From: hackbard Date: Thu, 9 Feb 2012 10:45:18 +0000 (+0100) Subject: vector stuff, still need: dual space and inner product X-Git-Url: https://hackdaworld.org/gitweb/?a=commitdiff_plain;h=8ef90b6c5d602acae2f9788f01b5fea7531c8fb5;p=lectures%2Flatex.git vector stuff, still need: dual space and inner product --- diff --git a/physics_compact/math_app.tex b/physics_compact/math_app.tex index edee2ff..001ed2b 100644 --- a/physics_compact/math_app.tex +++ b/physics_compact/math_app.tex @@ -44,8 +44,42 @@ The inner product ... \end{definition} \begin{definition} -If $\vec{u}\in U$ and $\vec{v}\in V$ are vectors within the respective vector spaces and $V^{\dagger}$ is the dual space of $V$, the outer product of $\vec{u}$ and $\vec{v}$ is defined as the tensor product ... +If $\vec{u}\in U$, $\vec{v}\in V$ and $\vec{v}^{\dagger}\in V^{\dagger}$ are vectors within the respective vector spaces and $V^{\dagger}$ is the dual space of $V$, +the outer product $\vec{u}\otimes\vec{v}$ is defined as the tensor product of $\vec{v}^{\dagger}$ and $\vec{u}$, +which constitutes a map $A:V\rightarrow U$ by +\begin{equation} +\vec{v}\mapsto\vec{v}^{\dagger}(\vec{v})\vec{u} +\text{ ,} +\end{equation} +where $\vec{v}^{\dagger}(\vec{v})$ denotes the linear functional $\vec{v}^{\dagger}\in V^{\dagger}$ on $V$ when evaluated at $\vec{v}\in V$, a scalar that in turn is multiplied with $\vec{u}\in U$. + +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 $\vec{v}=\sum_i^n\vec{e}'_iv_i$, +the outer product can be written as matrix $A$ as +\begin{equation} +\vec{u}\otimes\vec{v}=A=\left( +\begin{array}{c c c c} +u_1v_1 & u_1v_2 & \cdots & u_1v_n\\ +u_2v_1 & u_2v_2 & \cdots & u_2v_n\\ +\vdots & \vdots & \ddots & \vdots\\ +u_mv_1 & u_mv_2 & \cdots & u_mv_n\\ +\end{array} +\right) +\text{ .} +\end{equation} \end{definition} +\begin{remark} +The matrix can be equivalently obtained by 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: +\begin{equation} +(\vec{u}\otimes\vec{v})\vec{w}=\vec{u}(\vec{v},\vec{w}) +\end{equation} +\end{remark} \section{Spherical coordinates}