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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}
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+\usepackage{textcomp}
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+\usepackage{amsthm}
+\usepackage[svgnames]{xcolor}
+\usepackage{thmtools}
+\usepackage{lipsum}
+\usepackage{framed}
+\usepackage{parskip}
+\usepackage{titlesec}
+% \usepackage{inter}
+
+\renewcommand{\familydefault}{\sfdefault}
+% \renewcommand{\familydefault}{\sffamily}
+
+% figure support
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+\usepackage{xifthen}
+\pdfminorversion=7
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+
+\usepackage{LobsterTwo}
+\titleformat{\section}{\newpage\LobsterTwo \huge\bfseries}{\thesection.}{1em}{}
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+\renewenvironment{proof}{$\square$ \footnotesize\textit{Démonstration.}}{\begin{flushright}$\blacksquare$\end{flushright}}
+
+\title{TD du 6 mars}
+\author{William Hergès\thanks{Sorbonne Université - Faculté des Sciences, Faculté des Lettres}}
+
+\begin{document}
+	\maketitle
+	\section{Feuille du 20 février}
+	\subsection*{Exercice 4}
+	Soient $v_1,v_2$ dans $P$ et $\lambda$ dans $\mathbb{K}$. On a :
+	$$ v_1+\lambda v_2 = x+2y+4z+\lambda x'+\lambda2y'+\lambda4z' = 0 $$
+	car $0+0$ vaut $0$. De plus, $0_P\in P$, donc $P$ est un espace vectoriel.
+
+	On peut écrire $x+2y+4z=0$ comme $x+2y=-4z$. Ainsi, $z$ est lié, donc $P=\mathrm{Vect}\left( \small\begin{pmatrix} 1/4\\0\\0 \end{pmatrix} ,\small\begin{pmatrix} 0\\1/2\\0 \end{pmatrix} \right)$, ce qui est aussi la base de $P$.
+
+	Soient $v_1,v_2$ dans $D$ et $\lambda$ dans $\mathbb{K}$. On a :
+	$$ v_1+\lambda v_2 = \begin{pmatrix} x\\-y\\z \end{pmatrix} + \lambda\begin{pmatrix} x'\\-y'\\z' \end{pmatrix} \land\ldots $$
+	ce qui vaut bien $0$ car $0+0=0$. De plus, $O_D\in D$, donc $D$ est un ev.
+
+	On a :
+	$$ \left\{\begin{matrix} x&-y&+z&=&0\\ x&+y&-z&=&0 \end{matrix} \right.\iff \left\{\begin{matrix} x&-y&=&-z\\ x&+y&=&z \end{matrix} \right. $$
+	En additionnant $L_1$ et $L_2$, on obtient que $x=0$ et donc que $y=z$. Ainsi, $(0,1,1)$ est une base de $D$ et $\mathrm{dim}(D) = 1$.
+
+	\textit{Mutadis mutandis}, $H$ est un ev.
+
+	On a :
+	$$ x+y+z+t=0\iff x+y+z=-t $$
+	Donc $t$ est fixé. Une base de $H$ est $\{(1,0,0),(0,1,0),(0,0,1)\}$ et donc sa dimension vaut 3.
+	\subsection*{Exercice 6}
+	Trouver une équation de l'image signifie, trouver une équation dépendant uniquement des variables de l'image. Par exemple, quand on a un ensemble d'équation avec $x_n'$ dépendant de $(x_n)$ décrivant l'image, une équation de l'image dépendra uniquement de $(x_n')$.
+	\section{Feuille du 6 mars}
+	On a : $(1-\lambda)(3-\lambda)-8 = 0 \iff \lambda^2-4\lambda-5 = 0$.\\
+	$-1$ est une valeur propre (solution évidente).\\
+	$\Delta = 36 = 6^2$, donc $5$ est aussi une solution.
+	
+	Le sous-espace propre lié à $\lambda = 5$ est l'ensemble des $u$ tels que $(A-5I)u = 0$, i.e.
+	$$ \begin{pmatrix} -4&4\\-2&2 \end{pmatrix} u = 0 $$
+	donc, $u\in\{(a,a),a\in\mathbb{R}\}$.
+	
+	Le sous-espace propre lié à $\lambda = -1$ est l'ensemble des $u$ tels que $(A+1I)u = 0$, i.e.
+	$$ \begin{pmatrix} 2&4\\2&4 \end{pmatrix} u = 0 $$
+	donc, $u\in\{(-2a,a),a\in\mathbb{R}\}$.
+
+	Ainsi,
+	$$ \underset{D}{\underbrace{\begin{pmatrix} 5&0\\0&-1 \end{pmatrix}}} = P^{-1}A\underset{P}{\underbrace{\begin{pmatrix} 1&-2\\1&1 \end{pmatrix}}} $$
+	On a aussi que $P^{-1} = \frac{1}{3}\tiny\begin{pmatrix} 1&2\\-1&1 \end{pmatrix} $
+\end{document}