\documentclass[10pt,aspectratio=43,mathserif,table]{beamer}
%设置为 Beamer 文档类型，设置字体为 10pt，长宽比为16:9，数学字体为 serif 风格
\batchmode
\usepackage{graphicx}
\usepackage{animate}
\usepackage{hyperref}
%导入一些用到的宏包
\usepackage{amsmath,bm,amsfonts,amssymb,enumerate,epsfig,bbm,calc,color,ifthen,capt-of,multimedia,hyperref}
\usepackage{ctex} %导入中文包
% \setCJKmainfont{SimHei} %字体可采用黑体 Microsoft YaHei
\setCJKmainfont{FandolKai} %Overleaf中字体只能这条，采用楷体等见overleaf字体说明
\usetheme{Berlin} %主题
\usecolortheme{sustech} %主题颜色
\usepackage[ruled,linesnumbered]{algorithm2e}
\usepackage{fancybox}
\usepackage{xcolor}
\usepackage{times}
\usepackage{listings}
\usepackage{booktabs}
\usepackage{colortbl}
\usepackage{multicol}
\usepackage{tikz}
\usetikzlibrary{calc}
\usetikzlibrary{intersections,through}
\usetikzlibrary{decorations.pathreplacing}
% \setsansfont{Microsoft YaHei} %可设置非衬线字体采用黑体Microsoft YaHei
\setmainfont{Times New Roman} %设置英文字体采用 Times New Roman，overleaf中有效，其他需要将上方非衬线字体设置删除
\definecolor{mygreen}{rgb}{0,0.6,0}
\definecolor{mymauve}{rgb}{0.58,0,0.82}
\definecolor{mygray}{gray}{.9}
\definecolor{mypink}{rgb}{.99,.91,.95}
\definecolor{mycyan}{cmyk}{.3,0,0,0}
%题目，作者，学校，日期
\title{演示报告 \quad(内容为示例简述特征向量)}
\subtitle{\fontsize{9pt}{14pt}\textbf{标题小字 \quad(示例全英，可自行修改为中文)}}
\author{Yi Wang(姓名) \newline \newline \fontsize{6pt}{10pt}数学与统计学院}
\institute{\fontsize{6pt}{10pt}广东工业大学 \newline Guangdong University of Tech.}
\date{\today}
% 学校Logo
\logo{\includegraphics[height=1cm]{./fig/logo.png}\hspace*{0.1cm}}
\AtBeginSection[]
{
\begin{frame}<beamer>
\frametitle{\textbf{CONTENT}}
\tableofcontents[currentsection]
\end{frame}
}
\beamerdefaultoverlayspecification{<+->}
% -----------------------------------------------------------------------------
\begin{document}
% -----------------------------------------------------------------------------
\frame{\titlepage}
\section[CONTENT]{} %目录
\begin{frame}{CONTENT}
\tableofcontents
\end{frame}
% -----------------------------------------------------------------------------
\section{图片与枚举示例} %引言
\begin{frame}{Review of Higher Algebra}
\begin{multicols}{2}
\includegraphics[height=4cm]{fig/gaodai_cover.jpg}
\begin{itemize}
\item Full of matrix
\item No geometric graphics at all
\item Not intuitive
\end{itemize}
\end{multicols}
\end{frame}
\begin{frame}{Review of Higher Algebra}
\begin{multicols}{2}
\includegraphics[height=4cm]{fig/gaodai.jpg}
\includegraphics[height=4cm]{fig/gaodai_mat.jpg}
\end{multicols}
\end{frame}
\section{公式显示示例 Vector in Geometry}
\begin{frame}{Linear Simultaneous Equations}
Introduce a linear simultaneous equations
\begin{equation}
\begin{aligned}
& x & - & \quad 2y & = 1 \\
& 3x & + & \quad 2y & = 11
\end{aligned}
\end{equation}
\begin{multicols}{2} % 分两栏 若花括号中为3则是分三列
\begin{tikzpicture}[scale=0.7,domain=0:4]
\draw[very thin,color=gray] (-0.1,-1.1) grid (3.9,3.9);
\draw[->] (-0.2,0) -- (4.2,0) node[right] {$x$};
\draw[->] (0,-1) -- (0,1.8) node[above] {$y$};
\draw[color=blue] plot (\x,{\x/2 -0.5)}) node[right] {$x - 2y =1$};
\draw[color=orange] plot (\x,{-1.5*\x +5.5)}) node[right] {$3x+2y=11$};
\coordinate [label=S] (O) at (3,1);
\fill (O) circle(3pt);
\end{tikzpicture}
\hfill
\quad\\[0.5cm]
Row picture:\\[0.1cm]
\fbox{x - 2y =1}\\
\fbox{3x+2y=11}\\[0.1cm]
Point $S=(3,1)$ is the solution.
