%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% M99DT2BC.tex
% Version 1.02 du 29 mai 1999.
% Correction du probl\`eme IIB de la banque PT 1999
%
% R\'edacteur : Claude Mendes <csmendes@club-internet.fr>
% TeXnicien : Daniel Duparc <dduparc@club-internet.fr>
% Lyc\'ee Voltaire
% 101 av. de la R\'epublique
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% Merci \`a Yves Saladin <yves.saladin@wanadoo.fr>
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\title{\textbf{Banque PT 1999 : Math\'ematiques II-B }}

\begin{document}
\author{}
\date{}
\maketitle
%\pagestyle{empty}
%\pagestyle{headings}
%\pagestyle{myheadings}
%\markright{{\bf DUPARC 836}\ \ \ \ \ Correction de Claude {\sc Mendes}}
%\markright{{\bf Banque PT 1999 : Math\'ematiques II-B}}
%\thispagestyle{empty}

\newif\ifblackboard
\blackboardtrue

\ifblackboard
%  Macros de definition des ensembles usuels N,Z,Q,R,C
%  en utilisant les fontes de l'AMS
%  Pour agrandir les caracteres redefinir dans un groupe la 
%  fonte \bb . Par exemple {\font\bb=msbm10 scaled \magstep5 ...}
\newfont{\bb}{msbm10}
\def\R{\hbox{\bb R}}
\def\N{\hbox{\bb N}}
\def\C{\hbox{\bb C}}
\def\Z{\hbox{\bb Z}}
\def\Q{\hbox{\bb Q}}
\def\K{\hbox{\bb K}}

\else
%  Macros de definition des ensembles usuels N,Z,Q,R,C
%  d'apres Urban,Cornelis : Premiers pas en LaTeX page 80
%  Pour agrandir les caracteres redefinir dans un groupe la 
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\def\R{\hbox{\bb I\hskip -0.15em R}}
\def\N{\hbox{\bb I\hskip -0.15em N}}
\def\C{\hbox{\bb \ l\hskip -0.53em C\/}}
\def\Z{\hbox{\bb Z\hskip -0.3em Z}}
\def\Q{\hbox{\bb \ l\hskip -0.53em Q}}
\def\K{\hbox{\bb I\hskip -0.15em K}}

\fi

\font\cursiveonze=cmsy10 scaled 1100
\font\cursivedix=cmsy10 
  

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\centerline{\LARGE \textbf{ Premi\`ere partie}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question I.1
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{La droite vectorielle de base $\vec{e_3}$ et le plan de 
base $(\vec{e_1} , \vec{e_2} )$ sont stables par $f$}
$$E_1\times  {}^tE_2 = 
\pmatrix{1\cr
0\cr
0\cr
}\times
\pmatrix{0 & 1 & 0 \cr
}
=
\pmatrix{
0 & 1 & 0 \cr
0 & 0 & 0 \cr
0 & 0 & 0 \cr
}
$$
d'o\`u
$$F =
\pmatrix{
1 & 2\alpha & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
}
\ \ \ \mbox{et}\ \ \ 
{}^tF =
\pmatrix{
1 & 0 & 0 \cr
2\alpha & 1 & 0 \cr
0 & 0 & 1 \cr
}
= G
$$
$f(\vec{e_3}) =\vec{e_3}$ donc la droite vectorielle de base $\vec{e_3}$
est stable par $f$.

$f(\vec{e_1}) =\vec{e_1}$ et $f(\vec{e_2}) =2\alpha \vec{e_1} 
+  \vec{e_2}$, vecteurs qui constituent une famille g\'en\'eratrice
et libre (base !) du plan vectoriel image du plan de base
$(\vec{e_1} , \vec{e_2} )$, sont des vecteurs de ce plan.
Par suite, le plan vectoriel de base $(\vec{e_1} , \vec{e_2} )$
est stable par $f$.  
 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question I.2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Les valeurs propres $\{ \mu_i\}_{1\le i \le 3}$ 
de $g \circ f$ sont strictement positives 
}
$$G\times F =
\pmatrix{
1 & 0 & 0 \cr
2\alpha & 1 & 0 \cr
0 & 0 & 1 \cr
}
\times
\pmatrix{
1 & 2\alpha & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} =
\pmatrix{
1       & 2\alpha       & 0 \cr
2\alpha & 4\alpha ^2 +1 & 0 \cr
0       & 0             & 1 \cr
}
$$
$$\det (G\times F-\lambda I_3 )=
\left | 
\matrix{
1 -\lambda & 2\alpha                & 0          \cr
2\alpha    & 4\alpha ^2 +1 -\lambda & 0          \cr
0          & 0                      & 1 -\lambda \cr
}\right |
$$
$$\det (G\times F-\lambda I_3 )=
(1-\lambda ) [(1-\lambda )(4\alpha^2 +1-\lambda)-4\alpha^2]
=(1-\lambda )[\lambda^2 -2 (2\alpha^2 +1 )\lambda +1 ]
$$
Le discriminant r\'eduit du polyn\^ome entre crochets est :

\noindent $\Delta '= (2\alpha ^2 +1 )^2 -1 = 4\alpha^2 (\alpha^2 +1)$.

\noindent Comme $ \sqrt{\alpha^2} = | \alpha | $, l'ensemble 
des valeurs propres (r\'eelles) de $g\circ f$ 
est donc 

\noindent $ \{ 2\alpha^2 +1+2| \alpha |\sqrt{\alpha^2 +1} ,
 2\alpha^2 +1-2 | \alpha | \sqrt{\alpha^2 +1}, 1 \}$, 
ou encore :
$ \{ 2\alpha^2 +1+2 \alpha \sqrt{\alpha^2 +1} ,
 2\alpha^2 +1-2  \alpha  \sqrt{\alpha^2 +1 } , 1 \}$.

\noindent Le produit des 2 valeurs propres qui ne sont
pas \'egales \`a $1$ vaut $  1$ comme  produit des z\'eros
d'un polyn\^ome du second degr\'e, et sont donc de m\^eme signe.
Or, au moins une des deux est strictement positive (\'evident sur
les formules ci-dessus). Donc les 3 valeurs propres sont
strictement positives et on peut donc les num\'eroter de sorte
que $\mu_1 > \mu_2$. De plus $\mu_1 \mu_2 = 1 = \mu_3$.  

