2021-02-04 15:08:52 +01:00
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%% ********************************************************************************
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%% AUTHOR: Raj Dahya
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%% CREATED: November 2020
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%% EDITED: —
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%% TYPE: Notizen
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%% TITLE: Zusatzaufgaben
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%% DOI: —
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%% DEPARTMENT: Fakultät for Mathematik und Informatik
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%% INSTITUTE: Universität Leipzig
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%% ********************************************************************************
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%% ********************************************************************************
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%% DOCUMENT STRUCTURE:
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%% ~~~~~~~~~~~~~~~~~~~
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%%
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%% - root.tex;
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%% |
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%% ---- parameters.tex;
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%% |
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2021-03-09 13:03:51 +01:00
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%% ---- srclocal/index.tex;
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2021-02-04 15:08:52 +01:00
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%% |
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2021-03-09 13:03:51 +01:00
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%% ---- src/setup-type.tex;
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2021-02-04 15:08:52 +01:00
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%% |
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2021-03-09 13:03:51 +01:00
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%% ---- src/setup-packages.tex;
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2021-02-04 15:08:52 +01:00
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%% |
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2021-03-09 13:03:51 +01:00
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%% ---- src/setup-parameters.tex;
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2021-02-04 15:08:52 +01:00
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%% |
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2021-03-09 13:03:51 +01:00
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%% ---- src/setup-macros.tex;
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2021-02-04 15:08:52 +01:00
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%% |
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2021-03-09 13:03:51 +01:00
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%% ---- src/setup-environments.tex;
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2021-02-04 15:08:52 +01:00
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%% |
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2021-03-09 13:03:51 +01:00
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%% ---- src/setup-layout.tex;
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2021-02-04 15:08:52 +01:00
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%% |
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2021-03-09 13:03:51 +01:00
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%% ---- srclocal/setup-localmacros.tex;
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2021-02-04 15:08:52 +01:00
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%% |
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%% ---- front/index.tex;
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%% |
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%% ---- front/title.tex;
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%% |
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%% ---- front/foreword.tex;
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%% |
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%% ---- front/contents.tex;
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%% |
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%% ---- body/index.tex;
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%% |
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%% ---- body/linear-extensions.tex;
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2021-02-08 20:56:52 +01:00
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%% |
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%% ---- body/linear-systems.tex;
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2021-02-04 15:08:52 +01:00
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%% |
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%% ---- back/index.tex;
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%% |
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2021-02-09 17:32:10 +01:00
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%% ---- ########;
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2021-02-04 15:08:52 +01:00
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%%
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%% DOCUMENT-RANDOM-SEED: 5637845
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%% ********************************************************************************
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%% ********************************************************************************
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%% FILE: root.tex
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%% ********************************************************************************
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%% ********************************************************************************
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%% FILE: parameters.tex
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%% ********************************************************************************
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2021-03-09 13:03:51 +01:00
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%% ********************************************************************************
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%% FILE: srclocal/index.tex
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%% ********************************************************************************
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\makeatletter
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%% ********************************************************************************
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%% FILE: src/setup-type.tex
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%% ********************************************************************************
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\documentclass[
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12pt,
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a4paper,
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oneside,
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openright,
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center,
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chapterbib,
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crosshair,
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fleqn,
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headcount,
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headline,
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indent,
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indentfirst=false,
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portrait,
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phonetic,
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oldernstyle,
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onecolumn,
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sfbold,
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upper,
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]{scrbook}
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%% ********************************************************************************
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%% FILE: src/setup-packages.tex
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%% ********************************************************************************
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\PassOptionsToPackage{T2A,OT1}{fontenc} % T1,OT1,T2A,OT2
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\PassOptionsToPackage{utf8}{inputenc} % utf8
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\PassOptionsToPackage{british,english,ngerman,russian}{babel}
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\PassOptionsToPackage{
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english,
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ngerman,
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russian,
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capitalise,
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}{cleveref}
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\PassOptionsToPackage{
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bookmarks=true,
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bookmarksopen=false,
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bookmarksopenlevel=0,
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bookmarkstype=toc,
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colorlinks=false,
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raiselinks=true,
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hyperfigures=true,
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}{hyperref}
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\PassOptionsToPackage{
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reset,
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left=1in,
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right=1in,
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top=20mm,
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bottom=20mm,
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heightrounded,
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}{geometry}
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\PassOptionsToPackage{
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framemethod=TikZ,
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}{mdframed}
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\PassOptionsToPackage{normalem}{ulem}
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\PassOptionsToPackage{
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amsmath,
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thmmarks,
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}{ntheorem}
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\PassOptionsToPackage{table}{xcolor}
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\PassOptionsToPackage{
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all,
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color,
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curve,
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frame,
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import,
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knot,
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line,
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movie,
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rotate,
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textures,
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tile,
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tips,
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web,
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xdvi,
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}{xy}
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\usepackage{amsfonts}
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\usepackage{amsmath}
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\usepackage{amssymb}
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\usepackage{ntheorem} % <— muss nach den ams* Packages vorkommen!!
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\usepackage{array}
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\usepackage{babel}
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\usepackage{bbding}
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\usepackage{bbm}
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\usepackage{calc}
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\usepackage{sectsty}
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\usepackage{titlesec}
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\usepackage{fancyhdr}
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\usepackage{footmisc}
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\usepackage{geometry}
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\usepackage{graphicx}
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\usepackage{ifpdf}
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\usepackage{ifthen}
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\usepackage{ifnextok}
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\usepackage{longtable}
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\usepackage{multicol}
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\usepackage{multirow}
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\usepackage{nameref}
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\usepackage{nowtoaux}
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\usepackage{paralist}
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\usepackage{enumerate} %% nach [paralist]
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\usepackage{pgf}
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\usepackage{pgfplots}
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\usepackage{proof}
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\usepackage{refcount}
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\usepackage{relsize}
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\usepackage{savesym}
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\usepackage{stmaryrd}
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\usepackage{subfigure}
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\usepackage{yfonts} %% <— Altgotische Fonts
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\usepackage{tikz}
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\usepackage{xy}
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\usepackage{undertilde}
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\usepackage{ulem} %% <– f\"ur besseren \underline-Befehl (\ul)
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\usepackage{xcolor}
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\usepackage{xspace}
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\usepackage{xstring}
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\usepackage{hyperref}
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\usepackage{cleveref} % must vor hyperref geladen werden.
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\pgfplotsset{compat=newest}
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\usetikzlibrary{
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angles,
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arrows,
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automata,
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calc,
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decorations,
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decorations.pathmorphing,
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decorations.pathreplacing,
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math,
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positioning,
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patterns,
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quotes,
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snakes,
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}
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%% \var ≈ alter Befehl
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%% \xvar ≈ wie das neue Package \var interpretieren soll.
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\savesymbol{Diamond}
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\savesymbol{emptyset}
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\savesymbol{ggg}
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\savesymbol{int}
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\savesymbol{lll}
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\savesymbol{RectangleBold}
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\savesymbol{langle}
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\savesymbol{rangle}
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\savesymbol{hookrightarrow}
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\savesymbol{hookleftarrow}
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\savesymbol{Asterisk}
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\usepackage{mathabx}
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\usepackage{wasysym}
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\let\varemptyset=\emptyset
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\restoresymbol{x}{Diamond}
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\restoresymbol{x}{emptyset}
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\restoresymbol{x}{ggg}
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\restoresymbol{x}{int}
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\restoresymbol{x}{lll}
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\restoresymbol{x}{RectangleBold}
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\restoresymbol{x}{langle}
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\restoresymbol{x}{rangle}
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\restoresymbol{x}{hookrightarrow}
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\restoresymbol{x}{hookleftarrow}
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\restoresymbol{x}{Asterisk}
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\ifpdf
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\usepackage{pdfcolmk}
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\fi
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\usepackage{mdframed}
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%% Force-Import aus MnSymbol
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\DeclareFontFamily{U}{MnSymbolA}{}
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\DeclareFontShape{U}{MnSymbolA}{m}{n}{
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<-6> MnSymbolA5
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<6-7> MnSymbolA6
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<7-8> MnSymbolA7
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<8-9> MnSymbolA8
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<9-10> MnSymbolA9
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<10-12> MnSymbolA10
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<12-> MnSymbolA12
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}{}
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\DeclareFontShape{U}{MnSymbolA}{b}{n}{
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<-6> MnSymbolA-Bold5
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<6-7> MnSymbolA-Bold6
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<7-8> MnSymbolA-Bold7
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<8-9> MnSymbolA-Bold8
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<9-10> MnSymbolA-Bold9
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<10-12> MnSymbolA-Bold10
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<12-> MnSymbolA-Bold12
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}{}
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\DeclareSymbolFont{MnSyA}{U}{MnSymbolA}{m}{n}
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\DeclareMathSymbol{\lcirclearrowright}{\mathrel}{MnSyA}{252}
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\DeclareMathSymbol{\lcirclearrowdown}{\mathrel}{MnSyA}{255}
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\DeclareMathSymbol{\rcirclearrowleft}{\mathrel}{MnSyA}{250}
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\DeclareMathSymbol{\rcirclearrowdown}{\mathrel}{MnSyA}{251}
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\DeclareFontFamily{U}{MnSymbolC}{}
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\DeclareSymbolFont{MnSyC}{U}{MnSymbolC}{m}{n}
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\DeclareFontShape{U}{MnSymbolC}{m}{n}{
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<-6> MnSymbolC5
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<6-7> MnSymbolC6
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<7-8> MnSymbolC7
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<8-9> MnSymbolC8
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<9-10> MnSymbolC9
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<10-12> MnSymbolC10
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<12-> MnSymbolC12%
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}{}
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\DeclareMathSymbol{\powerset}{\mathord}{MnSyC}{180}
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%% ********************************************************************************
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%% FILE: src/setup-parameters.tex
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%% ********************************************************************************
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\def\boolwahr{true}
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\def\boolfalsch{false}
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\def\boolleer{}
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\let\documenttwosided\boolfalsch
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\let\boolinappendix\boolfalsch
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\let\boolinmdframed\boolfalsch
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\let\eqtagset\boolfalsch
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\let\eqtaglabel\boolleer
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\let\eqtagsymb\boolleer
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\newcount\bufferctr
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\newcount\bufferreplace
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\newcounter{columnanzahl}
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\newlength\rtab
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\newlength\gesamtlinkerRand
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\newlength\gesamtrechterRand
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\newlength\ownspaceabovethm
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\newlength\ownspacebelowthm
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\setlength{\rtab}{0.025\textwidth}
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\setlength{\ownspaceabovethm}{0.5\baselineskip}
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\setlength{\ownspacebelowthm}{0.5\baselineskip}
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\setlength{\gesamtlinkerRand}{0pt}
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\setlength{\gesamtrechterRand}{0pt}
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\def\secnumberingpt{$\cdot$}
|
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|
\def\secnumberingseppt{.}
|
