started working on eigenvalue
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Robert
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\subfile{sections/vector_spaces.tex}
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\subfile{sections/matrices.tex}
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\subfile{sections/determinant.tex}
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\subfile{sections/scalar_product.tex}
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\subfile{sections/eigenvalue.tex}
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\end{document}
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18
chapters/sections/eigenvalue.tex
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18
chapters/sections/eigenvalue.tex
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\documentclass[../../script.tex]{subfiles}
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% !TEX root = ../../script.tex
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\begin{document}
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\section{Eigenvalue problems}
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\begin{defi}
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Let $A \in \field^{n \times n}$. Then $\lambda \in \field$ is called an eigenvalue of $A$, if
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\[
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\exists v \in \field^n, ~v \ne 0: ~~Av = \lambda v
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\]
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Such a vector $v$ is called eigenvector. We call
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\[
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\set[Av = \lambda v]{v \in \field^n} =: E_{\lambda}
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\]
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eigenspace belonging to $\lambda$
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\end{defi}
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\end{document}
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@ -60,7 +60,7 @@ $\field^{n\times m}$ is the space of all $n \times m$-matrices. The following op
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\item Conjugate transposition
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\[
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\conj{A} = \left(\overline{A}\right)^T
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A^H = \left(\conj{A}\right)^T
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\]
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\end{enumerate}
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\end{defi}
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272
chapters/sections/scalar_product.tex
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272
chapters/sections/scalar_product.tex
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\documentclass[../../script.tex]{subfiles}
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% !TEX root = ../../script.tex
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\begin{document}
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\section{Scalar Product}
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In this section $V$ will always denote a vector space and $\field$ a field (either $\realn$ or $\cmpln$).
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\begin{defi}
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A scalar product is a mapping
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\[
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\innerproduct{\cdot}{\cdot}: V \times V \longrightarrow \field
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\]
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that fulfils the following conditions:
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$\forall v_1, v_2, w_1, w_2 \in V, ~~\lambda \in \field$
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\begin{align*}
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\text{Linearity} && &\innerproduct{v_1}{w_1 + \lambda w_2} = \innerproduct{w_1}{w_1} + \lambda\innerproduct{v_1}{w_2} \\
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\text{Conjugated symmetry} && &\innerproduct{v_1}{w_1} = \conj{\innerproduct{w_1}{v_1}} \\
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\text{Positivity} && &\innerproduct{v_1}{v_1} \ge 0 \\
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\text{Definedness} && &\innerproduct{v_1}{v_2} = 0 \implies v_1 = 0 \\
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\text{Conjugated linearity} && &\innerproduct{v_1 + \lambda v_2}{w_1} = \innerproduct{v_1}{w_1} + \conj{\lambda} \innerproduct{v_2}{w_1} \\
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\end{align*}
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The mapping
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\begin{align*}
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\norm{\cdot}: V &\longrightarrow \field \\
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v &\longmapsto \sqrt{\innerproduct{v}{v}}
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\end{align*}
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\end{defi}
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\begin{eg}
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On $\realn^n$ the following is a scalar product
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\[
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\innerproduct{(x_1, x_2, \cdots, x_n)^T}{(y_1, y_2, \cdots, y_n)^T} = \series[n]{k} x_ky_k
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\]
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The norm is then equivalent to the Pythagorean theorem
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\[
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\norm{v} = \sqrt{\innerproduct{v}{v}} = \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2}
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\]
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Analogously for $\cmpln^n$
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\[
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\innerproduct{(u_1, u_2, \cdots, u_n)^T}{(v_1, v_2, \cdots, v_n)^T} = \series[n]{k} \conj{u_k}v_k
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\]
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\end{eg}
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\begin{rem}
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\begin{itemize}
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\item The length of $v \in V$ is $\norm{v}$
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\item The distance between elements $v, w \in V$ is $\norm{v-w}$
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\item The angle $\phi$ between $v, w \in V$ is $\cos \phi = \frac{\innerproduct{v}{w}}{\norm{v}\cdot\norm{w}}$
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\end{itemize}
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\end{rem}
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\begin{thm}
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Let $v, w \in V$. Then
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\begin{align*}
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\text{Cauchy-Schwarz-Inequality} && &|\innerproduct{v}{w}| \le \norm{v}\norm{w} \\
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\text{Triangle Inequality} && &\norm{v + w} \le \norm{v} + \norm{w}
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\end{align*}
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\end{thm}
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\begin{proof}
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For $\lambda \in \field$ we know that
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\begin{equation}
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\begin{split}
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0 \le \innerproduct{v - \lambda w}{v - \lambda w} &= \innerproduct{v - \lambda w}{v} - \lambda\innerproduct{v - \lambda w}{w} \\
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&= \innerproduct{v}{v} - \conj{\lambda}\innerproduct{w}{v} - \lambda\innerproduct{v}{w} + \underbrace{\lambda\conj{\lambda}}_{|\lambda|^2}\innerproduct{w}{w}
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\end{split}
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\end{equation}
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Let $\lambda = \frac{\innerproduct{w}{v}}{\norm{w}^2}$. Then
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\begin{equation}
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\begin{split}
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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 \\
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&= \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}} \\
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&= \norm{v}^2 - \frac{|\innerproduct{w}{v}|^2}{\norm{w}^2}
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\end{split}
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\end{equation}
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Through the monotony of the square root this implies that
