Please wait
\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}
% -----------------------------------------------------------------------------
\end{document}
Overleaf is perfect for all types of projects — from papers and presentations to newsletters, CVs and much more! It's also a great way to learn how to use LaTeX and produce professional looking projects quickly.
Upload or create templates for journals you submit to and theses and presentation templates for your institution. Just create it as a project on Overleaf and use the publish menu. It's free! No sign-up required.
New template are added all the time. Follow us on twitter for the highlights!
Overleaf is a free online collaborative LaTeX editor. No sign up required.
Learn more