-The kernel is $\vec{v}=0$, structural identity (isomorphism) of $V$ and $V^{\dagger}$ is .
+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 $\vec{v}\in V$.
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 $V^{\dagger}$
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 $V^{\dagger}$
@@ -114,14+118,14 @@ In doing so, the conjugate transpose is associated with the dual vector.
\end{remark}
\begin{definition}
\end{remark}
\begin{definition}
-If $\vec{u}\in U$, $\vec{v}\in V$ are vectors within the respective vector spaces and $\vec{\varphi}^{\dagger}\in V^{\dagger}$ 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 $\vec{\varphi}^{\dagger}$ and $\vec{u}$,
+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 $\vec{u}$,
-where $\vec{\varphi}^{\dagger}(\vec{v})$ denotes the linear functional $\vec{\varphi}^{\dagger}\in V^{\dagger}$ on $V$ when evaluated at $\vec{v}\in V$, a scalar that in turn is multiplied with $\vec{u}\in U$.
+where $\vec{\varphi}(\vec{v})$ 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 $\vec{u}\in U$.
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 $\vec{v}=\sum_i^n\vec{e}'_iv_i$,
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 $\vec{v}=\sum_i^n\vec{e}'_iv_i$,
@@ -129,10+133,10 @@ the outer product can be written as matrix $A$ as