\documentclass[../../script.tex]{subfiles} % !TEX root = ../../script.tex \begin{document} \section{Scalar Product} In this section $V$ will always denote a vector space and $\field$ a field (either $\realn$ or $\cmpln$). \begin{defi} A scalar product is a mapping \[ \innerproduct{\cdot}{\cdot}: V \times V \longrightarrow \field \] that fulfils the following conditions: $\forall v_1, v_2, w_1, w_2 \in V, ~~\lambda \in \field$ \begin{align*} \text{Linearity} && &\innerproduct{v_1}{w_1 + \lambda w_2} = \innerproduct{w_1}{w_1} + \lambda\innerproduct{v_1}{w_2} \\ \text{Conjugated symmetry} && &\innerproduct{v_1}{w_1} = \conj{\innerproduct{w_1}{v_1}} \\ \text{Positivity} && &\innerproduct{v_1}{v_1} \ge 0 \\ \text{Definedness} && &\innerproduct{v_1}{v_2} = 0 \implies v_1 = 0 \\ \text{Conjugated linearity} && &\innerproduct{v_1 + \lambda v_2}{w_1} = \innerproduct{v_1}{w_1} + \conj{\lambda} \innerproduct{v_2}{w_1} \\ \end{align*} The mapping \begin{align*} \norm{\cdot}: V &\longrightarrow \field \\ v &\longmapsto \sqrt{\innerproduct{v}{v}} \end{align*} \end{defi} \begin{eg} On $\realn^n$ the following is a scalar product \[ \innerproduct{(x_1, x_2, \cdots, x_n)^T}{(y_1, y_2, \cdots, y_n)^T} = \series[n]{k} x_ky_k \] The norm is then equivalent to the Pythagorean theorem \[ \norm{v} = \sqrt{\innerproduct{v}{v}} = \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2} \] Analogously for $\cmpln^n$ \[ \innerproduct{(u_1, u_2, \cdots, u_n)^T}{(v_1, v_2, \cdots, v_n)^T} = \series[n]{k} \conj{u_k}v_k \] \end{eg} \begin{rem} \begin{itemize} \item The length of $v \in V$ is $\norm{v}$ \item The distance between elements $v, w \in V$ is $\norm{v-w}$ \item The angle $\phi$ between $v, w \in V$ is $\cos \phi = \frac{\innerproduct{v}{w}}{\norm{v}\cdot\norm{w}}$ \end{itemize} \end{rem} \begin{thm} Let $v, w \in V$. Then \begin{align*} \text{Cauchy-Schwarz-Inequality} && &|\innerproduct{v}{w}| \le \norm{v}\norm{w} \\ \text{Triangle Inequality} && &\norm{v + w} \le \norm{v} + \norm{w} \end{align*} \end{thm} \begin{proof} For $\lambda \in \field$ we know that \begin{equation} \begin{split} 0 \le \innerproduct{v - \lambda w}{v - \lambda w} &= \innerproduct{v - \lambda w}{v} - \lambda\innerproduct{v - \lambda w}{w} \\ &= \innerproduct{v}{v} - \conj{\lambda}\innerproduct{w}{v} - \lambda\innerproduct{v}{w} + \underbrace{\lambda\conj{\lambda}}_{|\lambda|^2}\innerproduct{w}{w} \end{split} \end{equation} Let $\lambda = \frac{\innerproduct{w}{v}}{\norm{w}^2}$. Then \begin{equation} \begin{split} 0 &\le \norm{v}^2 - \frac{\conj{\innerproduct{w}{v}}}{\norm{w}^2} \cdot \innerproduct{w}{v} - \frac{\innerproduct{w}{v}}{\norm{w}^2} \cdot \innerproduct{v}{w} + \frac{|\innerproduct{w}{v}|^2}{\norm{w}^4} \norm{w}^2 \\ &= \norm{v}^2 - \frac{|\innerproduct{w}{v}|^2}{\norm{w}^2} - \cancel{\frac{|\innerproduct{w}{v}|^2}{\norm{w}^2}} + \cancel{\frac{|\innerproduct{w}{v}|^2}{\norm{w}^2}} \\ &= \norm{v}^2 - \frac{|\innerproduct{w}{v}|^2}{\norm{w}^2} \end{split} \end{equation} Through the monotony of