Let $A \in \mathbb{R}^{n \times n}$ be a real $n \times n$-matrix. Consider the function
$$q: \mathbb{R}^n \to \mathbb{R}, x \mapsto x^t A x$$
where $x^t$ is the transposed vector $x$.
I now want to prove that $q$ has both a maximum and a minimum on the unit sphere $S = \{x \in \mathbb{R}^n, ||x||_2 = 1\}$.
Also, consider the case $n = 2$. What would be concrete examples for $A$, so that the set $\{x \in \mathbb{R}^n, q(x) = 1\}$ is compact or not compact?
Thanks in advance. I don't really know how to approach a function like this when it comes to properties like it's maximum and minimum.
If $f: \Bbb R^n \to \Bbb R$ is given by $f(x)=\|x\|^2$, then $f$ is continuous and so $S = f^{-1}(\{1\})$ is closed. Also, $S \subset B(0, 2)$ is bounded. So $S$ is compact and $q$ is continuous, so we're done by Weierstrass theorem.
Define $\langle \cdot,\cdot \rangle_L : \Bbb R^2 \times \Bbb R^2 \to \Bbb R$ by $\langle x,y \rangle_L = x_1y_1-x_2y_2$, and from here put $q(x) = \langle x,x\rangle_L$. The "unit positive sphere" according to $\langle\cdot,\cdot\rangle_L$ (also called the $2$-dimensional De Sitter space) is nothing more then the hyperbola $x^2-y^2 = 1$, which is not compact.