\subsection{Dual space}
+\begin{definition}
+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{ .}
+\end{equation}
+These type of linear maps are termed linear functionals.
+The dual space $V^{\dagger}$ becomes a vector space over $K$ itself by the following additional definitions
+\begin{eqnarray}
+(\varphi+\psi)(\vec{v}) & = & \varphi(\vec{v})+\psi(\vec{v}) \\
+(\lambda\varphi)(\vec{v}) & = & \lambda\varphi(\vec{v})
+\end{eqnarray}
+for all $\vec{v}\in V$, $\varphi,\psi\in V^{\dagger}$ and $\lambda\in K$.
+
+The map $V^{\dagger}\times V \rightarrow K: [\varphi,\vec{v}]=\varphi(\vec{v})$ is termed dual pairing of a functional $\varphi\in V^{\dagger}$ and an elemnt $\vec{v}\in V$.
+\end{definition}
+
\subsection{Inner and outer product}
\label{math_app:product}
\begin{definition}
-The inner product on a vector space $V$ over $K$ is a map $(\cdot,\cdot):V\times V \rightarrow K$, which satisfies
+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
\begin{itemize}
\item $(\vec{u},\vec{v})=(\vec{v},\vec{u})^*$
(conjugate symmetry, symmetric for $K=\mathbb{R}$)
(positive definite)
\end{itemize}
for $\vec{u},\vec{v}\in V$ and $\lambda\in K$.
+Taking the complex conjugate $(\cdot)^*$ is the map from $K\ni z=a+bi\mapsto a-bi=z^*\in K$.
\end{definition}
\begin{remark}
\lambda^*(\vec{v}',\vec{u})^*+\lambda^*(\vec{v}'',\vec{u})^*=
\lambda^*(\vec{u},\vec{v}')+\lambda^*(\vec{u},\vec{v}'')
\end{equation}
+
+The inner product $(\cdot,\cdot)$ provides a mapping
+\begin{equation}
+V\rightarrow V^{\dagger}:\vec{v}\mapsto \vec{v}^{\dagger}
+\end{equation}
+given by
+\begin{equation}
+v^{\dagger}()
+\end{equation}
+indicating structural identity (isomorphism) of $V$ and $V^{\dagger}$.
+
In physics and matrix algebra, the inner product is often defined with linearity in the second argument and conjugate linearity in the first argument.
-This allows to express the inner product $(\vec{u},\vec{v})$ as a product of vector $\vec{v}$ with the dual vector or linear functional of dual space $V^{\dagger}$
+This allows to express the inner product $(\vec{u},\vec{v})$ as a product of vector $\vec{v}$ with a dual vector or linear functional of dual space $V^{\dagger}$
\begin{equation}
(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}(\vec{u})\vec{v}
+\text{ CHECK ! ! !}
\end{equation}
or the conjugate transpose in matrix formalism
\begin{equation}
\end{remark}
\begin{definition}
-If $\vec{u}\in U$, $\vec{v}\in V$ are vectors within the respective vector spaces and $\vec{y}^{\dagger}\in V^{\dagger}$ 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 $\vec{y}^{\dagger}$ and $\vec{u}$,
+If $\vec{u}\in U$, $\vec{v}\in V$ are vectors within the respective vector spaces and $\vec{\varphi}^{\dagger}\in V^{\dagger}$ 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 $\vec{\varphi}^{\dagger}$ and $\vec{u}$,
which constitutes a map $A:V\rightarrow U$ by
\begin{equation}
-\vec{v}\mapsto\vec{y}^{\dagger}(\vec{v})\vec{u}
+\vec{v}\mapsto\vec{\varphi}^{\dagger}(\vec{v})\vec{u}
\text{ ,}
\end{equation}
-where $\vec{y}^{\dagger}(\vec{v})$ denotes the linear functional $\vec{y}^{\dagger}\in V^{\dagger}$ on $V$ when evaluated at $\vec{v}\in V$, a scalar that in turn is multiplied with $\vec{u}\in U$.
+where $\vec{\varphi}^{\dagger}(\vec{v})$ denotes the linear functional $\vec{\varphi}^{\dagger}\in V^{\dagger}$ on $V$ when evaluated at $\vec{v}\in V$, a scalar that in turn is multiplied with $\vec{u}\in U$.
In matrix formalism, with respect to a given basis ${\vec{e}_i}$ of $\vec{u}$ and ${\vec{e}'_i}$ of $\vec{v}$,
if $\vec{u}=\sum_i^m \vec{e}_iu_i$ and $\vec{v}=\sum_i^n\vec{e}'_iv_i$,