\end{multicols}
\end{frame}
\begin{frame}{Vector}
Column picture:\\
\begin{equation}
x\left[\begin{array}{c}
1 \\
3 \\
\end{array}\right]
+y\left[\begin{array}{r}
-2 \\
2 \\
\end{array}\right]
=\left[\begin{array}{r}
1 \\
11 \\
\end{array}\right]
\end{equation}
\begin{multicols}{3} % 分两栏 若花括号中为3则是分三列
\begin{tikzpicture}[scale=0.3,domain=-2:2]
\draw[very thin,color=gray,opacity=0.1] (-2,-1) grid (3,11);
\draw[->] (-2,0) -- (3,0) node[right] {$x$};
\draw[->] (0,-1) -- (0,11) node[above] {$y$};
\draw[->] (0,0) -- (-2,2) node[above] {\tiny{$\left[\begin{array}{r}
-2 \\
2 \\
\end{array}\right]$}};
\draw[->] (0,0) -- (1,3) node[right] {\tiny{$\left[\begin{array}{c}
1 \\
3 \\
\end{array}\right]$}};
\end{tikzpicture}
\begin{tikzpicture}[scale=0.3,domain=-2:2]
\draw[very thin,color=gray,opacity=0.1] (-2,-1) grid (3,11);
\draw[->] (-2,0) -- (3,0) node[right] {$x$};
\draw[->] (0,-1) -- (0,11) node[above] {$y$};
\draw[->] (0,0) -- (-2,2) node[above] {\tiny{$\left[\begin{array}{r}
-2 \\
2 \\
\end{array}\right]$}};
\draw[->] (0,0) -- (1,3) node[right] {\tiny{$\left[\begin{array}{c}
1 \\
3 \\
\end{array}\right]$}};
\draw[->] (1,3) -- (3,9) node[above]{} ;
\draw[dashed,opacity=0.5] (-2,2) -- (1,11);
\draw[dashed,opacity=0.5] (3,9) -- (1,11);
\draw[->] (0,0) -- (1,11) node[above]{$S$} ;
\end{tikzpicture}
\hfill
Where we take
\\[0.2cm]
$\left[\begin{array}{r}
-2 \\
2 \\
\end{array}\right]$ and $\left[\begin{array}{c}
1 \\
3 \\
\end{array}\right]$ \\[0.2cm]
as vectors,\\[0.2cm]
when $x=3$,\quad $y=1$ ,the $b = \left[\begin{array}{r}
1 \\
11 \\
\end{array}\right]$
\end{multicols}
\end{frame}
\section{公式显示示例 Determinant in Geometry}
\begin{frame}{Coefficient matrix}
$$ \left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
\end{array}\right]
\left[\begin{array}{r}
x \\
y \\
\end{array}\right]
=\left[\begin{array}{r}
\cdots \\
\cdots \\
\end{array}\right]
or
\left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
\end{array}\right]
\left[\begin{array}{r}
a \\
b \\
\end{array}\right]
=\left[\begin{array}{r}
\cdots \\
\cdots \\
\end{array}\right]
$$
Coefficient matrix
$ A = \left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
\end{array}\right]
$
is also a rectangular matrix.\\[0.2cm]
$det(A) = \left|\begin{array}{rr}
3 & 1 \\
0 & 2 \\
\end{array}\right| =6$\\[0.2cm]
Obvious matrix A has two vectors:$ \left[\begin{array}{r}
3 \\
0 \\
\end{array}\right]
$,
$ \left[\begin{array}{r}
1 \\
2 \\
\end{array}\right]
$
\end{frame}
\begin{frame}{公式显示示例 Linear transformations}
Unit vectors in the 2-dimensional plane are $\hat{i}=\left[\begin{array}{r}
1 \\
0 \\
\end{array}\right]$ , $\hat{j}=\left[\begin{array}{r}
0 \\
1 \\
\end{array}\right]$.\\[0.5cm]
\begin{multicols}{3}
$a\cdot i + b\cdot j $\\[0.1cm]
