\end{definition}
\begin{remark}
+\label{math_app:ip_remark}
Due to conjugate symmetry, linearity in the first argument results in conjugate linearity (also termed antilinearity) in the second argument.
\begin{equation}
(\vec{u},\lambda(\vec{v}'+\vec{v}''))=(\lambda(\vec{v}'+\vec{v}''),\vec{u})^*=
\text{ defined by }
\quad
\varphi_{\vec{v}}(\vec{u})=(\vec{v},\vec{u}) \text{ .}
+\label{eq:ip_mapping}
\end{equation}
Since the inner product is linear in the first argument, the same is true for the defined mapping.
\begin{equation}
The dual pairing $[\vec{v}^{\dagger},\vec{u}]=[\varphi_{\vec{v}},\vec{u}]=\varphi_{\vec{v}}(\vec{u})$ is associated with the inner product $(\vec{v},\vec{u})$.
Now, in physics and matrix algebra, the inner product is often defined with linearity in the second argument and conjugate linearity in the first argument.
-In this case, the antilinearity property is assigned to element $\varphi_{\vec{v}}=\vec{v}^{\dagger}$ of dual space indicating an isomorphism of $V$ to the conjugate complex of its dual space.
+In this case, the antilinearity property is assigned to the element $\varphi_{\vec{v}}=\vec{v}^{\dagger}$ of dual space
\begin{equation}
-[(\lambda\vec{v})^{\dagger},\vec{u}]=
-[\varphi_{\lambda\vec{v}},\vec{u}]=
\varphi_{\lambda\vec{v}}(\vec{u})=
-\lambda^*\varphi_{\vec{v}}(\vec{u})=
-\lambda^*(\vec{v},\vec{u})
+(\lambda\vec{v},\vec{u})=
+\lambda^*(\vec{v},\vec{u})=
+\lambda^*\varphi_{\vec{v}}(\vec{u})
\end{equation}
-According to this, in matrix formalism, the dual vector is associated with the conjugate transpose.
+and $V$ is found to be isomorphic to the conjugate complex of its dual space.
+Then, the inner product $(\vec{v},\vec{u})$ is associated with the dual pairing of element $\vec{u}$ of the vector space and $\vec{v}^{\dagger}$ of its conjugate complex dual space
\begin{equation}
-(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v}
+(\vec{v},\vec{u})\rightarrow
+[\varphi_{\vec{v}},\vec{u}]=
+[\vec{v}^{\dagger},\vec{u}]
+\text{ .}
\end{equation}
+
+The standard sesquilinear form $\langle\cdot,\cdot\rangle$, also called Hermitian form, on $\mathbb{C}^n$ and linearity in the second argument, is given by
+\begin{equation}
+\langle\vec{v},\vec{u}\rangle=\sum_i^nv_i^*u_i
+\text{ .}
+\end{equation}
+In this case, in matrix formalism, the inner product is reformulated
+\begin{equation}
+(\vec{v},\vec{u}) \rightarrow \vec{v}^{\dagger}\vec{u}
+\text{ ,}
+\end{equation}
+where the dual vector is associated with the conjugate transpose $\vec{v}^{\dagger}$ of the corresponding vector $\vec{v}$
+and the usual rules of matrix multiplication.
\end{remark}
\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 $\vec{u}$,
+If $\vec{u}\in U$, $\vec{v},\vec{w}\in V$ are vectors within the respective vector spaces and $\varphi_{\vec{v}}\in V^{\dagger}$ is a linear functional of the dual space $V^{\dagger}$ of $V$ determined in some way from $\vec{v}$ (e.g.\ as in \eqref{eq:ip_mapping}),
+the outer product $\vec{u}\otimes\vec{v}$ is defined as the tensor product of $\varphi_{\vec{v}}$ and $\vec{u}$,
which constitutes a map $A:V\rightarrow U$ by
\begin{equation}
-\vec{v}\mapsto\vec{\varphi}(\vec{v})\vec{u}
+\vec{w}\mapsto\varphi_{\vec{v}}(\vec{w})\vec{u}
\text{ ,}
\end{equation}
-where $\vec{\varphi}(\vec{v})$ denotes the linear functional $\vec{\varphi}\in V^{\dagger}$ on $V$ when evaluated at $\vec{v}\in V$, a scalar that in turn is multiplied with $\vec{u}\in U$.
+where $\varphi_{\vec{v}}(\vec{w})$ denotes the linear functional $\varphi_{\vec{v}}\in V^{\dagger}$ on $V$ when evaluated at some $\vec{w}\in V$, a scalar that in turn is multiplied with $\vec{u}\in U$.
+\end{definition}
+\begin{remark}
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 $\vec{v}=\sum_i^n\vec{e}'_iv_i$,
the outer product can be written as matrix $A$ as
\right)
\text{ .}
\end{equation}
-\end{definition}
-\begin{remark}
+
The matrix can be equivalently obtained by matrix multiplication:
\begin{equation}
\vec{u}\otimes\vec{v}=\vec{u}\vec{v}^{\dagger} \text{ ,}