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\title{TD du 13 février}
\author{William Hergès\thanks{Sorbonne Université - Faculté des Sciences, Faculté des Lettres}}

\begin{document}
	\maketitle
	\newpage
	\section*{Exercice 1}
	Trivial
	\section*{Exercice 2}
	Trivial
	\section*{Exercice 3}
	Trivial
	\section*{Exercice 4}
	On a :
	\begin{align*}
		2 &= a-b+c \\
		1 &= a+b-c \\
		3 &= -a+b+c
	\end{align*}
	où $(a,b,c)$ sont les coordonnées du vecteur dans la base.

	D'où :
	$$\begin{pmatrix} 1&-1&1&|&2\\1&1&-1&|&1\\-1&1&1&|&3 \end{pmatrix}\iff\begin{pmatrix} 1&-1&1&|&2\\0&2&-2&|&-1\\0&0&2&|&5\end{pmatrix}$$
	Ainsi,
	\begin{align*}
		c &= 5/2 \\
		b &= 2 \\
		a &= 3/2
	\end{align*}
	\section*{Exercice 5}
	Si $x=0$, alors $v = u + w$, donc ce n'est pas libre.\\
	Si $x\neq 0$, alors $(uvx)$ est échelonné, donc elle est libre.

	Il s'agit  d'une base si et seulement si $x\neq 0$.
	\section*{Exercice 6}
	\begin{lititle}
		Matrice de changement de base.
	\end{lititle}
	On a $v$ un vecteur dans la base canonique. $\lambda$ est dans la base $A$ et $\mu$ dans la base $B$.

	Ainsi, on a :
	\begin{align*}
		\mathrm{I}_n v &= A\lambda \\
		\mathrm{I}_n v &= B\mu
	\end{align*}
	d'où :
	$$ A\lambda = B\mu \iff \mu = B^{-1}A\lambda $$
	Ainsi, $B^{-1}A$ est la matrice de passage de $A$ à $B$.
	\section*{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.
\end{document}