$=a\left[\begin{array}{r}
1 \\
0 \\
\end{array}\right]
+b\left[\begin{array}{r}
0 \\
1 \\
\end{array}\right]$\\[0.1cm]
$=\left[\begin{array}{r}
a \\
b \\
\end{array}\right]$\quad
\begin{tikzpicture}[scale=0.7,domain=-2:2]
\draw[very thin,color=gray,opacity=0.2] (-2,-2) grid (2,2);
\draw[->] (-2,0) -- (2,0) node[right] {$x$};
\draw[->] (0,-2) -- (0,2) node[above] {$y$};
\draw[->,color=red] (0,0)--(1,0) node[below]{$\hat{i}$};
\draw[->,color=red] (0,0)--(0,1) node[left]{$\hat{j}$};
\draw[->,color=red] (0,0)--(1.5,1.8) node[left]{$v$};
\draw[dashed,color=blue] (1.5,0)--(1.5,1.8) node[right]{$b$};
\draw[dashed,color=blue] (0,1.8)--(1.5,1.8) node[above]{$a$};
\end{tikzpicture}\\[0.2cm]
\begin{tikzpicture}[scale=0.7,domain=-2:2]
\draw[very thin,color=gray,opacity=0.2] (-2,-2) grid (2,2);
\draw[->] (-2,0) -- (2,0) node[right] {$x$};
\draw[->] (0,-2) -- (0,2) node[above] {$y$};
\draw[->,color=red] (0,0)--(1,0) node[below]{$\hat{i}$};
\draw[->,color=red] (0,0)--(0,1) node[left]{$\hat{j}$};
\draw[->,color=red] (0,0)--(1,1) node[left]{$v$};
\draw[dashed,color=blue] (1,0)--(1,1) node[right]{$b$};
\draw[dashed,color=blue] (0,1)--(1,1) node[above]{$a$};
\filldraw[color=yellow,opacity=0.2] (0,0) -- (1,0) -- (1,1) -- (0,1);
\end{tikzpicture}\\[0.2cm]
\end{multicols}
\fbox{$a\cdot i + b\cdot j $} is a linear transformations.\quad $a=1,b=1 $,then area is $1 $.\\[0.2cm]
also $\left[\begin{array}{rr}
i & j \\
\end{array}\right]\left[\begin{array}{r}
a \\
b \\
\end{array}\right] = \left[\begin{array}{r}
a \\
b \\
\end{array}\right]$\quad $\Rightarrow$ \quad
$\left[\begin{array}{rr}
1 & 0 \\
0 & 1 \\
\end{array}\right]\left[\begin{array}{r}
a \\
b \\
\end{array}\right] = \left[\begin{array}{r}
a \\
b \\
\end{array}\right]$
\end{frame}
\begin{frame}
Hense, we can tell that \\[0.2cm]
$ \left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
\end{array}\right]
\left[\begin{array}{r}
a \\
b \\
\end{array}\right]
$
$
= \left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
\end{array}\right]
\left[\begin{array}{rr}
1 & 0 \\
0 & 1 \\
\end{array}\right]
\left[\begin{array}{r}
a \\
b \\
\end{array}\right]
$\\[0.2cm]
Which is actually the original two-dimensional space of the unit vector is linearly transformed. $\rightarrow \times \left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
\end{array}\right]$\\[0.2cm]
$\hat{i} = \left[\begin{array}{r}
1 \\
0 \\
\end{array}\right]
\rightarrow \left[\begin{array}{r}
3 \\
0 \\
\end{array}\right]
$\quad ,\quad
$\hat{j} = \left[\begin{array}{r}
0 \\
1 \\
\end{array}\right]\rightarrow \left[\begin{array}{r}
1 \\
2 \\
\end{array}\right]
$\\[0.5cm]
\begin{multicols}{3}
\begin{tikzpicture}[scale=0.7,domain=-2:2]
\draw[very thin,color=gray,opacity=0.2] (-2,-2) grid (2,2);