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\def\subsecnumberingseppt{}
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\def\thmnumberingpt{$\cdot$}
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\def\thmnumberingseppt{}
|
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|
|
\def\thmForceSepPt{.}
|
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\definecolor{leer}{gray}{1}
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\definecolor{hellgrau}{gray}{0.85}
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\definecolor{dunkelgrau}{gray}{0.5}
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\definecolor{maroon}{rgb}{0.6901961,0.1882353,0.3764706}
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\definecolor{dunkelgruen}{rgb}{0.015625,0.363281,0.109375}
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\definecolor{dunkelrot}{rgb}{0.5450980392,0,0}
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\definecolor{dunkelblau}{rgb}{0,0,0.5450980392}
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\definecolor{blau}{rgb}{0,0,1}
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\definecolor{newresult}{rgb}{0.6,0.6,0.6}
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\definecolor{improvedresult}{rgb}{0.9,0.9,0.9}
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\definecolor{hervorheben}{rgb}{0,0.9,0.7}
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\definecolor{starkesblau}{rgb}{0.1019607843,0.3176470588,0.8156862745}
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\definecolor{achtung}{rgb}{1,0.5,0.5}
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\definecolor{frage}{rgb}{0.5,1,0.5}
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\definecolor{schreibweise}{rgb}{0,0.7,0.9}
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\definecolor{axiom}{rgb}{0,0.3,0.3}
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|
|
%% ********************************************************************************
|
|
|
|
|
|
%% FILE: src/setup-macros.tex
|
|
|
|
|
|
%% ********************************************************************************
|
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|
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%% ****************************************************************
|
|
|
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|
|
%% TEX:
|
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|
|
%% ****************************************************************
|
|
|
|
|
|
|
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|
|
\def\let@name#1#2{\expandafter\let\csname #1\expandafter\endcsname\csname #2\endcsname\relax}
|
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|
|
|
\DeclareRobustCommand\crfamily{\fontfamily{ccr}\selectfont}
|
|
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|
|
\DeclareTextFontCommand{\textcr}{\crfamily}
|
|
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|
|
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|
|
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|
|
\def\nichtzeigen#1{\phantom{#1}}
|
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|
|
|
|
%% ****************************************************************
|
|
|
|
|
|
%% SPACING:
|
|
|
|
|
|
%% ****************************************************************
|
|
|
|
|
|
|
|
|
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|
|
\def\ifthenelseleer#1#2#3{\ifthenelse{\equal{#1}{}}{#2}{#1#3}}
|
|
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|
|
\def\bedingtesspaceexpand#1#2#3{\ifthenelseleer{\csname #1\endcsname}{#3}{#2#3}}
|
|
|
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|
|
\def\voritemise{\leavevmode\nvraum{1}}
|
|
|
|
|
|
\def\hraum{\null\hfill\null}
|
|
|
|
|
|
\def\vraum{\null\vfill\null}
|
|
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|
|
\def\nvraum{\@ifnextchar\bgroup{\nvraum@c}{\nvraum@bes}}
|
|
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|
|
\def\nvraum@c#1{\vspace*{-#1\baselineskip}}
|
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|
|
\def\nvraum@bes{\vspace*{-\baselineskip}}
|
|
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|
|
\def\erlaubeplatz{\relax\ifmmode\else\@\xspace\fi}
|
|
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|
|
\def\entferneplatz{\relax\ifmmode\else\expandafter\@gobble\fi}
|
|
|
|
|
|
|
|
|
|
|
|
%% ****************************************************************
|
|
|
|
|
|
%% TAGS / BEZEICHNUNGEN / LABELLING:
|
|
|
|
|
|
%% ****************************************************************
|
|
|
|
|
|
|
|
|
|
|
|
\def\send@toaux#1{\@bsphack\protected@write\@auxout{}{\string#1}\@esphack}
|
|
|
|
|
|
|
|
|
|
|
|
%% \rlabel{LABEL}[CTR]{CREF-SHORT}{CREF-LONG}{DISPLAYTEXT}
|
|
|
|
|
|
\def\rlabel#1[#2]#3#4#5{#5\rlabel@aux{#1}[#2]{#3}{#4}{#5}}
|
|
|
|
|
|
\def\rlabel@aux#1[#2]#3#4#5{%
|
|
|
|
|
|
\send@toaux{\newlabel{#1}{{\@currentlabel}{\thepage}{{\unexpanded{#5}}}{#2.\csname the#2\endcsname}{}}}\relax%
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
%% \tag@rawscheme{CREF-SHORT}{CREF-LONG}[CTR]{LEFT-BRKT}{RIGHT-BRKT} [LABEL]{DISPLAYTEXT}
|
|
|
|
|
|
\def\tag@rawscheme#1#2[#3]#4#5{\@ifnextchar[{\tag@rawscheme@{#1}{#2}[#3]{#4}{#5}}{\tag@rawscheme@{#1}{#2}[#3]{#4}{#5}[*]}}
|
|
|
|
|
|
\def\tag@rawscheme@#1#2[#3]#4#5[#6]{\@ifnextchar\bgroup{\tag@rawscheme@@{#1}{#2}[#3]{#4}{#5}[#6]}{\tag@rawscheme@@{#1}{#2}[#3]{#4}{#5}[#6]{}}}
|
|
|
|
|
|
\def\tag@rawscheme@@#1#2[#3]#4#5[#6]#7{%
|
|
|
|
|
|
\ifthenelse{\equal{#6}{*}}{%
|
|
|
|
|
|
\ifthenelse{\equal{#7}{\boolleer}}{\refstepcounter{#3}#4\csname the#3\endcsname#5}{#4#7#5}%
|
|
|
|
|
|
}{%
|
|
|
|
|
|
\refstepcounter{#3}#4%
|
|
|
|
|
|
\ifthenelse{\equal{#7}{\boolleer}}{\rlabel{#6}[#3]{#1}{#2}{\csname the#3\endcsname}}{\rlabel{#6}[#3]{#1}{#2}{#7}}%
|
|
|
|
|
|
#5%
|
|
|
|
|
|
}%
|
|
|
|
|
|
}
|
|
|
|
|
|
%% \tag@scheme{CREF-SHORT}{CREF-LONG}[CTR] [LABEL]{DISPLAYTEXT}
|
|
|
|
|
|
\def\tag@scheme#1#2[#3]{\tag@rawscheme{#1}{#2}[#3]{\upshape(}{\upshape)}}
|
|
|
|
|
|
|
|
|
|
|
|
%% \eqtag[LABEL]{DISPLAYTEXT}
|
|
|
|
|
|
\def\eqtag@post#1{\makebox[0pt][r]{#1}}
|
|
|
|
|
|
\def\eqtag@pre{\tag@scheme{Eq}{Equation}[Xe]}
|
|
|
|
|
|
\def\eqtag{\@ifnextchar[{\eqtag@}{\eqtag@[*]}}
|
|
|
|
|
|
\def\eqtag@[#1]{\@ifnextchar\bgroup{\eqtag@@[#1]}{\eqtag@@[#1]{}}}
|
|
|
|
|
|
\def\eqtag@@[#1]#2{\eqtag@post{\eqtag@pre[#1]{#2}}}
|
|
|
|
|
|
|
|
|
|
|
|
\def\eqcref#1{\text{(\ref{#1})}}
|
|
|
|
|
|
\def\ptcref#1{\ref{#1}}
|
|
|
|
|
|
\def\punktlabel#1{\label{it:#1:\beweislabel}}
|
|
|
|
|
|
\def\punktcref#1{\eqcref{it:#1:\beweislabel}}
|
|
|
|
|
|
\def\crefit#1#2{\cref{#1}~\eqcref{it:#2:#1}}
|
|
|
|
|
|
\def\Crefit#1#2{\Cref{#1}~\eqcref{it:#2:#1}}
|
|
|
|
|
|
|
|
|
|
|
|
%% UNDER/OVERSET BEFEHLE
|
|
|
|
|
|
\def\opfromto[#1]_#2^#3{\underset{#2}{\overset{#3}{#1}}}
|
|
|
|
|
|
\def\textoverset#1#2{\overset{\text{#1}}{#2}}
|
|
|
|
|
|
\def\textunderset#1#2{\underset{#2}{\text{#1}}}
|
|
|
|
|
|
\def\crefoverset#1#2{\textoverset{\cref{#1}}{#2}}
|
|
|
|
|
|
\def\Crefoverset#1#2{\textoverset{\Cref{#1}}{#2}}
|
|
|
|
|
|
\def\crefunderset#1#2{\textunderset{#2}{\cref{#1}}}
|
|
|
|
|
|
\def\Crefunderset#1#2{\textunderset{#2}{\Cref{#1}}}
|
|
|
|
|
|
\def\eqcrefoverset#1#2{\textoverset{\eqcref{#1}}{#2}}
|
|
|
|
|
|
\def\eqcrefunderset#1#2{\textunderset{#2}{\eqcref{#1}}}
|
|
|
|
|
|
\def\mathclap#1{#1}
|
|
|
|
|
|
\def\oberunterset#1{\@ifnextchar^{\oberunterset@oben{#1}}{\oberunterset@unten{#1}}}
|
|
|
|
|
|
\def\oberunterset@oben#1^#2_#3{\underset{\mathclap{#3}}{\overset{\mathclap{#2}}{#1}}}
|
|
|
|
|
|
\def\oberunterset@unten#1_#2^#3{\underset{\mathclap{#2}}{\overset{\mathclap{#3}}{#1}}}
|
|
|
|
|
|
\def\breitunderbrace#1_#2{\underbrace{#1}_{\mathclap{#2}}}
|
|
|
|
|
|
\def\breitoverbrace#1^#2{\overbrace{#1}^{\mathclap{#2}}}
|
|
|
|
|
|
\def\breitunderbracket#1_#2{\underbracket{#1}_{\mathclap{#2}}}
|
|
|
|
|
|
\def\breitoverbracket#1^#2{\overbracket{#1}^{\mathclap{#2}}}
|
|
|
|
|
|
|
|
|
|
|
|
\def\generatenestedsecnumbering#1#2#3{%
|
|
|
|
|
|
\expandafter\gdef\csname thelong#3\endcsname{%
|
|
|
|
|
|
\expandafter\csname the#2\endcsname%
|
|
|
|
|
|
\secnumberingpt%
|
|
|
|
|
|
\expandafter\csname #1\endcsname{#3}%
|
|
|
|
|
|
}%
|
|
|
|
|
|
\expandafter\gdef\csname theshort#3\endcsname{%
|
|
|
|
|
|
\expandafter\csname #1\endcsname{#3}%
|
|
|
|
|
|
}%
|
|
|
|
|
|
}
|
|
|
|
|
|
\def\generatenestedthmnumbering#1#2#3{%
|
|
|
|
|
|
\expandafter\gdef\csname the#3\endcsname{%
|
|
|
|
|
|
\expandafter\csname the#2\endcsname%
|
|
|
|
|
|
\thmnumberingpt%
|
|
|
|
|
|
\expandafter\csname #1\endcsname{#3}%
|
|
|
|
|
|
}%
|
|
|
|
|
|
\expandafter\gdef\csname theshort#3\endcsname{%
|
|
|
|
|
|
\expandafter\csname #1\endcsname{#3}%
|
|
|
|
|
|
}%
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
%% ****************************************************************
|
|
|
|
|
|
%% ALLG. MACROS:
|
|
|
|
|
|
%% ****************************************************************
|
|
|
|
|
|
|
|
|
|
|
|
\def\+#1{\addtocounter{#1}{1}}
|
|
|
|
|
|
\def\setcounternach#1#2{\setcounter{#1}{#2}\addtocounter{#1}{-1}}
|
|
|
|
|
|
\def\textsubscript#1{${}_{\textup{#1}}$}
|
|
|
|
|
|
\def\rome#1{\overline{\underline{#1}}}
|
|
|
|
|
|
\def\textTODO{\text{[{\large\textcolor{red}{More work needed!}}]}}
|
|
|
|
|
|
\def\hlineEIGENpt{\hdashline[0.5pt/5pt]}
|
|
|
|
|
|
\def\clineEIGENpt#1{\cdashline{#1}[0.5pt/5pt]}
|
|
|
|
|
|
|
|
|
|
|
|
\def\forcepunkt#1{#1\IfEndWith{#1}{.}{}{.}}
|
|
|
|
|
|
\def\lateinabkuerzung#1#2{%
|
|
|
|
|
|
\expandafter\gdef\csname #1\endcsname{\emph{#2}\@ifnextchar.{\entferneplatz}{\erlaubeplatz}}
|
|
|
|
|
|
}
|
|
|
|
|
|
\def\deutscheabkuerzung#1#2{%
|
|
|
|
|
|
\expandafter\gdef\csname #1\endcsname{{#2}\@ifnextchar.{\entferneplatz}{\erlaubeplatz}}
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
%% ****************************************************************
|
|
|
|
|
|
%% MATHE
|
|
|
|
|
|
%% ****************************************************************
|
|
|
|
|
|
|
|
|
|
|
|
\def\matrix#1{\left(\begin{array}{#1}}
|
|
|
|
|
|
\def\endmatrix{\end{array}\right)}
|
|
|
|
|
|
\def\smatrix{\left(\begin{smallmatrix}}
|
|
|
|
|
|
\def\endsmatrix{\end{smallmatrix}\right)}
|
|
|
|
|
|
|
|
|
|
|
|
\def\multiargrekursiverbefehl#1#2#3#4#5#6#7#8{%
|
|
|
|
|
|
\expandafter\gdef\csname#1\endcsname #2##1#4{\csname #1@anfang\endcsname##1#3\egroup}
|
|
|
|
|
|
\expandafter\def\csname #1@anfang\endcsname##1#3{#5##1\@ifnextchar\egroup{\csname #1@ende\endcsname}{#7\csname #1@mitte\endcsname}}
|
|
|
|
|
|
\expandafter\def\csname #1@mitte\endcsname##1#3{#6##1\@ifnextchar\egroup{\csname #1@ende\endcsname}{#7\csname #1@mitte\endcsname}}
|
|
|
|
|
|
\expandafter\def\csname #1@ende\endcsname##1{#8}
|
|
|
|
|
|
}
|
|
|
|
|
|
\multiargrekursiverbefehl{svektor}{[}{;}{]}{\begin{smatrix}}{}{\\}{\\\end{smatrix}}
|
|
|
|
|
|
\multiargrekursiverbefehl{vektor}{[}{;}{]}{\begin{matrix}{c}}{}{\\}{\\\end{matrix}}
|
|
|
|
|
|
\multiargrekursiverbefehl{vektorzeile}{}{,}{;}{}{&}{}{}
|
|
|
|
|
|
\multiargrekursiverbefehl{matlabmatrix}{[}{;}{]}{\begin{smatrix}\vektorzeile}{\vektorzeile}{;\\}{;\end{smatrix}}
|
|
|
|
|
|
|
|
|
|
|
|
\def\cases[#1]#2{\left\{\begin{array}[#1]{#2}}
|
|
|
|
|
|
\def\endcases{\end{array}\right.}
|
|
|
|
|
|
|
|
|
|
|
|
\def\BeweisRichtung[#1]{\@ifnextchar\bgroup{\@BeweisRichtung@c[#1]}{\@BeweisRichtung@bes[#1]}}
|
|
|
|
|
|
\def\@BeweisRichtung@bes[#1]{{\bfseries(#1).~}}
|
|
|
|
|
|
\def\@BeweisRichtung@c[#1]#2#3{{\bfseries(#2#1#3).~}}
|
|
|
|
|
|
\def\erzeugeBeweisRichtungBefehle#1#2{
|
|
|
|
|
|
\expandafter\gdef\csname #1text\endcsname##1##2{\BeweisRichtung[#2]{##1}{##2}}
|
|
|
|
|
|
\expandafter\gdef\csname #1\endcsname{%
|
|
|
|
|
|
\@ifnextchar\bgroup{\csname #1@\endcsname}{\csname #1text\endcsname{}{}}%
|
|
|
|
|
|
}
|
|
|
|
|
|
\expandafter\gdef\csname #1@\endcsname##1##2{%
|
|
|
|
|
|
\csname #1text\endcsname{\punktcref{##1}}{\punktcref{##2}}%
|
|
|
|
|
|
}
|
|
|
|
|
|
}
|
|
|
|
|
|
\erzeugeBeweisRichtungBefehle{hinRichtung}{$\Longrightarrow$}
|
|
|
|
|
|
\erzeugeBeweisRichtungBefehle{herRichtung}{$\Longleftarrow$}
|
|
|
|
|
|
\erzeugeBeweisRichtungBefehle{hinherRichtung}{$\Longleftrightarrow$}
|
|
|
|
|
|
|
|
|
|
|
|
\def\cal#1{\mathcal{#1}}
|
|
|
|
|
|
\def\brkt#1{\langle{}#1{}\rangle}
|
|
|
|
|
|
\def\mathfrak#1{\mbox{\usefont{U}{euf}{m}{n}#1}}
|
|
|
|
|
|
\def\kurs#1{\textit{#1}}
|
|
|
|
|
|
\def\rectangleblack{\text{\RectangleBold}}
|
|
|
|
|
|
\def\rectanglewhite{\text{\Rectangle}}
|
|
|
|
|
|
\def\squareblack{\blacksquare}
|
|
|
|
|
|
\def\squarewhite{\Box}
|
|
|
|
|
|
|
|
|
|
|
|
%% ********************************************************************************
|
|
|
|
|
|
%% FILE: src/setup-environments.tex
|
|
|
|
|
|
%% ********************************************************************************
|
|
|
|
|
|
|
|
|
|
|
|
%% **********************************************************************
|
|
|
|
|
|
%% CLEVEREF: ************************************************************
|
|
|
|
|
|
|
|
|
|
|
|
\def\crefname@full#1#2#3{\crefname{#1}{#2}{#3}\Crefname{#1}{#2}{#3}}
|
|
|
|
|
|
\crefname@full{chapter}{Kapitel}{Kapitel}
|
|
|
|
|
|
\crefname@full{section}{Abschnitt}{Abschnitte}
|
|
|
|
|
|
\crefname@full{figure}{Fig.}{Fig.}
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\crefname@full{subfigure}{Fig.}{Fig.}
|
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\crefname@full{proof}{Beweis}{Beweise}
|
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\crefname@full{thm}{Theorem}{Theoreme}
|
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|
\crefname@full{satz}{Satz}{Sätze}
|
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|
\crefname@full{claim}{Behauptung}{Behauptungen}
|
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|
|
\crefname@full{lemm}{Lemma}{Lemmata}
|
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|
\crefname@full{cor}{Korollar}{Korollarien}
|
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|
\crefname@full{folg}{Folgerung}{Folgerungen}
|
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|
\crefname@full{prop}{Proposition}{Propositionen}
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|
\crefname@full{defn}{Definition}{Definitionen}
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|
\crefname@full{conv}{Konvention}{Konventionen}
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\crefname@full{fact}{Fakt}{Fakten}
|
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\crefname@full{rem}{Bemerkung}{Bemerkungen}
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|
\crefname@full{qstn}{Frage}{Fragen}
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|
\crefname@full{e.g.}{Beipsiel}{Beipsiele}
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|
|
%% ****************************************************************
|
|
|
|
|
|
%% THEOREME:
|
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|
|
|
|
%% ****************************************************************
|
|
|
|
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|
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|
\def\qedEIGEN#1{\@ifnextchar[{\qedEIGEN@c{#1}}{\qedEIGEN@bes{#1}}}%]
|
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|
\def\qedEIGEN@bes#1{%
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\parfillskip=0pt% % so \par doesnt push \square to left
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\widowpenalty=10000% % so we dont break the page before \square
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\displaywidowpenalty=10000% % ditto
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\finalhyphendemerits=0% % TeXbook exercise 14.32
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\leavevmode% % \nobreak means lines not pages
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\unskip% % remove previous space or glue
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\nobreak% % don’t break lines
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\hfil% % ragged right if we spill over
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\penalty50% % discouragement to do so
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\hskip.2em% % ensure some space
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\null% % anchor following \hfill
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\hfill% % push \square to right
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#1% % the end-of-proof mark
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\par%
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}
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\def\qedEIGEN@c#1[#2]{%
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|
\parfillskip=0pt% % so \par doesnt push \square to left
|
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\widowpenalty=10000% % so we dont break the page before \square
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\displaywidowpenalty=10000% % ditto
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|
\finalhyphendemerits=0% % TeXbook exercise 14.32
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|
\leavevmode% % \nobreak means lines not pages
|
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|
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|
\unskip% % remove previous space or glue
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|
\nobreak% % don’t break lines
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|
\hfil% % ragged right if we spill over
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|
\penalty50% % discouragement to do so
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|
\hskip.2em% % ensure some space
|
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|
\null% % anchor following \hfill
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|
\hfill% % push \square to right
|
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|
{#1~{\smaller\bfseries\upshape (#2)}}%
|
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|
\par%
|
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|
}
|
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|
\def\qedVARIANT#1#2{
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|
\expandafter\def\csname ennde#1Sign\endcsname{#2}
|
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|
\expandafter\def\csname ennde#1\endcsname{\@ifnextchar[{\qedEIGEN@c{#2}}{\qedEIGEN@bes{#2}}} %]
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|
}
|
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|
|
\qedVARIANT{OfProof}{$\squareblack$}
|
|
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|
|
\qedVARIANT{OfWork}{\rectangleblack}
|
|
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|
|
\qedVARIANT{OfSomething}{$\dashv$}
|
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|
|
\qedVARIANT{OnNeutral}{$\lozenge$} % \lozenge \bigcirc \blacklozenge
|
|
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|
\def\qedsymbol{\enndeOfProofSign}
|
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|
|
\def\proofSymbol{\enndeOfProofSign}
|
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|
|
\def\ra@pretheoremwork{
|
|
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|
|
\setlength{\theorempreskipamount}{\ownspaceabovethm}
|
|
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|
|
}
|
|
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|
|
|
\def\rathmtransfer#1#2{
|
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|
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|
|
\expandafter\def\csname #2\endcsname{\csname #1\endcsname}
|
|
|
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|
|
\expandafter\def\csname end#2\endcsname{\csname end#1\endcsname}
|
|
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|
|
|
}
|
|
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|
|
\def\ranewthm#1#2#3[#4]{
|
|
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|
|
%% FOR \BEGIN{THM}
|
|
|
|
|
|
\theoremstyle{\current@theoremstyle}
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|
\theoremseparator{\current@theoremseparator}
|
|
|
|
|
|
\theoremprework{\ra@pretheoremwork}
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|
|
\@ifundefined{#1@basic}{\newtheorem{#1@basic}[#4]{#2}}{\renewtheorem{#1@basic}[#4]{#2}}
|
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|
|
%% FOR \BEGIN{THM}[...]
|
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|
|
\theoremstyle{\current@theoremstyle}
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|
\theoremseparator{\thmForceSepPt}
|
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|
|
\theoremprework{\ra@pretheoremwork}
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|
|
\@ifundefined{#1@withName}{\newtheorem{#1@withName}[#4]{#2}}{\renewtheorem{#1@withName}[#4]{#2}}
|
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|
|
|
%% FOR \BEGIN{THM*}
|
|
|
|
|
|
\theoremstyle{nonumberplain}
|
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|
\theoremseparator{\thmForceSepPt}
|
|
|
|
|
|
\theoremprework{\ra@pretheoremwork}
|
|
|
|
|
|
\@ifundefined{#1@star@basic}{\newtheorem{#1@star@basic}[Xdisplaynone]{#2}}{\renewtheorem{#1@star@basic}[Xdisplaynone]{#2}}
|
|
|
|
|
|
%% FOR \BEGIN{THM*}[...]
|
|
|
|
|
|
\theoremstyle{nonumberplain}
|
|
|
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|
|
\theoremseparator{\thmForceSepPt}
|
|
|
|
|
|
\theoremprework{\ra@pretheoremwork}
|
|
|
|
|
|
\@ifundefined{#1@star@withName}{\newtheorem{#1@star@withName}[Xdisplaynone]{#2}}{\renewtheorem{#1@star@withName}[Xdisplaynone]{#2}}
|
|
|
|
|
|
%% GENERATE ENVIRONMENTS:
|
|
|
|
|
|
\umbauenenv{#1}{#3}[#4]
|
|
|
|
|
|
\umbauenenv{#1@star}{#3}[Xdisplaynone]
|
|
|
|
|
|
%% TRANSFER *-DEFINITION
|
|
|
|
|
|
\rathmtransfer{#1@star}{#1*}
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
\def\umbauenenv#1#2[#3]{%
|
|
|
|
|
|
%% \BEGIN{THM}...
|
|
|
|
|
|
\expandafter\def\csname #1\endcsname{\relax%
|
|
|
|
|
|
\@ifnextchar[{\csname #1@\endcsname}{\csname #1@\endcsname[*]}%
|
|
|
|
|
|
}
|
|
|
|
|
|
%% \BEGIN{THM}[ANFANG]...
|
|
|
|
|
|
\expandafter\def\csname #1@\endcsname[##1]{\relax%
|
|
|
|
|
|
\@ifnextchar[{\csname #1@@\endcsname[##1]}{\csname #1@@\endcsname[##1][*]}%
|
|
|
|
|
|
}
|
|
|
|
|
|
%% \BEGIN{THM}[ANFANG][SCHLUSS]
|
|
|
|
|
|
\expandafter\def\csname #1@@\endcsname[##1][##2]{%
|
|
|
|
|
|
\ifx*##1%
|
|
|
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|
|
\def\enndeOfBlock{\csname end#1@basic\endcsname}
|
|
|
|
|
|
\csname #1@basic\endcsname%
|
|
|
|
|
|
\else%
|
|
|
|
|
|
\def\enndeOfBlock{\csname end#1@withName\endcsname}
|
|
|
|
|
|
\csname #1@withName\endcsname[##1]%
|
|
|
|
|
|
\fi%
|
|
|
|
|
|
\def\makelabel####1{%
|
|
|
|
|
|
\gdef\beweislabel{####1}%
|
|
|
|
|
|
\label{\beweislabel}%
|
|
|
|
|
|
}%
|
|
|
|
|
|
\ifx*##2%
|
|
|
|
|
|
\def\enndeSymbol{\qedEIGEN{#2}}
|
|
|
|
|
|
\else%
|
|
|
|
|
|
\def\enndeSymbol{\qedEIGEN{#2}[##2]}
|
|
|
|
|
|
\fi
|
|
|
|
|
|
}
|
|
|
|
|
|
%% \END{THM}
|
|
|
|
|
|
\expandafter\gdef\csname end#1\endcsname{\enndeSymbol\enndeOfBlock}
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
%% NEWTHEOREM EINSTELLUNGSOPTIONEN:
|
|
|
|
|
|
%% F\"UR \theoremstyle
|
|
|
|
|
|
%% plain Emulates original LATEX defin, except uses param \theorem...skipamount.
|
|
|
|
|
|
%% break Header followed by line break.
|
|
|
|
|
|
%% change Header, Number and Text are interchanged, without a line break.
|
|
|
|
|
|
%% changebreak =change, but with a line break after Header.
|
|
|
|
|
|
%% margin Number in left margin, without a line break.
|
|
|
|
|
|
%% marginbreak =margin, but with a line break after the header.
|
|
|
|
|
|
%% nonumberplain =plain, without number.
|
|
|
|
|
|
%% nonumberbreak =break, without number.
|
|
|
|
|
|
%% empty No number, no name. Only the optional argument is typeset.
|
|
|
|
|
|
%% \theoremclass \theoremnumbering
|
|
|
|
|
|
%% \theorempreskip \theorempostkip \theoremindent
|
|
|
|
|
|
%% \theoremprework \theorempostwork
|
|
|
|
|
|
|
|
|
|
|
|
\def\current@theoremstyle{plain}
|
|
|
|
|
|
\def\current@theoremseparator{\thmnumberingseppt}
|
|
|
|
|
|
\theoremstyle{\current@theoremstyle}
|
|
|
|
|
|
\theoremseparator{\current@theoremseparator}
|
|
|
|
|
|
\theoremsymbol{}
|
|
|
|
|
|
|
|
|
|
|
|
\newtheorem{X}{X}[chapter] % for most theorems
|
|
|
|
|
|
\newtheorem{Xe}{Xe}[chapter] % for equations
|
|
|
|
|
|
\newtheorem*{Xdisplaynone}{Xdisplaynone}[chapter] % a dummy counter, that will never be displayed.