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question I.3
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Vecteurs $\vec{u_1} = a \vec{e_1} + \vec{e_2}$ et
$\vec{u_2} = b \vec{e_1} + \vec{e_2}$ }
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question I.3 (a)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{a) Relations sur $a$ et $b$ pour que 
$\vec{u_1} \cdot \vec{u_2} = 0$
et $f( \vec{u_1} ) \cdot f ( \vec{u_2} ) = 0$
}
$\vec{u_1} = a \vec{e_1} + \vec{e_2}$ et 
$\vec{u_2} = b \vec{e_1} + \vec{e_2}$,
donc $\vec{u_1} \cdot \vec{u_2} = 0$ si et seulement si $ab+1 = 0$.

\noindent $f ( \vec{u_1} )= a f ( \vec{e_1} )+f (  \vec{e_2})=
(a+2\alpha ) \vec{e_1} + \vec{e_2}$

\noindent $f ( \vec{u_2} )= b f ( \vec{e_1} )+f (  \vec{e_2})=
(b+2\alpha )\vec{e_1} + \vec{e_2}$

\noindent $f( \vec{u_1} ) \cdot f ( \vec{u_2} ) = 0$ si et seulement si 
$ab + 2\alpha (a+b) +4\alpha^2 +1 = 0$

\noindent Donc $\vec{u_1} \cdot \vec{u_2} = 0$ 
et $f( \vec{u_1} ) \cdot f ( \vec{u_2} ) = 0$
si et seulement si :
$$\left \{ \matrix{
ab+1 & = & 0\cr
ab+2\alpha (a+b) +4\alpha^2 +1 & =&0\cr
}
\right .
$$
ce qui \'equivaut \`a :
$$\left \{ \matrix{
ab & = & -1\cr
a+b  & =&-2\alpha \cr
}
\right .
$$
\noindent Donc $\vec{u_1} \cdot \vec{u_2} = 0$ 
et $f( \vec{u_1} ) \cdot f ( \vec{u_2} ) = 0$
si et seulement si $a$ et $b$ sont racines de l'\'equation 
$x^2 +2\alpha x -1 =0$, soit :
$$ \{ a , b \} = \{ -\alpha +\sqrt{\alpha^2 +1},  -\alpha -\sqrt{\alpha^2 +1}\}
$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question I.3 (b)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{b) D\'etermination de $a$ et $b$ pour que $B$ et
$\bigl( f( \vec{u_i})\bigr)_{1\le i\le 3}$ soient des bases orthogonales
}
Si $\vec{u_3} = \vec{e_3}$, $\vec{u_1} = a \vec{e_1} + \vec{e_2}$ et
$\vec{u_2} = b \vec{e_1} + \vec{e_2}$, alors $\vec{u_3}$ est orthogonal
\`a $\vec{u_1}$ et $\vec{u_2}$. Pour que 
$\bigl( \vec{u_1} , \vec{u_2} , \vec{u_3} \bigr)$ soit une base 
orthogonale, il faut et il suffit d'avoir :
$$\vec{u_1} \cdot \vec{u_2} =0$$

\noindent De m\^eme $\bigl( f( \vec{u_i})\bigr)_{1\le i\le 3}$
est une base orthogonale si et seulement si : 
$$f( \vec{u_1}) \cdot f(\vec{u_2}) =0$$

\noindent Par suite les deux propri\'et\'es sont acquises simultan\'ement
si et seulement si $a$ et $b$ prennent les valeurs trouv\'ees dans
la question pr\'ec\'edente.

\noindent Nous avons alors le d\'eterminant des vecteurs de la base $B$ 
dans la base canonique $B_c$ :
$$\det{}_{B_c} ( B) =
\left| \matrix{
a & b & 0 \cr
1 & 1 & 0 \cr
0 & 0 & 1 \cr
}\right|
= a-b
$$
\noindent Donc $B$ sera de sens direct si et seulement si $a > b$,
donc si et seulement si $ a = -\alpha +\sqrt{\alpha^2 +1}$
et $b = -\alpha -\sqrt{\alpha^2 +1}$.
\noindent Dans ce cas, 
$$\left\{
\matrix{
\vec{u_1} & = & (-\alpha +\sqrt{\alpha^2 +1}) \vec{e_1} + \vec{e_2} \cr  
\vec{u_2} & = & (-\alpha -\sqrt{\alpha^2 +1}) \vec{e_1} + \vec{e_2} \cr
\vec{u_3} & = &  \vec{e_3} \cr
}
\right.
$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question I.4
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{$B$ est une base de vecteurs propres de $g \circ f$
}
$$\matrix{
 g \circ f ( \vec{u_1} ) &=&  
(-\alpha +\sqrt{\alpha^2 +1}) (\vec{e_1} + 2\vec{e_2})
+[2\alpha \vec{e_1}+ (4\alpha^2 +1) \vec{e_2}] \cr
&=&(\alpha +\sqrt{\alpha^2 +1}) \vec{e_1} 
+[2\alpha^2 +1+ 2\alpha\sqrt{\alpha^2 +1}] \vec{e_2} \cr
}
$$
Or $(2\alpha^2 +1+ 2\alpha\sqrt{\alpha^2 +1}) (-\alpha +\sqrt{\alpha^2 +1}) 
= \alpha +\sqrt{\alpha^2 +1}$, donc $\vec{u_1}$ est  un vecteur propre
de  $g \circ f$ de valeur propre 
$2\alpha^2 +1+ 2\alpha\sqrt{\alpha^2 +1}$.