\draw[->] (-2,0) -- (2,0) node[right] {$x$};
\draw[->] (0,-2) -- (0,2) node[above] {$y$};
\draw[->,color=red] (0,0)--(1,0) node[below]{$\hat{i}$};
\draw[->,color=red] (0,0)--(0,1) node[left]{$\hat{j}$};
\draw[->,color=red] (0,0)--(1,1) node[left]{$v$};
\draw[dashed,color=blue] (1,0)--(1,1) node[right]{};
\draw[dashed,color=blue] (0,1)--(1,1) node[above]{};
\filldraw[color=yellow,opacity=0.2] (0,0) -- (1,0) -- (1,1) -- (0,1);
\end{tikzpicture}\\[1cm]
\hfill
$$\Rightarrow$$
Area : $1 \rightarrow 6$\\[0.1cm]
\hfill
\begin{tikzpicture}[scale=0.7,domain=-2:2]
\draw[very thin,color=gray,opacity=0.2] (0,0) grid (4,4);
\draw[->] (0,0) -- (4,0) node[right] {$x$};
\draw[->] (0,0) -- (2,4) node[above] {$y$};
\draw[->,color=red] (0,0)--(3,0) node[right]{$\left[\begin{array}{r}
3 \\
0 \\
\end{array}\right]$};
\draw[->,color=red] (0,0)--(1,2) node[left]{$\left[\begin{array}{r}
1 \\
2 \\
\end{array}\right]$};
\draw[->,color=red] (0,0)--(4,2) node[left]{$v$};
\draw[dashed,color=blue] (3,0)--(4,2) node[right]{};
\draw[dashed,color=blue] (1,2)--(4,2) node[above]{};
\filldraw[color=yellow,opacity=0.2] (0,0) -- (3,0) -- (4,2) -- (1,2);
\end{tikzpicture}
\end{multicols}
\end{frame}
\begin{frame}{Determinant in Geometry}
Since
$\left|\begin{array}{rr}
3 & 1 \\
0 & 2 \\
\end{array}\right|
= 6
$ ,\quad
Area : $1 \rightarrow 6$\\[0.2cm]
\begin{multicols}{3}
\begin{tikzpicture}[scale=0.7,domain=-2:2]
\draw[very thin,color=gray,opacity=0.2] (-2,-2) grid (2,2);
\draw[->] (-2,0) -- (2,0) node[right] {$x$};
\draw[->] (0,-2) -- (0,2) node[above] {$y$};
\draw[->,color=red] (0,0)--(1,0) node[below]{$\hat{i}$};
\draw[->,color=red] (0,0)--(0,1) node[left]{$\hat{j}$};
\draw[->,color=red] (0,0)--(1,1) node[left]{$v$};
\draw[dashed,color=blue] (1,0)--(1,1) node[right]{};
\draw[dashed,color=blue] (0,1)--(1,1) node[above]{};
\filldraw[color=yellow,opacity=0.2] (0,0) -- (1,0) -- (1,1) -- (0,1);
\end{tikzpicture}\\[1cm]
\hfill
$$\Rightarrow$$
Area scaled \\[0.1cm]
by \textbf{6} times.\\[0.1cm]
\hfill
\begin{tikzpicture}[scale=0.7,domain=-2:2]
\draw[very thin,color=gray,opacity=0.2] (0,0) grid (4,4);
\draw[->] (0,0) -- (4,0) node[right] {$x$};
\draw[->] (0,0) -- (2,4) node[above] {$y$};
\draw[->,color=red] (0,0)--(3,0) node[right]{$\left[\begin{array}{r}
3 \\
0 \\
\end{array}\right]$};
\draw[->,color=red] (0,0)--(1,2) node[left]{$\left[\begin{array}{r}
1 \\
2 \\
\end{array}\right]$};
\draw[->,color=red] (0,0)--(4,2) node[left]{$v$};
\draw[dashed,color=blue] (3,0)--(4,2) node[right]{};
\draw[dashed,color=blue] (1,2)--(4,2) node[above]{};
\filldraw[color=yellow,opacity=0.2] (0,0) -- (3,0) -- (4,2) -- (1,2);
\end{tikzpicture}
\end{multicols}
We can conclude that :\\[0.2cm]
The Determinant in Geometry
is how much are areas scaled.
