\subsection{Vector space}
\label{math_app:vector_space}
-\begin{definition}
+\begin{definition}[Vector space]
A vector space $V$ over a field $(K,+,\cdot)$ is an additive abelian group $(V,+)$ and an additionally defined scalar multiplication of $\vec{v}\in V$ by $\lambda\in K$, which fullfills:
\begin{itemize}
\item $\forall \vec{v} \, \exists 1$ with: $\vec{v}1=\vec{v}$
\subsection{Dual space}
-\begin{definition}
+\begin{definition}[Dual space]
The dual space $V^{\dagger}$ of vector space $V$ over field $K$ is defined as the set of all linear maps from the vector space $V$ into its field $K$
\begin{equation}
\varphi:V\rightarrow K \text{ .}
\subsection{Inner and outer product}
\label{math_app:product}
-\begin{definition}
+\begin{definition}[Inner product]
The inner product on a vector space $V$ over $K$ is a map
-\begin{equation}
-(\cdot,\cdot):V\times V \rightarrow K
-\text{ ,}
-\end{equation}
-which satisfies
+$(\cdot,\cdot):V\times V \rightarrow K$, which satisfies
\begin{itemize}
\item $(\vec{u},\vec{v})=(\vec{v},\vec{u})^*$
(conjugate symmetry, symmetric for $K=\mathbb{R}$)
In doing so, the conjugate transpose is associated with the dual vector.
\end{remark}
-\begin{definition}
+\begin{definition}[Outer product]
If $\vec{u}\in U$, $\vec{v}\in V$ are vectors within the respective vector spaces and $\vec{\varphi}\in V^{\dagger}$ is a linear functional of the dual space $V^{\dagger}$ of $V$,
the outer product $\vec{u}\otimes\vec{v}$ is defined as the tensor product of $\vec{\varphi}$ and $\vec{u}$,
which constitutes a map $A:V\rightarrow U$ by