From: hackbard <hackbard@hackdaworld.org>
Date: Sat, 18 Feb 2012 22:44:16 +0000 (+0100)
Subject: fixed outer product, still remarks to do!
X-Git-Url: https://hackdaworld.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=437d458c00fa2ee5b6d18a0fa5b450fe5a04d563;p=lectures%2Flatex.git

fixed outer product, still remarks to do!
---

diff --git a/physics_compact/math_app.tex b/physics_compact/math_app.tex
index e3417bd..f8361e8 100644
--- a/physics_compact/math_app.tex
+++ b/physics_compact/math_app.tex
@@ -76,6 +76,7 @@ z=a+bi\mapsto z^*=a-bi \text{, } z,z^*\in K \text{.}
 \end{definition}
 
 \begin{remark}
+\label{math_app:ip_remark}
 Due to conjugate symmetry, linearity in the first argument results in conjugate linearity (also termed antilinearity) in the second argument.
 \begin{equation}
 (\vec{u},\lambda(\vec{v}'+\vec{v}''))=(\lambda(\vec{v}'+\vec{v}''),\vec{u})^*=
@@ -92,6 +93,7 @@ V\rightarrow V^{\dagger}:\vec{v}\mapsto \varphi_{\vec{v}}
 \text{ defined by }
 \quad
 \varphi_{\vec{v}}(\vec{u})=(\vec{v},\vec{u}) \text{ .}
+\label{eq:ip_mapping}
 \end{equation}
 Since the inner product is linear in the first argument, the same is true for the defined mapping.
 \begin{equation}
@@ -106,30 +108,48 @@ Vector $\vec{v}^{\dagger}\equiv \varphi_{\vec{v}}\in V^{\dagger}$ is said to be
 The dual pairing $[\vec{v}^{\dagger},\vec{u}]=[\varphi_{\vec{v}},\vec{u}]=\varphi_{\vec{v}}(\vec{u})$ is associated with the inner product $(\vec{v},\vec{u})$.
 
 Now, in physics and matrix algebra, the inner product is often defined with linearity in the second argument and conjugate linearity in the first argument.
-In this case, the antilinearity property is assigned to element $\varphi_{\vec{v}}=\vec{v}^{\dagger}$ of dual space indicating an isomorphism of $V$ to the conjugate complex of its dual space.
+In this case, the antilinearity property is assigned to the element $\varphi_{\vec{v}}=\vec{v}^{\dagger}$ of dual space
 \begin{equation}
-[(\lambda\vec{v})^{\dagger},\vec{u}]=
-[\varphi_{\lambda\vec{v}},\vec{u}]=
 \varphi_{\lambda\vec{v}}(\vec{u})=
-\lambda^*\varphi_{\vec{v}}(\vec{u})=
-\lambda^*(\vec{v},\vec{u})
+(\lambda\vec{v},\vec{u})=
+\lambda^*(\vec{v},\vec{u})=
+\lambda^*\varphi_{\vec{v}}(\vec{u})
 \end{equation}
-According to this, in matrix formalism, the dual vector is associated with the conjugate transpose.
+and $V$ is found to be isomorphic to the conjugate complex of its dual space.
+Then, the inner product $(\vec{v},\vec{u})$ is associated with the dual pairing of element $\vec{u}$ of the vector space and $\vec{v}^{\dagger}$ of its conjugate complex dual space
 \begin{equation}
-(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v}
+(\vec{v},\vec{u})\rightarrow 
+[\varphi_{\vec{v}},\vec{u}]=
+[\vec{v}^{\dagger},\vec{u}]
+\text{ .}
 \end{equation}
+
+The standard sesquilinear form $\langle\cdot,\cdot\rangle$, also called Hermitian form, on $\mathbb{C}^n$ and linearity in the second argument, is given by
+\begin{equation}
+\langle\vec{v},\vec{u}\rangle=\sum_i^nv_i^*u_i
+\text{ .}
+\end{equation}
+In this case, in matrix formalism, the inner product is reformulated
+\begin{equation}
+(\vec{v},\vec{u}) \rightarrow \vec{v}^{\dagger}\vec{u}
+\text{ ,}
+\end{equation}
+where the dual vector is associated with the conjugate transpose $\vec{v}^{\dagger}$ of the corresponding vector $\vec{v}$
+and the usual rules of matrix multiplication.
 \end{remark}
 
 \begin{definition}[Outer product]
-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 $\vec{u}$,
+If $\vec{u}\in U$, $\vec{v},\vec{w}\in V$ are vectors within the respective vector spaces and $\varphi_{\vec{v}}\in V^{\dagger}$ is a linear functional of the dual space $V^{\dagger}$ of $V$ determined in some way from $\vec{v}$ (e.g.\  as in \eqref{eq:ip_mapping}),
+the outer product $\vec{u}\otimes\vec{v}$ is defined as the tensor product of $\varphi_{\vec{v}}$ and $\vec{u}$,
 which constitutes a map $A:V\rightarrow U$ by
 \begin{equation}
-\vec{v}\mapsto\vec{\varphi}(\vec{v})\vec{u}
+\vec{w}\mapsto\varphi_{\vec{v}}(\vec{w})\vec{u}
 \text{ ,}
 \end{equation}
-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 $\vec{u}\in U$.
+where $\varphi_{\vec{v}}(\vec{w})$ denotes the linear functional $\varphi_{\vec{v}}\in V^{\dagger}$ on $V$ when evaluated at some $\vec{w}\in V$, a scalar that in turn is multiplied with $\vec{u}\in U$.
+\end{definition}
 
+\begin{remark}
 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
@@ -144,8 +164,7 @@ u_mv_1^* & u_mv_2^* & \cdots & u_mv_n^*\\
 \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{ ,}