From e1a0c997b7a088254ca1e2d05263cd03c7f1a681 Mon Sep 17 00:00:00 2001 From: Syndamia Date: Thu, 6 Apr 2023 20:03:17 +0300 Subject: [A2/H1/E1] Solved first exercise --- Year 2/Algebra/Homework1/Exercise1.pdf | Bin 0 -> 31733 bytes Year 2/Algebra/Homework1/Exercise1.tex | 189 +++++++++++++++++++++++++++++++++ 2 files changed, 189 insertions(+) create mode 100644 Year 2/Algebra/Homework1/Exercise1.pdf create mode 100644 Year 2/Algebra/Homework1/Exercise1.tex diff --git a/Year 2/Algebra/Homework1/Exercise1.pdf b/Year 2/Algebra/Homework1/Exercise1.pdf new file mode 100644 index 0000000..4b1a72e Binary files /dev/null and b/Year 2/Algebra/Homework1/Exercise1.pdf differ diff --git a/Year 2/Algebra/Homework1/Exercise1.tex b/Year 2/Algebra/Homework1/Exercise1.tex new file mode 100644 index 0000000..4311e01 --- /dev/null +++ b/Year 2/Algebra/Homework1/Exercise1.tex @@ -0,0 +1,189 @@ +% Document MUST be built with XeLaTeX! +\documentclass[11pt,a4paper]{article} + +\usepackage[margin=1.5cm]{geometry} + +% Поддръжка на български език +% Thanks to https://tex.stackexchange.com/a/136926 +\usepackage{fontspec} +\defaultfontfeatures{Ligatures=TeX} +\setmainfont{CMU Serif} +\setsansfont{CMU Sans Serif} + +\usepackage[bulgarian,english]{babel} + +\usepackage{amsmath} % Математически символи +\usepackage{amssymb} +\usepackage{mathtools} +\usepackage{amsfonts} % Още математически символи +\usepackage{cancel} % Зачертаване +\usepackage[dvipsnames]{xcolor} % Цветни кутийки + допълнителни цветове +\usepackage{indentfirst} % Индентация на параграф след секция + +\usepackage{enumitem} +\setlist[enumerate]{label=\arabic*.} + +\renewcommand{\thesection}{Решение} +\renewcommand{\thesubsection}{(\roman{subsection})} + +\newcommand{\im}{\operatorname{im}} + +\let\oldforall\forall +\renewcommand{\forall}{\; \oldforall} + +\let\oldexist\exists +\renewcommand{\exists}{\; \oldexist} + +\setlist[enumerate]{label=\Roman*.} + +\usepackage[pdfauthor={Камен Димитров Младенов, КН Летен семестър, 2023}, + pdftitle={Домашна работа 1 по ДАА}, + pdfsubject={}, + pdfkeywords={}, + colorlinks=true, + urlcolor=blue]{hyperref} + +\title{\vspace{-1.5cm}Домашно №1, Задача 1 по Алгебра 2 на Компютърни Науки} +\author{Камен Димитров Младенов, Група 6} +\date{7 Април 2023} + +\begin{document} + +\maketitle + +\subsection{Подточка} + +Ще докажем, че $G$ е подгрупа на $GL_2(\mathbb{R})$, съгласно +\begin{align*} + \forall A = \left( \begin{array}{cc} + a & c \\ + 0 & b + \end{array} \right) \in G, \text{ знаем че } a, b \in \mathbb{R}^* \implies \det(A) = a.b - c.0 = ab \neq 0 +\end{align*} + +Тоест ще докажем, че $\forall A,B \in G \Rightarrow A.B \in G, B^{-1} \in G$. Нека +\begin{align*} + A = \left( \begin{array}{cc} + a_1 & c_1 \\ + 0 & b_1 + \end{array} \right) \qquad + B = \left( \begin{array}{cc} + a_2 & c_2 \\ + 0 & b_2 + \end{array} \right) +\end{align*} + +\begin{align*} + A.B = + \left( \begin{array}{cc} + a_1 & c_1 \\ + 0 & b_1 + \end{array} \right) + \left( \begin{array}{cc} + a_2 & c_2 \\ + 0 & b_2 + \end{array} \right) = + \left( \begin{array}{cc} + a_1 a_2 & a_1 c_2 + c_1 b_2 \\ + 0 & b_1 b_2 + \end{array} \right) = C +\end{align*} +\begin{align*} + a_1 a_2, b_1 b_2 \in \mathbb{R}^* \implies C \in G \text{ и } \det(C) \neq 0 +\end{align*} +\begin{align*} + B^{-1} = \left( \begin{array}{cc} + a_2 & c_2 \\ + 0 & b_2 + \end{array} \right)^{-1} = \frac{1}{\det(B)} + \left( \begin{array}{cc} + B_{11} & B_{12} \\ + B_{21} & B_{22} + \end{array} \right)^\top = \frac{1}{\det(B)} + \left( \begin{array}{cc} + b_2 & 0 \\ + c_2 & a_2 + \end{array} \right)^\top = \frac{1}{\det(B)} + \left( \begin{array}{cc} + b_2 & c_2 \\ + 0 & a_2 + \end{array} \right) +\end{align*} +\begin{align*} + a_2, b_2 \in \mathbb{R}^* \text{ и } \det(B) \neq 0 \implies B^{-1} \in G \text{ и } \det(B^{-1}) \neq 0 +\end{align*} + +\subsection{Подточка} + +Ще докажем, че изображението $\varphi$ е хомоморфизъм на групи, като +\begin{align*} + \varphi : G \rightarrow (\mathbb{R}^*, .), \quad \forall + \left( \begin{array}{cc} + a & c \\ + 0 & b + \end{array} \right) \in G: \; + \varphi + \left( \begin{array}{cc} + a & c \\ + 0 & b + \end{array} \right) = a^2b +\end{align*} +\begin{align*} + \varphi \left( + \left( \begin{array}{cc} + a_1 & c_1 \\ + 0 & b_1 + \end{array} \right) + \left( \begin{array}{cc} + a_2 & c_2 \\ + 0 & b_2 + \end{array} \right) + \right) & = + \varphi \left( + \left( \begin{array}{cc} + a_1 a_2 & a_1 c_2 + c_1 b_2 \\ + 0 & b_1 b_2 + \end{array} \right) + \right) = (a_1a_2)^2(b_1b_2) = (a_1^2b_1)(a_2^2b_2) \\ + & = \varphi \left( + \left( \begin{array}{cc} + a_1 & c_1 \\ + 0 & b_1 + \end{array} \right) \right) + \varphi \left( + \left( \begin{array}{cc} + a_2 & c_2 \\ + 0 & b_2 + \end{array} \right) \right) \implies \varphi \text{ е хомоморфизъм} +\end{align*} +\begin{align*} + \forall a \in \mathbb{R}^*, \forall c \in \mathbb{R}: \; \varphi \left( + \left( \begin{array}{cc} + a & c \\ + 0 & a^{-1} + \end{array} \right) \right) = a^2a^{-1} = a \implies \varphi \text{ е върху} +\end{align*} + +Забелязваме, че +\begin{align*} + \forall a \in \mathbb{R}^* \forall c \in \mathbb{R}: \; \varphi \left( + \left( \begin{array}{cc} + a & c \\ + 0 & a^{-2} + \end{array} \right) \right) = a^2a^{-2} = 1 + \implies \ker(\varphi) = \left\{ \left( \left. \begin{array}{cc} + a & c \\ + 0 & a^{-2} + \end{array} \right) \right| a \in \mathbb{R}^*, c \in \mathbb{R} \right\} = H +\end{align*} +\begin{align*} + \forall a, b \in \mathbb{R}^*, \forall c \in \mathbb{R}: \; \varphi + \left( \begin{array}{cc} + a & c \\ + 0 & b + \end{array} \right) = a^2b \in (\mathbb{R}^*, .) \implies \im(\varphi) = (\mathbb{R}^*, .) +\end{align*} + +От теоремата за хомоморфизмите, следва че $G / H = G / \ker(\varphi) \simeq \im (\varphi) = (\mathbb{R}^*, .)$ + +\end{document} \ No newline at end of file -- cgit v1.3.1-23-gd82d