|
|
|
|
|
|
\newtheorem{Xsp}{Xsp}[chapter] % for special theorems
|
|
|
|
|
|
\generatenestedthmnumbering{arabic}{chapter}{X}
|
|
|
|
|
|
\generatenestedthmnumbering{arabic}{chapter}{Xe}
|
|
|
|
|
|
\generatenestedthmnumbering{Roman}{chapter}{Xsp}
|
|
|
|
|
|
\let\theXsp\theshortXsp
|
|
|
|
|
|
|
|
|
|
|
|
\theoremheaderfont{\upshape\bfseries}
|
|
|
|
|
|
\theorembodyfont{\slshape}
|
|
|
|
|
|
|
|
|
|
|
|
\ranewthm{thm}{Theorem}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
\ranewthm{satz}{Satz}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
\ranewthm{claim}{Behauptung}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
\ranewthm{lemm}{Lemma}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
\ranewthm{cor}{Korollar}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
\ranewthm{folg}{Folgerung}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
\ranewthm{prop}{Proposition}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
|
|
|
|
|
|
\theorembodyfont{\upshape}
|
|
|
|
|
|
|
|
|
|
|
|
\ranewthm{defn}{Definition}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
\ranewthm{conv}{Konvention}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
\ranewthm{obs}{Beobachtung}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
\ranewthm{e.g.}{Beipsiel}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
\ranewthm{fact}{Fakt}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
\ranewthm{rem}{Bemerkung}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
\ranewthm{qstn}{Frage}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
\ranewthm{exer}{Aufgabe}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
\ranewthm{soln}{Lösung}{\enndeOnNeutralSign}[X]
|
|
|
|
|
|
|
|
|
|
|
|
\theoremheaderfont{\itshape\bfseries}
|
|
|
|
|
|
\theorembodyfont{\upshape}
|
|
|
|
|
|
|
|
|
|
|
|
\ranewthm{proof@tmp}{Beweis}{\enndeOfProofSign}[Xdisplaynone]
|
|
|
|
|
|
\rathmtransfer{proof@tmp*}{proof}
|
|
|
|
|
|
|
|
|
|
|
|
\def\behauptungbeleg@claim{%
|
|
|
|
|
|
\iflanguage{british}{Claim}{%
|
|
|
|
|
|
\iflanguage{english}{Claim}{%
|
|
|
|
|
|
\iflanguage{ngerman}{Behauptung}{%
|
|
|
|
|
|
\iflanguage{russian}{Утверждение}{%
|
|
|
|
|
|
Claim%
|
|
|
|
|
|
}}}}%
|
|
|
|
|
|
}
|
|
|
|
|
|
\def\behauptungbeleg@pf@kurz{%
|
|
|
|
|
|
\iflanguage{british}{Pf}{%
|
|
|
|
|
|
\iflanguage{english}{Pf}{%
|
|
|
|
|
|
\iflanguage{ngerman}{Bew}{%
|
|
|
|
|
|
\iflanguage{russian}{Доказательство}{%
|
|
|
|
|
|
Pf%
|
|
|
|
|
|
}}}}%
|
|
|
|
|
|
}
|
|
|
|
|
|
\def\behauptungbeleg{\@ifnextchar\bgroup{\behauptungbeleg@c}{\behauptungbeleg@bes}}
|
|
|
|
|
|
\def\behauptungbeleg@c#1{\item[{\bfseries \behauptungbeleg@claim\erlaubeplatz #1.}]}
|
|
|
|
|
|
\def\behauptungbeleg@bes{\item[{\bfseries \behauptungbeleg@claim.}]}
|
|
|
|
|
|
\def\belegbehauptung{\item[{\bfseries\itshape\behauptungbeleg@pf@kurz.}]}
|
|
|
|
|
|
|
|
|
|
|
|
%% ****************************************************************
|
|
|
|
|
|
%% ALTE UMGEBUNGEN:
|
|
|
|
|
|
%% ****************************************************************
|
|
|
|
|
|
|
|
|
|
|
|
\newcolumntype{\RECHTS}[1]{>{\raggedleft}p{#1}}
|
|
|
|
|
|
\newcolumntype{\LINKS}[1]{>{\raggedright}p{#1}}
|
|
|
|
|
|
\newcolumntype{m}{>{$}l<{$}}
|
|
|
|
|
|
\newcolumntype{C}{>{$}c<{$}}
|
|
|
|
|
|
\newcolumntype{L}{>{$}l<{$}}
|
|
|
|
|
|
\newcolumntype{R}{>{$}r<{$}}
|
|
|
|
|
|
\newcolumntype{0}{@{\hspace{0pt}}}
|
|
|
|
|
|
\newcolumntype{\LINKSRAND}{@{\hspace{\@totalleftmargin}}}
|
|
|
|
|
|
\newcolumntype{h}{@{\extracolsep{\fill}}}
|
|
|
|
|
|
\newcolumntype{i}{>{\itshape}}
|
|
|
|
|
|
\newcolumntype{t}{@{\hspace{\tabcolsep}}}
|
|
|
|
|
|
\newcolumntype{q}{@{\hspace{1em}}}
|
|
|
|
|
|
\newcolumntype{n}{@{\hspace{-\tabcolsep}}}
|
|
|
|
|
|
\newcolumntype{M}[2]{%
|
|
|
|
|
|
>{\begin{minipage}{#2}\begin{math}}%
|
|
|
|
|
|
{#1}%
|
|
|
|
|
|
<{\end{math}\end{minipage}}%
|
|
|
|
|
|
}
|
|
|
|
|
|
\newcolumntype{T}[2]{%
|
|
|
|
|
|
>{\begin{minipage}{#2}}%
|
|
|
|
|
|
{#1}%
|
|
|
|
|
|
<{\end{minipage}}%
|
|
|
|
|
|
}
|
|
|
|
|
|
\setlength{\LTpre}{\baselineskip}
|
|
|
|
|
|
\setlength{\LTpost}{0pt}
|
|
|
|
|
|
\def\center{\centering}
|
|
|
|
|
|
\def\endcenter{}
|
|
|
|
|
|
|
|
|
|
|
|
\def\punkteumgebung@genbefehl#1#2#3{
|
|
|
|
|
|
\punkteumgebung@genbefehl@{#1}{#2}{#3}{}{}
|
|
|
|
|
|
\punkteumgebung@genbefehl@{multi#1}{#2}{#3}{
|
|
|
|
|
|
\setlength{\columnsep}{10pt}%
|
|
|
|
|
|
\setlength{\columnseprule}{0pt}%
|
|
|
|
|
|
\begin{multicols}{\thecolumnanzahl}%
|
|
|
|
|
|
}{\end{multicols}\nvraum{1}}
|
|
|
|
|
|
}
|
|
|
|
|
|
\def\punkteumgebung@genbefehl@#1#2#3#4#5{
|
|
|
|
|
|
\expandafter\gdef\csname #1\endcsname{
|
|
|
|
|
|
\@ifnextchar\bgroup{\csname #1@c\endcsname}{\csname #1@bes\endcsname}
|
|
|
|
|
|
}%]
|
|
|
|
|
|
\expandafter\def\csname #1@c\endcsname##1{
|
|
|
|
|
|
\@ifnextchar[{\csname #1@c@\endcsname{##1}}{\csname #1@c@\endcsname{##1}[\z@]}
|
|
|
|
|
|
}%]
|
|
|
|
|
|
\expandafter\def\csname #1@c@\endcsname##1[##2]{
|
|
|
|
|
|
\@ifnextchar[{\csname #1@c@@\endcsname{##1}[##2]}{\csname #1@c@@\endcsname{##1}[##2][\z@]}
|
|
|
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|
|
}%]
|
|
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|
|
|
\expandafter\def\csname #1@c@@\endcsname##1[##2][##3]{
|
|
|
|
|
|
\let\alterlinkerRand\gesamtlinkerRand
|
|
|
|
|
|
\let\alterrechterRand\gesamtrechterRand
|
|
|
|
|
|
\addtolength{\gesamtlinkerRand}{##2}
|
|
|
|
|
|
\addtolength{\gesamtrechterRand}{##3}
|
|
|
|
|
|
\advance\linewidth -##2%
|
|
|
|
|
|
\advance\linewidth -##3%
|
|
|
|
|
|
\advance\@totalleftmargin ##2%
|
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|
|
\parshape\@ne \@totalleftmargin\linewidth%
|
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|
|
|
|
#4
|
|
|
|
|
|
\begin{#2}[\upshape ##1]%
|
|
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|
|
\setlength{\parskip}{0.5\baselineskip}\relax%
|
|
|
|
|
|
\setlength{\topsep}{\z@}\relax%
|
|
|
|
|
|
\setlength{\partopsep}{\z@}\relax%
|
|
|
|
|
|
\setlength{\parsep}{\parskip}\relax%
|
|
|
|
|
|
\setlength{\itemsep}{#3}\relax%
|
|
|
|
|
|
\setlength{\listparindent}{\z@}\relax%
|
|
|
|
|
|
\setlength{\itemindent}{\z@}\relax%
|
|
|
|
|
|
}
|
|
|
|
|
|
\expandafter\def\csname #1@bes\endcsname{
|
|
|
|
|
|
\@ifnextchar[{\csname #1@bes@\endcsname}{\csname #1@bes@\endcsname[\z@]}
|
|
|
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|
}%]
|
|
|
|
|
|
\expandafter\def\csname #1@bes@\endcsname[##1]{
|
|
|
|
|
|
\@ifnextchar[{\csname #1@bes@@\endcsname[##1]}{\csname #1@bes@@\endcsname[##1][\z@]}
|
|
|
|
|
|
}%]
|
|
|
|
|
|
\expandafter\def\csname #1@bes@@\endcsname[##1][##2]{
|
|
|
|
|
|
\let\alterlinkerRand\gesamtlinkerRand
|
|
|
|
|
|
\let\alterrechterRand\gesamtrechterRand
|
|
|
|
|
|
\addtolength{\gesamtlinkerRand}{##1}
|
|
|
|
|
|
\addtolength{\gesamtrechterRand}{##2}
|
|
|
|
|
|
\advance\linewidth -##1%
|
|
|
|
|
|
\advance\linewidth -##2%
|
|
|
|
|
|
\advance\@totalleftmargin ##1%
|
|
|
|
|
|
\parshape\@ne \@totalleftmargin\linewidth%
|
|
|
|
|
|
#4
|
|
|
|
|
|
\begin{#2}%
|
|
|
|
|
|
\setlength{\parskip}{0.5\baselineskip}\relax%
|
|
|
|
|
|
\setlength{\topsep}{\z@}\relax%
|
|
|
|
|
|
\setlength{\partopsep}{\z@}\relax%
|
|
|
|
|
|
\setlength{\parsep}{\parskip}\relax%
|
|
|
|
|
|
\setlength{\itemsep}{#3}\relax%
|
|
|
|
|
|
\setlength{\listparindent}{\z@}\relax%
|
|
|
|
|
|
\setlength{\itemindent}{\z@}\relax%
|
|
|
|
|
|
}
|
|
|
|
|
|
\expandafter\gdef\csname end#1\endcsname{%
|
|
|
|
|
|
\end{#2}#5
|
|
|
|
|
|
\setlength{\gesamtlinkerRand}{\alterlinkerRand}
|
|
|
|
|
|
\setlength{\gesamtlinkerRand}{\alterrechterRand}
|
|
|
|
|
|
}
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
\def\ritempunkt{{\Large\textbullet}} % \textbullet, $\sqbullet$, $\blacktriangleright$
|
|
|
|
|
|
\setdefaultitem{\ritempunkt}{\ritempunkt}{\ritempunkt}{\ritempunkt}
|
|
|
|
|
|
\punkteumgebung@genbefehl{itemise}{compactitem}{\parskip}{}{}
|
|
|
|
|
|
\punkteumgebung@genbefehl{kompaktitem}{compactitem}{\z@}{}{}
|
|
|
|
|
|
\punkteumgebung@genbefehl{enumerate}{compactenum}{\parskip}{}{}
|
|
|
|
|
|
\punkteumgebung@genbefehl{kompaktenum}{compactenum}{\z@}{}{}
|
|
|
|
|
|
|
|
|
|
|
|
\let\ALTthebibliography\thebibliography
|
|
|
|
|
|
\renewenvironment{thebibliography}[1]{%
|
|
|
|
|
|
\begin{ALTthebibliography}{#1}
|
|
|
|
|
|
\addcontentsline{toc}{part}{\bibname}
|
|
|
|
|
|
}{%
|
|
|
|
|
|
\end{ALTthebibliography}
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
%% ****************************************************************
|
|
|
|
|
|
%% NEUE UMGEBUNGEN:
|
|
|
|
|
|
%% ****************************************************************
|
|
|
|
|
|
|
|
|
|
|
|
\def\matrix#1{\left(\begin{array}[mc]{#1}}
|
|
|
|
|
|
\def\endmatrix{\end{array}\right)}
|
|
|
|
|
|
\def\smatrix{\left(\begin{smallmatrix}}
|
|
|
|
|
|
\def\endsmatrix{\end{smallmatrix}\right)}
|
|
|
|
|
|
\def\vector{\begin{matrix}{c}}
|
|
|
|
|
|
\def\endvector{\end{matrix}}
|
|
|
|
|
|
\def\svector{\begin{smatrix}}
|
|
|
|
|
|
\def\endsvector{\end{smatrix}}
|
|
|
|
|
|
|
|
|
|
|
|
\def\multiargrekursiverbefehl#1#2#3#4#5#6#7#8{%
|
|
|
|
|
|
\expandafter\gdef\csname#1\endcsname #2##1#4{\csname #1@anfang\endcsname##1#3\egroup}
|
|
|
|
|
|
\expandafter\def\csname #1@anfang\endcsname##1#3{#5##1\@ifnextchar\egroup{\csname #1@ende\endcsname}{#7\csname #1@mitte\endcsname}}
|
|
|
|
|
|
\expandafter\def\csname #1@mitte\endcsname##1#3{#6##1\@ifnextchar\egroup{\csname #1@ende\endcsname}{#7\csname #1@mitte\endcsname}}
|
|
|
|
|
|
\expandafter\def\csname #1@ende\endcsname##1{#8}
|
|
|
|
|
|
}
|
|
|
|
|
|
\multiargrekursiverbefehl{svektor}{[}{;}{]}{\begin{smatrix}}{}{\\}{\\\end{smatrix}}
|
|
|
|
|
|
\multiargrekursiverbefehl{vektor}{[}{;}{]}{\begin{matrix}{c}}{}{\\}{\\\end{matrix}}
|
|
|
|
|
|
\multiargrekursiverbefehl{vektorzeile}{}{,}{;}{}{&}{}{}
|
|
|
|
|
|
\multiargrekursiverbefehl{matlabmatrix}{[}{;}{]}{\begin{smatrix}\vektorzeile}{\vektorzeile}{;\\}{;\end{smatrix}}
|
|
|
|
|
|
|
|
|
|
|
|
\def\underbracenodisplay#1{%
|
|
|
|
|
|
\mathop{\vtop{\m@th\ialign{##\crcr
|
|
|
|
|
|
$\hfil\displaystyle{#1}\hfil$\crcr
|
|
|
|
|
|
\noalign{\kern3\p@\nointerlineskip}%
|
|
|
|
|
|
\upbracefill\crcr\noalign{\kern3\p@}}}}\limits%
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
\def\mathe[#1]#2{%
|
|
|
|
|
|
\ifthenelse{\equal{\boolinmdframed}{\boolwahr}}{}{\begin{escapeeinzug}}
|
|
|
|
|
|
\noindent%
|
|
|
|
|
|
\let\eqtagset\boolfalsch
|
|
|
|
|
|
\let\eqtaglabel\boolleer
|
|
|
|
|
|
\let\eqtagsymb\boolleer
|
|
|
|
|
|
\let\alteqtag\eqtag
|
|
|
|
|
|
\def\eqtag{\@ifnextchar[{\eqtag@loc@}{\eqtag@loc@[*]}}%
|
|
|
|
|
|
\def\eqtag@loc@[##1]{\@ifnextchar\bgroup{\eqtag@loc@@[##1]}{\eqtag@loc@@[##1]{}}}%
|
|
|
|
|
|
\def\eqtag@loc@@[##1]##2{%
|
|
|
|
|
|
\gdef\eqtagset{\boolwahr}
|
|
|
|
|
|
\gdef\eqtaglabel{##1}
|
|
|
|
|
|
\gdef\eqtagsymb{##2}
|
|
|
|
|
|
}%
|
|
|
|
|
|
\def\verticalalign{}%
|
|
|
|
|
|
\IfBeginWith{#1}{t}{\def\verticalalign{t}}{}%
|
|
|
|
|
|
\IfBeginWith{#1}{m}{\def\verticalalign{c}}{}%
|
|
|
|
|
|
\IfBeginWith{#1}{b}{\def\verticalalign{b}}{}%
|
|
|
|
|
|
\def\horizontalalign{\null\hfill\null}%
|
|
|
|
|
|
\IfEndWith{#1}{l}{}{\null\hfill\null}%
|
|
|
|
|
|
\IfEndWith{#1}{r}{\def\horizontalalign{}}{}%
|
|
|
|
|
|
\begin{math}
|
|
|
|
|
|
\begin{array}[\verticalalign]{0#2}%
|
|
|
|
|
|
}
|
|
|
|
|
|
\def\endmathe{%
|
|
|
|
|
|
\end{array}
|
|
|
|
|
|
\end{math}\horizontalalign%
|
|
|
|
|
|
\let\eqtag\alteqtag
|
|
|
|
|
|
\ifthenelse{\equal{\eqtagset}{\boolwahr}}{\eqtag[\eqtaglabel]{\eqtagsymb}}{}
|
|
|
|
|
|
\ifthenelse{\equal{\boolinmdframed}{\boolwahr}}{}{\end{escapeeinzug}}%
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
\def\longmathe[#1]#2{\relax
|
|
|
|
|
|
\let\altarraystretch\arraystretch
|
|
|
|
|
|
\renewcommand\arraystretch{1.2}\relax
|
|
|
|
|
|
\begin{longtable}[#1]{\LINKSRAND #2}
|
|
|
|
|
|
}
|
|
|
|
|
|
\def\endlongmathe{
|
|
|
|
|
|
\end{longtable}
|
|
|
|
|
|
\renewcommand\arraystretch{\altarraystretch}
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
\def\einzug{\@ifnextchar[{\indents@}{\indents@[\z@]}}%]
|
|
|
|
|
|
\def\indents@[#1]{\@ifnextchar[{\indents@@[#1]}{\indents@@[#1][\z@]}}%]
|
|
|
|
|
|
\def\indents@@[#1][#2]{%
|
|
|
|
|
|
\begin{list}{}{\relax
|
|
|
|
|
|
\setlength{\topsep}{\z@}\relax
|
|
|
|
|
|
\setlength{\partopsep}{\z@}\relax
|
|
|
|
|
|
\setlength{\parsep}{\parskip}\relax
|
|
|
|
|
|
\setlength{\listparindent}{\z@}\relax
|
|
|
|
|
|
\setlength{\itemindent}{\z@}\relax
|
|
|
|
|
|
\setlength{\leftmargin}{#1}\relax
|
|
|
|
|
|
\setlength{\rightmargin}{#2}\relax
|
|
|
|
|
|
\let\alterlinkerRand\gesamtlinkerRand
|
|
|
|
|
|
\let\alterrechterRand\gesamtrechterRand
|
|
|
|
|
|
\addtolength{\gesamtlinkerRand}{#1}
|
|
|
|
|
|
\addtolength{\gesamtrechterRand}{#2}
|
|
|
|
|
|
}\relax
|
|
|
|
|
|
\item[]\relax
|
|
|
|
|
|
}
|
|
|
|
|
|
\def\endeinzug{%
|
|
|
|
|
|
\setlength{\gesamtlinkerRand}{\alterlinkerRand}
|
|
|
|
|
|
\setlength{\gesamtlinkerRand}{\alterrechterRand}
|
|
|
|
|
|
\end{list}%
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
\def\escapeeinzug{\begin{einzug}[-\gesamtlinkerRand][-\gesamtrechterRand]}
|
|
|
|
|
|
\def\endescapeeinzug{\end{einzug}}
|
|
|
|
|
|
|
|
|
|
|
|
\def\programmiercode{
|
|
|
|
|
|
\modulolinenumbers[1]
|
|
|
|
|
|
\begin{einzug}[\rtab][\rtab]%
|
|
|
|
|
|
\begin{linenumbers}%
|
|
|
|
|
|
\fontfamily{cmtt}\fontseries{m}\fontshape{u}\selectfont%
|
|
|
|
|
|
\setlength{\parskip}{1\baselineskip}%
|
|
|
|
|
|
\setlength{\parindent}{0pt}%
|
|
|
|
|
|
}
|
|
|
|
|
|
\def\endprogrammiercode{
|
|
|
|
|
|
\end{linenumbers}
|
|
|
|
|
|
\end{einzug}
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
\def\schattiertebox@genbefehl#1#2#3{
|
|
|
|
|
|
\expandafter\gdef\csname #1\endcsname{%
|
|
|
|
|
|
\@ifnextchar[{\csname #1@args\endcsname}{\csname #1@args\endcsname[#3]}%]%
|
|
|
|
|
|
}
|
|
|
|
|
|
\expandafter\def\csname #1@args\endcsname[##1]{%
|
|
|
|
|
|
\@ifnextchar[{\csname #1@args@l\endcsname[##1]}{\csname #1@args@n\endcsname[##1]}%]%
|
|
|
|
|
|
}
|
|
|
|
|
|
\expandafter\def\csname #1@args@l\endcsname[##1][##2]{%
|
|
|
|
|
|
\@ifnextchar[{\csname #1@args@l@r\endcsname[##1][##2]}{\csname #1@args@l@n\endcsname[##1][##2]}%]%
|
|
|
|
|
|
}
|
|
|
|
|
|
\expandafter\def\csname #1@args@n\endcsname[##1]{%
|
|
|
|
|
|
\let\boolinmdframed\boolwahr
|
|
|
|
|
|
\begin{mdframed}[#2leftmargin=0,rightmargin=0,outermargin=0,innermargin=0,##1]
|
|
|
|
|
|
}
|
|
|
|
|
|
\expandafter\def\csname #1@args@l@n\endcsname[##1][##2]{%
|
|
|
|
|
|
\let\boolinmdframed\boolwahr
|
|
|
|
|
|
\begin{mdframed}[#2leftmargin=##2/2,rightmargin=##2/2,outermargin=##2/2,innermargin=##2/2,##1]
|
|
|
|
|
|
}
|
|
|
|
|
|
\expandafter\def\csname #1@args@l@r\endcsname[##1][##2][##3]{%
|
|
|
|
|
|
\let\boolinmdframed\boolwahr
|
|
|
|
|
|
\begin{mdframed}[#2leftmargin=##2,rightmargin=##3,outermargin=##2,innermargin=##3,##1]
|
|
|
|
|
|
}
|
|
|
|
|
|
\expandafter\gdef\csname end#1\endcsname{%
|
|
|
|
|
|
\end{mdframed}
|
|
|
|
|
|
\let\boolinmdframed\boolfalsch
|
|
|
|
|
|
}
|
|
|
|
|
|
}
|
|
|
|
|
|
\schattiertebox@genbefehl{schattiertebox}{
|
|
|
|
|
|
splittopskip=0,%
|
|
|
|
|
|
splitbottomskip=0,%
|
|
|
|
|
|
frametitleaboveskip=0,%
|
|
|
|
|
|
frametitlebelowskip=0,%
|
|
|
|
|
|
skipabove=1\baselineskip,%
|
|
|
|
|
|
skipbelow=1\baselineskip,%
|
|
|
|
|
|
linewidth=2pt,%
|
|
|
|
|
|
linecolor=black,%
|
|
|
|
|
|
roundcorner=4pt,%
|
|
|
|
|
|
}{
|
|
|
|
|
|
backgroundcolor=leer,%
|
|
|
|
|
|
nobreak=true,%
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
\schattiertebox@genbefehl{schattierteboxdunn}{
|
|
|
|
|
|
splittopskip=0,%
|
|
|
|
|
|
splitbottomskip=0,%
|
|
|
|
|
|
frametitleaboveskip=0,%
|
|
|
|
|
|
frametitlebelowskip=0,%
|
|
|
|
|
|
skipabove=1\baselineskip,%
|
|
|
|
|
|
skipbelow=1\baselineskip,%
|
|
|
|
|
|
linewidth=1pt,%
|
|
|
|
|
|
linecolor=black,%
|
|
|
|
|
|
roundcorner=2pt,%
|
|
|
|
|
|
}{
|
|
|
|
|
|
backgroundcolor=leer,%
|
|
|
|
|
|
nobreak=true,%
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
\def\algorithm{\schattiertebox[backgroundcolor=hellgrau,nobreak=false]}
|
|
|
|
|
|
\def\endalgorithm{\endschattiertebox}
|
|
|
|
|
|
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\def\tikzsetzenode#1{%
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\tikz[remember picture,baseline,overlay]{\node #1;}%
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}
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\def\tikzsetzepfeil#1{%
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\begin{tikzpicture}[remember picture,overlay,>=latex]%
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\draw #1;%
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\end{tikzpicture}%
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}
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\def\tikzsetzeoverlay#1{%
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\begin{tikzpicture}[remember picture,overlay,>=latex]%
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#1%
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\end{tikzpicture}%
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}
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\def\tikzsetzekreise[#1]#2#3{%
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\tikzsetzepfeil{%
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[rounded corners,#1]%
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([shift={(-\tabcolsep,0.75\baselineskip)}]#2)%
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rectangle%
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([shift={(\tabcolsep,-0.5\baselineskip)}]#3)
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}%
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}
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\tikzset{
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>=stealth,
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auto,
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thick,
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main node/.style={
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circle,draw,font=\sffamily\Large\bfseries,minimum size=0pt
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},
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}
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%% ********************************************************************************
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%% FILE: src/setup-layout.tex
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%% ********************************************************************************
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\pagestyle{fancyplain}
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\@ifundefined{setcitestyle}{%
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%% do nothing
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}{%
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\setcitestyle{numeric-comp,open={[},close={]}}
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}
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\def\crefpairconjunction{ und }
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\def\crefmiddleconjunction{, }
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\def\creflastconjunction{, und }
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\raggedbottom %% <- pushes footers up
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\sloppy
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\def\headrulewidth{0pt}
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\def\footrulewidth{0pt}
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\setlength{\columnsep}{20pt}
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\setlength{\columnseprule}{1pt}
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\setlength{\headheight}{11pt}
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\setlength{\partopsep}{0pt}
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\setlength{\topsep}{\baselineskip}
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\setlength{\topskip}{0.5\baselineskip}
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\setlength{\footskip}{-1\baselineskip}
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\setlength{\maxdepth}{0pt}
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\renewcommand{\baselinestretch}{1}
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\renewcommand{\arraystretch}{1}
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\setcounter{LTchunksize}{\infty}
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\setlength{\abovedisplayskip}{0pt}
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\setlength{\parskip}{1\baselineskip}
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\def\firstparagraph{\noindent}
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\def\continueparagraph{\noindent}
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\hypersetup{
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hidelinks=true,
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}
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\@addtoreset{chapter}{part} %% nötig für Hyperref.
