The inner product $(\cdot,\cdot)$ provides a mapping
\begin{equation}
-V\rightarrow V^{\dagger}:\vec{v}\mapsto \vec{v}^{\dagger}
+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{ .}
\end{equation}
-given by
+Since the inner product is linear in the first argument, the same is true for the defined mapping.
\begin{equation}
-v^{\dagger}()
+\lambda(\vec{u}+\vec{v}) \mapsto
+\varphi_{\lambda(\vec{u}+\vec{v})}=
+\lambda\varphi_{\vec{u}}+\lambda\varphi_{\vec{v}}\\
\end{equation}
-indicating structural identity (isomorphism) of $V$ and $V^{\dagger}$.
+The kernel is $\vec{v}=0$, structural identity (isomorphism) of $V$ and $V^{\dagger}$ is .
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 a dual vector or linear functional of dual space $V^{\dagger}$
\begin{equation}
(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \text{ .}
\end{equation}
-In doing so, conjugacy is associated with duality.
+In doing so, the conjugate transpose is associated with the dual vector.
\end{remark}
\begin{definition}