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\subsection{Spans, linear independence and the
Steinitz exchange lemma}

\begin{definition*}[Spand of a family of vectors]
    Let $V$ be an $F$ vector space. Let $S \subset
    V$ be a subset ($S =$ collection of vectors).
    We define:
    \begin{align*}
        \ub{\langle S \rangle}_{\text{``span of
        $S$''}}
        &= \{\text{finite linear combination of
        elements of $S$}\} \\
        &= \left\{ \sum_{s \in J} \lambda_s v_s,
        v_s \in S, \lambda_s \in F, \text{$J$ is
        finite} \right\}
    \end{align*}
\end{definition*}

\noindent
Convention: $\langle \emptyset \rangle = \{0\}$.

\begin{remark*}
    $\langle S \rangle$ is the smallest vector
    subspace which contains $S$.
\end{remark*}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item $V = \RR^3$
        \[ S = \left\{ \left|
        \begin{matrix}
            1 \\
            0 \\
            0
        \end{matrix}
        \right. , \left|
        \begin{matrix}
            0 \\
            1 \\
            2
        \end{matrix}
        \right. , \left|
        \begin{matrix}
            2 \\
            -2 \\
            -4
        \end{matrix}
        \right. \right\} \]
        \[ \implies \langle S \rangle = \left\{
        \left|
        \begin{matrix}
            a \\
            b \\
            2b
        \end{matrix}
        \right. , (a, b) \in \RR^2 \right\} \]
    \item \[ V = \RR^n = \left\{ \left|
        \begin{matrix}
            x_1 \\
            \vdots \\
            x_n
        \end{matrix}
        \right. , x_i \in \RR, 1 \le i \le n\right\} \]
        Define:
        \[ e_i = \left|
        \begin{matrix}
            0 \\
            \vdots \\
            1 \quad \text{(in position $i$)} \\
            \vdots \\
            0
        \end{matrix}
        \right. \]
        \[ \implies V = \langle e_1, \dots, e_n
        \rangle \]
    \item $X$ is a set, $V = \RR^X = \{f : X \to
        \RR\}$.
        \[ S_x : X \to \RR \]
        \[ y \mapsto \begin{cases}
            1 & \text{if $x = y$} \\
            0 & \text{otherwise}
        \end{cases} \]
        \begin{align*}
            \langle (S_x)_{x \in X} \rangle
            &= \Span ((S_x)_{x \in X}) \\
            &= \{f \in \RR^X : f \text{ has finite
            support}\}
        \end{align*}
        (Support of $f$ is $\{x \in X : f(x) \neq
        0\}$)
\end{enumerate}

\begin{definition*}
    Let $V$ be an $F$ vector space. Let $S$ be a
    subset of $V$. We say that $S$ \emph{spans}
    $V$ if:
    \[ \langle S \rangle = V .\]
\end{definition*}

\begin{example*}
    $V = \RR^2$
    \begin{center}
    \begin{tsqx}
        ! size(2.5cm);
        (0,0)->(1,1)
        (0,0)->(1.5,0.5)
        label $v_1$ @ 1.2*(1,1)
        label $v_2$ @ 1.2*(1.5,0.5)
    \end{tsqx}
    \end{center}
    $\{v_1, v_2\}$ spans $V$.
\end{example*}

\begin{definition*}[Finite dimension]
    Let $V$ be an $F$ vector space. We say that
    $V$ is finite dimensional if it is spanned by
    a finite set.
\end{definition*}

\begin{example*}
    Let $V_1 = P[x]$ be the set of polynomials
    over $\RR$, and let $V = P_n[x]$ be the set of
    polynomials over $\RR$ with degree $\le n$.
    Then $\{1, x, \dots, x^n\}$ spans $P_n[x]$, so
    $P_n[x] = \langle 1, x, \dots, x^n \rangle$.
    So $P_n[x]$ is finite dimensional. \\
    On the other hand, $P[x]$ is \emph{not} finite
    dimensional: it is infinite dimensional,
    because there is no family of $V$ with
    finitely many elements which spans $V$.
\end{example*}

\noindent
Question: If $V$ is finite dimensional, is there a
\emph{minimal} number of vectors in the family so
that they span $V$.