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\def\partfont{\documentfont\fontseries{bx}\Huge\selectfont}
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\def\chapterfont{\documentfont\fontseries{bx}\huge\selectfont}
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\def\sectionfont{\documentfont\fontseries{bx}\Large\selectfont}
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\def\subsectionfont{\documentfont\fontseries{bx}\large\selectfont}
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\def\thepart{\Roman{part}}
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\generatenestedsecnumbering{arabic}{part}{chapter}
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\generatenestedsecnumbering{arabic}{chapter}{section}
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\generatenestedsecnumbering{arabic}{section}{subsection}
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\generatenestedsecnumbering{arabic}{subsection}{subsubsection}
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\def\theunitnamepart{\thepart}
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\def\theunitnamechapter{\theshortchapter}
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\def\theunitnamesection{\thelongsection}
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\def\theunitnamesubsection{\thelongsubsection}
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\def\theunitnamesubsubsection{\thelongsubsubsection}
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\def\partname{Teil\erlaubeplatz}
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\def\chaptername{Kapitel\erlaubeplatz}
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\def\sectionname{\S\erlaubeplatz}
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\def\subsectionname{}
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\def\subsubsectionname{}
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\let\appendix@orig\appendix
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\def\appendix{%
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\appendix@orig%
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\let\boolinappendix\boolwahr
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\addcontentsline{toc}{part}{\appendixname}%
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\addtocontents{toc}{\protect\setcounter{tocdepth}{0}}
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\def\sectionname{Appendix}%
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\def\theunitnamesection{\Alph{section}}%
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}
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\def\notappendix{%
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\let\boolinappendix\boolfalse
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|
\addtocontents{toc}{\protect\setcounter{tocdepth}{1 }}
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\def\sectionname{}%
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\def\theunitnamesection{\arabic{section}}%
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}
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%% \titlespacing{<sectionclassname>}
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%% {linker einzug}{platz oberhalb}{platz unterhalb}[rechter einzug]
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\titlespacing{\section}{0pt}{\baselineskip}{\baselineskip}
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\titlespacing{\subsection}{0pt}{\baselineskip}{\baselineskip}
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\titlespacing{\subsubsection}{0pt}{\baselineskip}{\baselineskip}
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\titlespacing{\paragraph}{0pt}{0pt}{1em}
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\titleformat{\part}[display]
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|
{\normalfont\headingfont\bfseries\Huge\centering}
|
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{%
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|
\ifthenelse{\equal{\partname}{}}{%
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|
\theunitnamepart%
|
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|
}{%
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|
\MakeUppercase{\partname}~\theunitnamepart%
|
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|
}%
|
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|
}{0pt}{%
|
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|
}[\thispagestyle{empty}]
|
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|
\titleformat{\chapter}[frame]
|
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|
{\normalfont\headingfont\bfseries\Large}
|
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|
{%
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|
|
\bedingtesspaceexpand{chaptername}{~}{\theunitnamechapter}%
|
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|
}{0.5em}{%
|
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|
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|
|
}[\thispagestyle{empty}]%\titlerule%[2pt]%
|
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|
|
\titleformat{\section}[hang]
|
|
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|
|
{\normalfont\headingfont\bfseries\flushleft\large}
|
|
|
|
|
|
{%
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|
|
\bedingtesspaceexpand{sectionname}{~}{\theunitnamesection}%
|
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|
}{0.5em}
|
|
|
|
|
|
{%
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|
}
|
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[%
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|
|
\nvraum{0.25}%
|
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|
]
|
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|
|
\titleformat{\subsection}[hang]
|
|
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|
|
{\normalfont\headingfont\bfseries\flushleft\large}
|
|
|
|
|
|
{%
|
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|
|
\bedingtesspaceexpand{subsectionname}{~}{\theunitnamesubsection}%
|
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|
|
}{0.5em}
|
|
|
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|
|
{%
|
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|
}
|
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|
[%
|
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|
\nvraum{0.25}%
|
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|
]
|
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|
|
\titleformat{\subsubsection}[hang]
|
|
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|
|
{\normalfont\headingfont\bfseries\flushleft\large}
|
|
|
|
|
|
{%
|
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|
|
\bedingtesspaceexpand{subsubsectionname}{~}{\theunitnamesubsubsection}%
|
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|
}{0.5em}
|
|
|
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{%
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|
}
|
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[%
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|
\nvraum{0.25}%
|
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]
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|
\def\rafootnotectr{20}
|
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|
\def\incrftnotectr#1{%
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|
|
\addtocounter{#1}{1}%
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|
\ifnum\value{#1}>\rafootnotectr\relax
|
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|
\setcounter{#1}{0}%
|
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|
|
\fi%
|
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|
|
}
|
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|
\def\footnoteref[#1]{\protected@xdef\@thefnmark{\ref{#1}}\@footnotemark}
|
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|
|
\let\altfootnotetext\footnotetext
|
|
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|
|
\def\footnotetext[#1]#2{\incrftnotectr{footnote}\altfootnotetext[\value{footnote}]{\label{#1}#2}}
|
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|
|
\let\altfootnotemark\footnotemark
|
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|
|
%% Undesirable solution, as the text is not hyperlinked.
|
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|
|
\def\footnotemark[#1]{\text{\textsuperscript{\getrefnumber{#1}}}}
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|
|
\DefineFNsymbols*{custom}{abcdefghijklmnopqrstuvwxyz}
|
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|
|
\setfnsymbol{custom}
|
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|
|
\def\footnotelayout{\documentfont\scriptsize}
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|
\def\thefootnote{\fnsymbol{footnote}}
|
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|
\def\kopfzeileleer{
|
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|
|
\lhead[]{}
|
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|
|
|
\chead[]{}
|
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|
|
|
\rhead[]{}
|
|
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|
|
|
\lfoot[]{}
|
|
|
|
|
|
\cfoot[]{}
|
|
|
|
|
|
\rfoot[]{}
|
|
|
|
|
|
}
|
|
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|
|
|
\def\kopfzeiledefault{
|
|
|
|
|
|
\lhead[]{}
|
|
|
|
|
|
\lhead[]{}
|
|
|
|
|
|
\chead[]{}
|
|
|
|
|
|
\rhead[]{}
|
|
|
|
|
|
\lfoot[]{}
|
|
|
|
|
|
\cfoot{\footnotesize\thepage}
|
|
|
|
|
|
\rfoot[]{}
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
\DeclareRobustCommand\crfamily{\fontfamily{pcr}\selectfont}
|
|
|
|
|
|
\def\headingfont{\fontfamily{cmss}\selectfont}
|
|
|
|
|
|
\def\documentfancyfont{%
|
|
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|
|
\gdef\headingfont{\crfamily}%
|
|
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|
|
|
\fontfamily{ccr}\fontseries{m}\selectfont%
|
|
|
|
|
|
}
|
|
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|
|
|
\def\documentfont{%
|
|
|
|
|
|
\gdef\headingfont{\fontfamily{cmss}\selectfont}%
|
|
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|
|
|
\fontfamily{cmss}\fontseries{m}\selectfont%
|
|
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|
|
|
\renewcommand{\sfdefault}{phv}%
|
|
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|
|
\renewcommand{\ttdefault}{pcr}%
|
|
|
|
|
|
\renewcommand{\rmdefault}{cmr}% <— funktionieren nicht mit {ptm}
|
|
|
|
|
|
\renewcommand{\bfdefault}{bx}%
|
|
|
|
|
|
\renewcommand{\itdefault}{it}%
|
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|
|
\renewcommand{\sldefault}{sl}%
|
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|
|
\renewcommand{\scdefault}{sc}%
|
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|
|
\renewcommand{\updefault}{n}%
|
|
|
|
|
|
}
|
|
|
|
|
|
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|
|
\allowdisplaybreaks
|
|
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|
|
\let\altcleardoublepage\cleardoublepage
|
|
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|
|
\let\cleardoublepage\clearpage
|
|
|
|
|
|
|
|
|
|
|
|
\def\startdocumentlayoutoptions{
|
|
|
|
|
|
\selectlanguage{ngerman}
|
|
|
|
|
|
\setlength{\parskip}{0.5\baselineskip}
|
|
|
|
|
|
\setlength{\parindent}{0pt}
|
|
|
|
|
|
\kopfzeiledefault
|
|
|
|
|
|
\documentfont
|
|
|
|
|
|
\normalsize
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
\def\highlightTerm#1{\emph{#1}}
|
|
|
|
|
|
|
|
|
|
|
|
%% ********************************************************************************
|
|
|
|
|
|
%% FILE: srclocal/setup-localmacros.tex
|
|
|
|
|
|
%% ********************************************************************************
|
|
|
|
|
|
|
|
|
|
|
|
%% ****************************************************************
|
|
|
|
|
|
%% MATHE:
|
|
|
|
|
|
%% ****************************************************************
|
|
|
|
|
|
|
|
|
|
|
|
\def\cal#1{\mathcal{#1}}
|
|
|
|
|
|
\def\reell{\mathbb{R}}
|
|
|
|
|
|
\def\kmplx{\mathbb{C}}
|
|
|
|
|
|
\def\Torus{\mathbb{T}}
|
|
|
|
|
|
\def\rtnl{\mathbb{Q}}
|
|
|
|
|
|
\def\intgr{\mathbb{Z}}
|
|
|
|
|
|
|
|
|
|
|
|
\def\ntrl{\mathbb{N}}
|
|
|
|
|
|
\def\ntrlpos{\mathbb{N}}
|
|
|
|
|
|
\def\ntrlzero{\mathbb{N}_{0}}
|
|
|
|
|
|
\def\reellNonNeg{\reell_{+}}
|
|
|
|
|
|
|
|
|
|
|
|
\def\imageinh{\imath}
|
|
|
|
|
|
\def\ReTeil{\mathop{\mathfrak{R}\text{\upshape e}}}
|
|
|
|
|
|
\def\ImTeil{\mathop{\mathfrak{I}\text{\upshape m}}}
|
|
|
|
|
|
|
|
|
|
|
|
\def\leer{\emptyset}
|
|
|
|
|
|
\def\restr#1{\vert_{#1}}
|
|
|
|
|
|
\def\ohne{\mathbin{\setminus}}
|
|
|
|
|
|
\def\Pot{\mathop{\mathcal{P}}}
|
|
|
|
|
|
\def\einser{\mathbf{1}}
|
|
|
|
|
|
\def\supp{\mathop{\mathrm{supp}}}
|
|
|
|
|
|
|
|
|
|
|
|
\def\brkt#1{\langle{}#1{}\rangle}
|
|
|
|
|
|
\def\lsim{\mathop{\sim}}
|
|
|
|
|
|
\def\lneg{\mathop{\neg}}
|
|
|
|
|
|
\def\land{\mathop{\wedge}}
|
|
|
|
|
|
\def\lor{\mathop{\vee}}
|
|
|
|
|
|
|
|
|
|
|
|
\def\eps{\varepsilon}
|
|
|
|
|
|
\let\altphi\phi
|
|
|
|
|
|
\let\altvarphi\varphi
|
|
|
|
|
|
\def\phi{\altvarphi}
|
|
|
|
|
|
\def\varphi{\altphi}
|
|
|
|
|
|
|
|
|
|
|
|
\def\vectorspacespan{\mathop{\text{\upshape Lin}}}
|
|
|
|
|
|
\def\dim{\mathop{\text{\upshape dim}}}
|
|
|
|
|
|
\def\rank{\mathop{\text{\upshape Rang}}}
|
|
|
|
|
|
\def\onematrix{\text{\upshape\bfseries I}}
|
|
|
|
|
|
\def\zeromatrix{\text{\upshape\bfseries 0}}
|
|
|
|
|
|
\def\zerovector{\text{\upshape\bfseries 0}}
|
|
|
|
|
|
|
|
|
|
|
|
\def\graph{\mathop{\text{\upshape Gph}}}
|
|
|
|
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\def\domain{\mathop{\text{\upshape dom}}}
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\def\range{\mathop{\text{\upshape Bild}}}
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\def\ker{\mathop{\text{\upshape Kern}}}
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\def\functionspace{\mathop{\text{\upshape Abb}}}
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\def\id{\text{\upshape id}}
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\def\modfn{\mathop{\text{\upshape mod}}}
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\def\divides{\mathbin{\mid}}
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\def\ndivides{\mathbin{\nmid}}
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\def\ggT{\mathop{\text{\upshape ggT}}}
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\def\choose#1#2{\begin{smatrix}#1\\#2\\\end{smatrix}}
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\makeatother
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2021-02-04 15:08:52 +01:00
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\begin{document}
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\startdocumentlayoutoptions
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%% FRONTMATTER:
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\thispagestyle{plain}
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%% ********************************************************************************
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%% FILE: front/index.tex
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%% ********************************************************************************
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%% ********************************************************************************
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%% FILE: front/title.tex
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%% ********************************************************************************
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\begin{titlepage}
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\null
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\vraum
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\noindent\rule{\linewidth}{2pt}
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{\hraum\LARGE Lineare Algebra I\hraum}\\
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{\hraum\LARGE $\oast$\,\rule[0.175\baselineskip]{0.65\linewidth}{1pt}\,$\oast$ \hraum}\\
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{\hraum\Large Zusatzaufgaben aus der Übungsgruppe\hraum}
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\noindent\rule{\linewidth}{2pt}
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\vraum
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\noindent
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\hraum{\footnotesize Raj Dahya}\hraum\\
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\hraum{\small \itshape Fakultät für Mathematik und Informatik}\hraum\\
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\hraum{\small \itshape Universität Leipzig.}\hraum\\
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\hraum{\small Wintersemester 2020/2021 }\hraum
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\end{titlepage}
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%% ********************************************************************************
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%% FILE: front/foreword.tex
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%% ********************************************************************************
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\chapter*{Vorwort}
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Dieses Dokument enthält zusätzliche Aufgaben und Themen,
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die in den Übungsgruppen erörtert wurden.
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%% ********************************************************************************
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%% FILE: front/contents.tex
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%% ********************************************************************************
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\kopfzeiledefault
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\footnotesize
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\setcounter{tocdepth}{1}
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\def\contentsname{Inhaltsverzeichnis}
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\tableofcontents
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%% HAUPTTEXT:
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%% ********************************************************************************
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%% FILE: body/index.tex
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%% ********************************************************************************
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\def\chaptername{}
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%% ********************************************************************************
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%% FILE: body/linear-extensions.tex
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%% ********************************************************************************
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\chapter[Lineare Ausdehnung]{Lineare Ausdehnung}
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\label{ch:lin-ext}
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In der Übungsgruppe in Woche 12 (am 3.2.2021) diskutierten wir
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verzwickte Situationen und Fragentypen, die zum Thema linearer Ausdehnung vorkommen können.
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Wir hatten das größtenteils theoretisch ausgelegt.
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Hier wollen wir ein paar Aufgaben komplett durchrechnen.
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\textbf{Beachte!} Hier geht es niemals darum,
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2021-02-10 12:19:33 +01:00
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eine lineare Ausdehnung \emph{explizit darzustellen},
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2021-02-04 15:08:52 +01:00
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sondern vielmehr
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(1) \cite[Satz~6.1.13]{sinn2020} als zentrales Resultat anzuwenden,
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(2) eine Basis aus den Inputvektoren zu generieren
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(ggf. durch Entfernung von „linear abhängigen“ Vektoren, ggf. durch Basiserweiterung, ggf. durch beides!)
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(3) die Input und Outputvektoren in der partielldefinierten Funktion
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zu untersuchen, und \uline{rein aufgrund dessen} ein Urteil zu treffen,
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ob
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(3a) eine lineare Ausdehnung überhaupt möglich ist,
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(3b) eine injektive/nicht injektive lineare Ausdehnung möglich ist,
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(3c) eine surjektive/nicht surjektive lineare Ausdehnung möglich ist,
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(3d) eine Isomorphismus (=Bijektion)/nicht-Isomorphismus als lineare Ausdehnung möglich ist.
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Nun, im Falle von Funktionen ${\phi:U\to V}$, wobei $U,V$ Vektorräume mit $\dim(U)=\dim(V)$,
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sind wegen \cite[Korollar~6.1.11]{sinn2020} die Nebenfragen (3a)–(3c) alle äquivalent.
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Im Falle $\dim(U)\neq\dim(V)$ machen wir von folgender Beobachtung Gebrauch:
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\begin{obs*}
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Seien $U$, $V$ (endlich dimensionale) Vektorräume über einem Körper $K$
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und sei ${\phi:U\to V}$ linear.
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Da $\range(\phi)\subseteq V$ gilt offensichtlich $\dim(\range(\phi))\leq\dim(V)$.
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Und wenn wir eine Basis ${\{u_{1},u_{2}\ldots,u_{n}\}\subseteq U}$
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für $U$ fixieren, mit $n=\dim(U)$,
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so gilt wegen Linearität
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${\range(\phi)=\vectorspacespan\{\phi(u_{1}),\phi(u_{2})\ldots,\phi(u_{n})\}}$.
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Das heißt, $\{\phi(u_{1}),\phi(u_{2})\ldots,\phi(u_{n})\}$
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ist ein Erzeugendensystem für $\range(\phi)$.
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Folglich gilt $\dim(\range(\phi))\leq n=\dim(U)$.
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Da per Definition $\rank(\phi)=\dim(\range(\phi))$,
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haben wir gezeigt,
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dass
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${\rank(\phi)\leq\dim(V)}$
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und ${\rank(\phi)\leq\dim(U)}$
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\uline{stets gelten}.
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Kürzer formuliert:
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\begin{mathe}[mc]{rcl}
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\eqtag[eq:lin-abb-leq:ch:lin-ext]
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\rank(\phi) &\leq &\min\{\dim(U),\dim(V)\}\\
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\end{mathe}
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gilt immer für alle lineare Abbildungen ${\phi:U\to V}$
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und alle Vektorräume $U, V$.
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\end{obs*}
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Aus dieser Beobachtung können wir über (3b–3d) folgende Urteile generell treffen, wenn $\dim(U)\neq\dim(V)$:
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\begin{kompaktitem}
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\item
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Falls $\dim(U)>\dim(V)$ kann es bei offensichtlich höchstens nicht-injektive lineare Ausdehnungen geben,
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weil für ${\phi:U\to V}$ linear gilt $\rank(\phi)\leq\dim(V)<\dim(U)$,
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sodass laut \cite[Korollar~6.3.15(1)]{sinn2020} $\phi$ niemals injektiv sein kann.
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\item[]
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Darum lautet die Antwort zu (3b/3d) \emph{Gibt es injektive/bijektive...?} immer nein.
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Die Fragen (3b/3d) \emph{Gibt es nicht-injektive/nicht-bijektive...?} sind dann äquivalent zu (3a).
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\item
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Falls $\dim(U)<\dim(V)$ kann es bei (3c) offensichtlich höchstens nicht-surjektive lineare Ausdehnungen geben,
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weil für ${\phi:U\to V}$ linear gilt $\rank(\phi)\leq\dim(U)<\dim(V)$,
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sodass laut \cite[Korollar~6.3.15(2)]{sinn2020} $\phi$ niemals surjektiv sein kann.
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\item[]
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Darum lautet die Antwort zu (3c/3d) \emph{Gibt es surjektive/bijektive...?} immer nein.
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Die Fragen (3b/3d) \emph{Gibt es nicht-surjektive/nicht-bijektive...?} sind dann äquivalent zu (3a).
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\end{kompaktitem}
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Daher können wir die Fragentypen in den Aufgaben immer teilweise sofort beantworten
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und zum Teil vereinfachen,
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je nachdem, ob $\dim(U)=\dim(V)$, oder $\dim(U)<\dim(V)$, oder $\dim(U)>\dim(V)$
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gelten.
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%% AUFGABE 1
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\clearpage
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\let\altsectionname\sectionname
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\def\sectionname{Aufgabe}
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\section[Aufgabe 1]{}
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\label{sec:1}
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\let\sectionname\altsectionname
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2021-02-04 20:37:34 +01:00
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Betrachten wir die $\reell$-Vektorräume $U:=\reell^{4}$ und $V:=\reell^{2}$
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und die Vektoren $u_{1},u_{2},u_{3}\in U$ und $v_{1},v_{2},v_{3}\in V$
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2021-02-04 15:08:52 +01:00
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\begin{mathe}[mc]{ccc}
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u_{1} = \begin{svector} 0\\ 0\\ 1\\ 1\\\end{svector},
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&u_{2} = \begin{svector} 3\\ 0\\ 1\\ 1\\\end{svector},
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&u_{3} = \begin{svector} 30\\ 0\\ 0\\ 0\\\end{svector},\\
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\end{mathe}
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\begin{mathe}[mc]{ccc}
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v_{1} = \begin{svector} 5\\ 8\\\end{svector},
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&v_{2} = \begin{svector} -9\\ 11\\\end{svector},
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&v_{3} = \begin{svector} -140\\ 30\\\end{svector}.\\
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\end{mathe}
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\begin{qstn}
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Gibt es eine lineare Abbildung, ${\phi:U\to V}$,
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so dass
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\setcounter{columnanzahl}{3}
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\begin{multikompaktenum}{\bfseries i)}[\rtab][\rtab]
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\item
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$\phi(u_{1})=v_{1}$
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\item
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$\phi(u_{2})=v_{2}$
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\item
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$\phi(u_{3})=v_{3}$
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\end{multikompaktenum}
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\uline{alle} erfüllt sind?
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\end{qstn}
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\begin{soln*}
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Wir beachten zuerst, dass $\{u_{1},u_{2}\}$ linear unabhängig sind\footnote{
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ich lasse hier die Beweise weg,
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aber man sollte die zeigen,
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z.\,B. durch das Gaußverfahren.
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}
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und dass $u_{3}\in\vectorspacespan\{u_{1},_{2}\}$,
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da ${u_{3}=10u_{2}-10u_{1}}$.
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Wir beachten auch, dass
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\begin{mathe}[mc]{rcccl}
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10v_{2}-10v_{1} &= &\begin{svector} -140\\ 30\\\end{svector} &= &v_{3}\\
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\end{mathe}
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gilt.
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Darum können wir die Frage auf Bedingungen {i) + ii)} reduzieren:
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existiert eine lineare Abbildung, die {i) + ii)} erfüllt,
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dann wird wegen Linearität Bedingung iii) automatisch mit erfüllt.
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Existiert keine lineare Abbildung, die {i) + ii)} erfüllt,
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dann existiert natürlich auch keine, die i)--iii) erfüllt.
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Wir \uline{erweitern} nun die lineare unabhängige Menge
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$\{u_{1},u_{2}\}$
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zu einer Basis
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$\{u_{1},u_{2},u'_{3},u'_{4}\}$
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von $U$.
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Wähle außerdem beliebige Vektoren, $v'_{3},v'_{4}\in V$.
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Da $\{u_{1},u_{2},u'_{3},u'_{4}\}$ eine Basis von $U$
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ist und $v_{1},v_{2},v'_{3},v'_{4}\in V$,
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existiert laut \cite[Satz~6.1.13]{sinn2020}
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eine lineare Ausdehnung,
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${\phi:U\to V}$,
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so dass
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\begin{mathe}[mc]{cccc}
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\phi(u_{1})=v_{1},
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&\phi(u_{2})=v_{2},
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&\phi(u'_{3})=v'_{3},
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&\phi(u'_{4})=v'_{4},
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\end{mathe}
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gelten. Insbesondere sind Bedingungen {i) + ii)} erfüllt.
