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\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}