From: hackbard <hackbard@hackdaworld.org>
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/cgi-bin/gitweb.cgi?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}