From: hackbard Date: Fri, 10 Feb 2012 16:57:54 +0000 (+0100) Subject: more X-Git-Url: https://hackdaworld.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=f74a53b2acb28533ef22dbf3adb0c97dbb5dd958;p=lectures%2Flatex.git more --- diff --git a/physics_compact/math_app.tex b/physics_compact/math_app.tex index d097e70..687f928 100644 --- a/physics_compact/math_app.tex +++ b/physics_compact/math_app.tex @@ -86,13 +86,19 @@ This is called a sesquilinear form. The inner product $(\cdot,\cdot)$ provides a mapping \begin{equation} -V\rightarrow V^{\dagger}:\vec{v}\mapsto \vec{v}^{\dagger} +V\rightarrow V^{\dagger}:\vec{v}\mapsto \varphi_{\vec{v}} +\quad +\text{ defined by } +\quad +\varphi_{\vec{v}}(\vec{u})=(\vec{v},\vec{u}) \text{ .} \end{equation} -given by +Since the inner product is linear in the first argument, the same is true for the defined mapping. \begin{equation} -v^{\dagger}() +\lambda(\vec{u}+\vec{v}) \mapsto +\varphi_{\lambda(\vec{u}+\vec{v})}= +\lambda\varphi_{\vec{u}}+\lambda\varphi_{\vec{v}}\\ \end{equation} -indicating structural identity (isomorphism) of $V$ and $V^{\dagger}$. +The kernel is $\vec{v}=0$, structural identity (isomorphism) of $V$ and $V^{\dagger}$ is . 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 $V^{\dagger}$ @@ -104,7 +110,7 @@ or the conjugate transpose in matrix formalism \begin{equation} (\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \text{ .} \end{equation} -In doing so, conjugacy is associated with duality. +In doing so, the conjugate transpose is associated with the dual vector. \end{remark} \begin{definition}