Stationary points with matrix I have an exercise but I do not even know where I should start.
Consider the normalised quadratic form $\frac{x^T Ax}{x^T x}$ where $x\in\mathbb{R}^2$, $A$ is a general $2\times 2$ matrix. Find the minima, maxima or saddle points.
Any ideas?
 A: Assume $A\in M_n(\mathbb{R}),x\in \mathbb{R}$. As JeanMarie wrote $2R(x)=\dfrac{x^T(A+A^T)x}{x^Tx}=\dfrac{x^TBx}{x^Tx}$ where $B$ is real symmetric. The critical points of the function $R$ defined on the surface $g(x)=x^Tx-1=0$, are given by the Lagrange method: there is $\lambda\in\mathbb{R}$ s.t. $D_xR-\lambda D_xg=0$, that is, for every $h\in\mathbb{R}^n$, $2x^TBh-\lambda 2x^Th=0$. That is equivalent to $Bx-\lambda x=0$. Finally, the critical points are the unitary eigenvectors $x_0$ of $B$, associated to $\lambda_0$, with $R(x_0)=\lambda_0$ (since $B$ is real symmetric, $x_0$ and $\lambda_0$ are real). Let $spectrum(B)=\{\lambda_1\geq\cdots\geq\lambda_n\}$. Then $\sup(R)$ is $\lambda_1$ and is reached in $x_1$ (s.t. $Ax_1=\lambda_1x_1$) and $\inf(R)$  is $\lambda_n$ and is reached in $x_n$. 
The hessian of the Lagrangian is $B-\lambda I_n$. Assume that $\lambda_1>\lambda_p=\cdots=\lambda_q>\lambda_n$. Then, in a critical point $x$ associated to the eigenvalue $\lambda_p=\cdots=\lambda_q$ of multiplicity $q-p+1$, the signature is $(p-1)\times +,(q-p+1)\times 0,(n-q)\times -$. In particular, such a point is a saddle point.
