From: hackbard Date: Fri, 10 Feb 2012 22:18:45 +0000 (+0100) Subject: faster! X-Git-Url: https://hackdaworld.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=52293c68b82f2108ec3346435de0561e0236e466;p=lectures%2Flatex.git faster! --- diff --git a/physics_compact/math_app.tex b/physics_compact/math_app.tex index 687f928..5f2b0ca 100644 --- a/physics_compact/math_app.tex +++ b/physics_compact/math_app.tex @@ -77,12 +77,13 @@ Taking the complex conjugate $(\cdot)^*$ is the map from $K\ni z=a+bi\mapsto a-b \begin{remark} Due to conjugate symmetry, linearity in the first argument results in conjugate linearity (also termed antilinearity) in the second argument. -This is called a sesquilinear form. \begin{equation} (\vec{u},\lambda(\vec{v}'+\vec{v}''))=(\lambda(\vec{v}'+\vec{v}''),\vec{u})^*= \lambda^*(\vec{v}',\vec{u})^*+\lambda^*(\vec{v}'',\vec{u})^*= \lambda^*(\vec{u},\vec{v}')+\lambda^*(\vec{u},\vec{v}'') \end{equation} +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 \begin{equation} @@ -98,7 +99,10 @@ Since the inner product is linear in the first argument, the same is true for th \varphi_{\lambda(\vec{u}+\vec{v})}= \lambda\varphi_{\vec{u}}+\lambda\varphi_{\vec{v}}\\ \end{equation} -The kernel is $\vec{v}=0$, structural identity (isomorphism) of $V$ and $V^{\dagger}$ is . +If the inner product is nondegenerate, i.e.\ $\forall\vec{u}\, (\vec{v},\vec{u})=0 \Leftrightarrow \vec{v}=0$, as it applies for the scalar product for instance, the mapping is injective. +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 $\vec{v}\in V$. 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 $V^{\dagger}$ @@ -114,14 +118,14 @@ In doing so, the conjugate transpose is associated with the dual vector. \end{remark} \begin{definition} -If $\vec{u}\in U$, $\vec{v}\in V$ are vectors within the respective vector spaces and $\vec{\varphi}^{\dagger}\in V^{\dagger}$ is a linear functional of the dual space $V^{\dagger}$ of $V$, -the outer product $\vec{u}\otimes\vec{v}$ is defined as the tensor product of $\vec{\varphi}^{\dagger}$ and $\vec{u}$, +If $\vec{u}\in U$, $\vec{v}\in V$ are vectors within the respective vector spaces and $\vec{\varphi}\in V^{\dagger}$ is a linear functional of the dual space $V^{\dagger}$ of $V$, +the outer product $\vec{u}\otimes\vec{v}$ is defined as the tensor product of $\vec{\varphi}$ and $\vec{u}$, which constitutes a map $A:V\rightarrow U$ by \begin{equation} -\vec{v}\mapsto\vec{\varphi}^{\dagger}(\vec{v})\vec{u} +\vec{v}\mapsto\vec{\varphi}(\vec{v})\vec{u} \text{ ,} \end{equation} -where $\vec{\varphi}^{\dagger}(\vec{v})$ denotes the linear functional $\vec{\varphi}^{\dagger}\in V^{\dagger}$ on $V$ when evaluated at $\vec{v}\in V$, a scalar that in turn is multiplied with $\vec{u}\in U$. +where $\vec{\varphi}(\vec{v})$ denotes the linear functional $\vec{\varphi}\in V^{\dagger}$ 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 $\vec{v}=\sum_i^n\vec{e}'_iv_i$, @@ -129,10 +133,10 @@ 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\\ +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\\ +u_mv_1^* & u_mv_2^* & \cdots & u_mv_n^*\\ \end{array} \right) \text{ .}