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2021-02-04 20:37:34 +01:00
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Also lautet wie oben argumentiert, die Antwort auf die Originalfrage: \fbox{Ja}!
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2021-02-04 15:08:52 +01:00
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\end{soln*}
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\begin{qstn}
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Gibt es eine
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\setcounter{columnanzahl}{3}
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\begin{multikompaktenum}
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\item[] b) injektive
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\item[] b') nicht-injektive
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\item[] c) surjektive
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\item[] c') nicht-surjektive
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\item[] d) bijektive\footnote{
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also einen »Isomorphismus«
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}
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\item[] d') nicht-bijektive
|
|
|
|
|
|
\end{multikompaktenum}
|
|
|
|
|
|
|
|
|
|
|
|
lineare Abbildung, ${\phi:U\to V}$,
|
|
|
|
|
|
so dass i)--iii) erfüllt sind?
|
|
|
|
|
|
\end{qstn}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{soln*}
|
|
|
|
|
|
Da $\dim(U)>\dim(V)$,
|
|
|
|
|
|
kann es generell keine injektiven linearen Abbildungen
|
|
|
|
|
|
von $U$ nach $V$ geben.
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Also lauten die Antworten auf \textbf{b)}, \textbf{d)} \fbox{Nein},
|
2021-02-10 12:23:34 +01:00
|
|
|
|
und da mindestens eine lineare Ausdehnung existiert,
|
2021-02-04 20:37:34 +01:00
|
|
|
|
lautet die Antwort auf \textbf{b')} und \textbf{d')} \fbox{Ja}.
|
2021-02-04 15:08:52 +01:00
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Es bleiben nur noch \textbf{c)} und \textbf{c')} zu bestimmen.
|
2021-02-04 15:08:52 +01:00
|
|
|
|
Sei ${\phi:U\to V}$ eine lineare Ausdehnung von i)--iii).
|
|
|
|
|
|
Dann wegen Bedingungen {i) + ii)} und Linearität gilt
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcccccl}
|
|
|
|
|
|
\range(\phi)
|
|
|
|
|
|
&\supseteq
|
|
|
|
|
|
&\vectorspacespan\{\phi(u_{1}),\phi(u_{2})\}
|
|
|
|
|
|
&= &\vectorspacespan\{v_{1},v_{2}\}
|
|
|
|
|
|
&= &V.\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
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|
|
|
Die letzte Gleichung gilt, weil $\{v_{1},v_{2}\}$
|
2021-02-04 20:37:34 +01:00
|
|
|
|
linear unabhängig ist,\footnote{
|
2021-02-04 15:08:52 +01:00
|
|
|
|
ich lasse wieder den Beweis weg,
|
|
|
|
|
|
aber man sollte das machen
|
2021-02-04 20:37:34 +01:00
|
|
|
|
}
|
2021-02-04 15:08:52 +01:00
|
|
|
|
und somit eine Basis von dem $2$-dimensionalen Raum, $V$, ist.
|
|
|
|
|
|
Darum ist $\range(\phi)$ surjektiv.
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Da $\phi$ beliebig war,
|
|
|
|
|
|
haben wir tatsächlich gezeigt,
|
2021-02-04 15:08:52 +01:00
|
|
|
|
dass alle lineare Ausdehnungen von i)--iii) surjektiv sind.
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Darum lautet die Antwort auf \textbf{c)} \fbox{Ja} und auf \textbf{c')} \fbox{Nein}.
|
2021-02-04 15:08:52 +01:00
|
|
|
|
\end{soln*}
|
|
|
|
|
|
|
|
|
|
|
|
%% AUFGABE 2
|
|
|
|
|
|
\clearpage
|
|
|
|
|
|
\let\altsectionname\sectionname
|
|
|
|
|
|
\def\sectionname{Aufgabe}
|
|
|
|
|
|
\section[Aufgabe 2]{}
|
|
|
|
|
|
\label{sec:2}
|
|
|
|
|
|
\let\sectionname\altsectionname
|
|
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Betrachten wir die $\reell$-Vektorräume $U:=\reell^{4}$ und $V:=\reell^{2}$
|
|
|
|
|
|
und die Vektoren $u_{1},u_{2},u_{3}\in U$ und $v_{1},v_{2},v_{3}\in V$
|
2021-02-04 15:08:52 +01:00
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{cccc}
|
|
|
|
|
|
u_{1} = \begin{svector} 0\\ 0\\ 1\\ 1\\\end{svector},
|
|
|
|
|
|
&u_{2} = \begin{svector} 3\\ 0\\ 1\\ 1\\\end{svector},
|
|
|
|
|
|
&u_{3} = \begin{svector} 30\\ 0\\ 0\\ 0\\\end{svector},
|
|
|
|
|
|
&u_{4} = \begin{svector} 3\\ 0\\ 2\\ 2\\\end{svector},\\
|
|
|
|
|
|
\end{mathe}
|
|
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|
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|
|
|
|
\begin{mathe}[mc]{cccc}
|
|
|
|
|
|
v_{1} = \begin{svector} 5\\ 8\\\end{svector},
|
|
|
|
|
|
&v_{2} = \begin{svector} 25\\ 40\\\end{svector},
|
2021-02-06 18:54:31 +01:00
|
|
|
|
&v_{3} = \begin{svector} 200\\ 320\\\end{svector},
|
2021-02-04 15:08:52 +01:00
|
|
|
|
&v_{4} = \begin{svector} 30\\ 48\\\end{svector}.\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{qstn}
|
|
|
|
|
|
Gibt es eine lineare Abbildung, ${\phi:U\to V}$,
|
|
|
|
|
|
so dass
|
|
|
|
|
|
|
|
|
|
|
|
\setcounter{columnanzahl}{2}
|
|
|
|
|
|
\begin{multikompaktenum}{\bfseries i)}[\rtab][\rtab]
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{1})=v_{1}$
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{2})=v_{2}$
|
|
|
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|
|
\item
|
|
|
|
|
|
$\phi(u_{3})=v_{3}$
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{4})=v_{4}$
|
|
|
|
|
|
\end{multikompaktenum}
|
|
|
|
|
|
|
|
|
|
|
|
\uline{alle} erfüllt sind?
|
|
|
|
|
|
\end{qstn}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{soln*}
|
|
|
|
|
|
Wir beachten zuerst, dass $\{u_{1},u_{2}\}$ linear unabhängig sind
|
|
|
|
|
|
und dass $u_{3},u_{4}\in\vectorspacespan\{u_{1},_{2}\}$,
|
|
|
|
|
|
da ${u_{3}=10u_{2}-10u_{1}}$ und ${u_{4}=u_{1}+u_{2}}$.
|
|
|
|
|
|
Wir beachten auch, dass
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcccl}
|
2021-02-06 18:54:31 +01:00
|
|
|
|
10v_{2}-10v_{1} &= &\begin{svector} 200\\ 320\\\end{svector} &= &v_{3},\\
|
2021-02-04 15:08:52 +01:00
|
|
|
|
v_{1}+v_{2} &= &\begin{svector} 30\\ 48\\\end{svector} &= &v_{4}\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
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|
|
gelten.
|
|
|
|
|
|
Darum können wir die Frage auf Bedingungen {i) + ii)} reduzieren,
|
|
|
|
|
|
weil wegen der o.\,s. Verhältnisse {iii) + iv)}
|
|
|
|
|
|
für lineare Abbildungen automatisch mit erfüllt werden.
|
|
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Wir \uline{erweitern} nun die lineare unabhängige Menge
|
|
|
|
|
|
$\{u_{1},u_{2}\}$
|
2021-02-04 15:08:52 +01:00
|
|
|
|
zu einer Basis
|
|
|
|
|
|
$\{u_{1},u_{2},u'_{3},u'_{4}\}$
|
|
|
|
|
|
von $U$.
|
|
|
|
|
|
Wähle außerdem beliebige Vektoren, $v'_{3},v'_{4}\in V$.
|
|
|
|
|
|
Da $\{u_{1},u_{2},u'_{3},u'_{4}\}$ eine Basis von $U$
|
|
|
|
|
|
ist und $v_{1},v_{2},v'_{3},v'_{4}\in V$,
|
|
|
|
|
|
existiert laut \cite[Satz~6.1.13]{sinn2020}
|
|
|
|
|
|
eine lineare Ausdehnung,
|
|
|
|
|
|
${\phi:U\to V}$,
|
|
|
|
|
|
so dass
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{cccc}
|
|
|
|
|
|
\phi(u_{1})=v_{1},
|
|
|
|
|
|
&\phi(u_{2})=v_{2},
|
|
|
|
|
|
&\phi(u'_{3})=v'_{3},
|
|
|
|
|
|
&\phi(u'_{4})=v'_{4},
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
gelten. Insbesondere sind Bedingungen {i) + ii)} erfüllt.
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Also lautet wie oben argumentiert, die Antwort auf die Originalfrage: \fbox{Ja}!
|
2021-02-04 15:08:52 +01:00
|
|
|
|
\end{soln*}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{qstn}
|
|
|
|
|
|
Gibt es eine
|
|
|
|
|
|
|
|
|
|
|
|
\setcounter{columnanzahl}{3}
|
|
|
|
|
|
\begin{multikompaktenum}
|
|
|
|
|
|
\item[] b) injektive
|
|
|
|
|
|
\item[] b') nicht-injektive
|
|
|
|
|
|
\item[] c) surjektive
|
|
|
|
|
|
\item[] c') nicht-surjektive
|
|
|
|
|
|
\item[] d) bijektive
|
|
|
|
|
|
\item[] d') nicht-bijektive
|
|
|
|
|
|
\end{multikompaktenum}
|
|
|
|
|
|
|
|
|
|
|
|
lineare Abbildung, ${\phi:U\to V}$,
|
|
|
|
|
|
so dass i)--iv) erfüllt sind?
|
|
|
|
|
|
\end{qstn}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{soln*}
|
|
|
|
|
|
Da $\dim(U)>\dim(V)$,
|
|
|
|
|
|
kann es generell keine injektiven linearen Abbildungen
|
|
|
|
|
|
von $U$ nach $V$ geben.
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Also lauten die Antworten auf \textbf{b)}, \textbf{d)} \fbox{Nein},
|
2021-02-10 12:23:34 +01:00
|
|
|
|
und da mindestens eine lineare Ausdehnung existiert,
|
2021-02-04 20:37:34 +01:00
|
|
|
|
lautet die Antwort auf \textbf{b')} und \textbf{d')} \fbox{Ja}.
|
2021-02-04 15:08:52 +01:00
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Es bleiben nur noch \textbf{c)} und \textbf{c')} zu bestimmen.
|
2021-02-04 15:08:52 +01:00
|
|
|
|
Beachte, dass in der Konstruktion von $\phi$ im o.\,s. Beweis
|
|
|
|
|
|
wir $v'_{3},v'_{4}$ beliebig auswählen konnten.
|
|
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Zu \textbf{c)} wähle bspw. $v'_{3}:=\begin{svector} 1\\ 0\\\end{svector}$
|
2021-02-04 15:08:52 +01:00
|
|
|
|
und $v'_{4}:=\zerovector$
|
|
|
|
|
|
und sei ${\phi_{1}:U\to V}$ die lineare Abbildung
|
|
|
|
|
|
im o.\,s. Beweis mit diesen Vektoren in der Konstruktion.
|
|
|
|
|
|
Da $\{u_{1},u_{2},u'_{3},u'_{4}\}$ eine Basis für $U$ ist,
|
|
|
|
|
|
gilt wegen Linearität von $\phi_{1}$
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcl}
|
|
|
|
|
|
\range(\phi_{1})
|
|
|
|
|
|
&= &\phi_{1}(\vectorspacespan\{u_{1},u_{2},u'_{3},u'_{4}\})\\
|
|
|
|
|
|
&= &\vectorspacespan\{\phi(u_{1}),\phi(u_{2}),\phi(u'_{3}),\phi(u'_{4})\}\\
|
|
|
|
|
|
&= &\vectorspacespan\{v_{1},v_{2},v'_{3},v'_{4}\}\\
|
2021-02-04 20:37:34 +01:00
|
|
|
|
&\supseteq &\vectorspacespan\{v_{1},v'_{3}\}\\
|
|
|
|
|
|
&= &V\\
|
2021-02-04 15:08:52 +01:00
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Die letzte Gleichung gilt,
|
|
|
|
|
|
weil \uline{per Wahl} $\{v_{1},v'_{3}\}$ linear unabhängig ist
|
|
|
|
|
|
und somit eine Basis des $2$-dimenionalen Vektorraums, $V$ ist.
|
|
|
|
|
|
Da $\range(\phi_{1})\supseteq V$, ist $\phi_{1}$ surjektiv.
|
|
|
|
|
|
Die Antwort auf \textbf{c)} lautet also \fbox{Ja}.
|
2021-02-04 15:08:52 +01:00
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Zu \textbf{c')} wähle $v'_{3},v'_{4}:=\zerovector$
|
2021-02-04 15:08:52 +01:00
|
|
|
|
und sei ${\phi_{2}:U\to V}$ die lineare Abbildung
|
|
|
|
|
|
im o.\,s. Beweis mit diesen Vektoren in der Konstruktion.
|
|
|
|
|
|
Wie oben gilt
|
|
|
|
|
|
$%
|
|
|
|
|
|
\rank(\phi_{2})
|
|
|
|
|
|
\textoverset{Defn}{=}\dim(\range(\phi_{2}))
|
|
|
|
|
|
=\dim(\vectorspacespan\{v_{1},v_{2},v'_{3},v'_{4}\})
|
|
|
|
|
|
=\dim(\vectorspacespan\{v_{1}\})
|
|
|
|
|
|
\leq 1
|
|
|
|
|
|
$,
|
2021-02-04 20:37:34 +01:00
|
|
|
|
da \uline{per Wahl} $v_{2},v_{3},v_{4}\in\vectorspacespan\{v_{1}\}$
|
2021-02-04 15:08:52 +01:00
|
|
|
|
und $v_{1}\neq\zerovector$.
|
|
|
|
|
|
Also, $\rank(\phi_{2})<2=\dim(V)$.
|
|
|
|
|
|
Folglich ist $\phi_{2}$
|
|
|
|
|
|
laut \cite[Korollar~6.3.15(1)]{sinn2020}
|
|
|
|
|
|
nicht-surjektiv.
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Die Antwort auf \textbf{c')} lautet also \fbox{Ja}.
|
2021-02-04 15:08:52 +01:00
|
|
|
|
\end{soln*}
|
|
|
|
|
|
|
|
|
|
|
|
%% AUFGABE 3
|
|
|
|
|
|
\clearpage
|
|
|
|
|
|
\let\altsectionname\sectionname
|
|
|
|
|
|
\def\sectionname{Aufgabe}
|
|
|
|
|
|
\section[Aufgabe 3]{}
|
|
|
|
|
|
\label{sec:3}
|
|
|
|
|
|
\let\sectionname\altsectionname
|
|
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Betrachten wir die $\reell$-Vektorräume $U:=\reell^{2}$ und $V:=\reell^{4}$
|
|
|
|
|
|
und die Vektoren $u_{1},u_{2},u_{3}\in U$ und $v_{1},v_{2},v_{3}\in V$
|
2021-02-04 15:08:52 +01:00
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{ccc}
|
|
|
|
|
|
u_{1} = \begin{svector} 1\\ 1\\\end{svector},
|
|
|
|
|
|
&u_{2} = \begin{svector} 0\\ 2\\\end{svector},
|
|
|
|
|
|
&u_{3} = \begin{svector} 1\\ 3\\\end{svector}\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{ccc}
|
|
|
|
|
|
v_{1} = \begin{svector} -9\\ 0\\ 0\\ 1\\\end{svector},
|
|
|
|
|
|
&v_{2} = \begin{svector} 4\\ 0\\ 0\\ 2\\\end{svector},
|
|
|
|
|
|
&v_{3} = \begin{svector} 5\\ 1\\ 0\\ 3\\\end{svector}.\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{qstn}
|
|
|
|
|
|
Gibt es eine lineare Abbildung, ${\phi:U\to V}$,
|
|
|
|
|
|
so dass
|
|
|
|
|
|
|
|
|
|
|
|
\setcounter{columnanzahl}{3}
|
|
|
|
|
|
\begin{multikompaktenum}{\bfseries i)}[\rtab][\rtab]
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{1})=v_{1}$
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{2})=v_{2}$
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{3})=v_{3}$
|
|
|
|
|
|
\end{multikompaktenum}
|
|
|
|
|
|
|
|
|
|
|
|
\uline{alle} erfüllt sind?
|
|
|
|
|
|
\end{qstn}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{soln*}
|
|
|
|
|
|
Beachte, dass $u_{3}=u_{1}+u_{2}$, aber
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcccl}
|
|
|
|
|
|
v_{1}+v_{2}
|
|
|
|
|
|
&= &\begin{svector} -5\\ 0\\ 0\\ 3\\\end{svector}
|
|
|
|
|
|
&\neq &v_{3}.\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
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|
|
|
|
Angenommen, es gebe eine lineare Ausdehnung ${\phi:U\to V}$,
|
|
|
|
|
|
die i)--iii) erfüllt.
|
|
|
|
|
|
Dann muss
|
|
|
|
|
|
$v_{3}=\phi(u_{3})=\phi(u_{1}+u_{2})=\phi(u_{1})+\phi(u_{2})=v_{1}+v_{2}$
|
|
|
|
|
|
gelten. Laut der o.\,s. Gleichung kann dies aber nicht gelten.
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Darum lautet die Antwort \fbox{Nein}.
|
2021-02-04 15:08:52 +01:00
|
|
|
|
Es gibt keine lineare Ausdehnung.
|
|
|
|
|
|
\end{soln*}
|
|
|
|
|
|
|
|
|
|
|
|
%% AUFGABE 4
|
|
|
|
|
|
\clearpage
|
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|
|
|
\let\altsectionname\sectionname
|
|
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|
\def\sectionname{Aufgabe}
|
|
|
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|
|
\section[Aufgabe 4]{}
|
|
|
|
|
|
\label{sec:4}
|
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|
|
\let\sectionname\altsectionname
|
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|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Betrachten wir die $\reell$-Vektorräume $U:=\reell^{2}$ und $V:=\reell^{4}$
|
|
|
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|
|
und die Vektoren $u_{1},u_{2},u_{3}\in U$ und $v_{1},v_{2},v_{3}\in V$
|
2021-02-04 15:08:52 +01:00
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{ccc}
|
|
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|
|
u_{1} = \begin{svector} 1\\ 1\\\end{svector},
|
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|
|
&u_{2} = \begin{svector} 2\\ 3\\\end{svector},
|
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|
&u_{3} = \begin{svector} 0\\ 1\\\end{svector}\\
|
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|
\end{mathe}
|
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|
|
\begin{mathe}[mc]{ccc}
|
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|
|
v_{1} = \begin{svector} 8\\ 0\\ 0\\ 4\\\end{svector},
|
2021-02-06 18:54:31 +01:00
|
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|
&v_{2} = \begin{svector} 18\\ 0\\ 0\\ 9\\\end{svector},
|
2021-02-04 15:08:52 +01:00
|
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|
|
&v_{3} = \begin{svector} 2\\ 0\\ 0\\ 1\\\end{svector}.\\
|
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|
\end{mathe}
|
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|
|
\begin{qstn}
|
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|
|
Gibt es eine lineare Abbildung, ${\phi:U\to V}$,
|
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|
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|
|
so dass
|
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|
|
\setcounter{columnanzahl}{3}
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|
\begin{multikompaktenum}{\bfseries i)}[\rtab][\rtab]
|
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|
|
\item
|
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|
|
$\phi(u_{1})=v_{1}$
|
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|
\item
|
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|
|
$\phi(u_{2})=v_{2}$
|
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|
\item
|
|
|
|
|
|
$\phi(u_{3})=v_{3}$
|
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|
|
|
\end{multikompaktenum}
|
|
|
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|
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|
|
\uline{alle} erfüllt sind?
|
|
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|
|
\end{qstn}
|
|
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|
|
|
|
\begin{soln*}
|
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|
|
|
|
Wir beachten zuerst, dass $\{u_{1},u_{3}\}$ linear unabhängig ist
|
|
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|
|
|
und dass $u_{2}\in\vectorspacespan\{u_{1}\}$,
|
|
|
|
|
|
da ${u_{2}=2u_{1}+u_{3}}$.
|
|
|
|
|
|
Wir beachten auch, dass
|
|
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|
|
|
|
|
|
|
\begin{mathe}[mc]{rcccl}
|
|
|
|
|
|
2v_{1}+u_{3} &= &\begin{svector} 18\\ 0\\ 0\\ 9\\\end{svector} &= &v_{2}\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
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|
|
gilt.
|
|
|
|
|
|
Darum können wir die Frage auf Bedingung {i) + iii)} reduzieren,
|
|
|
|
|
|
weil wegen der o.\,s. Verhältnisse {ii)}
|
|
|
|
|
|
für lineare Abbildungen automatisch mit erfüllt wird.
|
|
|
|
|
|
|
|
|
|
|
|
Wegen linearer Unabhängigkeit ist $\{u_{1},u_{3}\}$
|
|
|
|
|
|
bereits eine Basis des $2$-dimensionalen Raums, $U$.
|
|
|
|
|
|
Deswegen brauchen wir in dieser Aufgabe keine Erweiterung zu machen.
|
|
|
|
|
|
Da $\{u_{1},u_{3}\}$ eine Basis für $U$
|
|
|
|
|
|
und $\{v_{1},v_{3}\}\subseteq V$,
|
|
|
|
|
|
existiert laut \cite[Satz~6.1.13]{sinn2020}
|
|
|
|
|
|
eine lineare Abbildung,
|
|
|
|
|
|
${\phi:U\to V}$,
|
|
|
|
|
|
so dass
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{cc}
|
|
|
|
|
|
\phi(u_{1})=v_{1}, &\phi(u_{3})=v_{3},
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
gelten. Insbesondere sind Bedingung {i) + iii)} erfüllt.
|
|
|
|
|
|
Also lautet wie oben argumentiert,
|
2021-02-04 20:37:34 +01:00
|
|
|
|
die Antwort auf die Originalfrage: \fbox{Ja}!
|
2021-02-04 15:08:52 +01:00
|
|
|
|
\end{soln*}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{qstn}
|
|
|
|
|
|
Gibt es eine
|
|
|
|
|
|
|
|
|
|
|
|
\setcounter{columnanzahl}{3}
|
|
|
|
|
|
\begin{multikompaktenum}
|
|
|
|
|
|
\item[] b) injektive
|
|
|
|
|
|
\item[] b') nicht-injektive
|
|
|
|
|
|
\item[] c) surjektive
|
|
|
|
|
|
\item[] c') nicht-surjektive
|
|
|
|
|
|
\item[] d) bijektive
|
|
|
|
|
|
\item[] d') nicht-bijektive
|
|
|
|
|
|
\end{multikompaktenum}
|
|
|
|
|
|
|
|
|
|
|
|
lineare Abbildung, ${\phi:U\to V}$,
|
|
|
|
|
|
so dass i)--iii) erfüllt sind?
|
|
|
|
|
|
\end{qstn}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{soln*}
|
|
|
|
|
|
Da $\dim(U)<\dim(V)$,
|
|
|
|
|
|
kann es generell keine surjektive linearen Abbildungen
|
|
|
|
|
|
von $U$ nach $V$ geben.