\noindent De m\^eme :
$$\matrix{
 g \circ f ( \vec{u_2} ) &=&  
(-\alpha -\sqrt{\alpha^2 +1}) (\vec{e_1} + 2\vec{e_2})
+[2\alpha \vec{e_1}+ (4\alpha^2 +1) \vec{e_2}] \cr
&=&(\alpha -\sqrt{\alpha^2 +1}) \vec{e_1} 
+[2\alpha^2 +1- 2\alpha\sqrt{\alpha^2 +1}] \vec{e_2} \cr
}
$$
Or $(2\alpha^2 +1- 2\alpha\sqrt{\alpha^2 +1}) (-\alpha -\sqrt{\alpha^2 +1}) 
= \alpha -\sqrt{\alpha^2 +1}$, donc $\vec{u_2}$ est  un vecteur propre
de  $g \circ f$ de valeur propre 
$2\alpha^2 +1- 2\alpha\sqrt{\alpha^2 +1}$.

\noindent \'Evidemment, $\vec{u_3}$ est  un vecteur propre
de  $g \circ f$ de valeur propre $1$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question I.5
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Valeur de $\theta$ telle que $s = r_\theta^{-1}\circ f$
soit sym\'etrique, de valeurs propres $>0$
}
Avec les conventions de l'\'enonc\'e :
$$R_\theta =
\pmatrix{
\cos \theta & -\sin \theta & 0 \cr
\sin \theta &  \cos \theta & 0 \cr
0           &  0           & 1 \cr
}
$$
$$S=
\pmatrix{
\cos \theta & \sin \theta & 0 \cr
-\sin \theta &  \cos \theta & 0 \cr
0           &  0           & 1 \cr
}
\pmatrix{
1 & 2\alpha & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
}
=
\pmatrix{
\cos\theta  & 2\alpha\cos \theta+\sin\theta  & 0 \cr
-\sin\theta & -2\alpha\sin \theta+\cos\theta & 0 \cr
0           & 0                              & 1 \cr
}
$$
$S$ est sym\'etrique si et seulement si 
$2\alpha \cos\theta +\sin\theta = -\sin\theta$ c'est-\`a-dire si et seulement si
$ \sin\theta = -\alpha \cos\theta $. Alors :
$$S=
\pmatrix{
\cos\theta        & \alpha\cos \theta              & 0 \cr
\alpha\cos \theta & (2\alpha^2+1)\cos\theta        & 0 \cr
0                 & 0                              & 1 \cr
}
$$
L'\'equation caract\'eristique associ\'ee \`a $s$ est :
$$\left|
\matrix{
\cos\theta -\lambda & \alpha\cos \theta                & 0 \cr
\alpha\cos \theta   & (2\alpha^2+1)\cos\theta -\lambda & 0 \cr
0                 & 0                                  & 1-\lambda \cr
}\right| =0
$$
soit :
$$(1-\lambda ) 
\Bigl[ \Bigl( (2\alpha^2 +1)\cos\theta -\lambda \Bigr)
(\cos\theta -\lambda )-\alpha^2 \cos^2 \theta\Bigr] =0
$$
$$(1-\lambda )
\Bigl( \lambda^2 -2 (\alpha^2 +1) \cos\theta\  \lambda
+ (\alpha^2 +1 ) \cos\theta  
\Bigr)=0
$$
R\'esolvons l'\'equation 
$\lambda^2 -2 (\alpha^2 +1) \cos\theta\  \lambda
+ (\alpha^2 +1 ) \cos\theta =0$.

\noindent $\Delta' = (\alpha^2 +1)^2 \cos^2 \theta -
(\alpha^2 +1 )\cos^2 \theta = \alpha^2 (\alpha^2 +1)\cos^2 \theta >0$.

\noindent Les valeurs propres autres que $1$ ont le signe de la somme
des racines soit de $\cos \theta$. Pour qu'elles soient positives,
il faut choisir $\theta$ dans 
$\displaystyle \Bigl] -{\pi \over 2} , {\pi \over 2}\Bigr[$,
soit :
$$ \theta = - \hbox{arctan}(\alpha)
$$

\vskip 5mm

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\centerline{\LARGE \textbf{ Deuxi\`eme partie}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question II.1
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\setcounter{section}{0}
\section{\'Etude de suites
}
$\vec{Op}_0 = f( \vec{Om}_0)$, $\vec{Om}_{n+1} = g( \vec{Op}_n)$  et
$\vec{Op}_{n+1} = f( \vec{Om}_{n+1})$, avec :
$$F=
\pmatrix{
1 & 2 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
}
\hbox{,\ \ \ \ \ \ \ \ }
G=
\pmatrix{
1 & 0 & 0 \cr
2 & 1 & 0 \cr
0 & 0 & 1 \cr
}
\hbox{\ \ \ \ et\ \ \ \ \ \ }
G F=
\pmatrix{
1 & 2 & 0 \cr
2 & 5 & 0 \cr
0 & 0 & 1 \cr
}.
$$
$\mu_1 =3+2\sqrt{2}$, $\mu_2 =3-2\sqrt{2}$  et $\mu_3 = 1$.

\noindent $\vec{u_1} = (\sqrt{2} - 1)\vec{e_1} +\vec{e_2}$ est 
vecteur propre de $g \circ f$ de valeur propre $\mu_1$. 
\noindent $\vec{u_2} = (-\sqrt{2} - 1)\vec{e_1} +\vec{e_2}$ est 
vecteur propre de $g \circ f$ de valeur propre $\mu_2$. 
\noindent $\vec{u_3} = \vec{e_3} $
est vecteur propre de $g \circ f$ de valeur propre $\mu_3 = 1$. 


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question II.1 (a)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{a) \'Etude des suites $(z_n)_{n\in \N}$ et
$(\tilde z_n)_{n\in \N}$
}
$\tilde z_0 =  z_0$, les deux suites $(z_n)_{n\in \N}$ et
$(\tilde z_n)_{n\in \N}$ sont constantes, tous leurs termes
sont \'egaux \`a $z_0$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question II.1 (b)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{b) \'Etude conjointe des suites $(m_n)_{n\in \N}$ et
$( p_n)_{n\in \N}$
}
$\vec{Op}_{n} = f( \vec{Om}_{n})$, donc pour tout entier $n \ge 0$,
$\tilde y_n = y_n$. 

\noindent $\vec{Om}_{n+1} = g( \vec{Op}_{n})$, donc pour 
tout entier $n \ge 0$,
$ x_{n+1} = \tilde x_n$.

