author hackbard Sun, 19 Feb 2012 17:28:00 +0000 (18:28 +0100) committer hackbard Sun, 19 Feb 2012 17:28:00 +0000 (18:28 +0100)

index f8361e8..f1ac777 100644 (file)
@@ -139,7 +139,7 @@ and the usual rules of matrix multiplication.
\end{remark}

\begin{definition}[Outer product]
-If $\vec{u}\in U$, $\vec{v},\vec{w}\in V$ are vectors within the respective vector spaces and $\varphi_{\vec{v}}\in V^{\dagger}$ is a linear functional of the dual space $V^{\dagger}$ of $V$ determined in some way from $\vec{v}$ (e.g.\  as in \eqref{eq:ip_mapping}),
+If $\vec{u}\in U$, $\vec{v},\vec{w}\in V$ are vectors within the respective vector spaces and $\varphi_{\vec{v}}\in V^{\dagger}$ is a linear functional of the dual space $V^{\dagger}$ of $V$ (determined in some way by $\vec{v}$),
the outer product $\vec{u}\otimes\vec{v}$ is defined as the tensor product of $\varphi_{\vec{v}}$ and $\vec{u}$,
which constitutes a map $A:V\rightarrow U$ by
\begin{equation}
@@ -150,9 +150,10 @@ where $\varphi_{\vec{v}}(\vec{w})$ denotes the linear functional $\varphi_{\vec{ \end{definition} \begin{remark} -In matrix formalism, with respect to a given basis${\vec{e}_i}$of$\vec{u}$and${\vec{e}'_i}$of$\vec{v}$, + +In matrix formalism, if$\varphi_{\vec{v}}$is defined as in \eqref{eq:ip_mapping} and if$\vec{u}=\sum_i^m \vec{e}_iu_i$and$\vec{v}=\sum_i^n\vec{e}'_iv_i$, -the outer product can be written as matrix$A$as +the standard form of the outer product can be written as the matrix \begin{equation} \vec{u}\otimes\vec{v}=A=\left( \begin{array}{c c c c} @@ -162,19 +163,19 @@ u_2v_1^* & u_2v_2^* & \cdots & u_2v_n^*\\ u_mv_1^* & u_mv_2^* & \cdots & u_mv_n^*\\ \end{array} \right) -\text{ .} +\text{ ,} \end{equation} - -The matrix can be equivalently obtained by matrix multiplication: +which can be equivalently obtained by the rulrs of matrix multiplication \begin{equation} \vec{u}\otimes\vec{v}=\vec{u}\vec{v}^{\dagger} \text{ ,} \end{equation} if$\vec{u}$and$\vec{v}$are represented as$m\times 1$and$n\times 1$column vectors, respectively. -Here,$\vec{v}^{\dagger}$represents the conjugate transpose of$\vec{v}$. -By definition, and as can be easily seen in the matrix representation, the following identity holds: +Here, again,$\vec{v}^{\dagger}$represents the conjugate transpose of$\vec{v}\$.
+By definition, and as can be easily seen in matrix representation, the identity
\begin{equation}
(\vec{u}\otimes\vec{v})\vec{w}=\vec{u}(\vec{v},\vec{w})
\end{equation}
+holds.
\end{remark}

\section{Spherical coordinates}