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Also lauten die Antworten auf \textbf{c)}, \textbf{d)} \fbox{Nein},
|
2021-02-10 12:23:34 +01:00
|
|
|
|
und da mindestens eine lineare Ausdehnung existiert,
|
2021-02-04 20:37:34 +01:00
|
|
|
|
lautet die Antwort auf \textbf{c')} und \textbf{d')} \fbox{Ja}.
|
2021-02-04 15:08:52 +01:00
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Es bleiben nur noch \textbf{b)} und \textbf{b')} zu bestimmen.
|
|
|
|
|
|
Sei $\phi$ eine lineare Ausdehnung, die i)--iii) erfüllt.
|
|
|
|
|
|
Dann wegen Linearität von $\phi$
|
2021-02-04 15:08:52 +01:00
|
|
|
|
und da $\{u_{1},u_{3}\}$ eine Basis von $U$ ist,
|
|
|
|
|
|
gilt
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcccccl}
|
2021-02-04 20:37:34 +01:00
|
|
|
|
\range(\phi)
|
|
|
|
|
|
&= &\phi(\vectorspacespan\{u_{1},u_{3}\})
|
|
|
|
|
|
&= &\vectorspacespan\{\phi(u_{1}),\phi(u_{3})\}
|
2021-02-04 15:08:52 +01:00
|
|
|
|
&= &\vectorspacespan\{v_{1},v_{3}\}
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
und damit
|
|
|
|
|
|
$%
|
|
|
|
|
|
\rank(\phi)
|
|
|
|
|
|
\textoverset{Defn}{=}\dim(\range(\phi))
|
|
|
|
|
|
=\dim(\vectorspacespan\{v_{1},v_{3}\})
|
|
|
|
|
|
=\dim(\vectorspacespan\{v_{3}\})
|
|
|
|
|
|
=1
|
|
|
|
|
|
$,
|
|
|
|
|
|
da $v_{1}\in\vectorspacespan\{v_{3}\}$
|
|
|
|
|
|
und $v_{3}\neq\zerovector$.
|
|
|
|
|
|
Also, $\rank(\phi)<2=\dim(U)$.
|
|
|
|
|
|
Folglich ist $\phi$
|
|
|
|
|
|
laut \cite[Korollar~6.3.15(1)]{sinn2020}
|
|
|
|
|
|
nicht injektiv.
|
|
|
|
|
|
Da hier $\phi$ beliebig gewählt wurde,
|
|
|
|
|
|
sind alle linearen Ausdehnungen von i)--iii)
|
|
|
|
|
|
immer nicht-injektiv.
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Darum lautet die Antwort auf \textbf{b)} \fbox{Nein}
|
|
|
|
|
|
und auf \textbf{b')} \fbox{Ja}.
|
2021-02-04 15:08:52 +01:00
|
|
|
|
\end{soln*}
|
|
|
|
|
|
|
|
|
|
|
|
%% AUFGABE 5
|
|
|
|
|
|
\clearpage
|
|
|
|
|
|
\let\altsectionname\sectionname
|
|
|
|
|
|
\def\sectionname{Aufgabe}
|
|
|
|
|
|
\section[Aufgabe 5]{}
|
|
|
|
|
|
\label{sec:5}
|
|
|
|
|
|
\let\sectionname\altsectionname
|
|
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Betrachten wir die $\reell$-Vektorräume $U:=\reell^{2}$ und $V:=\reell^{4}$
|
|
|
|
|
|
und die Vektoren $u_{1},u_{2},u_{3}\in U$ und $v_{1},v_{2},v_{3}\in V$
|
2021-02-04 15:08:52 +01:00
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{ccc}
|
|
|
|
|
|
u_{1} = \begin{svector} 1\\ 1\\\end{svector},
|
|
|
|
|
|
&u_{2} = \begin{svector} 2\\ 2\\\end{svector},
|
|
|
|
|
|
&u_{3} = \begin{svector} 3\\ 3\\\end{svector}\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{ccc}
|
|
|
|
|
|
v_{1} = \begin{svector} 8\\ 0\\ 0\\ 4\\\end{svector},
|
|
|
|
|
|
&v_{2} = \begin{svector} 16\\ 0\\ 0\\ 8\\\end{svector},
|
|
|
|
|
|
&v_{3} = \begin{svector} 24\\ 0\\ 0\\ 12\\\end{svector}.\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{qstn}
|
|
|
|
|
|
Gibt es eine lineare Abbildung, ${\phi:U\to V}$,
|
|
|
|
|
|
so dass
|
|
|
|
|
|
|
|
|
|
|
|
\setcounter{columnanzahl}{3}
|
|
|
|
|
|
\begin{multikompaktenum}{\bfseries i)}[\rtab][\rtab]
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{1})=v_{1}$
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{2})=v_{2}$
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{3})=v_{3}$
|
|
|
|
|
|
\end{multikompaktenum}
|
|
|
|
|
|
|
|
|
|
|
|
\uline{alle} erfüllt sind?
|
|
|
|
|
|
\end{qstn}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{soln*}
|
|
|
|
|
|
Wir beachten zuerst, dass $\{u_{1}\}$ linear unabhängig ist
|
|
|
|
|
|
und dass $u_{2},u_{3}\in\vectorspacespan\{u_{1}\}$,
|
|
|
|
|
|
da $u_{2}=2u_{1}$ und $u_{3}=3u_{1}$.
|
|
|
|
|
|
Wir beachten auch, dass
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcccl}
|
|
|
|
|
|
2v_{1} &= &\begin{svector} 16\\ 0\\ 0\\ 8\\\end{svector} &= &v_{2},\\
|
|
|
|
|
|
3v_{1} &= &\begin{svector} 24\\ 0\\ 0\\ 12\\\end{svector} &= &v_{3}\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
gelten.
|
|
|
|
|
|
Darum können wir die Frage auf Bedingung {i)} reduzieren,
|
|
|
|
|
|
weil wegen der o.\,s. Verhältnisse {ii) + iii)}
|
|
|
|
|
|
für lineare Abbildungen automatisch mit erfüllt werden.
|
|
|
|
|
|
|
|
|
|
|
|
Erweitere $\{u_{1}\}$ zu einer Basis $\{u_{1},u'_{2}\}$
|
|
|
|
|
|
des $2$-dimensionalen Raums, $U$,
|
|
|
|
|
|
und wähle einen Vektor $v'_{2}\in V$.
|
|
|
|
|
|
Da $\{u_{1},u'_{2}\}$ eine Basis für $U$
|
|
|
|
|
|
und $\{v_{1},v'_{2}\}\subseteq V$,
|
|
|
|
|
|
existiert laut \cite[Satz~6.1.13]{sinn2020}
|
|
|
|
|
|
eine lineare Abbildung,
|
|
|
|
|
|
${\phi:U\to V}$,
|
|
|
|
|
|
so dass
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{cc}
|
|
|
|
|
|
\phi(u_{1})=v_{1}, &\phi(u'_{2})=v'_{2},
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
gelten. Insbesondere sind Bedingung {i)} erfüllt.
|
|
|
|
|
|
Also lautet wie oben argumentiert,
|
2021-02-04 20:37:34 +01:00
|
|
|
|
die Antwort auf die Originalfrage: \fbox{Ja}!
|
2021-02-04 15:08:52 +01:00
|
|
|
|
\end{soln*}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{qstn}
|
|
|
|
|
|
Gibt es eine
|
|
|
|
|
|
|
|
|
|
|
|
\setcounter{columnanzahl}{3}
|
|
|
|
|
|
\begin{multikompaktenum}
|
|
|
|
|
|
\item[] b) injektive
|
|
|
|
|
|
\item[] b') nicht-injektive
|
|
|
|
|
|
\item[] c) surjektive
|
|
|
|
|
|
\item[] c') nicht-surjektive
|
|
|
|
|
|
\item[] d) bijektive
|
|
|
|
|
|
\item[] d') nicht-bijektive
|
|
|
|
|
|
\end{multikompaktenum}
|
|
|
|
|
|
|
|
|
|
|
|
lineare Abbildung, ${\phi:U\to V}$,
|
|
|
|
|
|
so dass i)--iii) erfüllt sind?
|
|
|
|
|
|
\end{qstn}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{soln*}
|
|
|
|
|
|
Da $\dim(U)<\dim(V)$,
|
|
|
|
|
|
kann es generell keine surjektive linearen Abbildungen
|
|
|
|
|
|
von $U$ nach $V$ geben.
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Also lauten die Antworten auf \textbf{c)}, \textbf{d)} \fbox{Nein},
|
2021-02-10 12:23:34 +01:00
|
|
|
|
und da mindestens eine lineare Ausdehnung existiert,
|
2021-02-04 20:37:34 +01:00
|
|
|
|
lautet die Antwort auf \textbf{c')} und \textbf{d')} \fbox{Ja}.
|
2021-02-04 15:08:52 +01:00
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Es bleiben nur noch \textbf{b)} und \textbf{b')} zu bestimmen.
|
2021-02-04 15:08:52 +01:00
|
|
|
|
Beachte, dass in der Konstruktion von $\phi$ im o.\,s. Beweis
|
|
|
|
|
|
wir $v'_{2}$ beliebig auswählen konnten.
|
|
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Zu \textbf{b)} wähle $v'_{2}:=\begin{svector} 0\\ 1\\ 0\\ 0\\\end{svector}$
|
2021-02-04 15:08:52 +01:00
|
|
|
|
und sei ${\phi_{1}:U\to V}$ die lineare Abbildung
|
|
|
|
|
|
im o.\,s. Beweis mit diesen Vektoren in der Konstruktion.
|
|
|
|
|
|
Da $\{u_{1},u'_{2}\}$ eine Basis für $U$ ist,
|
|
|
|
|
|
gilt wegen Linearität von $\phi_{1}$
|
|
|
|
|
|
|
2021-02-07 14:39:05 +01:00
|
|
|
|
\begin{mathe}[mc]{rcccccl}
|
2021-02-04 15:08:52 +01:00
|
|
|
|
\range(\phi_{1})
|
|
|
|
|
|
&= &\phi_{1}(\vectorspacespan\{u_{1},u'_{2}\})
|
|
|
|
|
|
&= &\vectorspacespan\{\phi_{1}(u_{1}),\phi_{1}(u'_{2})\}
|
2021-02-07 14:39:05 +01:00
|
|
|
|
&= &\vectorspacespan\{v_{1},v'_{2}\},
|
2021-02-04 15:08:52 +01:00
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
und damit
|
|
|
|
|
|
$%
|
|
|
|
|
|
\rank(\phi_{1})
|
|
|
|
|
|
\textoverset{Defn}{=}\dim(\range(\phi_{1}))
|
|
|
|
|
|
=\dim(\vectorspacespan\{v_{1},v'_{2}\})
|
|
|
|
|
|
=2
|
|
|
|
|
|
$,
|
|
|
|
|
|
da \uline{per Wahl} $\{v_{1},v'_{2}\}$ linear unabhängig ist.
|
|
|
|
|
|
Also, $\rank(\phi_{1})\geq 2=\dim(U)$.
|
|
|
|
|
|
Folglich ist $\phi_{1}$
|
|
|
|
|
|
laut \cite[Korollar~6.3.15(1)]{sinn2020}
|
|
|
|
|
|
injektiv.
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Die Antwort auf \textbf{b)} lautet also \fbox{Ja}.
|
2021-02-04 15:08:52 +01:00
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Zu \textbf{b')} wähle $v'_{2}:=\zerovector$
|
2021-02-04 15:08:52 +01:00
|
|
|
|
und sei ${\phi_{2}:U\to V}$ die lineare Abbildung
|
|
|
|
|
|
im o.\,s. Beweis mit diesen Vektoren in der Konstruktion.
|
|
|
|
|
|
Wie oben gilt
|
|
|
|
|
|
$%
|
|
|
|
|
|
\rank(\phi_{2})
|
|
|
|
|
|
\textoverset{Defn}{=}\dim(\range(\phi_{2}))
|
|
|
|
|
|
=\dim(\vectorspacespan\{v_{1},v'_{2}\})
|
|
|
|
|
|
=\dim(\vectorspacespan\{v_{1}\})
|
2021-02-04 20:37:34 +01:00
|
|
|
|
\leq 1
|
2021-02-04 15:08:52 +01:00
|
|
|
|
$,
|
2021-02-04 20:37:34 +01:00
|
|
|
|
da \uline{per Wahl} $v'_{2}\in\vectorspacespan\{v_{1}\}$.
|
2021-02-04 15:08:52 +01:00
|
|
|
|
Also, $\rank(\phi_{2})<2=\dim(U)$.
|
|
|
|
|
|
Folglich ist $\phi_{2}$
|
|
|
|
|
|
laut \cite[Korollar~6.3.15(1)]{sinn2020}
|
|
|
|
|
|
nicht-injektiv.
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Die Antwort auf \textbf{b')} lautet also \fbox{Ja}.
|
2021-02-04 15:08:52 +01:00
|
|
|
|
\end{soln*}
|
|
|
|
|
|
|
|
|
|
|
|
%% AUFGABE 6
|
|
|
|
|
|
\clearpage
|
|
|
|
|
|
\let\altsectionname\sectionname
|
|
|
|
|
|
\def\sectionname{Aufgabe}
|
|
|
|
|
|
\section[Aufgabe 6]{}
|
|
|
|
|
|
\label{sec:6}
|
|
|
|
|
|
\let\sectionname\altsectionname
|
|
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Betrachten wir die $\reell$-Vektorräume $U=V:=\reell^{4}$
|
|
|
|
|
|
und die Vektoren $u_{1},u_{2},u_{3},u_{4}\in U$ und $v_{1},v_{2},v_{3},v_{4}\in V$
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{cccc}
|
|
|
|
|
|
u_{1} = \begin{svector} 0\\ 0\\ 1\\ 1\\\end{svector},
|
|
|
|
|
|
&u_{2} = \begin{svector} 3\\ 0\\ 1\\ 1\\\end{svector},
|
|
|
|
|
|
&u_{3} = \begin{svector} 30\\ 0\\ 0\\ 0\\\end{svector},
|
|
|
|
|
|
&u_{4} = \begin{svector} 3\\ 0\\ 2\\ 2\\\end{svector},\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{cccc}
|
|
|
|
|
|
v_{1} = \begin{svector} 0\\ 3\\ 0\\ 0\\\end{svector},
|
|
|
|
|
|
&v_{2} = \begin{svector} 1\\ 3\\ 0\\ 0\\\end{svector},
|
|
|
|
|
|
&v_{3} = \begin{svector} 10\\ 0\\ 0\\ 0\\\end{svector},
|
|
|
|
|
|
&v_{4} = \begin{svector} 1\\ 6\\ 0\\ 0\\\end{svector}.\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{qstn}
|
|
|
|
|
|
\makelabel{qstn:6:ch:lin-ext}
|
|
|
|
|
|
Gibt es eine lineare Abbildung, ${\phi:U\to V}$,
|
|
|
|
|
|
so dass
|
|
|
|
|
|
|
|
|
|
|
|
\setcounter{columnanzahl}{2}
|
|
|
|
|
|
\begin{multikompaktenum}{\bfseries i)}[\rtab][\rtab]
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{1})=v_{1}$
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{2})=v_{2}$
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{3})=v_{3}$
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{4})=v_{4}$
|
|
|
|
|
|
\end{multikompaktenum}
|
|
|
|
|
|
|
|
|
|
|
|
\uline{alle} erfüllt sind?
|
|
|
|
|
|
\end{qstn}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{soln*}
|
|
|
|
|
|
Zunächst beobachte, dass $\{u_{1},u_{2}\}$ linear unabhängig ist,
|
|
|
|
|
|
und dass $u_{3},u_{4}\in\vectorspacespan\{u_{1},u_{2}\}$,
|
|
|
|
|
|
da ${u_{3}=10u_{2}-10u_{1}}$ und ${u_{4}=u_{1}+u_{2}}$.
|
|
|
|
|
|
Beachte auch, dass sich diese Verhältnisse in den Outputvektoren wiederspiegeln:
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcccl}
|
|
|
|
|
|
10v_{2}-10v_{1}
|
|
|
|
|
|
&= &\begin{svector} 10\\ 0\\ 0\\ 0\\\end{svector} &= &v_{3},\\
|
|
|
|
|
|
v_{1}+v_{2}
|
|
|
|
|
|
&= &\begin{svector} 1\\ 6\\ 0\\ 0\\\end{svector} &= &v_{4}.\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
Darum können wir die Frage auf Bedingungen {i) + ii)} reduzieren,
|
|
|
|
|
|
weil wegen der o.\,s. Verhältnisse {iii) + iv)}
|
|
|
|
|
|
für lineare Abbildungen automatisch mit erfüllt werden.
|
|
|
|
|
|
|
|
|
|
|
|
\uline{Erweitere} nun die linear unabhängige Menge
|
|
|
|
|
|
$\{u_{1},u_{2}\}$
|
|
|
|
|
|
zu einer Basis
|
|
|
|
|
|
$\{u_{1},u_{2},u'_{3},u'_{4}\}$
|
|
|
|
|
|
von $U$.
|
|
|
|
|
|
Wähle außerdem beliebige Vektoren, $v'_{3},v'_{4}\in V$.
|
|
|
|
|
|
Da $\{u_{1},u_{2},u'_{3},u'_{4}\}$ eine Basis von $U$
|
|
|
|
|
|
ist und $v_{1},v_{2},v'_{3},v'_{4}\in V$,
|
|
|
|
|
|
existiert laut \cite[Satz~6.1.13]{sinn2020}
|
|
|
|
|
|
eine lineare Ausdehnung,
|
|
|
|
|
|
${\phi:U\to V}$,
|
|
|
|
|
|
so dass
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{cccc}
|
|
|
|
|
|
\phi(u_{1})=v_{1},
|
|
|
|
|
|
&\phi(u_{2})=v_{2},
|
|
|
|
|
|
&\phi(u'_{3})=v'_{3},
|
|
|
|
|
|
&\phi(u'_{4})=v'_{4},
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
gelten. Insbesondere sind Bedingungen {i) + ii)} erfüllt.
|
|
|
|
|
|
Also lautet wie oben argumentiert, die Antwort auf die Originalfrage: \fbox{Ja}!
|
|
|
|
|
|
\end{soln*}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{qstn}
|
|
|
|
|
|
Gibt es eine
|
|
|
|
|
|
|
|
|
|
|
|
\setcounter{columnanzahl}{3}
|
|
|
|
|
|
\begin{multikompaktenum}
|
|
|
|
|
|
\item[] b) injektive
|
|
|
|
|
|
\item[] b') nicht-injektive
|
|
|
|
|
|
\item[] c) surjektive
|
|
|
|
|
|
\item[] c') nicht-surjektive
|
|
|
|
|
|
\item[] d) bijektive
|
|
|
|
|
|
\item[] d') nicht-bijektive
|
|
|
|
|
|
\end{multikompaktenum}
|
|
|
|
|
|
|
|
|
|
|
|
lineare Abbildung, ${\phi:U\to V}$,
|
|
|
|
|
|
so dass i)--iv) erfüllt sind?
|
|
|
|
|
|
\end{qstn}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{soln*}
|
|
|
|
|
|
Da $\dim(U)=\dim(V)$,
|
|
|
|
|
|
sind \textbf{b)}, \textbf{c)}, \textbf{d)} äquivalent
|
|
|
|
|
|
und genauso sind \textbf{b')}, \textbf{c')}, \textbf{d)} äquivalent,
|
|
|
|
|
|
da für lineare Abbildung zwischen endlichdimensionalen Räumen gleicher Dimensionen
|
|
|
|
|
|
Injektivität, Surjektivität, und Bijektivität
|
|
|
|
|
|
äquivalent sind.
|
|
|
|
|
|
Darum reicht es aus, nur \textbf{c)} und \textbf{c')} zu behandeln.
|
|
|
|
|
|
|
|
|
|
|
|
Zu \textbf{c)}, da $\{v_{1},v_{2}\}\subseteq V$ linear unabhängig sind,
|
|
|
|
|
|
wähle $v'_{3},v'_{4}\in V$ so,
|
|
|
|
|
|
dass $\{v_{1},v_{2},v'_{3},v'_{4}\}$ eine Basis
|
|
|
|
|
|
des $4$-dimensionalen Raums, $V$, bildet.
|
|
|
|
|
|
Sei ${\phi_{1}:U\to V}$ die lineare Abbildung
|
|
|
|
|
|
im o.\,s. Beweis mit diesen Vektoren in der Konstruktion.
|
|
|
|
|
|
Da $\{u_{1},u_{2},u'_{3},u'_{4}\}$ eine Basis für $U$ ist,
|
|
|
|
|
|
gilt wegen Linearität von $\phi_{1}$
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcl}
|
|
|
|
|
|
\range(\phi_{1})
|
|
|
|
|
|
&= &\phi_{1}(\vectorspacespan\{u_{1},u_{2},u'_{3},u'_{4}\})\\
|
|
|
|
|
|
&= &\vectorspacespan\{\phi(u_{1}),\phi(u_{2}),\phi(u'_{3}),\phi(u'_{4})\}\\
|
|
|
|
|
|
&= &\vectorspacespan\{v_{1},v_{2},v'_{3},v'_{4}\}\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
und damit $\range(\phi_{1})=V$,
|
|
|
|
|
|
da \uline{per Wahl} $\{v_{1},v_{2},v'_{3},v'_{4}\}$ eine Basis von $V$ ist.
|
|
|
|
|
|
Folglich ist $\phi_{1}$ surjektiv.
|
|
|
|
|
|
Die Antwort auf \textbf{c)} (und \textbf{b)} und \textbf{d)}), lautet also \fbox{Ja}.