\noindent Par suite $\Bigl( x_n \Bigr)_{n\in \N}$ converge
si et seulement si $\Bigl( \tilde x_n \Bigr)_{n\in \N}$
converge et $\Bigl( y_n \Bigr)_{n\in \N}$ converge si et seulement
si $\Bigl( \tilde y_n \Bigr)_{n\in \N}$ converge.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question II.2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{\'Etude de la suite convergente $\Bigl( m_n \Bigr)_{n\in \N}$
}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question II.2 (a)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{a) $\vec{Om}$ est un vecteur invariant par $g\circ f$
}
Pour tout entier naturel $n$, $g\circ f ( \vec{Om}_n ) = \vec{Om}_{n+1}$,
d'o\`u :
$$\left\{ \matrix{
x_{n+1} & = & x_n    & + & 2 y_n \cr
y_{n+1} & = & 2 x_n  & + & 5 y_n \cr
}\right.
$$
Par suite si $\Bigl( m_n \Bigr)_{n\in \N}$ converge vers $m$ 
de coordonn\'ees $(x,y,z)$, on aura : 
$$\left\{ \matrix{
x & = & x    & + & 2 y \cr
y & = & 2 x  & + & 5 y \cr
z & = & z    & & (= z_0) \cr
}\right.
$$
Donc $\vec{Om} $ est invariant par $g \circ f$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question II.2 (b)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{b) D\'efinition g\'eom\'etrique de $m$
}
On en d\'eduit que $\vec{Om}$ est un vecteur propre de $g \circ f$
de valeur propre 1. En r\'esolvant le syst\`eme pr\'ec\'edent, on trouve
que $ m$ est n\'ecessairement le point de coordonn\'ees 
$(0,0,z_0)$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question II.3
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{\'Etude de la matrice $M'_n$ 
}
$$M'_n
=
\pmatrix{
x'_n\cr
y'_n\cr
z'_n\cr
}
=
Q^{-1}
\pmatrix{
x_n\cr
y_n\cr
z_n\cr
}
$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question II.3 (a)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{a) Identification de $M'_n$
}
$M'_n$ repr\'esente la matrice de $\vec{Om}_n$ dans la base $B$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question II.3 (b)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{b) Relation de r\'ecurrence liant $M'_{n+1}$ et $M'_n$
}
Comme $\vec{Om}_{n+1} = g \circ f ( \vec{Om}_n)$,
$$M'_{n+1} =
\hbox{\cursiveonze M}_B (g \circ f) M'_n
$$
c'est-\`a-dire :
$$M'_{n+1} =
\pmatrix{
\mu_1 &   0   &   0   \cr
0     & \mu_2 &   0   \cr
0     &   0   & \mu_3 \cr
}
M'_n
$$
$$M'_{n+1} =
\pmatrix{
3+2\sqrt{2} &   0   &   0   \cr
0     & 3-2\sqrt{2} &   0   \cr
0     &   0   & 1 \cr
}
M'_n
$$
Par suite $x'_{n+1} = (3+2\sqrt{2} ) x'_n$, 
$y'_{n+1} = (3-2\sqrt{2} ) y'_n$, et 
$z'_{n+1} =  z'_n$, et donc 
$$ \left\{
\matrix{
x'_{n} = (3+2\sqrt{2} )^n x'_0 \cr
y'_{n} = (3-2\sqrt{2} )^n y'_0 \cr 
z'_{n} =  z'_0 \cr
}
\right.
$$ 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question II.3 (c)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{c) Condition n\'ecessaire et suffisante pour que la suite
$\Bigl( M'_n \Bigr)_{n\in \N}$ soit convergente 
}
Comme $3+2\sqrt{2} > 1$, pour que $\Bigl( M'_n \Bigr)_{n\in \N}$
converge, il faut et il suffit que $x'_0 = 0$. 
$\Bigl( y'_n \Bigr)_{n\in \N} $
tend vers $0$ et $\Bigl( z'_n \Bigr)_{n\in \N}=
\Bigl( z'_0 \Bigr)_{n\in \N}$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question II.4
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{\'Etude de la suite $\Bigl( m_n \Bigr)_{n\in \N}$
}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question II.4 (a)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{a) Condition n\'ecessaire et suffisante
pour que la suite  $\Bigl( m_n \Bigr)_{n\in \N}$  converge
} 
La suite  $\Bigl( m_n \Bigr)_{n\in \N}$ converge si et seulement
si $x'_0 = 0$ donc si et seulement si $m_0$ appartient au
plan $\Pi$ engendr\'e par $\vec{u_2}$ et $\vec{u_3}$.

Son \'equation est donc :
 
$$
\left| \matrix{
x & -\sqrt{2}-1   & 0   \cr
y & 1             & 0   \cr
z          &  0              & 1   \cr
}
\right|
= 0$$
c'est-\`a-dire :
$$ x + (\sqrt{2}+1) y =0
$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question II.4 (b)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{b) Les points $m_n$ et $p_n$ appartiennent \`a
des droites 
}
Si $m_0$ appartient \`a $\Pi$ sans appartenir \`a la droite 
vectorielle de base $(\vec{e}_3 )$,
alors $ x_0 + (\sqrt{2}+1) y_0 =0$ sans que $(x_0 , y_0) = (0 , 0)$.

Alors pour tout entier naturel $n$, $x'_n = 0$ donc  
$ x_n + (\sqrt{2}+1) y_n =0$ et $z_n = z_0$. Donc les
points $m_n$ appartiennent  \`a la droite $(\Delta )$ dont les
\'equations sont :
$$(\Delta )
\left\{ \matrix{
x + (\sqrt{2}+1) y &=&0 \cr
z &=& z_0 \cr   
}\right.
$$ 
Quant aux points $p_n$, ils appartiennent \`a la droite 
$(\Delta ') = f(\Delta  )$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question II.4 (c)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{c) Construction g\'eom\'etrique des points $m_n$ et $p_n$
}

% Dessin :

\vskip -9cm
\hskip -2cm
\epsfxsize=16.0cm  \epsffile{M99DT2Fc.eps}

\vskip -0.5cm
Si $m_0$ est un point de $\Pi$ tel que $z_0 = 0$,
$\vec{Om}_0  = y'_0\  \vec{u}_2$ ($y'_0\neq 0$) et pour tout entier
naturel $n$, $\vec{Om}_n  = (3-2\sqrt{2})^n y_0\  \vec{u}_2
= (3-2\sqrt{2})^n \vec{Om}_0$ 

Donc $m_n $ se d\'eduit de $m_0$ par une homoth\'etie de centre $O$
et de rapport $(3-2\sqrt{2})^n$. Bien entendu,
$m_{n+1} $ se d\'eduit de $m_n$ par une homoth\'etie de centre $O$
et de rapport $(3-2\sqrt{2})$.

