\label{math_app:product}
\begin{definition}
-The inner product ...
+The inner product on a vector space $V$ over $K$ is a map $(\cdot,\cdot):V\times V \rightarrow K$, which satisfies
+\begin{itemize}
+\item $(\vec{u},\vec{v})=(\vec{v},\vec{u})^*$
+ (conjugate symmetry, symmetric for $K=\mathbb{R}$)
+\item $(\lambda\vec{u},\vec{v})=\lambda(\vec{u},\vec{v})$ and
+ $(\vec{u}'+\vec{u}'',\vec{v})=(\vec{u}',\vec{v})+(\vec{u}'',\vec{v})$
+ (linearity in first argument)
+\item $(\vec{u},\vec{u})\geq 0 \text{, } ``=" \Leftrightarrow \vec{u}=0$
+ (positive definite)
+\end{itemize}
+for $\vec{u},\vec{v}\in V$ and $\lambda\in K$.
\end{definition}
+\begin{remark}
+Due to conjugate symmetry, linearity in the first argument results in conjugate linearity (also termed antilinearity) in the second argument.
+This is called a sesquilinear form.
+\begin{equation}
+(\vec{u},\lambda(\vec{v}'+\vec{v}''))=(\lambda(\vec{v}'+\vec{v}''),\vec{u})^*=
+\lambda^*(\vec{v}',\vec{u})^*+\lambda^*(\vec{v}'',\vec{u})^*=
+\lambda^*(\vec{u},\vec{v}')+\lambda^*(\vec{u},\vec{v}'')
+\end{equation}
+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 $V^{\dagger}$
+\begin{equation}
+(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}(\vec{u})\vec{v}
+\end{equation}
+or the conjugate transpose in matrix formalism
+\begin{equation}
+(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \text{ .}
+\end{equation}
+In doing so, conjugacy is associated with duality.
+\end{remark}
+
\begin{definition}
If $\vec{u}\in U$, $\vec{v}\in V$ and $\vec{v}^{\dagger}\in V^{\dagger}$ are vectors within the respective vector spaces and $V^{\dagger}$ is the dual space of $V$,
the outer product $\vec{u}\otimes\vec{v}$ is defined as the tensor product of $\vec{v}^{\dagger}$ and $\vec{u}$,