(positive definite)
\end{itemize}
for $\vec{u},\vec{v}\in V$ and $\lambda\in K$.
-Taking the complex conjugate $(\cdot)^*$ is the map from $K\ni z=a+bi\mapsto a-bi=z^*\in K$.
+Taking the complex conjugate $(\cdot)^*$ is the map from
+\begin{equation}
+z=a+bi\mapsto z^*=a-bi \text{, } z,z^*\in K \text{.}
+\end{equation}
\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})^*=
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
+Furtermore, the inner product $(\cdot,\cdot)$ provides a mapping
\begin{equation}
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{ .}
+\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}
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 $\vec{v}\in V$.
+The dual pairing $[\vec{v}^{\dagger},\vec{u}]=[\varphi_{\vec{v}},\vec{u}]=\varphi_{\vec{v}}(\vec{u})$ is associated with the inner product $(\vec{v},\vec{u})$.
-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 $V^{\dagger}$
+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 the element $\varphi_{\vec{v}}=\vec{v}^{\dagger}$ of dual space
\begin{equation}
-(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}(\vec{u})\vec{v}
-\text{ CHECK ! ! !}
+\varphi_{\lambda\vec{v}}(\vec{u})=
+(\lambda\vec{v},\vec{u})=
+\lambda^*(\vec{v},\vec{u})=
+\lambda^*\varphi_{\vec{v}}(\vec{u})
\end{equation}
-or the conjugate transpose in matrix formalism
+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} \text{ .}
+(\vec{v},\vec{u})\rightarrow
+[\varphi_{\vec{v}},\vec{u}]=
+[\vec{v}^{\dagger},\vec{u}]
+\text{ .}
\end{equation}
-In doing so, the conjugate transpose is associated with the dual vector.
+
+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 by $\vec{v}$),
+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 $\vec{v}$,
+In matrix formalism, if $\varphi_{\vec{v}}$ is defined as in \eqref{eq:ip_mapping} and
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
+the standard form of the outer product can be written as the matrix
\begin{equation}
\vec{u}\otimes\vec{v}=A=\left(
\begin{array}{c c c c}
u_mv_1^* & u_mv_2^* & \cdots & u_mv_n^*\\
\end{array}
\right)
-\text{ .}
+\text{ ,}
\end{equation}
-\end{definition}
-\begin{remark}
-The matrix can be equivalently obtained by matrix multiplication:
+which can be equivalently obtained by the rulrs of matrix multiplication
\begin{equation}
\vec{u}\otimes\vec{v}=\vec{u}\vec{v}^{\dagger} \text{ ,}
\end{equation}
if $\vec{u}$ and $\vec{v}$ are represented as $m\times 1$ and $n\times 1$ column vectors, respectively.
-Here, $\vec{v}^{\dagger}$ represents the conjugate transpose of $\vec{v}$.
-By definition, and as can be easily seen in the matrix representation, the following identity holds:
+Here, again, $\vec{v}^{\dagger}$ represents the conjugate transpose of $\vec{v}$.
+By definition, and as can be easily seen in matrix representation, the identity
\begin{equation}
(\vec{u}\otimes\vec{v})\vec{w}=\vec{u}(\vec{v},\vec{w})
\end{equation}
+holds.
\end{remark}
\section{Spherical coordinates}
projects / lectures / latex.git / blobdiff