Finding the maximizer Let be $ A\in \mathbb{R}^{2,2} $ a symmetrical and positive definite matrix with different eigenvalues $ \lambda_1>\lambda_2>0 $. Further we have the rayleigh quotient $$ \max_{x\neq 0}\frac{\overline{x}^T\cdot A\cdot x}{\overline{x}^T\cdot x}=\max_{\|x\|_2=1}\overline{x}^T\cdot A\cdot x $$.
For how many points $ x\in \mathbb{R}^2 $ the maximum $$ \max_{\|x\|_2=1}\overline{x}^T\cdot A\cdot x $$ is reached?
My idea:
Let $$ A:=\begin{pmatrix}a&b\\b&c\end{pmatrix},\quad x:=\begin{pmatrix}x_1\\x_2\end{pmatrix} ,\quad \|x\|_2^2=1=x_1^2+x_2^2$$
and define
$$ F(x)=x^T\cdot A\cdot x=(x_1, x_2)\cdot \begin{pmatrix}a&b\\b&c\end{pmatrix} \cdot \begin{pmatrix}x_1\\x_2\end{pmatrix}\\=(x_1, \sqrt{1-x_1^2})\cdot \begin{pmatrix}a&b\\b&c\end{pmatrix} \cdot \begin{pmatrix}x_1\\\sqrt{1-x_1^2}\end{pmatrix}=:F(x_1) $$.
Derivative:
$$ \nabla F(x_1)=\left(1,\frac{-x_1}{\sqrt{1-x_1^2}}\right)\cdot \begin{pmatrix}a&b\\b&c\end{pmatrix} \cdot \begin{pmatrix}1\\\frac{-x_1}{\sqrt{1-x_1^2}}\end{pmatrix}\stackrel{!}{=}0 $$.
From here I get stuck and I don't see how can I use the information about $ A $, the eigenvalues and the rayleigh qoutient.

Note that the set $\|x\|=1$ is compact, so a $\max$ exists.
Solve $\max_{ \|x\|^2 =1 } {1 \over 2}x^T A x$ using Lagrange multipliers to get
$Ax = \lambda x$, so we see that the Lagrange multiplier must be an eigenvalue.
Hence $\max_{ \|x\|^2 =1 } x^T A x = \lambda_\max$ and in the given example,
since the eigenvalues are distinct we see that $x$ must be an eigenvector and hence the two solutions are $\pm x$.
