faster!

diff --git a/physics_compact/math_app.tex b/physics_compact/math_app.tex
index 687f928..5f2b0ca 100644 (file)
--- 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{ .}