\end{frame}
\section{公式显示示例 Eigenvalue in Geometry}
\begin{frame}{Vectors remain on their own span}
\hspace{-2cm}{
\begin{multicols}{3}
\begin{tikzpicture}[scale=0.7,domain=-2:2]
\draw[very thin,color=gray,opacity=0.2] (-2,-2) grid (2,2);
\draw[->] (-2,0) -- (2,0) node[right] {$x$};
\draw[->] (0,-2) -- (0,2) node[above] {$y$};
\draw[->,color=red] (0,0)--(1,0) node[below]{$\hat{i}$};
\draw[->,color=red] (0,0)--(0,1) node[left]{$\hat{j}$};
\draw[->,color=red] (0,0)--(1,1) node[left]{$v$};
\draw[dashed,color=blue] (1,0)--(1,1) node[right]{};
\draw[dashed,color=blue] (0,1)--(1,1) node[above]{};
\filldraw[color=yellow,opacity=0.2] (0,0) -- (1,0) -- (1,1) -- (0,1);
\end{tikzpicture}\\[1cm]
\hfill
linearly transformed.\\[0.2cm]
$\rightarrow \times \left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
\end{array}\right]
\rightarrow$\\[0.2cm]
\hfill
\hspace{-1.8cm}
\begin{tikzpicture}[scale=0.5,domain=-2:2]
\draw[very thin,color=gray,opacity=0.2] (-4,-4) grid (4,4);
\draw[very thin,color=gray,opacity=0.2] (-4,0) -- (4,0) ;
\draw[very thin,color=gray,opacity=0.2] (-4,2) -- (4,2) ;
\draw[very thin,color=gray,opacity=0.2] (-4,4) -- (4,4) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-2) -- (4,-2) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-4) -- (4,-4) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-0) -- (-2,4) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-4) -- (0,4) ;
\draw[very thin,color=gray,opacity=0.2] (-2,-4) -- (2,4) ;
\draw[very thin,color=gray,opacity=0.2] (0,-4) -- (4,4) ;
\draw[very thin,color=gray,opacity=0.2] (2,-4) -- (4,0) ;
\draw[->] (-4,0) -- (4,0) node[right] {$x$};
\draw[->] (-2,-4) -- (2,4) node[above] {$y$};
\draw[->,color=red] (0,0)--(3,0) node[below]{$\left[\begin{array}{r}
3 \\
0 \\
\end{array}\right]$};
\draw[->,color=red] (0,0)--(1,2) node[above]{$\left[\begin{array}{r}
1 \\
2 \\
\end{array}\right]$};
\draw[->,color=red] (0,0)--(4,2) node[left]{$v$};
\draw[dashed,color=blue] (3,0)--(4,2) node[right]{};
\draw[dashed,color=blue] (1,2)--(4,2) node[above]{};
\filldraw[color=yellow,opacity=0.2] (0,0) -- (3,0) -- (4,2) -- (1,2);
\end{tikzpicture}
\end{multicols}}
\end{frame}
\begin{frame}{Vectors remain on their own span}
\begin{multicols}{3}
\begin{tikzpicture}[scale=0.7,domain=-2:2]
\draw[very thin,color=gray,opacity=0.2] (-2,-2) grid (2,2);
\draw[->] (-2,0) -- (2,0) node[right] {$x$};
\draw[->] (0,-2) -- (0,2) node[above] {$y$};
\draw[very thick,->,color=red] (0,0)--(1,0) node[below]{$\hat{i}$};
\draw[very thick,->,color=blue] (0,0)--(-1,0) node[below]{$\alpha$};
\draw[decorate,decoration={brace,raise=5pt},blue] (0,0)--(-1,0);
\node at (-0.5,-1) {1};
\draw[->,color=red] (0,0)--(0,1) node[left]{$\hat{j}$};
\filldraw[color=yellow,opacity=0.2] (0,0) -- (1,0) -- (1,1) -- (0,1);
\end{tikzpicture}\\[1cm]
\hfill
linearly transformed.\\[0.2cm]
