+This is called a sesquilinear form.
+If $K=\mathbb{R}$, conjugate symmetry reduces to symmetry and the sesquilinear form gets a bilinear form.
+
+Furtermore, the inner product $(\cdot,\cdot)$ provides a mapping
+\begin{equation}
+V\rightarrow V^{\dagger}:\vec{v}\mapsto \varphi_{\vec{v}}
+\quad
+\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}
+\lambda(\vec{u}+\vec{v}) \mapsto
+\varphi_{\lambda(\vec{u}+\vec{v})}=
+\lambda\varphi_{\vec{u}}+\lambda\varphi_{\vec{v}}\\
+\end{equation}
+If the inner product is nondegenerate, i.e.\ $\forall\vec{u}\, (\vec{v},\vec{u})=0 \Leftrightarrow \vec{v}=0$, as it applies for the scalar product for instance, the mapping is injective.
+Since the dimension of $V$ and $V^{\dagger}$ is equal, it is additionally surjective.
+Then, $V$ is isomorphic to $V^{\dagger}$.
+Vector $\vec{v}^{\dagger}\equiv \varphi_{\vec{v}}\in V^{\dagger}$ is said to be the dual vector of $\vec{v}\in V$.
+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 the element $\varphi_{\vec{v}}=\vec{v}^{\dagger}$ of dual space
+\begin{equation}
+\varphi_{\lambda\vec{v}}(\vec{u})=
+(\lambda\vec{v},\vec{u})=
+\lambda^*(\vec{v},\vec{u})=
+\lambda^*\varphi_{\vec{v}}(\vec{u})
+\end{equation}
+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{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