We introduce the new field $\field$ which will stand for any field. It can be either $\realn$, $\cmpln$ or any other set that fulfils the field axioms.
\begin{defi}
A vector space is a set $V$ with the operations
\noindent\begin{minipage}[t]{.5\linewidth}
\[
\text{Addition}
\]
\[
\begin{split}
+: V \times V &\longrightarrow V \\
(x, y) &\longmapsto x + y
\end{split}
\]
\end{minipage}
\begin{minipage}[t]{.5\linewidth}
\[
\text{Scalar Multiplication}
\]
\[
\begin{split}
\cdot: \field\times V &\longrightarrow V \\
(\alpha, y) &\longmapsto\alpha x
\end{split}
\]
\end{minipage}
We require the following conditions for these operations
\begin{enumerate}[(i)]
\item$\exists0\in V ~\forall x \in V: ~~x +0= x$
\item$\forall x \in V ~\exists(-x)\in V: ~~x +(-x)=0$
\item$\forall x, y \in V: ~~x + y = y + x$
\item$\forall x, y, z \in V: ~~(x + y)+ z = x +(y + z)$
\item$\forall\alpha\in\field ~\forall x, y \in V: ~~\alpha(x + y)=\alpha x +\alpha y$
\item$\forall\alpha, \beta\in\field ~\forall x \in V: ~~(\alpha+\beta)x =\alpha x +\beta x$
\item$\forall\alpha, \beta\in\field ~\forall x \in V: ~~(\alpha\beta)x =\alpha(\beta x)$
\item$\forall x \in V: ~~1\cdot x = x$
\end{enumerate}
Elements from $V$ are called vectors, elements from $\field$ are called scalars.
\end{defi}
\begin{rem}
We now have two different addition operations that are denoted the same way:
\begin{enumerate}[(i)]
\item$+: V \times V \rightarrow V$
\item$+: \field\times\field\rightarrow\field$
\end{enumerate}
Analogously there are two neutral elements and two multiplication operations.
\end{rem}
\begin{eg}\leavevmode
\begin{enumerate}[(i)]
\item$\field$ is already a vector space
\item$V =\field^2$. In the case that $\field=\realn$ this vector space is the two-dimensional Euclidean space. The neutral element is $(0, 0)$, and the inverse is $(\chi_1, \chi_2)\rightarrow(-\chi_1, -\chi_2)$. This can be extended to $\field^n$.
\item$\field$-valued sequences:
\[
V = \set[\chi \in \field ~~\forall n \in \natn]{\seq{\chi}_{n \in\natn}}
\]
\item Let $M$ be a set. Then the set of all $\field$-valued functions on $M$ is a vector space
\[
V = \set[f: M \rightarrow \field]{f}
\]
\end{enumerate}
\end{eg}
\begin{defi}
Let $V$ be a vector space, let $x, x_1, \cdots, x_n \in V$ and let $M \subset V$.
\begin{enumerate}[(i)]
\item$x$ is said to be a linear combination of $x_1, \cdots, x_n$ if $\exists\alpha_1, \cdots, \alpha_n \in\field$ such that
\[
x = \sum_{k=1}^n \alpha_k x_k
\]
\item The set of all linear combinations of elements from $M$ is called the \textit{span}, or the \textit{linear hull} of $M$
implies $\alpha_0=\alpha_1=\cdots=\alpha_n =0$. The span of the $f_k$ is the set of all polynomials of $(\le n)$-th degree. The function $x \mapsto(x-1)^3$ is a linear combination of $f_0, \cdots, f_3$:
\[
(x-1)^3 = x^3 - 3x^2 + 3x - 1
\]
\end{enumerate}
\end{eg}
\begin{rem}
Let $V$ be a vector space, $y \in V$ a linear combination of $y_1, \cdots, y_n$, and each of those a linear combination of $x_1, \cdots, x_n$. I.e.
Let $V$ be a finite-dimensional vector space, and let $x_1, \cdots, x_n \in V$. Then the following are equivalent
\begin{enumerate}[(i)]
\item$x_1, \cdots, x_n$ is a basis.
\item$x_1, \cdots, x_n$ is a minimal generator (Minimal means that no subset is a generator).
\item$x_1, \cdots, x_n$ is a maximal linearly independent system (Maximal means that $x_1, \cdots, x_n, y$ is not linearly independent).