$\rightarrow \times \left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
\end{array}\right]
\rightarrow$\\[0.2cm]
\hfill
\hspace{-1.8cm}
\begin{tikzpicture}[scale=0.5,domain=-2:2]
\draw[very thin,color=gray,opacity=0.2] (-4,-4) grid (4,4);
\draw[very thin,color=gray,opacity=0.2] (-4,0) -- (4,0) ;
\draw[very thin,color=gray,opacity=0.2] (-4,2) -- (4,2) ;
\draw[very thin,color=gray,opacity=0.2] (-4,4) -- (4,4) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-2) -- (4,-2) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-4) -- (4,-4) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-0) -- (-2,4) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-4) -- (0,4) ;
\draw[very thin,color=gray,opacity=0.2] (-2,-4) -- (2,4) ;
\draw[very thin,color=gray,opacity=0.2] (0,-4) -- (4,4) ;
\draw[very thin,color=gray,opacity=0.2] (2,-4) -- (4,0) ;
\draw[->] (-4,0) -- (4,0) node[right] {$x$};
\draw[->] (-2,-4) -- (2,4) node[above] {$y$};
\draw[very thick,->,color=red] (0,0)--(3,0) node[below]{$\left[\begin{array}{r}
3 \\
0 \\
\end{array}\right]$};
\draw[very thick,->,color=blue] (0,0)--(-3,0) node[below]{$\alpha$};
\draw[decorate,decoration={brace,mirror,raise=5pt},blue] (0,0)--(-3,0);
\node at (-1.6,1) {3};
\draw[->,color=red] (0,0)--(1,2) node[above]{};
\filldraw[color=yellow,opacity=0.2] (0,0) -- (3,0) -- (4,2) -- (1,2);
\end{tikzpicture}
\end{multicols}
\textbf{$\overrightarrow{\alpha}$} remains on the line of the x-axis,stretched by a factor of \textbf{3}.
\end{frame}
\begin{frame}{Vectors remain on their own span}
\begin{multicols}{3}
\begin{tikzpicture}[scale=0.7,domain=-2:2]
\draw[very thin,color=gray,opacity=0.2] (-2,-2) grid (2,2);
\draw[->] (-2,0) -- (2,0) node[right] {$x$};
\draw[->] (0,-2) -- (0,2) node[above] {$y$};
\draw[very thick,->,color=brown] (0,0)--(1,-1) node[left]{$\gamma$};
\draw[very thick,->,color=purple] (-1,1)--(-2,2) node[left]{$\theta$};
\draw[very thick,->,color=gray] (1,-1)--(2,-2) node[left]{$\omega$};
\draw[very thick,->,color=blue] (0,0)--(-1,1) node[left]{$\beta$};
\draw[decorate,decoration={brace,raise=5pt},blue] (0,0)--(-1,1);
\node at (-1,0) {1};
\filldraw[color=yellow,opacity=0.2] (0,0) -- (1,0) -- (1,1) -- (0,1);
\end{tikzpicture}\\[1cm]
\hfill
linearly transformed.\\[0.2cm]
$\rightarrow \times \left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
\end{array}\right]
\rightarrow$\\[0.2cm]
\hfill
\hspace{-1.8cm}
\begin{tikzpicture}[scale=0.5,domain=-2:2]
\draw[very thin,color=gray,opacity=0.2] (-4,-4) grid (4,4);
\draw[very thin,color=gray,opacity=0.2] (-4,0) -- (4,0) ;
\draw[very thin,color=gray,opacity=0.2] (-4,2) -- (4,2) ;
\draw[very thin,color=gray,opacity=0.2] (-4,4) -- (4,4) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-2) -- (4,-2) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-4) -- (4,-4) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-0) -- (-2,4) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-4) -- (0,4) ;