Pour tout entier naturel $n$, les points $m_n$ 
(not\'es {\og}m(n){\fg} sur le dessin)
appartiennent \`a la droite $\Delta$ (not\'ee {\og}(Delta){\fg} sur le dessin)
du plan $z=0$, dirig\'ee par le vecteur 
$\vec{t} = (\sqrt{2} + 1) \vec{e}_1 -\vec{e}_2$,
passant par $O$.

La droite $\Delta '$ (not\'ee {\og}(f(Delta)){\fg} sur le dessin),
ensemble des points $p_n$ (not\'es {\og}p(n){\fg} sur le dessin), 
est une droite du plan d'\'equation $z=0$
dirig\'ee par le vecteur
$f( \vec{t}) = (\sqrt{2} + 1) \vec{e}_1 -(2\vec{e}_1 +\vec{e}_2)
=(\sqrt{2} - 1) \vec{e}_1 -\vec{e}_2$,
passant pas $O$.

Pour chaque valeur de $n$, $\tilde y_n = y_n$ car la 
deuxi\`eme ligne de $F$ est $[ 0 , 1 , 0 ]$. $p_n$
est le point de $\Delta '$ ayant la m\^eme ordonn\'ee
que $m_n$.

D'autre part, les abscisses de $p_n$ et $ m_{n+1}$
sont \'egales, donc la droite $p_n  m_{n+1}$ est parall\`ele
\`a l'axe des $y$. On peut donc obtenir $ m_{n+1}$
\`a partir de $p_n$ en cherchant l'intersection de 
$\Delta $  avec la parall\`ele \`a $Oy$
passant par $ p_{n}$ (voir le dessin).

%\vskip 8mm
\newpage

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\centerline{\LARGE \textbf{ Troisi\`eme partie}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question III.1
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\setcounter{section}{0}
\section{\'Etude des endomorphismes d\'efinis positifs
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question III.1 (a)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{a) Si un endomorphisme est d\'efini positif 
alors ses valeurs propres sont strictement positives
}
$\lambda$ \'etant une valeur propre de $\phi $ et $\vec{x}$ \'etant 
un vecteur propre de $\phi $ de valeur propre $\lambda $,
$$ \vec{x} \cdot \phi (\vec{x} ) =
\vec{x} \cdot  (\lambda \vec{x} ) =
  \lambda (\vec{x}\cdot \vec{x} ) >0
$$ 
Comme $\vec{x}\cdot \vec{x}  >0$ (car $\vec{x}$ \'etant un vecteur
propre est diff\'erent de $\vec{0}$,  on a bien $\lambda >0$).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question III.1 (b)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{b) Si les valeurs propres d'un endomorphisme $\phi$
sont strictement positives alors il est d\'efini positif
}
$\phi$ \'etant un endomorphisme sym\'etrique, il poss\`ede une 
base propre orthonormale $(\vec{v}_1 , \vec{v}_2 , \vec{v}_3)$
associ\'ee (respectivement) aux valeurs propres r\'eelles 
$ ( \lambda_1 , \lambda_2 , \lambda_3 )$ suppos\'ees ici
strictement positives.

$\vec{x}$ \'etant un vecteur non nul de 
{\font\bb=msbm10 scaled 1100 $\R^3$}, il existe trois r\'eels
$x_1$, $x_2$ et  $x_3$ non tous nuls tels que
$ \vec{x}= x_1 \vec{v}_1 + x_2 \vec{v}_2 + x_3 \vec{v}_3$.

Alors 
$$\phi (\vec{x} )= 
\phi ( x_1 \vec{v}_1 + x_2 \vec{v}_2 + x_3 \vec{v}_3 )
=  \lambda_1 x_1 \vec{v}_1 + \lambda_2 x_2 \vec{v}_2 + 
\lambda_3 x_3 \vec{v}_3  
$$ 
et
$$\vec{x} \cdot \phi ( \vec{x} ) =
\lambda_1 x_1^2 + \lambda_2 x_2^2  + \lambda_3 x_3^2  >0
$$  

Donc $\phi$ est d\'efini positif.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question III.1 (c)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{c) $g\circ f$ est sym\'etrique et d\'efini positif
}
$\hbox{\cursiveonze M}_{B_c} (g\circ f)=
G\times F = {}^tF \times F$.

Alors $
{}^t(G \times  F)= {}^tF \times {}^tG =
G\times F = \hbox{\cursiveonze M}_{B_c} (g\circ f) $

Donc $g\circ f$est un endomorphisme sym\'etrique.

Soit $\vec{x}$ un \'el\'ement quelconque de 
{\font\bb=msbm10 scaled 1100 $\R^3$} :
$\vec{x} = x_1 \vec{e_1 } + x_2 \vec{e_2 } + x_3 \vec{e_3 } $.

