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+%%=====================================================================================
+%%
+%%       Filename:  cours.tex
+%%
+%%    Description:  
+%%
+%%        Version:  1.0
+%%        Created:  03/06/2024
+%%       Revision:  none
+%%
+%%         Author:  YOUR NAME (), 
+%%   Organization:  
+%%      Copyright:  Copyright (c) 2024, YOUR NAME
+%%
+%%          Notes:  
+%%
+%%=====================================================================================
+\documentclass[a4paper, titlepage]{article}
+
+\usepackage[utf8]{inputenc}
+\usepackage[T1]{fontenc}
+\usepackage{textcomp}
+\usepackage[french]{babel}
+\usepackage{amsmath, amssymb}
+\usepackage{amsthm}
+\usepackage[svgnames]{xcolor}
+\usepackage{thmtools}
+\usepackage{lipsum}
+\usepackage{framed}
+\usepackage{parskip}
+\usepackage{titlesec}
+
+\renewcommand{\familydefault}{\sfdefault}
+
+% figure support
+\usepackage{import}
+\usepackage{xifthen}
+\pdfminorversion=7
+\usepackage{pdfpages}
+\usepackage{transparent}
+\newcommand{\incfig}[1]{%
+	\def\svgwidth{\columnwidth}
+	\import{./figures/}{#1.pdf_tex}
+}
+
+\pdfsuppresswarningpagegroup=1
+
+\colorlet{defn-color}{DarkBlue}
+\colorlet{props-color}{Blue}
+\colorlet{warn-color}{Red}
+\colorlet{exemple-color}{Green}
+\colorlet{corol-color}{Orange}
+\newenvironment{defn-leftbar}{%
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+  \MakeFramed {\advance\hsize-\width \FrameRestore}}%
+ {\endMakeFramed}
+\newenvironment{warn-leftbar}{%
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+\newenvironment{exemple-leftbar}{%
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+  \MakeFramed {\advance\hsize-\width \FrameRestore}}%
+ {\endMakeFramed}
+\newenvironment{corol-leftbar}{%
+  \def\FrameCommand{{\color{corol-color}\vrule width 3pt} \hspace{10pt}}%
+  \MakeFramed {\advance\hsize-\width \FrameRestore}}%
+ {\endMakeFramed}
+
+\def \freespace {1em}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
+ notefont=\sffamily\bfseries,%
+ notebraces={}{},%
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+ preheadhook=\vspace{\freespace}\begin{defn-leftbar},%
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+]{better-defn}
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+]{better-exemple}
+\declaretheoremstyle[headfont=\sffamily\bfseries,%
+ notefont=\sffamily\bfseries,%
+ notebraces={}{},%
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+ preheadhook=\vspace{\freespace}\begin{props-leftbar},%
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+ preheadhook=\vspace{\freespace}\begin{corol-leftbar},%
+ postfoothook=\end{corol-leftbar},%
+]{better-corol}
+
+\declaretheorem[style=better-defn]{defn}
+\declaretheorem[style=better-warn]{warn}
+\declaretheorem[style=better-exemple]{exemple}
+\declaretheorem[style=better-corol]{corol}
+\declaretheorem[style=better-props, numberwithin=defn]{props}
+\declaretheorem[style=better-thm, sibling=props]{thm}
+\newtheorem*{lemme}{Lemme}%[subsection]
+%\newtheorem{props}{Propriétés}[defn]
+
+\newenvironment{system}%
+{\left\lbrace\begin{align}}%
+{\end{align}\right.}
+
+\newenvironment{AQT}{{\fontfamily{qbk}\selectfont AQT}}
+
+\usepackage{LobsterTwo}
+\titleformat{\section}{\newpage\LobsterTwo \huge\bfseries}{\thesection.}{1em}{}
+\titleformat{\subsection}{\vspace{2em}\LobsterTwo \Large\bfseries}{\thesubsection.}{1em}{}
+\titleformat{\subsubsection}{\vspace{1em}\LobsterTwo \large\bfseries}{\thesubsubsection.}{1em}{}
+
+\newenvironment{lititle}%
+{\vspace{7mm}\LobsterTwo \large}%
+{\\}
+
+\renewenvironment{proof}{$\square$ \footnotesize\textit{Démonstration.}}{\begin{flushright}$\blacksquare$\end{flushright}}
+
+\title{TD du 20 février}
+\author{William Hergès\thanks{Sorbonne Université - Faculté des Sciences, Faculté des Lettres}}
+
+\begin{document}
+	\maketitle
+	\newpage
+	\section{Fin de la feuille du 13}
+	\subsection*{Exercice 7}
+	$\lambda v_1+v_2 = (\lambda x_1+x_2,\lambda y_1 + y_2)$\\
+	$\lambda y_1+y_2=\lambda x_1+x_2+\lambda+1 \neq \lambda x_1+x_2+1$\\
+	donc $A$ n'est pas un sev.
+
+	$\lambda v_1+v_2 = (\lambda x_1 + x_2, \lambda x_1 + x_2)$\\
+	donc $A_1$ est un sev.
+
+	$\lambda v_1+v_2 = (\lambda x_1+x_2,\lambda y_1 + y_2)$\\
+	$\lambda y_1 = \lambda^2 x^2 \neq x^2$\\
+	donc $A_2$ n'est pas un sev.
+	\section{Feuille du 20}
+	\subsection*{Exercice 2}
+	$g\circ f (x,y,z) = (z-2y-x, 2x-y+z)$
+
+	$BA = \begin{pmatrix} -1&-2&1\\2&-1&1 \end{pmatrix} $
+	\subsection*{Exercice 3}
+	Par le théorème du rang, on a que la dimension du Ker est 1. La représentation matricielle de $f$ est :
+	$$ \begin{pmatrix} 1&-1&0\\0&1&-1 \end{pmatrix}  $$
+	$$x-y = 0 \quad\land\quad y-z = 0 \iff x = y = z$$
+	Donc $\mathrm{Ker}(f) = \mathrm{Vect}((1,1,1)) = \{(a,a,a)|a\in\mathbb{R}\}$
+
+	Comme $f$ est de rang $2$, alors $f$ est surjective (tous les éléments sont atteints), mais pas injective.
+\end{document}