\draw[very thin,color=gray,opacity=0.2] (-2,-4) -- (2,4) ;
\draw[very thin,color=gray,opacity=0.2] (0,-4) -- (4,4) ;
\draw[very thin,color=gray,opacity=0.2] (2,-4) -- (4,0) ;
\draw[->] (-4,0) -- (4,0) node[right] {$x$};
\draw[->] (-2,-4) -- (2,4) node[above] {$y$};
\draw[very thick,->,color=brown] (0,0)--(2,-2) node[left]{$\gamma$};
\draw[very thick,->,color=purple] (-2,2)--(-4,4) node[left]{$\theta$};
\draw[very thick,->,color=gray] (2,-2)--(4,-4) node[left]{$\omega$};
\draw[very thick,->,color=blue] (0,0)--(-2,2) node[left]{$\beta$};
\draw[decorate,decoration={brace,raise=5pt},blue] (0,0)--(-2,2);
\node at (-1.5,0.5) {2};
\filldraw[color=yellow,opacity=0.2] (0,0) -- (3,0) -- (4,2) -- (1,2);
\end{tikzpicture}
\end{multicols}
\textbf{$\overrightarrow{\beta}$} remains on the line of the x-axis,stretched by a factor of \textbf{2}.\\[0.2cm]
The other vectors($\gamma,\theta,\omega$) on the line are also stretched by a factor of \textbf{2}
\end{frame}
\begin{frame}{Eigenvalue \& Eigenvector}
\begin{multicols}{2}
\begin{tikzpicture}[scale=0.5,domain=-2:2]
\draw[very thin,color=gray,opacity=0.2] (-4,-4) grid (4,4);
\draw[very thin,color=gray,opacity=0.2] (-4,0) -- (4,0) ;
\draw[very thin,color=gray,opacity=0.2] (-4,2) -- (4,2) ;
\draw[very thin,color=gray,opacity=0.2] (-4,4) -- (4,4) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-2) -- (4,-2) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-4) -- (4,-4) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-0) -- (-2,4) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-4) -- (0,4) ;
\draw[very thin,color=gray,opacity=0.2] (-2,-4) -- (2,4) ;
\draw[very thin,color=gray,opacity=0.2] (0,-4) -- (4,4) ;
\draw[very thin,color=gray,opacity=0.2] (2,-4) -- (4,0) ;
\draw[->,opacity=0.2] (-4,0) -- (4,0) node[right] {$x$};
\draw[->,opacity=0.2] (-2,-4) -- (2,4) node[above] {$y$};
\draw[very thick,->,color=brown] (0,0)--(2,-2) node[left]{};
\draw[very thick,->,color=purple] (-2,2)--(-4,4) node[left]{};
\draw[very thick,->,color=gray] (2,-2)--(4,-4) node[left]{};
\draw[very thick,->,color=blue] (0,0)--(-2,2) node[left]{$\beta$};
\draw[decorate,decoration={brace,mirror,raise=5pt},blue] (0,0)--(-2,2);
\node at (1,2) {Scaled by 2};
\draw[very thick,->,color=gray] (0,0)--(3,0) node[below]{};
\draw[very thick,->,color=red] (0,0)--(-3,0) node[below]{$\alpha$};
\draw[decorate,decoration={brace,raise=5pt},blue] (0,0)--(-3,0);
\node at (-1.6,-1) {Scaled by 3};
\filldraw[color=yellow,opacity=0.2] (0,0) -- (3,0) -- (4,2) -- (1,2);
\end{tikzpicture}
The vector representing these lines are
$$\left[\begin{array}{r}
1 \\
0 \\
\end{array}\right],\left[\begin{array}{r}
-1 \\
1 \\
\end{array}\right]$$
\end{multicols}
\end{frame}
\begin{frame}{Eigenvalue \& Eigenvector}
\begin{multicols}{2}
\begin{tikzpicture}[scale=0.5,domain=-2:2]
