vector stuff, still need: dual space and inner product

[lectures/latex.git] / physics_compact / math_app.tex

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 \chapter{Mathematical tools}
 
+\section{Vector algebra}
+
+\subsection{Vector space}
+\label{math_app:vector_space}
+
+\begin{definition}
+A vector space $V$ over a field $(K,+,\cdot)$ is an additive abelian group $(V,+)$ and an additionally defined scalar multiplication of $\vec{v}\in V$ by $\lambda\in K$, which fullfills:
+\begin{itemize}
+\item $\forall \vec{v} \, \exists 1$ with: $\vec{v}1=\vec{v}$
+      (identity element of scalar multiplication)
+\item $\vec{v}(\lambda_1+\lambda_2)=\vec{v}\lambda_1+\vec{v}\lambda_2$
+      (distributivity of scalar multiplication)
+\item $(\vec{v}_1+\vec{v}_2)\lambda=\vec{v}_1\lambda + \vec{v}_2\lambda$
+      (distributivity of scalar multiplication)
+\item $(\vec{v}\lambda_1)\lambda_2=\vec{v}(\lambda_1\lambda_2)$
+      (compatibility of scalar multiplication with field multiplication)
+\end{itemize}
+The elements $\vec{v}\in V$ are called vectors.
+\end{definition}
+\begin{remark}
+Due to the additive abelian group, the following properties are additionally valid:
+\begin{itemize}
+\item $\vec{u}+\vec{v}=\vec{v}+\vec{u}$ (commutativity of addition)
+\item $\vec{u}+(\vec{v}+\vec{w})=(\vec{u}+\vec{v})+\vec{w}$
+      (associativity of addition)
+\item $\forall \vec{v} \, \exists \vec{0}$ with:
+      $\vec{0}+\vec{v}=\vec{v}+\vec{0}=\vec{v}$
+      (identity elemnt of addition)
+\item $\forall \vec{v} \, \exists -\vec{v}$ with: $\vec{v}+(-\vec{v})=0$
+      (inverse element of addition)
+\end{itemize}
+The addition of two vectors is called vector addition.
+\end{remark}
+
+\subsection{Dual space}
+
+\subsection{Inner and outer product}
+\label{math_app:product}
+
+\begin{definition}
+The inner product ...
+\end{definition}
+
+\begin{definition}
+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}
 
 \section{Fourier integrals}