author hackbard Fri, 10 Feb 2012 22:18:45 +0000 (23:18 +0100) committer hackbard Fri, 10 Feb 2012 22:18:45 +0000 (23:18 +0100)

index 687f928..5f2b0ca 100644 (file)
@@ -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{ .}