the square root this implies that \begin{equation} |\innerproduct{w}{v}| \le \norm{v} \norm{w} \end{equation} To prove the triangle inequality, consider \begin{equation} \begin{split} ||v + w||^2 &= \innerproduct{v+w}{v+w} \\ &= \underbrace{\innerproduct{v}{v}}_{\norm{v}^2} + \innerproduct{v}{w} + \underbrace{\innerproduct{w}{v}}_{\conj{\innerproduct{v}{w}}} + \underbrace{\innerproduct{w}{w}}_{\norm{w}^2} \\ &\le \norm{v}^2 + 2 \cdot \Re\innerproduct{v}{w} + \norm{w}^2 \\ &\le \norm{v}^2 + 2\norm{v}\norm{w} + \norm{w}^2 \\ &= (\norm{v} + \norm{w})^2 \end{split} \end{equation} Using the same argument as above, this implies \begin{equation} \norm{v + w} \le \norm{v} + \norm{w} \end{equation} \end{proof} \begin{defi} $v, w \in V$ are called orthogonal if \[ \innerproduct{v}{w} = 0 \] The elements $v_1, \cdots, v_m \in V$ are called an orthogonal set if they are non-zero and they are pairwise orthogonal. I.e. \[ \forall i,j \in \set{1, \cdots, m}: \innerproduct{v_i}{v_j} = 0 \] If $\norm{v_i} = 1$, then the $v_i$ are called an orthonormal set. If their span is $V$ they are an orthonormal basis. \end{defi} \begin{thm} If $v_1, \cdots, v_n$ are an orthonormal set, they are linearly independent. \end{thm} \begin{proof} Let $\alpha_1, \cdots, \alpha_n \in \field$, such that \begin{equation} 0 = \alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n \end{equation} Then \begin{equation} \begin{split} 0 &= \innerproduct{v_i}{0} = \innerproduct{v_i}{\alpha_1v_1 + \alpha_2v_2 + \cdots + \alpha_nv_n} \\ &= \alpha_1\innerproduct{v_i}{v_1} + \alpha_2\innerproduct{v_i}{v_2} + \cdots + \alpha_n\innerproduct{v_i}{v_n} \\ &= \alpha_i \innerproduct{v_i}{v_i} ~~i \in \set{1, \cdots, n} \end{split} \end{equation} Since $v_i$ is not a zero vector, $\innerproduct{v_i}{v_i} \ne 0$, and thus $\alpha_i = 0$. Since $i$ is arbitrary, the $v_i$ are linearly independent. \end{proof} \begin{eg} \begin{enumerate}[(i)] \item The canonical basis in $\realn^n$ is an orthonormal basis regarding the canonical scalar product. \item Let $\phi \in \realn$. Then \begin{align*} v_1 = (\cos\phi, \sin\phi)^T && v_2 = (-\sin\phi, \cos\phi)^T \end{align*} are an orthonormal basis for $\realn^2$ \end{enumerate} \end{eg} \begin{thm} Let $v_1, \cdots, v_n$ be an orthonormal basis of $V$. Then for $v \in V$: \[ v = \series[n]{i} \innerproduct{v_i}{v} v_i \] \end{thm} \begin{proof} Since $v_1, \cdots, v_n$ is a basis, \begin{equation} \exists \alpha_1, \cdots, \alpha_n \in \field: ~~v = \series[n]{i} \alpha_i v_i \end{equation} And therefore, for $j \in \set{1, \cdots, n}$ \begin{equation} \innerproduct{v_j}{v} = \series[n]{i} \alpha_i \innerproduct{v_j}{v_i} = \alpha_j \underbrace{\innerproduct{v_j}{v_j}}_{\norm{v_j}^2 = 1} \end{equation} \end{proof} \begin{thm} Let $A \in \field^{m \times n}$ and $\innerproduct{\cdot}{\cdot}$ the canonical scalar product on $\field^n$. Then \[ \innerproduct{v}{Aw} = \innerproduct{A^Hv}{w} \] \end{thm} \begin{proof} First consider \begin{multicols}{2} \begin{subequations} \noindent \begin{equation} (Aw)_i = \series[n]{j} A_{ij} w_i \end{equation} \begin{equation} (A^H w)_j = \series[n]{i} A_{ji} v_i \end{equation} \end{subequations} \end{multicols} Now we can compute \begin{equation} \begin{split} \innerproduct{v}{Aw} &= \series[n]{i} \conj{v_i} (Aw)_i = \series[n]{i}\left(\conj{v_i} \cdot \series[n]{j} A_{ij} w_j \right) = \series[n]{i}\series[n]{j} A_{ij} \conj{v_i}w_j \\ &= \series[n]{j} \left(\series[n]{i} A{ij} \conj{v_i} \right) w_j = \series[n]{j} \left(\conj{\series[n]{i} \conj{A_{ij}} v_i}\right) w_j \\ &= \series[n]{j} \conj{(A^H v)_j} \cdot w_j \\ &= \innerproduct{A^H v}{w} \end{split} \end{equation} \end{proof} \begin{defi} A matrix $A \in \realn^{n \times n}$ is called orthogonal if \[ A^T A = AA^T = I \] or \[ A^T = A^{-1} \] The set of all orthogonal matrices \[ O(n) := \set[A^T A = I]{A \in \realn{n \times n}} \] is called the orthogonal group. \[ SO(n) = \set[A^TA = I \wedge \det A = 1]{A = \realn{n \times n}} \subset O(n) \] is called the special orthogonal group.6 \end{defi} \begin{eg} Let $\phi \in [0, 2\pi]$, then \[ A = \begin{pmatrix} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{pmatrix} \] is orthogonal. \end{eg} \begin{rem} \begin{enumerate}[(i)] \item Let $A, B \in \field^{n \times n}$, then \[ AB = I \implies BA = I \] \item \[ 1 = \det I = \det A^TA = \det A^T \cdot \det A = {\det}^2 A \] \item The $i$-$j$-component of $A^TA$ is equal to the canonical scalar product of the $i$-th row of $A^T$ and the $j$-th column of $A$. Since the rows of $A^T$ are the columns of $A$, we can conclude that \[ A \text{ orthogonal} \iff \innerproduct{r_i}{r_j} = \delta_{ij} \] where the $r_i$ are the columns of $A$. In this case, the $r_i$ are an orthonormal basis on $\realn^n$. This works analogously for the rows. \item Let $A$ be orthogonal, and $x, y \in \realn^n$ \begin{align*} \innerproduct{Ax}{Ay} &= \innerproduct{A^TAx}{y} = \innerproduct{x}{y} \\ \norm{Ax} &= \sqrt{\innerproduct{Ax}{Ax}} = \sqrt{\innerproduct{x}{x}} = \norm{x} \end{align*} $A$ perserves scalar products, lengths, distances and angles. These kinds of operations are called mirroring and rotation. \item Let $A, B \in O(n)$ \[ (AB)^T \cdot (AB) = B^TA^TAB = B^TIB = I \] This implies $(AB) \in O(n)$. It also implies $I \in O(n)$. Now consider $A \in O(n)$. Then \[ (\inv{A})^T \inv{A} = (A^T)^T \cdot A^T = AA^T = I \] This implies $\inv{A} \in O(T)$. Such a structure (a set with a multiplication operation, neutral element and multiplicative inverse) is called a group. \end{enumerate} \end{rem} \begin{eg} $O(n)$, $SO(n)$, $\realn \setminus \set{0}$, $\cmpln \setminus \set{0}$, $Gl(n)$ (set of invertible matrices) and $S_n$ are all groups. \end{eg} \begin{defi} A matrix $U \in \cmpln^{n \times n}$ is called unitary if \[ U^H U = I = UU^H \] We also introduce \[ \set[U^HU = I]{U \in \cmpln{n \times n}} \] the unitary group, and \[ \set[U^HU = I \wedge \det U = 1]{U \in \cmpln{n \times n}} \] the special unitary group. \end{defi} \end{document}