end{itemize}
    Hence, using $\theta$, $L(V, W) \simeq
    \mathcal{M}_{m, n}(F)$.
\end{proof}

\begin{remark*}
    Let $\mathcal{B}, \mathcal{C}$ be bases of $V,
    W$. Let $\eps_\mathcal{B} : V \to F^n$ be
    defined such that $v \mapsto
    [\alpha]_{\mathcal{B}}$, and similarly define
    $\eps_{\mathcal{C}} : W \to F^m$ such that $w
    \mapsto [w]_{\mathcal{C}}$. Then the following
    diagram commutes:
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/b3088fea4e7311ed.png}
    \end{center}
\end{remark*}