\draw[very thin,color=gray,opacity=0.2] (-4,-4) grid (4,4);
\draw[very thin,color=gray,opacity=0.2] (-4,0) -- (4,0) ;
\draw[very thin,color=gray,opacity=0.2] (-4,2) -- (4,2) ;
\draw[very thin,color=gray,opacity=0.2] (-4,4) -- (4,4) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-2) -- (4,-2) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-4) -- (4,-4) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-0) -- (-2,4) ;
\draw[very thin,color=gray,opacity=0.2] (-4,-4) -- (0,4) ;
\draw[very thin,color=gray,opacity=0.2] (-2,-4) -- (2,4) ;
\draw[very thin,color=gray,opacity=0.2] (0,-4) -- (4,4) ;
\draw[very thin,color=gray,opacity=0.2] (2,-4) -- (4,0) ;
\draw[->,opacity=0.2] (-4,0) -- (4,0) node[right] {$x$};
\draw[->,opacity=0.2] (-2,-4) -- (2,4) node[above] {$y$};
\draw[very thick,->,color=brown] (0,0)--(2,-2) node[left]{};
\draw[very thick,->,color=purple] (-2,2)--(-4,4) node[left]{};
\draw[very thick,->,color=gray] (2,-2)--(4,-4) node[left]{};
\draw[very thick,->,color=blue] (0,0)--(-2,2) node[left]{$\left[\begin{array}{r}
1 \\
0 \\
\end{array}\right]$};
\draw[decorate,decoration={brace,mirror,raise=5pt},blue] (0,0)--(-2,2);
\node at (1,2) {Scaled by 2};
\draw[very thick,->,color=gray] (0,0)--(3,0) node[below]{};
\draw[very thick,->,color=red] (0,0)--(-3,0) node[below]{$\left[\begin{array}{r}
-1 \\
1 \\
\end{array}\right]$};
\draw[decorate,decoration={brace,raise=5pt},blue] (0,0)--(-3,0);
\node at (0,-1) {Scaled by 3};
\filldraw[color=yellow,opacity=0.2] (0,0) -- (3,0) -- (4,2) -- (1,2);
\end{tikzpicture}
\hfill
$$A = \left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
\end{array}\right]$$\\[0.2cm]
The vector representing the line is called the eigenvector of the matrix A.\\[0.2cm]
特征向量:$\left[\begin{array}{r}
1 \\
0 \\
\end{array}\right],\left[\begin{array}{r}
-1 \\
1 \\
\end{array}\right]$\\[0.2cm]
The eigenvalue of the matrix A is just the factor by which it stretched or squashed during the transformation.\\[0.2cm]
特征值:$2,3$
\end{multicols}
\end{frame}
\begin{frame}{Eigenvalue \& Eigenvector}
So maybe you can tell why we can get eigenvalue of matrix from this equation:
$$Ax = \lambda x$$
\end{frame}
\section{参考文献示例 Refference}
\begin{frame}{Refference}
\begin{multicols}{2}
\begin{itemize}
\item Introduction to Linear Algebra(Strang)
\item Essense of Linear Algebra @3Blue1Brown
\item Linear algebra and its applications 4th
\end{itemize}
\includegraphics[scale=0.3]{fig/ref1.png}
\end{multicols}
\end{frame}
\begin{frame}{Acknowledgements}
\begin{center}
\begin{minipage}{1\textwidth}
\setbeamercolor{mybox}{fg=white, bg=black!50!blue}
\begin{beamercolorbox}[wd=0.70\textwidth, rounded=true, shadow=true]{mybox}
\LARGE \centering Thank you for listening! %结束语
\end{beamercolorbox}
\end{minipage}
\end{center}
\end{frame}
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\end{document}
