author hackbard Fri, 17 Feb 2012 22:43:04 +0000 (23:43 +0100) committer hackbard Fri, 17 Feb 2012 22:43:04 +0000 (23:43 +0100)

index f327750..e3417bd 100644 (file)
@@ -69,7 +69,10 @@ $(\cdot,\cdot):V\times V \rightarrow K$, which satisfies
(positive definite)
\end{itemize}
for $\vec{u},\vec{v}\in V$ and $\lambda\in K$.
-Taking the complex conjugate $(\cdot)^*$ is the map from $K\ni z=a+bi\mapsto a-bi=z^*\in K$.
+Taking the complex conjugate $(\cdot)^*$ is the map from
+\begin{equation}
+z=a+bi\mapsto z^*=a-bi \text{, } z,z^*\in K \text{.}
+\end{equation}
\end{definition}

\begin{remark}
@@ -82,7 +85,7 @@ Due to conjugate symmetry, linearity in the first argument results in conjugate
This is called a sesquilinear form.
If $K=\mathbb{R}$, conjugate symmetry reduces to symmetry and the sesquilinear form gets a bilinear form.

-The inner product $(\cdot,\cdot)$ provides a mapping
+Furtermore, the inner product $(\cdot,\cdot)$ provides a mapping
\begin{equation}
V\rightarrow V^{\dagger}:\vec{v}\mapsto \varphi_{\vec{v}}
@@ -100,18 +103,21 @@ If the inner product is nondegenerate, i.e.\  $\forall\vec{u}\, (\vec{v},\vec{u} Since the dimension of$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})$. -In physics and matrix algebra, the inner product is often defined with linearity in the second argument and conjugate linearity in the first argument. -This allows to express the inner product$(\vec{u},\vec{v})$as a product of vector$\vec{v}$with a dual vector or linear functional of dual space$V^{\dagger}$+Now, in physics and matrix algebra, the inner product is often defined with linearity in the second argument and conjugate linearity in the first argument. +In this case, the antilinearity property is assigned to element$\varphi_{\vec{v}}=\vec{v}^{\dagger}$of dual space indicating an isomorphism of$V\$ to the conjugate complex of its dual space.
\begin{equation}
-(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}(\vec{u})\vec{v}
-\text{ CHECK ! ! !}
+[(\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})
\end{equation}
-or the conjugate transpose in matrix formalism
+According to this, in matrix formalism, the dual vector is associated with the conjugate transpose.
\begin{equation}
-(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \text{ .}
+(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v}
\end{equation}
-In doing so, the conjugate transpose is associated with the dual vector.
\end{remark}

\begin{definition}[Outer product]