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\begin{equation}
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|\innerproduct{w}{v}| \le \norm{v} \norm{w}
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\end{equation}
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To prove the triangle inequality, consider
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\begin{equation}
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\begin{split}
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||v + w||^2 &= \innerproduct{v+w}{v+w} \\
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&= \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} \\
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&\le \norm{v}^2 + 2 \cdot \Re\innerproduct{v}{w} + \norm{w}^2 \\
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&\le \norm{v}^2 + 2\norm{v}\norm{w} + \norm{w}^2 \\
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&= (\norm{v} + \norm{w})^2
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\end{split}
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\end{equation}
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Using the same argument as above, this implies
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\begin{equation}
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\norm{v + w} \le \norm{v} + \norm{w}
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\end{equation}
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\end{proof}
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\begin{defi}
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$v, w \in V$ are called orthogonal if
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\[
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\innerproduct{v}{w} = 0
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\]
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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.
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\[
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\forall i,j \in \set{1, \cdots, m}: \innerproduct{v_i}{v_j} = 0
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\]
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If $\norm{v_i} = 1$, then the $v_i$ are called an orthonormal set. If their span is $V$ they are an orthonormal basis.
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\end{defi}
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\begin{thm}
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If $v_1, \cdots, v_n$ are an orthonormal set, they are linearly independent.
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\end{thm}
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\begin{proof}
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Let $\alpha_1, \cdots, \alpha_n \in \field$, such that
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\begin{equation}
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0 = \alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n
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\end{equation}
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Then
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\begin{equation}
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\begin{split}
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0 &= \innerproduct{v_i}{0} = \innerproduct{v_i}{\alpha_1v_1 + \alpha_2v_2 + \cdots + \alpha_nv_n} \\
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&= \alpha_1\innerproduct{v_i}{v_1} + \alpha_2\innerproduct{v_i}{v_2} + \cdots + \alpha_n\innerproduct{v_i}{v_n} \\
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&= \alpha_i \innerproduct{v_i}{v_i} ~~i \in \set{1, \cdots, n}
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\end{split}
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\end{equation}
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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.
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\end{proof}
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\begin{eg}
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\begin{enumerate}[(i)]
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\item The canonical basis in $\realn^n$ is an orthonormal basis regarding the canonical scalar product.
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\item Let $\phi \in \realn$. Then
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\begin{align*}
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v_1 = (\cos\phi, \sin\phi)^T && v_2 = (-\sin\phi, \cos\phi)^T
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\end{align*}
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are an orthonormal basis for $\realn^2$
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\end{enumerate}
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\end{eg}
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\begin{thm}
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Let $v_1, \cdots, v_n$ be an orthonormal basis of $V$. Then for $v \in V$:
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\[
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v = \series[n]{i} \innerproduct{v_i}{v} v_i
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\]
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\end{thm}
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\begin{proof}
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Since $v_1, \cdots, v_n$ is a basis,
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\begin{equation}
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\exists \alpha_1, \cdots, \alpha_n \in \field: ~~v = \series[n]{i} \alpha_i v_i
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\end{equation}
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And therefore, for $j \in \set{1, \cdots, n}$
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\begin{equation}
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\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}
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\end{equation}
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\end{proof}
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\begin{thm}
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Let $A \in \field^{m \times n}$ and $\innerproduct{\cdot}{\cdot}$ the canonical scalar product on $\field^n$. Then
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\[
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\innerproduct{v}{Aw} = \innerproduct{A^Hv}{w}
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\]
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\end{thm}
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\begin{proof}
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First consider
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\begin{multicols}{2}
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\begin{subequations}
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\noindent
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\begin{equation}
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(Aw)_i = \series[n]{j} A_{ij} w_i
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\end{equation}
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\begin{equation}
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(A^H w)_j = \series[n]{i} A_{ji} v_i
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\end{equation}
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\end{subequations}
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\end{multicols}
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Now we can compute
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\begin{equation}
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\begin{split}
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\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 \\
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&= \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 \\
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&= \series[n]{j} \conj{(A^H v)_j} \cdot w_j \\
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&= \innerproduct{A^H v}{w}
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\end{split}
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\end{equation}
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\end{proof}
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\begin{defi}
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A matrix $A \in \realn^{n \times n}$ is called orthogonal if
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\[
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A^T A = AA^T = I
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\]
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or
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\[
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A^T = A^{-1}
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\]
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The set of all orthogonal matrices
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\[
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O(n) := \set[A^T A = I]{A \in \realn{n \times n}}
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\]
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is called the orthogonal group.
