+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}