\begin{definition*}[Independence]
    We say that $(v_1, \dots, v_n)$ elements of
    $V$ are \emph{linearly independent} if:
    \[ \sum_{i = 1}^n \lambda_i v_i = 0, ~
    \lambda_i \in F \implies \lambda_i = 0
    ~\forall i \]
\end{definition*}

\begin{remark*}
    \begin{enumerate}[(1)]
        \item We also say that the family $(v_1,
            \dots, v_n)$ is \emph{free}.
        \item Equivalently, $(v_1, \dots, v_n)$
            are \emph{not} linearly independent if
            one of these vectors is a linear
            combination of the remaining $(n - 1)$
            ones. Indeed, $\exists (\lambda_1,
            \dots, \lambda_n)$ \emph{not all
            zero} (that is, there exists $j$ such
            that $\lambda_j \neq 0$), such that
            \[ \sum_{i = 1}^n \lambda_i v_i = 0
            \implies v_j = -\frac{1}{\lambda_j}
            \sum_{\substack{i = 1 \\ i \neq j}}^n
            \lambda_i v_i \]
    \end{enumerate}
\end{remark*}

\begin{example*}
    $V = \RR^3$. If $(v_1, v_2)$ free, and $v_3$
    is coplanar with both, then $(v_1, v_2, v_3$
    is not free.
\end{example*}

\begin{remark*}
    $(v_i)_{1 \le i \le n}$ free family (linearly
    independent) then $\forall 1 \le i \le n$,
    $v_i \neq 0$.
\end{remark*}

\begin{definition*}[Basis]
    A sub set $S$ of $V$ is a \emph{basis} of $V$
    if and only if:
    \begin{itemize}
        \item $\langle S \rangle = V$ (generating
            family)
        \item $S$ linearly independent / free
    \end{itemize}
\end{definition*}

\begin{remark*}
    When $S$ spans $V$, we say that $S$ is a
    generating family. So a basis is a free
    generating family.
\end{remark*}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item \[ V = \RR^n = \left\{ \left|
        \begin{matrix}
            x_1 \\
            \vdots \\
            x_n
        \end{matrix}
        \right. , x_i \in \RR, 1 \le i \le n\right\} \]
        \[ e_i = \left|
        \begin{matrix}
            0 \\
            \vdots \\
            1 \quad \text{(in position $i$)} \\
            \vdots \\
            0
        \end{matrix}
        \right. \]
        Then $(e_i)_{1 \le i \le n}$ is a basis of
        $V$.
    \item $V = \CC$. If $F = \CC$ then $\{1\}$ is
        a basis of $V$. If $F = \RR$ then $\{1,
        i\}$ is a basis of $V$.
    \item $V = P[x] = \{\text{polynomials over
        $\RR$}\}$, $S = \{x^n : n \ge 0\}$. Then
        $S$ is a basis for $V$.
\end{enumerate}

\begin{lemma*}
    Let $V$ be an $F$ vector space. Then $(v_1,
    \dots, v_n)$ is a basis of $V$ if and only if
    any vector $v \in V$ has a \emph{unique}
    decomposition:
    \[  v = \sum_{i = 1}^n \lambda_i v_i , \quad
    \lambda_i \in F \]
\end{lemma*}

\begin{notation*}
    $(\lambda_1, \dots, \lambda_n)$ are the
    \emph{coordinates} of $v$ in the basis $(v_1,
    \dots, v_n)$.
\end{notation*}

\begin{proof}
    By assumption, $\langle v_1, \dots, v_n
    \rangle = V$ so
    \[ \forall v \in V, \exists (\lambda_1, \dots,
    \lambda_n) \in F^n \qquad v = \sum_{i = 1}^n
    \lambda_i v_i \]
    Uniqueness: let
    \[ v = \sum_{i = 1}^n \lambda_i v_i = \sum_{i
    = 1}^n \lambda_i' v_i \]
    \[ \implies \sum_{i = 1}^n (\lambda_i -
    \lambda_i') v_i = 0 \]
    \[ \implies \forall 1 \le i \le n, ~\lambda_i
    = \lambda_i' \]
\end{proof}

\begin{lemma*}
    If $(v_1, \dots, v_n)$ spans $V$, then some
    subset of this family is a basis of $V$.
\end{lemma*}

\begin{proof}
    If $(v_1, \dots, v_n)$ are linearly
    independent then done. Let's assume they are
    not independent. Then by possible reordering
    the vectors,
    \[ v_n \in \langle v_1, \dots, v_{n - 1}
    \rangle \]
    ($v_n$ is a linear combination of $v_1, \dots,
    v_{n - 1}$) so
    \[ V = \langle v_1, \dots, v_n \rangle =
    \langle v_1, \dots, v_{n - 1} \rangle \]
    Now we can iterate until the resulting set is
    a basis of $V$. (We only have to iterate
    finitely many times since $n$ is finite).
\end{proof}

\begin{flashcard}
\begin{theorem*}[Steinitz exchange lemma]
    Let $V$ be a \emph{finite dimensional} vector
    space over $F$. Take:
    \begin{enumerate}[(i)]
        \item \cloze{$(v_1, \dots, v_m)$ free}
        \item \cloze{$(w_1, \dots, w_n)$ generating
            ($\langle w_1, \dots, w_n \rangle =
            V$).}
    \end{enumerate}
    Then \cloze{$m \le n$}, and \cloze{up to
    reordering,}
    \[ \cloze{(v_1, \dots, v_m, w_{m + 1}, \dots,
    w_n)} \]
    \cloze{spans $V$}.
\end{theorem*}
\end{flashcard}