Alors $(g\circ f)(\vec{x}) $ a pour matrice 
$$({}^tFF)
\pmatrix{
x_1\cr
x_2\cr
x_3\cr
}
$$ 
et donc
$$\matrix{  
\vec{x} \cdot (g\circ f)(\vec{x})& =&
\pmatrix{
x_1 & x_2 &x_3\cr 
}
\cdot
{}^tFF
\pmatrix{
x_1\cr
x_2\cr
x_3\cr
}\cr
&=&
\Bigl(
\pmatrix{
x_1 & x_2 &x_3\cr 
}
{}^tF\Bigr)
\Bigl(F
\pmatrix{
x_1\cr
x_2\cr
x_3\cr
}\Bigr)
\cr
&=&
{}^t\Bigl(F
\pmatrix{
x_1\cr
x_2\cr
x_3\cr
}\Bigr)
\cdot
\Bigl(F
\pmatrix{
x_1\cr
x_2\cr
x_3\cr
}\Bigr)
\cr
&=&
f(\vec{x}) \cdot f(\vec{x} )
\cr
}
$$
(Produit scalaire de $f(\vec{x})$ avec $f(\vec{x})$)

Comme $\vec{x} \neq \vec{0 }$ et $\det (f) > 0$,
$f(\vec{x}) \neq \vec{0}$ et donc :
$$f(\vec{x}) \cdot f(\vec{x} ) >0$$
Finalement,
$$\vec{x} \cdot (g\circ f)(\vec{x}) >o
$$
et donc $g\circ f$ est un endomorphisme sym\'etrique
d\'efini positif.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question III.2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{\'Etude d'une famille orthogonale 
$\Bigl( \vec{u}_i \Bigr)_{1\le i \le 3}$
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question III.2 (a)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{a) Si $\Bigl( \vec{u}_i \Bigr)_{1\le i \le 3}$
est une base orthogonale de vecteurs propres de $g\circ f$, 
$\Bigl( f(\vec{u}_i )\Bigr)_{1\le i \le 3}$
est aussi une base orthogonale
}
On d\'esigne respectivement par $\lambda_1$, $\lambda_2$
et $\lambda_3$ les valeurs propres de $\vec{u}_1$,
$\vec{u}_2$ et $\vec{u}_3$. Les r\'eels $\lambda_1$, $\lambda_2$
et $\lambda_3$ sont strictement positifs.

Pour tout couple $(i,j)$ tel que $1\le i \le 3$, $1\le j \le 3$
et $i\neq j$, 
$$
\matrix{
f(\vec{u_i}) \cdot f(\vec{u_j}) &=& 
{}^t\Bigl( FU_i\Bigr) \times \Bigl( FU_j\Bigr)\cr
 &=& {}^tU_i \times {}^tF \times F \times U_j \cr
 &=& {}^tU_i (\lambda_j U_j)\cr
 &=& \lambda_j {}^tU_iU_j\cr
 &=& 0\cr 
}
$$ 
Donc $\Bigl( f(\vec{u}_i )\Bigr)_{1\le i \le 3}$
est une base orthogonale.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question III.2 (b)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{b) Si $\Bigl( \vec{u}_i \Bigr)_{1\le i \le 3}$
et $\Bigl( f(\vec{u}_i )\Bigr)_{1\le i \le 3}$
sont  des  bases orthogonales, alors 
$\Bigl( \vec{u}_i \Bigr)_{1\le i \le 3}$
est une base de vecteurs propres de $g\circ f$
}
Pour tout couple $(i,j)$ tel que $1\le i \le 3$, $1\le j \le 3$
et $i\neq j$,
$$\matrix{
\vec{u}_i  \cdot (g\circ f)(\vec{u}_j )&=& 
{}^tU_i \cdot (G  F)  U_j \cr
&=& ({}^tU_i {}^tF)\cdot (  F  U_j) \cr
&=&{}^t ({}FU_i)\cdot (  F  U_j) \cr
&=& f(\vec{u}_i )\cdot f(\vec{u}_j ) \cr
&=& 0\cr
}
$$
Par suite $(g\circ f)(\vec{u}_j )$ a des composantes nulles sur les
$\vec{u}_i$ pour $i\neq j$. Donc $\vec{u}_j$ est un vecteur propre
de $g\circ f$ et $\Bigl( \vec{u}_i \Bigr)_{1\le i \le 3}$
est une base de vecteurs propres de $g\circ f$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question III.3
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{\'Etude d'une base orthonormale de vecteurs propres 
$B=\Bigl( \vec{u}_i \Bigr)_{1\le i \le 3}$ de $g\circ f$, de valeurs 
propres $\mu_i$
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question III.3 (a)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{a) D\'emonstration de 
$\displaystyle {}^tFF= \sum_{i=1}^{3} \mu_i U_i{}^tU_i$
}
Soit $h$ l'endomorphisme de {{\font\bb=msbm10 scaled 1100 $\R^3$} 
de matrice dans la base canonique :
$$\hbox{\cursiveonze M}_{B_c}(h)=
\displaystyle \sum_{i=1}^{3} \mu_i U_i{}^tU_i$$
Calculons la matrice de $h(\vec{u}_k )$ 
dans la base canonique 
pour tout entier $k$, $1\le k \le 3$.
$$\hbox{\cursiveonze M}_{B_c}(h(\vec{u}_k ))=
\sum_{i=1}^{3} \mu_i (U_i{}^tU_i) U_k =
\sum_{i=1}^{3} \mu_i U_i({}^tU_i U_k)=
\mu_k U_k
$$
Donc 
$$h(\vec{u}_k )= \mu_k \vec{u}_k = (g\circ f)(\vec{u}_k)
$$
$h$ co\"incidant avec $(g\circ f)$ sur une base de 
{{\font\bb=msbm10 scaled 1100 $\R^3$} est \'egal \`a
$(g\circ f)$. Donc :
$$\displaystyle {}^tFF= \sum_{i=1}^{3} \mu_i U_i{}^tU_i$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question III.3 (b) d\'ebut
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{b) Calcul de $S^2$ avec 
$\displaystyle S=\sum_{i=1}^{3} \lambda_i U_i{}^tU_i$ et
$\lambda_i  = \sqrt{\mu_i}$ 
}
Posons $I=\{ 1 , 2 , 3\}$. 
$$ S^2 = \sum_{(i,j)\in I^2}\lambda_i \lambda_j U_i {}^tU_i U_j {}^tU_j 
$$
Pour $j=i$, ${}^tU_i U_i = 1$ et pour $j\neq i$, ${}^tU_i U_j = 0$
 puique la base 
$\Bigl( \vec{u}_i \Bigr)_{1\le i \le 3}$ est orthonormale.
Donc 
$$ S^2 = \sum_{(i,j)\in I^2}\lambda_i \lambda_j U_i {}^tU_i U_j {}^tU_j 
= {}^tFF
$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question III.3 (b) suite
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{ L'endomorphisme $s$ de matrice $S$ est un endomorphisme
sym\'etrique, d\'efini positif.
}
$$
\matrix{
 {}^tS &=& 
{}^t\Bigl( \displaystyle\sum_{i=1}^{3} \lambda_i\  U_i{}^tU_i  \Bigr) \cr
&=& \displaystyle\sum_{i=1}^{3} \lambda_i\  {}^t\Bigl( U_i{}^tU_i \Bigr) \cr
&=& \displaystyle\sum_{i=1}^{3} \lambda_i\  U_i {}^tU_i  \cr
 {}^tS &=& S \cr
}
$$
Donc $s$ est un endomorphisme sym\'etrique.

