Given that the elements of a $2 \times 2$ symmetric matrix are real, show that

  (i)     the eigenvalues are real;

  (ii)     the eigenvectors are orthogonal if the eigenvalues are distinct.

The matrix $\boldsymbol{A}$ is given by $$\boldsymbol{A} = \left( {\begin{array}{*{20}{c}} {11}&{\sqrt 3 }\\ {\sqrt 3 }&9 \end{array}} \right) .$$ Find the eigenvalues and eigenvectors of $\boldsymbol{A}$.

The ellipse $E$ has equation ${{\boldsymbol{X}}^T}{\boldsymbol{AX}} = 24$ where $\boldsymbol{X} = \left( \begin{array}{l} x\\ y \end{array} \right)$ and $\boldsymbol{A}$ is as defined in part (b).

   (i)     Show that $E$ can be rotated about the origin onto the ellipse $E'$ having equation $2{x^2} + 3{y^2} = 6$ .

   (ii)     Find the acute angle through which $E$ has to be rotated to coincide with $E'$ .

(i)     let $\boldsymbol{M} = \left( {\begin{array}{*{20}{c}} a&b\\ b&c \end{array}} \right)$     (M1)

the eigenvalues satisfy

$\det (\boldsymbol{M} - \lambda \boldsymbol{I}) = 0$     (M1)

$(a - \lambda )(c - \lambda ) - {b^2} = 0$     (A1)

${\lambda ^2} - \lambda (a + c) + ac - {b^2} = 0$     A1

discriminant $= {(a + c)^2} - 4(ac - {b^2})$     M1

$= {(a - c)^2} + 4{b^2} \ge 0$     A1

this shows that the eigenvalues are real     AG

(ii)     let the distinct eigenvalues be ${\lambda _1},{\lambda _2}$  , with eigenvectors ${{\boldsymbol{X}}_1}$, ${{\boldsymbol{X}}_2}$

then

${\lambda _1}{{\boldsymbol{X}}_1} = {\boldsymbol{M}}{{\boldsymbol{X}}_1}$ and ${\lambda _2}{{\boldsymbol{X}}_2} = {\boldsymbol{M}}{{\boldsymbol{X}}_1}$     M1

transpose the first equation and postmultiply by ${{\boldsymbol{X}}_2}$ to give

${\lambda _1}{\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = {\boldsymbol{X}}_1^T{\boldsymbol{M}}{{\boldsymbol{X}}_2}$     A1

premultiply the second equation by ${\boldsymbol{X}}_1^T$

${\lambda _2}{\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = {\boldsymbol{X}}_1^T{\boldsymbol{M}}{{\boldsymbol{X}}_2}$     A1

it follows that

$(\lambda 1 - {\lambda _2}){\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = 0$     A1

since $\lambda 1 \ne {\lambda _2}$ , it follows that ${\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = 0$ so that the eigenvectors are orthogonal     R1

[11 marks]

the eigenvalues satisfy $\left| \begin{array}{l} 11 - \lambda \\ \sqrt 3 \end{array} \right.\left. \begin{array}{l} \sqrt 3 \\ 9 - \lambda \end{array} \right| = 0$     M1A1

${\lambda ^2} - 20\lambda  + 96 = 0$    A1

$\lambda  = 8,12$     A1

first eigenvector satisfies

$\left( {\begin{array}{*{20}{c}} 3&{\sqrt 3 }\\ {\sqrt 3 }&1 \end{array}} \right)\left( \begin{array}{l} x\\ y \end{array} \right) = \left( \begin{array}{l} 0\\ 0 \end{array} \right)$     M1

$\left( \begin{array}{l} x\\ y \end{array} \right) =$ (any multiple of) $\left( {\begin{array}{*{20}{c}} 1\\ { - \sqrt 3 } \end{array}} \right)$     A1

second eigenvector satisfies

$\left( {\begin{array}{*{20}{c}} { - 1}&{\sqrt 3 }\\ {\sqrt 3 }&{ - 3} \end{array}} \right)\left( \begin{array}{l} x\\ y \end{array} \right) = \left( \begin{array}{l} 0\\ 0 \end{array} \right)$

$\left( \begin{array}{l} x\\ y \end{array} \right) =$ (any multiple of ) $\left( {\begin{array}{*{20}{c}} {\sqrt 3 }\\ 1 \end{array}} \right)$     A1

[7 marks]

(i)     consider the rotation in which $(x,y)$ is transformed onto $(x',y')$ defined by

$\left( \begin{array}{l} {x'}\\ {y'} \end{array} \right) = \left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&{ - \frac{{\sqrt 3 }}{2}}\\ {\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}} \end{array}} \right)\left( \begin{array}{l} x\\ y \end{array} \right)$ so that $\left( \begin{array}{l} x\\ y \end{array} \right) = \left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}\\ { - \frac{{\sqrt 3 }}{2}}&{\frac{1}{2}} \end{array}} \right)\left( \begin{array}{l} {x'}\\ {y'} \end{array} \right)$     M1A1

the ellipse $E$ becomes

$\left( {\begin{array}{*{20}{c}} {x'}&{y'} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&{ - \frac{{\sqrt 3 }}{2}}\\ {\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {11}&{\sqrt 3 }\\ {\sqrt 3 }&9 \end{array}} \right)\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}\\ { - \frac{{\sqrt 3 }}{2}}&{\frac{1}{2}} \end{array}} \right)\left( \begin{array}{l} {x'}\\ {y'} \end{array} \right) = 24$     M1A1

$\left( {\begin{array}{*{20}{c}} {x'}&{y'} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 8&0\\ 0&{12} \end{array}} \right)\left( \begin{array}{l} {x'}\\ {y'} \end{array} \right) = 24$     A1

$2{(x')^2} + 3{(y')^2} = 6$    AG

(ii)     the angle of rotation is given by $\cos \theta  = \frac{1}{2},\sin \theta  = \frac{{\sqrt 3 }}{2}$     M1

since a rotational matrix has the form $\left( {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta }\\ {\sin \theta }&{\cos \theta } \end{array}} \right)$

so $\theta  = {60^ \circ }$ (anticlockwise)     A1

[7 marks]