|
|
|
|
|
|
|
|
|
|
|
|
Zu \textbf{c')} wähle $v'_{3},v'_{4}:=\zerovector$
|
|
|
|
|
|
und sei ${\phi_{2}:U\to V}$ die lineare Abbildung
|
|
|
|
|
|
im o.\,s. Beweis mit diesen Vektoren in der Konstruktion.
|
|
|
|
|
|
Wie oben gilt
|
|
|
|
|
|
$%
|
|
|
|
|
|
\rank(\phi_{2})
|
|
|
|
|
|
\textoverset{Defn}{=}\dim(\range(\phi_{2}))
|
|
|
|
|
|
=\dim(\vectorspacespan\{v_{1},v_{2},v'_{3},v'_{4}\})
|
|
|
|
|
|
=\dim(\vectorspacespan\{v_{1},v_{2}\})
|
|
|
|
|
|
\leq 2
|
|
|
|
|
|
$,
|
|
|
|
|
|
da \uline{per Wahl} $v'_{3},v'_{4}\in\vectorspacespan\{v_{1},v_{2}\}$.
|
|
|
|
|
|
Also, $\rank(\phi_{2})<4=\dim(V)$.
|
|
|
|
|
|
Folglich ist $\phi_{2}$
|
|
|
|
|
|
laut \cite[Korollar~6.3.15(2)]{sinn2020}
|
|
|
|
|
|
nicht-surjektiv.
|
|
|
|
|
|
Die Antwort auf \textbf{c')} (und \textbf{b')} und \textbf{d')}) lautet also \fbox{Ja}.
|
|
|
|
|
|
\end{soln*}
|
2021-02-04 15:08:52 +01:00
|
|
|
|
|
|
|
|
|
|
%% AUFGABE 7
|
2021-02-04 20:37:34 +01:00
|
|
|
|
\clearpage
|
2021-02-04 15:08:52 +01:00
|
|
|
|
\let\altsectionname\sectionname
|
|
|
|
|
|
\def\sectionname{Aufgabe}
|
|
|
|
|
|
\section[Aufgabe 7]{}
|
|
|
|
|
|
\label{sec:7}
|
|
|
|
|
|
\let\sectionname\altsectionname
|
|
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Betrachten wir die $\reell$-Vektorräume $U=V:=\reell^{4}$
|
|
|
|
|
|
und die Vektoren $u_{1},u_{2},u_{3},u_{4}\in U$ und $v_{1},v_{2},v_{3},v_{4}\in V$
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{cccc}
|
|
|
|
|
|
u_{1} = \begin{svector} 0\\ 0\\ 1\\ 1\\\end{svector},
|
|
|
|
|
|
&u_{2} = \begin{svector} 3\\ 0\\ 1\\ 1\\\end{svector},
|
|
|
|
|
|
&u_{3} = \begin{svector} 30\\ 0\\ 0\\ 0\\\end{svector},
|
|
|
|
|
|
&u_{4} = \begin{svector} 3\\ 0\\ 2\\ 2\\\end{svector},\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{cccc}
|
|
|
|
|
|
v_{1} = \begin{svector} 0\\ 3\\ 0\\ 0\\\end{svector},
|
|
|
|
|
|
&v_{2} = \begin{svector} 0\\ 6\\ 0\\ 0\\\end{svector},
|
|
|
|
|
|
&v_{3} = \begin{svector} 0\\ 30\\ 0\\ 0\\\end{svector},
|
|
|
|
|
|
&v_{4} = \begin{svector} 0\\ 9\\ 0\\ 0\\\end{svector}.\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{qstn}
|
|
|
|
|
|
Gibt es eine lineare Abbildung, ${\phi:U\to V}$,
|
|
|
|
|
|
so dass
|
|
|
|
|
|
|
|
|
|
|
|
\setcounter{columnanzahl}{2}
|
|
|
|
|
|
\begin{multikompaktenum}{\bfseries i)}[\rtab][\rtab]
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{1})=v_{1}$
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{2})=v_{2}$
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{3})=v_{3}$
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{4})=v_{4}$
|
|
|
|
|
|
\end{multikompaktenum}
|
|
|
|
|
|
|
|
|
|
|
|
\uline{alle} erfüllt sind?
|
|
|
|
|
|
\end{qstn}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{soln*}
|
|
|
|
|
|
Zunächst beobachte, dass $\{u_{1},u_{2}\}$ linear unabhängig ist,
|
|
|
|
|
|
und dass $u_{3},u_{4}\in\vectorspacespan\{u_{1},u_{2}\}$,
|
|
|
|
|
|
da ${u_{3}=10u_{2}-10u_{1}}$ und ${u_{4}=u_{1}+u_{2}}$.
|
|
|
|
|
|
Beachte auch, dass sich diese Verhältnisse in den Outputvektoren wiederspiegeln:
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcccl}
|
|
|
|
|
|
10v_{2}-10v_{1}
|
|
|
|
|
|
&= &\begin{svector} 0\\ 30\\ 0\\ 0\\\end{svector} &= &v_{3},\\
|
|
|
|
|
|
v_{1}+v_{2}
|
|
|
|
|
|
&= &\begin{svector} 0\\ 9\\ 0\\ 0\\\end{svector} &= &v_{4}.\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
Der Rest dieser Aufgabe lässt sich nun genauso wie
|
|
|
|
|
|
bei \Cref{qstn:6:ch:lin-ext} erledigen.
|
|
|
|
|
|
Die Antwort hier lautet also wieder: \fbox{Ja},
|
|
|
|
|
|
es gibt eine lineare Ausdehnung, die i)--iv) erfüllt.
|
|
|
|
|
|
\end{soln*}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{qstn}
|
|
|
|
|
|
Gibt es eine
|
|
|
|
|
|
|
|
|
|
|
|
\setcounter{columnanzahl}{3}
|
|
|
|
|
|
\begin{multikompaktenum}
|
|
|
|
|
|
\item[] b) injektive
|
|
|
|
|
|
\item[] b') nicht-injektive
|
|
|
|
|
|
\item[] c) surjektive
|
|
|
|
|
|
\item[] c') nicht-surjektive
|
|
|
|
|
|
\item[] d) bijektive
|
|
|
|
|
|
\item[] d') nicht-bijektive
|
|
|
|
|
|
\end{multikompaktenum}
|
|
|
|
|
|
|
|
|
|
|
|
lineare Abbildung, ${\phi:U\to V}$,
|
|
|
|
|
|
so dass i)--iv) erfüllt sind?
|
|
|
|
|
|
\end{qstn}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{soln*}
|
|
|
|
|
|
Da $\dim(U)=\dim(V)$,
|
|
|
|
|
|
sind \textbf{b)}, \textbf{c)}, \textbf{d)} äquivalent
|
|
|
|
|
|
und genauso sind \textbf{b')}, \textbf{c')}, \textbf{d)} äquivalent,
|
|
|
|
|
|
da für lineare Abbildung zwischen endlichdimensionalen Räumen gleicher Dimensionen
|
|
|
|
|
|
Injektivität, Surjektivität, und Bijektivität
|
|
|
|
|
|
äquivalent sind.
|
|
|
|
|
|
Darum reicht es aus, nur \textbf{b)} und \textbf{b')} zu behandeln.
|
|
|
|
|
|
|
|
|
|
|
|
Sei nun ${\phi:U\to V}$ eine beliebige lineare Abbildung,
|
|
|
|
|
|
die i)--iv) erfüllt (laut der letzten Aufgabe existiert mindestens eine).
|
|
|
|
|
|
Da $\{u_{1},u_{2}\}$ linear unabhängig ist,
|
|
|
|
|
|
können wir dies zu einer Basis
|
|
|
|
|
|
$\{u_{1},u_{2},u'_{3},u'_{4}\}$
|
|
|
|
|
|
von $U$ erweitern.
|
|
|
|
|
|
Da $\phi$ eine Ausdehnung und linear ist,
|
|
|
|
|
|
gilt
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcl}
|
|
|
|
|
|
\range(\phi)
|
|
|
|
|
|
&= &\phi(\vectorspacespan\{u_{1},u_{2},u'_{3},u'_{4}\})\\
|
|
|
|
|
|
&= &\vectorspacespan\{\phi(u_{1}),\phi(u_{2}),\phi(u'_{3}),\phi(u'_{4})\}\\
|
|
|
|
|
|
&= &\vectorspacespan\{v_{1},v_{2},\phi(u'_{3}),\phi(u'_{4})\}.\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
Darum gilt
|
|
|
|
|
|
$%
|
|
|
|
|
|
\rank(\phi)
|
|
|
|
|
|
\textoverset{Defn}{=}\dim(\range(\phi))
|
|
|
|
|
|
=\dim(\vectorspacespan\{v_{1},v_{2},\phi(u'_{3}),\phi(u'_{4})\})
|
|
|
|
|
|
=\dim(\vectorspacespan\{v_{1},\phi(u'_{3}),\phi(u'_{4})\})
|
|
|
|
|
|
\leq 3
|
|
|
|
|
|
$,
|
|
|
|
|
|
da $v_{2}\in\vectorspacespan\{v_{1}\}$.
|
|
|
|
|
|
Also, $\rank(\phi)<4=\dim(V)$.
|
|
|
|
|
|
Folglich ist $\phi$
|
|
|
|
|
|
laut \cite[Korollar~6.3.15(2)]{sinn2020}
|
|
|
|
|
|
nicht-surjektiv.
|
|
|
|
|
|
Da $\phi$ beliebig war,
|
|
|
|
|
|
haben wir tatsächlich gezeigt,
|
|
|
|
|
|
dass alle lineare Ausdehnungen von i)--iv) nicht-surjektiv sind.
|
|
|
|
|
|
Darum lautet die Antwort auf
|
|
|
|
|
|
\textbf{c)} (und \textbf{b)} und \textbf{d)}) \fbox{Nein}
|
|
|
|
|
|
und auf \textbf{c')} (und \textbf{b')} und \textbf{d')}) \fbox{Ja}.
|
|
|
|
|
|
\end{soln*}
|
2021-02-04 15:08:52 +01:00
|
|
|
|
|
|
|
|
|
|
%% AUFGABE 8
|
2021-02-04 20:37:34 +01:00
|
|
|
|
\clearpage
|
2021-02-04 15:08:52 +01:00
|
|
|
|
\let\altsectionname\sectionname
|
|
|
|
|
|
\def\sectionname{Aufgabe}
|
|
|
|
|
|
\section[Aufgabe 8]{}
|
|
|
|
|
|
\label{sec:8}
|
|
|
|
|
|
\let\sectionname\altsectionname
|
|
|
|
|
|
|
2021-02-04 20:37:34 +01:00
|
|
|
|
Betrachten wir die $\reell$-Vektorräume $U=V:=\reell^{3}$
|
|
|
|
|
|
und die Vektoren $u_{1},u_{2},u_{3},u_{4}\in U$ und $v_{1},v_{2},v_{3},v_{4}\in V$
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{cccc}
|
|
|
|
|
|
u_{1} = \begin{svector} 0\\ 0\\ 1\\\end{svector},
|
|
|
|
|
|
&u_{2} = \begin{svector} 3\\ 0\\ 1\\\end{svector},
|
|
|
|
|
|
&u_{3} = \begin{svector} 30\\ 2\\ 0\\\end{svector},
|
|
|
|
|
|
&u_{4} = \begin{svector} 0\\ 2\\ 0\\\end{svector},\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{cccc}
|
|
|
|
|
|
v_{1} = \begin{svector} 0\\ 3\\ 0\\\end{svector},
|
|
|
|
|
|
&v_{2} = \begin{svector} 1\\ 3\\ 0\\\end{svector},
|
2021-02-06 18:54:31 +01:00
|
|
|
|
&v_{3} = \begin{svector} 11\\ 1\\ 1\\\end{svector},
|
2021-02-04 20:37:34 +01:00
|
|
|
|
&v_{4} = \begin{svector} 1\\ 1\\ 1\\\end{svector}.\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{qstn}
|
|
|
|
|
|
\makelabel{ex:8:ch:lin-ext}
|
|
|
|
|
|
Gibt es eine lineare Abbildung, ${\phi:U\to V}$,
|
|
|
|
|
|
so dass
|
|
|
|
|
|
|
|
|
|
|
|
\setcounter{columnanzahl}{2}
|
|
|
|
|
|
\begin{multikompaktenum}{\bfseries i)}[\rtab][\rtab]
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{1})=v_{1}$
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{2})=v_{2}$
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{3})=v_{3}$
|
|
|
|
|
|
\item
|
|
|
|
|
|
$\phi(u_{4})=v_{4}$
|
|
|
|
|
|
\end{multikompaktenum}
|
|
|
|
|
|
|
|
|
|
|
|
\uline{alle} erfüllt sind?
|
|
|
|
|
|
\end{qstn}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{soln*}
|
|
|
|
|
|
Zunächst beobachte, dass $\{u_{1},u_{2},u_{4}\}$ linear unabhängig ist,
|
|
|
|
|
|
und dass $u_{3}\in\vectorspacespan\{u_{1},u_{2},u_{4}\}$,
|
|
|
|
|
|
da ${u_{3}=10u_{2}-10u_{1}+u_{4}}$.
|
|
|
|
|
|
Beachte auch, dass sich dieses Verhältnis in den Outputvektoren wiederspiegelt:
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcccl}
|
|
|
|
|
|
10v_{2}-10v_{1}+v_{4}
|
|
|
|
|
|
&= &\begin{svector} 11\\ 1\\ 1\\\end{svector} &= &v_{3}.\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
Darum können wir die Frage auf Bedingungen {i)--iii)} reduzieren,
|
|
|
|
|
|
weil wegen des o.\,s. Verhältnisses {iv)}
|
|
|
|
|
|
für lineare Abbildungen automatisch mit erfüllt wird.
|
|
|
|
|
|
|
|
|
|
|
|
Hier müssen wir nun im Gegensatz zu den anderen Aufgaben \uline{nichts hinzufügen}!
|
|
|
|
|
|
Da $\{u_{1},u_{2},u_{3}\}$ linear unabhängig ist,
|
|
|
|
|
|
bildet dies bereits eine Basis des $3$-dimenionalen Raums $U$.
|
|
|
|
|
|
Da $\{u_{1},u_{2},u_{3}\}$ eine Basis von $U$ ist
|
|
|
|
|
|
und $v_{1},v_{2},v_{3}\in V$,
|
|
|
|
|
|
existiert laut \cite[Satz~6.1.13]{sinn2020}
|
|
|
|
|
|
eine lineare Ausdehnung,
|
|
|
|
|
|
${\phi:U\to V}$,
|
|
|
|
|
|
so dass
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{cccc}
|
|
|
|
|
|
\phi(u_{1})=v_{1},
|
|
|
|
|
|
&\phi(u_{2})=v_{2},
|
|
|
|
|
|
&\phi(u_{3})=v_{3}
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
gelten. Insbesondere sind Bedingungen {i)--iii)} erfüllt.
|
|
|
|
|
|
Also lautet wie oben argumentiert, die Antwort auf die Originalfrage: \fbox{Ja}!
|
|
|
|
|
|
\end{soln*}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{qstn}
|
|
|
|
|
|
Gibt es eine
|
|
|
|
|
|
|
|
|
|
|
|
\setcounter{columnanzahl}{3}
|
|
|
|
|
|
\begin{multikompaktenum}
|
|
|
|
|
|
\item[] b) injektive
|
|
|
|
|
|
\item[] b') nicht-injektive
|
|
|
|
|
|
\item[] c) surjektive
|
|
|
|
|
|
\item[] c') nicht-surjektive
|
|
|
|
|
|
\item[] d) bijektive
|
|
|
|
|
|
\item[] d') nicht-bijektive
|
|
|
|
|
|
\end{multikompaktenum}
|
|
|
|
|
|
|
|
|
|
|
|
lineare Abbildung, ${\phi:U\to V}$,
|
|
|
|
|
|
so dass i)--iv) erfüllt sind?
|
|
|
|
|
|
\end{qstn}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{soln*}
|
|
|
|
|
|
Da $\dim(U)=\dim(V)$,
|
|
|
|
|
|
sind \textbf{b)}, \textbf{c)}, \textbf{d)} äquivalent
|
|
|
|
|
|
und genauso sind \textbf{b')}, \textbf{c')}, \textbf{d)} äquivalent,
|
|
|
|
|
|
da für lineare Abbildung zwischen endlichdimensionalen Räumen gleicher Dimensionen
|
|
|
|
|
|
Injektivität, Surjektivität, und Bijektivität
|
|
|
|
|
|
äquivalent sind.
|
|
|
|
|
|
Darum reicht es aus, nur \textbf{c)} und \textbf{c')} zu behandeln.
|
|
|
|
|
|
|
|
|
|
|
|
Sei nun ${\phi:U\to V}$ eine beliebige lineare Abbildung,
|
|
|
|
|
|
die i)--iv) erfüllt (laut der letzten Aufgabe existiert mindestens eine).
|
|
|
|
|
|
Da $\phi$ eine Ausdehnung und linear ist
|
|
|
|
|
|
und da $\{u_{1},u_{2},u_{3}\}$ eine Basis von $U$ ist,
|
|
|
|
|
|
gilt
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcl}
|
|
|
|
|
|
\range(\phi)
|
|
|
|
|
|
&= &\phi(\vectorspacespan\{u_{1},u_{2},u_{3}\})\\
|
|
|
|
|
|
&= &\vectorspacespan\{\phi(u_{1}),\phi(u_{2}),\phi(u_{3})\}\\
|
|
|
|
|
|
&= &\vectorspacespan\{v_{1},v_{2},v_{3}\}.\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
Nun ist $\{v_{1},v_{2},v_{3}\}\subseteq V$ linear unabhängig
|
|
|
|
|
|
und bildet somit eine Basis des $3$-dimenionalen Vektorraums, $V$.
|
|
|
|
|
|
Darum gilt $\range(\phi)\supseteq V$,
|
|
|
|
|
|
sodass $\phi$ surjektiv ist.
|
|
|
|
|
|
Da $\phi$ beliebig war,
|
|
|
|
|
|
haben wir tatsächlich gezeigt,
|
|
|
|
|
|
dass alle lineare Ausdehnungen von i)--iv) surjektiv sind.
|
|
|
|
|
|
Darum lautet die Antwort auf
|
|
|
|
|
|
\textbf{c)} (und \textbf{b)} und \textbf{d)}) \fbox{Ja}
|
|
|
|
|
|
und auf \textbf{c')} (und \textbf{b')} und \textbf{d')}) \fbox{Nein}.
|
|
|
|
|
|
\end{soln*}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{rem}
|
|
|
|
|
|
Laut \cite[Satz~6.1.13]{sinn2020}
|
|
|
|
|
|
die für \Cref{ex:8:ch:lin-ext} konstruierte lineare Abbildung, $\phi$,
|
|
|
|
|
|
eindeutig.
|
|
|
|
|
|
Da wir keine freie Wahl trafen, gibt es also \uline{exakt eine}
|
|
|
|
|
|
lineare Abbildung, die i)--iv) erfüllt.
|
|
|
|
|
|
Darum ist die letzte Frage eigentlich
|
|
|
|
|
|
»Ist \uline{die} lineare Ausdehnung injektiv/nicht-injektiv/...?«.
|
|
|
|
|
|
\end{rem}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{rem}
|
|
|
|
|
|
Wir hätten die o.\,s. Aufgabe so aufstellen können,
|
|
|
|
|
|
dass $\{v_{1},v_{2},v_{3}\}$ linear abhängig wäre.
|
|
|
|
|
|
Dann hätte die lineare Ausdehnung den Rang $<3$.
|
2021-03-09 13:03:51 +01:00
|
|
|
|
Die Antworten auf \textbf{c)}, \textbf{b)}, \textbf{d)} würden dann »Nein« lauten,
|
2021-02-04 20:37:34 +01:00
|
|
|
|
und auf \textbf{c')}, \textbf{b')}, \textbf{d')} »Ja«.
|
|
|
|
|
|
\end{rem}
|
2021-02-04 15:08:52 +01:00
|
|
|
|
|
2021-02-08 20:56:52 +01:00
|
|
|
|
%% ********************************************************************************
|
|
|
|
|
|
%% FILE: body/linear-systems.tex
|
|
|
|
|
|
%% ********************************************************************************
|
|
|
|
|
|
|
|
|
|
|
|
\chapter[Lineare Gleichungssysteme]{Lineare Gleichungssysteme}
|
|
|
|
|
|
\label{ch:lgs}
|
|
|
|
|
|
|
|
|
|
|
|
%% AUFGABE 1
|
|
|
|
|
|
\let\altsectionname\sectionname
|
|
|
|
|
|
\def\sectionname{Aufgabe}
|
|
|
|
|
|
\section[Aufgabe 1]{}
|
|
|
|
|
|
\label{sec:9}
|
|
|
|
|
|
\let\sectionname\altsectionname
|
|
|
|
|
|
|
|
|
|
|
|
\begin{exer*}
|
|
|
|
|
|
Bestimmen Sie eine Basis des Lösungsraums von dem Gleichungssystem,
|
|
|
|
|
|
${A\mathbf{x}=\zerovector}$,
|
|
|
|
|
|
wobei $A$ die Matrix $\begin{smatrix}
|
|
|
|
|
|
2 &-4 &0 &6 &2\\
|
|
|
|
|
|
3 &-6 &1 &13 &2\\
|
|
|
|
|
|
-7 &14 &-1 &-32 &-9\\
|
|
|
|
|
|
\end{smatrix}$
|
2021-02-09 20:16:34 +01:00
|
|
|
|
über dem Körper $\reell$ ist.
|
2021-02-08 20:56:52 +01:00
|
|
|
|
\end{exer*}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{soln*}
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|
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Um die \uline{allgemeine Form} der Lösungen
|
|
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|
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zu ${A\mathbf{x}=\zerovector}$ zu bestimmen,
|
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reduzieren wir zunächst $A$ auf Zeilenstufenform
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und normalisieren:
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\begin{mathe}[mc]{rcl}
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A &\xrightarrow{\substack{
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Z_{2} \mapsfrom 2\cdot Z_{2} - 3\cdot Z_{1}\\
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Z_{3} \mapsfrom 2\cdot Z_{3} + 7\cdot Z_{1}\\
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}} &
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\begin{matrix}{rrrrr}
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2 &-4 &0 &6 &2\\
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0 &0 &2 &8 &-2\\
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0 &0 &-2 &-22 &-4\\
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\end{matrix}
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\\
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&\xrightarrow{\substack{
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Z_{3} \mapsfrom Z_{3} + Z_{2}\\
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}} &
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\begin{matrix}{rrrrr}
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2 &-4 &0 &6 &2\\
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0 &0 &2 &8 &-2\\
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0 &0 &0 &-14 &-6\\
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\end{matrix}
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\\
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&\xrightarrow{\substack{
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Z_{1} \mapsfrom 7\cdot Z_{1} + 3\cdot Z_{3}\\
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Z_{2} \mapsfrom 7\cdot Z_{2} + 4\cdot Z_{3}\\
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}} &
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\begin{matrix}{rrrrr}
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14 &-28 &0 &0 &-4\\
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0 &0 &14 &0 &-38\\
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0 &0 &0 &-14 &-6\\
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\end{matrix}
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\\
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&\xrightarrow{\substack{
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Z_{1} \mapsfrom Z_{1}:2\\
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Z_{2} \mapsfrom Z_{2}:2\\
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Z_{3} \mapsfrom Z_{3}:-2\\
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}} &
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\begin{matrix}{ccccc}
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\boxed{7} &-14 &0 &0 &-2\\
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0 &0 &\boxed{7} &0 &-19\\
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0 &0 &0 &\boxed{7} &3\\
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\end{matrix}
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.\\
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\end{mathe}
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Aus der Zeilenstufenform ergibt sich,
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in
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${A\mathbf{x}=\zerovector}$
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die Variablen $x_{2}, x_{5}$ frei sind
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und
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\begin{mathe}[mc]{rcr0lcr0l}
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\eqtag[eq:1:ex:1:ch:lgs]
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x_{4} &= & && &-(3/7)&x_{5}\\
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x_{3} &= & && &(19/7)&x_{5}\\
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x_{1} &= &2&x_{2} &+ &(2/7)&x_{5}.\\
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\end{mathe}
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Eine Basis des Lösungsraums,
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${\ker(A)=\{\mathbf{x}\in\reell^{5}\mid A\mathbf{x}=\zerovector\}}$,
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lässt sich nach \cite[Satz~5.3.8]{sinn2020} finden,
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indem wir die Lösungen berechnen,
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für die genau eine der freien Unbekannten auf einen Wert ungleich $0$
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und alle anderen auf $0$ gesetzt werden.\footnote{
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Im \cite[Satz~5.3.8]{sinn2020} wird beschrieben, dass man $1$ verwendet,
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aber man kann die Elemente einer Basis
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mit beliebigen Werten ungleich $0$ multiplizieren,
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ohne zu ändern,
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dass die Menge eine Basis ist.