\item$\forall x \in V$ there exists a unique $\alpha_1, \cdots, \alpha_n \in\field$
\[
x = \series[n]{k}\alpha_k x_k
\]
\end{enumerate}
\end{thm}
\begin{proof}
First we prove "(i) $\implies$ (ii)". Let $x_1, \cdots, x_n$ be a basis of $V$. By definition $x_1, \cdots, x_n$ is a generator. Assume that $x_2, \cdots, x_n$ is still a generator, then
However this contradicts the linear independence of the basis. Next, to prove "(ii) $\implies$ (iii)" let $x_1, \cdots, x_n$ be a minimal generator. Let $\alpha_1, \cdots, \alpha_n \in\field$ such that
\begin{equation}
0 = \series[n]{k}\alpha_k x_k
\end{equation}
Assume that one coefficient is $\ne0$ (w.l.o.g. $\alpha_1=0$). Then
\begin{equation}
x_1 = \sum_{k=2}^n -\frac{\alpha_k}{\alpha_1} x_k
\end{equation}
$x_1, \cdots, x_n$ is a generator, i.e. for $x \in V$
So $x_1, \cdots, x_n, y$ is linearly dependent, and therefore $x_1, \cdots, x_n$ is maximal. To prove "(iii) $\implies$ (iv)" let $x_1, \cdots, x_n$ be a maximal linearly independent system. If $y \in V$, then
This is a contradiction, so therefore $\beta\ne0$:
\begin{equation}
y = \series[n]{k} -\frac{\alpha_k}{\beta} x_k
\end{equation}
The uniqueness of these coefficients are left as an exercise for the reader. Finally, to finish the proof we need to show "(iv) $\implies$ (i)". By definition
\begin{equation}
V = \spn\set{x_1, \cdots, x_n}
\end{equation}
Hence, $\set{x_1, \cdots, x_n}$ is a generator. In case
\begin{equation}
0 = \series[n]{k}\alpha_k x_k
\end{equation}
holds, then $\alpha_1=\cdots=\alpha_n =0$ follows from the uniqueness.
\end{proof}
\begin{cor}
Every finite-dimensional vector space has a basis.
\end{cor}
\begin{proof}
By condition, there is a generator $x_1, \cdots, x_n$. Either this generator is minimal (then it would be a basis), or we remove elements until it is minimal.
\end{proof}
\begin{lem}\label{lem:steinitz}
Let $V$ be a vector space and $x_1, \cdots, x_k \in V$ a linearly independent set of elements. Let $y \in V$, then
\[
x_1, \cdots, x_k, y \text{ linearly independent}\iff y \notin\spn\set{x_1, \cdots, x_k}
\]
\end{lem}
\begin{proof}
To prove "$\impliedby$", assume $y \ne\spn\set{x_1,\cdots,x_k}$. Therefore $x_1, \cdots, x_k, y$ must be linearly independent. To see this, consider
Then $\beta=0$, otherwise we could solve the above for $y$, and that would contradict our assumption. The argument works in the other direction as well.
\end{proof}
\begin{thm}[Steinitz exchange lemma]
Let $V$ be a finite-dimensional vector space. If $x_1, \cdots, x_m$ is a generator and $y_1, \cdots, y_n$ a linear independent set of vectors, then $n \le m$. In case $x_1, \cdots, x_m$ and $y_1, \cdots, y_n$ are both bases, then $n=m$.
\end{thm}
\begin{hproof}
Let $K \in\set{0, \cdots, \min\set{m, n}-1}$ and let
W.l.o.g. $x:i = x_{K+1}$. By \Cref{lem:steinitz}, $x_1, \cdots, x_{K+1}, y_{K+2}, \cdots y_n$ is linearly independent. We can now sequentially replace $y_i$ with $x_i$ without losing the linear independence. Assume $n > m$, then this process leads to a linear independent system $x_1, \cdots, x_m, y_{m+1}, \cdots, y_n$. But since $x_1, \cdots, x_m$ is a generator, $y_{m+1}$ is a linear combination of $x_1, \cdots, x_m$. If $x_1, \cdots, x_m$ and $y_1, \cdots, y_n$ are both bases, then we cannot change the roles and therefore $m = n$.
\end{hproof}
\begin{defi}
The amount of elements in a basis is said to be the dimension of $V$, and is denoted as $\dim V$ .
Then $e_1, \cdots, e_n$ is a basis, in fact, it is the standard basis of $\realn^n$ ($\cmpln^n$).
\item Let $V$ be the vector space of polynomials
\[
V = \set[n \in \natn, ~\alpha_1, \cdots, \alpha_n \in \realn, ~~f(x) = \sum_{k=1}^n \alpha_k x^k ~~\forall x \in \realn]{f:\realn\longrightarrow\realn}
\]
This space has the basis
\[
\set[n \in \natn_0]{x \longmapsto x^n}
\]
\end{enumerate}
\end{eg}
\begin{cor}
In an $n$-dimensional vector space, every generator has at least $n$ elements, and every linearly independent system has at most $n$ elements.
\end{cor}
\begin{proof}
Let $M \subset\spn\set{x_1, \cdots, x_n}$. Then
\begin{equation}
V = \spn M \subset\spn{x_1, \cdots, x_n}
\end{equation}
Hence, $x_1, \cdots, x_n$ is a generator. On the other hand, assume
\begin{equation}
\exists y \in M \setminus\spn\set{x_1, \cdots, x_n}
\end{equation}
Then $x_1, \cdots, x_n, y$ is linearly independent (\Cref{lem:steinitz}), and we can sequentially add elements from $M$ until $x_1, \cdots, x_n, y_{n+1}, \cdots, y_{n+m}$ is a generator.
\end{proof}
\begin{defi}[Vector subspace]
Let $V$ be a vector space. A non-empty set $W \subset V$ is called a vector subspace if
\[
\forall x, y \in W ~\forall\alpha\in\field: ~~x + \alpha y \in W
