La matrice de $Q$ dans la base canonique est $A=\left( \begin{array}{ccc} 4&\sqrt{6}&5\sqrt{2}\\ \sqrt{6}&9&\sqrt{3}\\ 5\sqrt{2}&\sqrt{3}&-1 \end{array} \right)$ .

$\begin{align*}\ensuremath \chi_A&=\left| \begin{array}{ccc} 4-X&\sqrt{6}&5\sqrt{2}\\ \sqrt{6}&9-X&\sqrt{3}\\ 5\sqrt{2}&\sqrt{3}&-1-X \end{array} \right|= (4-X)(X^2-8X-12)-\sqrt{6}(-\sqrt{6}X-6\sqrt{6})+5\sqrt{2}(5\sqrt{2}X-42\sqrt{2})\\ &= -X^3+12X^2+36X-432 =-(X-6)(X+6)(X-12). \end{align*}$

Ensuite,

$\begin{align*}\ensuremath \left( \begin{array}{c} x\\ y\\ z \end{array} \right)\in\text{Ker}(A-6I_3)&\Leftrightarrow\left\{ \begin{array}{l} -2x+\sqrt{6}y+5\sqrt{2}z=0\\ \sqrt{6}x+3y+\sqrt{3}z=0\\ 5\sqrt{2}x+\sqrt{3}y-7z=0 \end{array} \right.\Leftrightarrow\left\{ \begin{array}{l} z=-\sqrt{2}x-\sqrt{3}y\\ -2x+\sqrt{6}y+5\sqrt{2}(-\sqrt{2}x-\sqrt{3}y)=0\\ 5\sqrt{2}x+\sqrt{3}y-7(-\sqrt{2}x-\sqrt{3}y)=0 \end{array} \right.\\ &\Leftrightarrow\left\{ \begin{array}{l} z=-\sqrt{2}x-\sqrt{3}y\\ -12x-4\sqrt{6}y=0\\ 12\sqrt{2}x+8\sqrt{3}y=0 \end{array} \right.\Leftrightarrow\left\{ \begin{array}{l} y=-\sqrt{\frac{3}{2}}x\\ z=-\sqrt{2}x-\sqrt{3}(-\sqrt{\frac{3}{2}}x) \end{array} \right.\Leftrightarrow\left\{ \begin{array}{l} y=-\sqrt{\frac{3}{2}}x\\ z=\frac{\sqrt{2}}{2}x \end{array} \right.\Leftrightarrow\left\{ \begin{array}{l} x=\sqrt{2}z\\ y=-\sqrt{3}{z} \end{array} \right. \end{align*}$

et $\text{Ker}(A-6I_3) =\text{Vect}(e_1)$ où $e_1=\frac{1}{\sqrt{6}}(\sqrt{2},-\sqrt{3},1)$ . De même,

$\begin{align*}\ensuremath \left( \begin{array}{c} x\\ y\\ z \end{array} \right)\in\text{Ker}(A+6I_3)&\Leftrightarrow\left\{ \begin{array}{l} 10x+\sqrt{6}y+5\sqrt{2}z=0\\ \sqrt{6}x+15y+\sqrt{3}z=0\\ 5\sqrt{2}x+\sqrt{3}y+5z=0 \end{array} \right.\Leftrightarrow\left\{ \begin{array}{l} z=-\sqrt{2}x-5\sqrt{3}y\\ 10x+\sqrt{6}y+5\sqrt{2}(-\sqrt{2}x-5\sqrt{3}y)=0\\ 5\sqrt{2}x+\sqrt{3}y+5(-\sqrt{2}x-5\sqrt{3}y)=0 \end{array} \right.\\ &\Leftrightarrow\left\{ \begin{array}{l} z=-\sqrt{2}x-\sqrt{3}y\\ y=0 \end{array} \right.\Leftrightarrow\left\{ \begin{array}{l} y=0\\ z=-\sqrt{2}x \end{array} \right. \end{align*}$

et $\text{Ker}(A+6I_3) =\text{Vect}(e_2)$ où $e_2=\frac{1}{\sqrt{3}}(1,0,-\sqrt{2})$ .

Enfin $\text{Ker}(A-12I_3) =\text{Vect}(e_3)$ où

$e_3 =e_1\wedge e_2=\frac{1}{3\sqrt{2}}\left( \begin{array}{c} \sqrt{2}\\ -\sqrt{3}\\ 1 \end{array} \right)\wedge\left( \begin{array}{c} 1\\ 0\\ -\sqrt{2} \end{array} \right)=\frac{1}{3\sqrt{2}}\left( \begin{array}{c} \sqrt{6}\\ 3\\ \sqrt{3} \end{array} \right)$ .

Ainsi, si on pose $P=\left( \begin{array}{ccc} \frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}}\\ -\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{6}}&-\frac{2}{\sqrt{6}}&\frac{1}{\sqrt{6}} \end{array}\right)$ alors $PA{^t}P=\text{diag}(6,-6,12)$ ou encore

$Q(Xe_1+Ye_2+Ze_3) =6X^2-6Y^2+12Z^2$ où $\text{Mat}_{(i,j,k)}(e_1,e_2,e_3)=\left( \begin{array}{ccc} \frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}}\\ -\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{6}}&-\frac{2}{\sqrt{6}}&\frac{1}{\sqrt{6}} \end{array}\right)$ .

exo7 5811