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}
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Hier gilt
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\begin{mathe}[mc]{rcrcl}
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x_{2}=1,\,x_{5}=0
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&\Longrightarrow
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&\mathbf{x}
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&\eqcrefoverset{eq:1:ex:1:ch:lgs}{=}
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&\begin{svector} 2\\ 1\\ 0\\ 0\\ 0\\\end{svector}\\
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x_{2}=0,\,x_{5}=7
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&\Longrightarrow
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&\mathbf{x}
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&\eqcrefoverset{eq:1:ex:1:ch:lgs}{=}
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&\begin{svector} 2\\ 0\\ 19\\ -3\\ 7\\\end{svector}\\
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\end{mathe}
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Darum ist \fbox{$\bigg\{\begin{svector} 2\\ 1\\ 0\\ 0\\ 0\\\end{svector}, \begin{svector} 2\\ 0\\ 19\\ -3\\ 7\\\end{svector}\bigg\}$}
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eine Basis des Lösungsraums.
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\end{soln*}
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\begin{rem*}
|
2021-02-09 20:16:34 +01:00
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Es wird hier empfohlen zu verifizieren,
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dass $A\mathbf{x}$ wirklich gleich $\zerovector$ für alle Basiselemente gilt,
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um zu überprüfen, dass unsere Lösung \emph{nicht offensichtlich falsch} ist.
|
2021-02-08 20:56:52 +01:00
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\end{rem*}
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|
\begin{exer*}
|
2021-02-09 20:16:34 +01:00
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Bestimmen Sie den Spaltenraum von $A$ aus der letzten Aufgabe (noch über $\reell$).
|
2021-02-08 20:56:52 +01:00
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|
\end{exer*}
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|
\begin{soln*}
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|
Bezeichne mit $a^{(j)}\in\reell^{3}$ für $j\in\{1,2,\ldots,5\}$
|
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die Spalten von $A$.
|
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|
Aus der o.\,s. Zeilenstufenform,
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|
|
\begin{mathe}[mc]{c}
|
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|
|
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|
\begin{matrix}{ccccc}
|
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|
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|
|
\boxed{7} &-14 &0 &0 &-2\\
|
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|
|
0 &0 &\boxed{7} &0 &-19\\
|
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|
|
0 &0 &0 &\boxed{7} &3\\
|
|
|
|
|
|
\end{matrix},
|
|
|
|
|
|
\end{mathe}
|
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|
|
|
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|
|
geht hervor, dass Spalten $1$, $3$, $4$ aus $A$ eine Basis
|
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|
|
des Spaltenraums bilden.\footnote{
|
|
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|
|
|
Die allgemeine Begründung ist wie folgt:
|
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|
|
|
Aus der Zeilenstufenform folgt, dass Spalten $1$, $3$, $4$ aus $A$
|
|
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|
|
|
linear unabhängig sind und dass die übrigen von diesen linear abhängig sind.
|
|
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|
|
D.\,h. $\cal{A}:=\{a^{(1)},\,a^{(3)},\,a^{(4)}\}$ ist linear unabhängig
|
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|
|
und $a^{(2)},a^{(5)}\in\vectorspacespan\cal{A}$.
|
|
|
|
|
|
Da $\range(A)=\vectorspacespan\{a^{(1)},\,a^{(2)},\,a^{(3)},\,a^{(4)},\,a^{(5)}\}$,
|
|
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|
|
|
ist folglich $\cal{A}$ eine Basis für $\range(A)$.
|
|
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|
|
|
}
|
|
|
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|
|
Also ist
|
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|
|
\begin{mathe}[mc]{c}
|
|
|
|
|
|
\boxed{\bigg\{
|
|
|
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|
|
\begin{vector} 2\\ 3\\ -7\\\end{vector},
|
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|
|
|
|
\begin{vector} 0\\ 1\\ -1\\\end{vector},
|
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|
|
\begin{vector} 6\\ 13\\ -32\\\end{vector}
|
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|
|
\bigg\}}
|
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|
|
|
\end{mathe}
|
|
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|
|
|
eine Basis des Spaltenraums.
|
|
|
|
|
|
\end{soln*}
|
|
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|
|
|
|
|
|
|
|
|
\textbf{Zur Kontrolle:} Aus der letzten Teilaufgabe erhielten wir
|
|
|
|
|
|
eine Basis des Lösungsraums der Länge $2$,
|
|
|
|
|
|
d.\,h. $\dim(\ker(A))=2$,
|
2021-02-09 20:16:34 +01:00
|
|
|
|
und hier wurde eine Basis des Spaltenraums der Länge $3$ gefunden,
|
2021-02-08 21:08:31 +01:00
|
|
|
|
d.\,h. $\dim(\range(A))=3$.
|
|
|
|
|
|
Wir sehen dass $\dim(\ker(A))+\dim(\range(A))=5=\dim(\reell^{5})$,
|
2021-02-08 20:56:52 +01:00
|
|
|
|
sodass die Dimensionsformel für lineare Abbildungen erfüllt ist.\footnote{
|
2021-02-09 20:16:34 +01:00
|
|
|
|
Das heißt nicht, dass unsere berechneten Basen deswegen richtig sind.
|
2021-02-08 20:56:52 +01:00
|
|
|
|
Dies ist lediglich zu kontrollieren,
|
2021-02-09 20:16:34 +01:00
|
|
|
|
dass unsere Berechnungen \emph{nicht offensichtlich falsch} sind.
|
2021-02-08 20:56:52 +01:00
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{rem*}
|
|
|
|
|
|
Falls in der letzten Aufgabe der Körper etwas anderes wäre,
|
|
|
|
|
|
so hätten wir die Zeilenstufenform erneut für diesen Körper berechnen müssen.
|
|
|
|
|
|
\end{rem*}
|
|
|
|
|
|
|
|
|
|
|
|
%% AUFGABE 2
|
|
|
|
|
|
\clearpage
|
|
|
|
|
|
\let\altsectionname\sectionname
|
|
|
|
|
|
\def\sectionname{Aufgabe}
|
|
|
|
|
|
\section[Aufgabe 2]{}
|
|
|
|
|
|
\label{sec:10}
|
|
|
|
|
|
\let\sectionname\altsectionname
|
|
|
|
|
|
|
|
|
|
|
|
\begin{exer*}
|
|
|
|
|
|
Bestimmen Sie eine Basis des Lösungsraums von dem Gleichungssystem,
|
|
|
|
|
|
${A\mathbf{x}=\zerovector}$,
|
|
|
|
|
|
wobei $A$ die Matrix $\begin{smatrix}
|
|
|
|
|
|
2 &-4 &0 &6 &2\\
|
|
|
|
|
|
3 &-6 &1 &13 &2\\
|
|
|
|
|
|
-7 &14 &-1 &-32 &-9\\
|
|
|
|
|
|
\end{smatrix}$
|
2021-02-09 20:16:34 +01:00
|
|
|
|
über dem Körper $\mathbb{F}_{7}$ ist.
|
2021-02-08 20:56:52 +01:00
|
|
|
|
\end{exer*}
|
|
|
|
|
|
|
|
|
|
|
|
\begin{soln*}
|
|
|
|
|
|
Um die \uline{allgemeine Form} der Lösungen
|
|
|
|
|
|
zu ${A\mathbf{x}=\zerovector}$ zu bestimmen,
|
|
|
|
|
|
reduzieren wir $A$ auf Zeilenstufenform
|
|
|
|
|
|
und normalisieren.
|
|
|
|
|
|
Bei/vor jedem Schritt ersetzen wir Werte durch kanonische Darstellungen von Zahlen modulo $7$.
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcl}
|
|
|
|
|
|
A &= &
|
|
|
|
|
|
\begin{matrix}{rrrrr}
|
|
|
|
|
|
2 &3 &0 &6 &2\\
|
|
|
|
|
|
3 &1 &1 &6 &2\\
|
|
|
|
|
|
0 &0 &6 &3 &5\\
|
|
|
|
|
|
\end{matrix}
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
Die Zeilenstufenform wird wie folgt berechnet:
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcl}
|
|
|
|
|
|
A &\xrightarrow{\substack{
|
|
|
|
|
|
Z_{2} \mapsfrom Z_{2}+2\cdot Z_{1}\\
|
|
|
|
|
|
}} &
|
|
|
|
|
|
\begin{matrix}{rrrrr}
|
|
|
|
|
|
2 &3 &0 &6 &2\\
|
|
|
|
|
|
0 &0 &1 &4 &6\\
|
|
|
|
|
|
0 &0 &6 &3 &5\\
|
|
|
|
|
|
\end{matrix}
|
|
|
|
|
|
\\
|
|
|
|
|
|
&\xrightarrow{\substack{
|
|
|
|
|
|
Z_{3} \mapsfrom Z_{3} + Z_{2}
|
|
|
|
|
|
}} &
|
|
|
|
|
|
\begin{matrix}{rrrrr}
|
|
|
|
|
|
2 &3 &0 &6 &2\\
|
|
|
|
|
|
0 &0 &1 &4 &6\\
|
|
|
|
|
|
0 &0 &0 &0 &4\\
|
|
|
|
|
|
\end{matrix}
|
|
|
|
|
|
\\
|
|
|
|
|
|
&\xrightarrow{\substack{
|
|
|
|
|
|
Z_{1} \mapsfrom 2\cdot Z_{1} - Z_{3}\\
|
|
|
|
|
|
Z_{2} \mapsfrom 2\cdot Z_{2} - 3\cdot Z_{3}\\
|
|
|
|
|
|
}} &
|
|
|
|
|
|
\begin{matrix}{rrrrr}
|
|
|
|
|
|
4 &6 &0 &5 &0\\
|
|
|
|
|
|
0 &0 &2 &1 &0\\
|
|
|
|
|
|
0 &0 &0 &0 &4\\
|
|
|
|
|
|
\end{matrix}
|
|
|
|
|
|
\\
|
|
|
|
|
|
&\xrightarrow{\substack{
|
|
|
|
|
|
Z_{1} \mapsfrom 2\cdot Z_{1}\\
|
|
|
|
|
|
Z_{2} \mapsfrom 4\cdot Z_{2}\\
|
|
|
|
|
|
Z_{3} \mapsfrom 2\cdot Z_{3}\\
|
|
|
|
|
|
}} &
|
|
|
|
|
|
\begin{matrix}{rrrrr}
|
|
|
|
|
|
\boxed{1} &5 &0 &3 &0\\
|
|
|
|
|
|
0 &0 &\boxed{1} &4 &0\\
|
|
|
|
|
|
0 &0 &0 &0 &\boxed{1}\\
|
|
|
|
|
|
\end{matrix}
|
|
|
|
|
|
.\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
(Beachte, dass $2^{-1}=4$ und $4^{-1}=2$ in $\mathbb{F}_{7}$, weil $2\cdot 4=8\equiv 1\mod 7$.
|
|
|
|
|
|
Darum wurden im letzten Schritt die Zeilen jeweils mit $2$ oder $4$ multipliziert.)
|
|
|
|
|
|
|
|
|
|
|
|
Aus der Zeilenstufenform ergibt sich,
|
|
|
|
|
|
in
|
|
|
|
|
|
${A\mathbf{x}=\zerovector}$
|
|
|
|
|
|
die Variablen $x_{2},x_{4}$ frei sind
|
|
|
|
|
|
und
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcr0lcr0l}
|
|
|
|
|
|
x_{5} &= &0\\
|
|
|
|
|
|
x_{3} &= & && &-4&x_{4}\\
|
|
|
|
|
|
x_{1} &= &-5&x_{2} &- &3&x_{4}.\\
|
|
|
|
|
|
\end{mathe}
|
|
|
|
|
|
|
|
|
|
|
|
Eine Basis des Lösungsraums,
|
|
|
|
|
|
${\ker(A)=\{\mathbf{x}\in\mathbb{F}_{7}^{5}\mid A\mathbf{x}=\zerovector\}}$,
|
|
|
|
|
|
lässt sich nach \cite[Satz~5.3.8]{sinn2020} finden,
|
|
|
|
|
|
indem wir die Lösungen berechnen,
|
|
|
|
|
|
für die genau eine der freien Unbekannten auf einen Wert ungleich $0$
|
|
|
|
|
|
und alle anderen auf $0$ gesetzt werden.
|
|
|
|
|
|
Hier gilt
|
|
|
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcrclcl}
|
|
|
|
|
|
x_{2}=1,\,x_{4}=0
|
|
|
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&\Longrightarrow
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&\mathbf{x}
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&= &\begin{svector} -5\\ 1\\ 0\\ 0\\ 0\\\end{svector}
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&= &\begin{svector} 2\\ 1\\ 0\\ 0\\ 0\\\end{svector}\\
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x_{2}=0,\,x_{4}=1
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&\Longrightarrow
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&\mathbf{x}
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&= &\begin{svector} -3\\ 0\\ -4\\ 1\\ 0\\\end{svector}
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&= &\begin{svector} 4\\ 0\\ 3\\ 1\\ 0\\\end{svector}\\
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\end{mathe}
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Darum ist \fbox{$\bigg\{\begin{svector} 2\\ 1\\ 0\\ 0\\ 0\\\end{svector}, \begin{svector} 4\\ 0\\ 3\\ 1\\ 0\\\end{svector}\bigg\}$}
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eine Basis des Lösungsraums.
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\end{soln*}
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\begin{rem*}
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Wiederum wird empfohlen, zu kontrollieren,
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dass $A\mathbf{x}=\zerovector$ für alle Basiselemente gilt.
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Beachte hier, dass wir modulo $7$ berechnen sollen.\footnote{
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2021-02-09 22:36:15 +01:00
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Falls man \textbf{octave} zum Berechnen benutzt, gibt man bspw.
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{\ttfamily mod(A{\char42}[2, 1, 0, 0, 0].{\char13}, 7)}
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für Moduloberechnungen ein. Und in \textbf{python} gibt man
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{\ttfamily np.matmul(A, [2, 1, 0, 0, 0])\%7}
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2021-02-09 20:16:34 +01:00
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ein (solange man konventionsgemäß numpy mittels \texttt{import numpy as np;} importiert).
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2021-02-08 20:56:52 +01:00
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}
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Bei den o.\,s. Lösungen kommen
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2021-02-09 20:16:34 +01:00
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\begin{mathe}[mc]{rclqcqrclcl}
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2021-02-08 20:56:52 +01:00
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A\cdot \begin{svector} 2\\ 1\\ 0\\ 0\\ 0\\\end{svector}
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2021-02-09 20:16:34 +01:00
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&= &\begin{svector} 0\\ 0\\ 0\\\end{svector}
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&\text{und}
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&A\cdot \begin{svector} 4\\ 0\\ 3\\ 1\\ 0\\\end{svector}
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&= &\begin{svector} 14\\ 28\\ -63\\\end{svector}
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&= &\begin{svector} 0\\ 0\\ 0\\\end{svector}
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\end{mathe}
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2021-02-08 20:56:52 +01:00
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raus, sodass wir erleichtert sein können,
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dass unsere Basiselemente richtig sind.
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2021-02-09 20:16:34 +01:00
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Ob die Größe der Basis stimmt, ist aber nicht damit überprüft.
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2021-02-08 20:56:52 +01:00
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Da müssen wir einfach prüfen, dass die Zeilenstufenform richtig berechnet wurde,
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2021-02-09 20:16:34 +01:00
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um die Anzahl der Stufen und freien Variablen zu bestätigen.
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2021-02-08 20:56:52 +01:00
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\end{rem*}
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\begin{exer*}
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2021-02-09 20:16:34 +01:00
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Bestimmen Sie den Spaltenraum von $A$ aus der letzten Aufgabe (noch über $\mathbb{F}_{7}$).
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2021-02-08 20:56:52 +01:00
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\end{exer*}
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\begin{soln*}
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Bezeichne mit $a^{(j)}\in\reell^{3}$ für $j\in\{1,2,\ldots,5\}$
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die Spalten von $A$.
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Aus der o.\,s. Zeilenstufenform,
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\begin{mathe}[mc]{c}
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\begin{matrix}{ccccc}
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\boxed{1} &5 &0 &3 &0\\
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0 &0 &\boxed{1} &4 &0\\
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0 &0 &0 &0 &\boxed{1}\\
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\end{matrix},
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\end{mathe}
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geht hervor, dass Spalten $1$, $3$, $5$ aus $A$ eine Basis
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des Spaltenraums bilden.
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Also ist
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\begin{mathe}[mc]{c}
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\boxed{\bigg\{
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\begin{vector} 2\\ 3\\ -7\\\end{vector},
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\begin{vector} 0\\ 1\\ -1\\\end{vector},
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\begin{vector} 2\\ 2\\ -9\\\end{vector}
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\bigg\}}
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\end{mathe}
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eine Basis des Spaltenraums.
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\end{soln*}
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\textbf{Zur Kontrolle:}
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|
Aus der letzten Teilaufgabe erhielten wir $\dim(\ker(A))=2$,
|
2021-02-08 21:08:31 +01:00
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|
und hier wurde nebenbei gezeigt, dass $\dim(\range(A))=3$.
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|
Also gilt $\dim(\ker(A))+\dim(\range(A))=5=\dim(\reell^{5})$,
|
2021-02-08 21:06:12 +01:00
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|
sodass die Dimensionsformel für lineare Abbildungen erfüllt ist.
|
2021-02-08 20:56:52 +01:00
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|
\begin{rem*}
|
2021-02-08 21:06:12 +01:00
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|
Die hier verwendete Matrix, $A$, war »die gleiche« wie in Aufgabe 1,
|
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|
nur mit einem anderen Körper.
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Wir sehen, dass wir die über $\reell$ berechnete Zeilenstufenform
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|
in dieser Aufgabe nicht einfach so übernehmen durften.
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|
D.\,h. wenn im 1. Teil einer Aufgabe man eine Basis des Lösungsraums von $A$ über $\reell$ bestimmen soll,
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|
und dann im 2. Teil eine Basis des Spaltenraums von $A$ über $\reell$
|
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|
und dann im 3. Teil eine Basis des Spaltenraums von $A$ über bspw. $\mathbb{F}_{7}$,
|
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|
dann kann man im 1.+2. Teil dieselbe Zeilenstufenform gebrauchen,
|
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|
aber im 3. Teil berechnet man die Zeilenstufenform am liebsten ganz von vorne in dem neuen Körper,
|
|
|
|
|
|
um einer Nummer sicher zu gehen.
|
2021-02-08 20:56:52 +01:00
|
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|
\end{rem*}
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|
2021-02-04 15:08:52 +01:00
|
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|
%% ********************************************************************************
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|
%% FILE: back/index.tex
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|
%% ********************************************************************************
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|
\bibliographystyle{alpha}
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|
\def\bibname{Literaturverzeichnis}
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|
\nocite{*}
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|
\bgroup
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|
\footnotesize
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|
%% ********************************************************************************
|
2021-02-09 17:32:10 +01:00
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|
%% FILE: ########
|
2021-02-04 15:08:52 +01:00
|
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|
%% ********************************************************************************
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|
\begin{thebibliography}{Wal16}
|
|
|
|
|
|
|
|
|
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|
|
\bibitem[Jec97]{jech1997}
|
|
|
|
|
|
Thomas Jech.
|
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|
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|
|
\newblock {\em {Set Theory}}.
|
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|
|
\newblock Springer-Verlag, 1997.
|
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|
\bibitem[Sin20]{sinn2020}
|
|
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|
|
Rainer Sinn.
|
|
|
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|
|
\newblock {Lineare Algebra I: Skript zur Veranstaltung Universit\"at Leipzig}.
|
|
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|
|
\newblock Vorlesungsskript, 2020.
|
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|
\bibitem[Wal16]{waldmann2016}
|
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|
|
Stefan Waldmann.
|
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|
|
\newblock {\em {Lineare Algebra 1: Die Grundlagen f\"ur Studierende der
|
|
|
|
|
|
Mathematik und Physik}}.
|
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|
\newblock Springer Berlin Heidelberg, 2016.
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|
\end{thebibliography}
|
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|
\egroup
|
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\end{document}
|