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\[
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SO(n) = \set[A^TA = I \wedge \det A = 1]{A = \realn{n \times n}} \subset O(n)
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\]
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is called the special orthogonal group.6
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\end{defi}
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\begin{eg}
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Let $\phi \in [0, 2\pi]$, then
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\[
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A = \begin{pmatrix}
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\cos \phi & -\sin \phi \\
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\sin \phi & \cos \phi
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\end{pmatrix}
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\]
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is orthogonal.
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\end{eg}
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\begin{rem}
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\begin{enumerate}[(i)]
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\item Let $A, B \in \field^{n \times n}$, then
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\[
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AB = I \implies BA = I
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\]
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\item \[
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1 = \det I = \det A^TA = \det A^T \cdot \det A = {\det}^2 A
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\]
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\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 $A$.
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Since the rows of $A^T$ are the columns of $A$, we can conclude that
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\[
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A \text{ orthogonal} \iff \innerproduct{r_i}{r_j} = \delta_{ij}
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\]
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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.
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\item Let $A$ be orthogonal, and $x, y \in \realn^n$
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\begin{align*}
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\innerproduct{Ax}{Ay} &= \innerproduct{A^TAx}{y} = \innerproduct{x}{y} \\
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\norm{Ax} &= \sqrt{\innerproduct{Ax}{Ax}} = \sqrt{\innerproduct{x}{x}} = \norm{x}
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\end{align*}
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$A$ perserves scalar products, lengths, distances and angles. These kinds of operations are called mirroring and rotation.
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\item Let $A, B \in O(n)$
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\[
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(AB)^T \cdot (AB) = B^TA^TAB = B^TIB = I
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\]
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This implies $(AB) \in O(n)$. It also implies $I \in O(n)$. Now consider $A \in O(n)$. Then
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\[
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(\inv{A})^T \inv{A} = (A^T)^T \cdot A^T = AA^T = I
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\]
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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.
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\end{enumerate}
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\end{rem}
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\begin{eg}
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$O(n)$, $SO(n)$, $\realn \setminus \set{0}$, $\cmpln \setminus \set{0}$, $Gl(n)$ (set of invertible matrices) and $S_n$ are all groups.
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\end{eg}
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\begin{defi}
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A matrix $U \in \cmpln^{n \times n}$ is called unitary if
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\[
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U^H U = I = UU^H
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\]
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We also introduce
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\[
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\set[U^HU = I]{U \in \cmpln{n \times n}}
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\]
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the unitary group, and
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\[
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\set[U^HU = I \wedge \det U = 1]{U \in \cmpln{n \times n}}
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\]
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the special unitary group.
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\end{defi}
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\end{document}
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BIN
script.pdf
BIN
script.pdf
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script.tex
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script.tex
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linkcolor=black,
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urlcolor=black
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}
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\usepackage[
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type={CC},
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modifier={by-sa},
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version={4.0},
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]{doclicense}
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\usepackage{subfiles}
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\newcommand{\rcseqdef}[1]{\seq{#1} \subset \realn \text{ (or } \cmpln \text{)}}
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\newcommand{\series}[2][\infty]{\sum_{#2 = 1}^{#1}}
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\newcommand{\finite}{\text{ finite}}
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\newcommand{\conj}[1]{#1^{\ast}}
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\newcommand{\conj}[1]{\overline{#1}}
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\newcommand{\conv}[1]{\xrightarrow{\makebox[2em][c]{$\scriptstyle#1$}}}
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\newcommand{\convinf}{\conv{n \rightarrow \infty}}
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{\,\middle|\, #1}%
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\right\}%
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}
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\renewcommand{\innerproduct}[2]{\langle#1,#2\rangle}
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\newcommand{\equalexpl}[1]{%
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\underset{\substack{\big\uparrow\\\mathrlap{\text{\hspace{-1.5em}#1}}}}{=}}
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\Crefname{cor}{Corollary}{Corollaries}
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\crefname{cor}{Cor.}{Cors.}
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\usepackage[
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type={CC},
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modifier={by-sa},
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version={4.0},
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]{doclicense}
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\begin{document}
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\title{Mathematics for Physicists}
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\author{https://www.github.com/Lauchmelder23/Mathematics}
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