Soit $\vec{x}$ un \'el\'ement quelconque de 
{\font\bb=msbm10 scaled 1100 $\R^3$} :
$\vec{x} = x_1 \vec{u_1 } + x_2 \vec{u_2 } + x_3 \vec{u_3 } $.

La matrice de $s (\vec{x} )$ dans la base canonique est :
$$\hbox{\cursiveonze M}_{B_c}(s (\vec{x} )  )=
\displaystyle \Bigl( \sum_{i=1}^{3} \lambda_i U_i{}^tU_i \Bigr)
\Bigl(\sum_{k=1}^{3} x_k {U_k }\Bigr)
= \sum_{i=1}^{3} \lambda_i x_i U_i
$$
car pour $k\neq i$, ${}^tU_i U_k = 0$ et pour $k=i$,
${}^tU_i U_i = 1$, comme ci-dessus.

Alors :
$$\vec{x} \cdot s (\vec{x} )= \Bigl(\sum_{k=1}^{3} x_k {U_k }\Bigr)
\times \sum_{i=1}^{3} \lambda_i x_i U_i
= \sum_{i=1}^{3} \lambda_i x_i^2
$$
Donc $s$ est un endomorphisme sym\'etrique d\'efini positif.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question III.4
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{D\'ecomposition de $f$ en la composition d'un endomorphisme
sym\'etrique d\'efini positif et d'une rotation
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question III.4 (a)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{a) $R = F S^{-1}$ est une matrice orthogonale
}
$R = F S^{-1}$ donc ${}^tR = {}^t(S^{-1}) \ {}^tF = S^{-1}\ {}^t F$ 
car $S$ 
est sym\'etrique.

Donc :
$$\matrix{R\  {}^tR & = & F\ S^{-1}\  S^{-1}\ {}^tF \cr
                  & = & F\ (S^{-1})^2\ {}^tF \cr
                  & = & F\ (S^{2})^{-1}\ {}^tF \cr
}$$
Comme $S^2 = {}^tF\  F$,
$$\matrix{R\  {}^tR & =  & F\ ({}^tF\  F )^{-1}\ {}^tF \cr
                    & = &  F\ F^{-1}\  ({}^tF )^{-1}\ {}^tF \cr
                    & = & I_3 \cr
}$$
Donc ${}^tR = R^{-1}$ : $R$ est orthogonale.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question III.4 (b)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{b) D\'ecomposition de $f$ en la composition d'un endomorphisme
sym\'etrique d\'efini positif et d'une rotation
}
$F = RS$ est \'equivalent \`a $f = r\circ s$.

$R$ \'etant une matrice orthogonale, $r$ est une isom\'etrie
de {{\font\bb=msbm10 scaled 1100 $\R^3$}.
$$ \det (r) = \det (R)= \det (F) \times \det (S^{-1})=
{1 \over \lambda_1  \lambda_2 \lambda_3} \det (F) > 0
$$
Donc $r$ est une isom\'etrie positive de  
{{\font\bb=msbm10 scaled 1100 $\R^3$}, donc une rotation
vectorielle.

(On a d\'eja montr\'e dans la question 3.b de la troisi\`eme 
partie que $s$ \'etait un endomorphisme sym\'etrique 
d\'efini positif.)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question III.5
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{ Expression de $R$ \`a l'aide de $Q$ et $Q'$
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question III.5 (a) 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{a) $B'$ est une base orthonormale de
{\font\bb=msbm10 scaled 1100 $\R^3$}
}

Pour $1\le i \le 3$, 
$$r (\vec{u}_i ) =  \vec{u'}_i $$
$r$ \'etant une isom\'etrie, transforme une base 
orthonormale en une base orthonormale.
 
Donc $B'$ est une base orthonormale de de
{\font\bb=msbm10 scaled 1100 $\R^3$}.  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% sous question III.5 (b) 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{b) Expression de $R$ \`a l'aide de $Q$ et $Q'$
}
$R$ peut \^etre interpr\'et\'e comme la matrice de 
passage de la base $B$ \`a la base $B'$.


$Q$ est la matrice de $Id_{\R^3}$ (application identique
{\font\bb=msbm10 scaled 1100 $\R^3 \mapsto \R^3$}),
l'espace de d\'epart {\font\bb=msbm10 scaled 1100 $\R^3$}
\'etant rapport\'e \`a la base $B$, 
l'espace d'arriv\'ee {\font\bb=msbm10 scaled 1100 $\R^3$}
\'etant rapport\'e \`a la base $B_c$, puisque les 
colonnes de $Q$ contiennent les composantes de
$(\vec{u}_1 , \vec{u}_2 , \vec{u}_3)$ dans la base 
$B_c$.


$(Q')^{-1}$ est la matrice de $Id_{\R^3}$, 
l'espace de d\'epart {\font\bb=msbm10 scaled 1100 $\R^3$}
\'etant rapport\'e \`a la base $B_c$, 
l'espace d'arriv\'ee {\font\bb=msbm10 scaled 1100 $\R^3$}
\'etant rapport\'e \`a la base $B'$, puisque les 
colonnes de $Q'$ contiennent les composantes de
$(\vec{u'}_1 , \vec{u'}_2 , \vec{u'}_3)$ dans la base 
$B_c$.

Donc 
$$ R = Q'\times{}^tQ
$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% question III.6
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{ Image par $f$ d'une sph\`ere centr\'ee en $O$ et de rayon $\rho$ 
} 
$f$ transforme le vecteur 
$\vec{x} = x_1 \vec{u}_1 + x_2 \vec{u}_2 + x_3 \vec{u}_3$
en le vecteur 
$\vec{y} = y_1 \vec{u'}_1 + y_2 \vec{u'}_2 + y_3 \vec{u'}_3$
tel que
$y_1 = \lambda_1 x_1$, $y_2 = \lambda_2 x_2$, $y_3 = \lambda_3 x_3$, 
donc :
$$ x_1 = {y_1\over \lambda_1 }\hbox{\ \ \ \ ,\ \ \ \ }
x_2 = {y_2\over \lambda_2 }\hbox{\ \ \ \ et\ \ \ \ }
x_3 = {y_3\over \lambda_3 }
$$
$m$ tel que $\vec{Om} = \vec{x}$ appartient \`a la sph\`ere de centre
$O$ et de rayon $\rho$ si et seulement si :
$$ x_1^2 + x_2^2 + x_3^2 = \rho^2
$$
donc si et seulement si :
$$ {y_1^2 \over (\lambda_1 \rho)^2 } +
{y_2^2 \over (\lambda_2 \rho)^2 } +
{y_3^2 \over (\lambda_3 \rho)^2 } =1
$$
dans la base $B'$.

Donc l'image par $f$ d'une sph\`ere centr\'ee en $O$ et de rayon $\rho$
est un ellipso\"ide de centre $O$ dont les axes sont port\'es par
les axes de coordonn\'ees du rep\`ere $(O , B')$, les longueurs de 
demi-axes \'etant $\lambda_1 \rho$, $\lambda_2 \rho$ et $\lambda_3 \rho$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Appendice : source du dessin pour Maple V
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Appendice : proc\'edure en langage Maple V pour le dessin de la 
question 4.c de la deuxi\`eme partie
}
\begin{verbatim}

#########################################################################
fig1:=proc()
#########################################################################
# Programme MAPLE V pour d\'essiner une figure pour la 
# question II.4.c de l'\'epreuve II.B de la banque PT 1999.
#  Mode d'emploi : sous Maple V (pour MSDOS et MS Windows *) :
#      read `T:\\monrep\\M99DT2Fc.txt`;
#  o\`u T:\monrep\ est le chemin d'acc\`es MSDOS du fichier, puis :
#      fig1();
#  Mode d'emploi sous Maple V (unix, Linux, FreeBSD,...) :
#      read `monrep/M99DT2Fc.txt`;
#  o\`u monrep est le chemin d'acc\`es unix du fichier, comme
#  /home/daniel/tex/d836 puis :
#      fig1();
#########################################################################
local g_1,g_2,g_3,g_4,g_5,g_m_n,g_m_np1,g_p_n,g_p_np1,g_delta,g_deltap,\
      x_m_n,y_m_n,x_m_np1,y_m_np1,x_p_n,y_p_n,x_p_np1,y_p_np1,\
      x_delta,y_delta,x_deltap,y_deltap;

 
with(plots):

# Si on veut faire des essais sur 'ecran, commenter interface,
# cad les trois lignes suivantes. Pour obtenir un fichier  
# PostScript, d\'ecommenter ! Pour modifier le dessin, etc...
# changer \`a votre convenance suivant l'OS, les chemins...
#interface(plotdevice=postscript,\
#plotoutput=`/home/daniel/tex/d836/M99DT2Fc.eps`,\
#plotoptions=`noborder`):

# les 2 droites Delta et Delta' :
g_1:=plot((-1/(sqrt(2)+1))*x,x=-1..10,y=-5..1,scaling=CONSTRAINED);
g_2:=plot((1/(-sqrt(2)+1))*x,x=-1..10,y=-5..1,scaling=CONSTRAINED);

# \'Etiquette pour le point m(n)
x_m_n:=8.0:
y_m_n:=evalf(x_m_n*(-1/(sqrt(2)+1))):
g_m_n:=PLOT(TEXT([x_m_n , y_m_n-0.50],`m(n)`));


# \'Etiquette pour le point m(n+1)
x_m_np1:=evalf(x_m_n*(3-2*sqrt(2))):
y_m_np1:=evalf(y_m_n*(3-2*sqrt(2))):
g_m_np1:=PLOT(TEXT([x_m_np1+0.80 , y_m_np1],`m(n+1)`));

# \'Etiquette pour le point p(n)
y_p_n:=y_m_n:
x_p_n:=evalf(y_p_n*((-sqrt(2)+1))):
g_p_n:=PLOT(TEXT([x_p_n + 1.0 , y_p_n-0.50],`p(n)`));

# \'Etiquette pour le point p(n+1)
x_p_np1:=evalf(x_p_n*(3-2*sqrt(2))):
y_p_np1:=evalf(y_p_n*(3-2*sqrt(2))):
g_p_np1:=PLOT(TEXT([x_p_np1 + 0.7 , y_p_np1-0.40],`p(n+1)`));

# \'Etiquette pour la droite Delta
x_delta:=x_m_n*0.6+1.0:
y_delta:=y_m_n*0.6:
g_delta:=PLOT(TEXT([x_delta , y_delta],`(Delta)`));

# \'Etiquette pour la droite Delta'
x_deltap:=x_p_n*1.4+0.70:
y_deltap:=y_p_n*1.4:
g_deltap:=PLOT(TEXT([x_deltap , y_deltap],`(f(Delta))`));

# trac\'e de la droite m(n)p(n)
g_3:=PLOT(CURVES([[x_m_n,y_m_n],[x_p_n,y_p_n]])):

# trac\'e de la droite m(n+1)p(n+1)
g_4:=PLOT(CURVES([[x_m_np1,y_m_np1],[x_p_np1,y_p_np1]])):

# trac\'e de la droite m(n+1)p(n)
g_5:=PLOT(CURVES([[x_m_np1,y_m_np1],[x_p_n,y_p_n]])):


display([g_1,g_2,g_3,g_4,g_5,g_delta,g_deltap,g_m_n,g_m_np1,g_p_n,g_p_np1]):

end;
#########################################################################


\end{verbatim}
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\end{document}
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