Verify that if $P$ is an orthogonal matrix and $x = Py$ then $y^Ty = x^Tx$. Let $A$ be a real symmetric $n × n$ matrix. Then we know that there exists a real orthogonal matrix $P$ such that $P^TAP$ is diagonal. By using the transformation $x = Py$, or otherwise, prove that for every $x\in {\mathbb R}^{n}$,
$$mx^Tx ≤ x^TAx ≤ Mx^Tx$$ where $m$ and $M$ are the smallest and greatest eigenvalues of $A$ respectively. For which $x$ is it true that $x^TAx = Mx^Tx$?
$\begin{bmatrix}5 & 1&\sqrt2\\1 & 5&\sqrt2\\ \sqrt2 &\sqrt2 &6\end{bmatrix}$
Find the maximum and minimum values of $x^Tx$ for those $x$ for which $x^TAx = 1$. Giving no heed to orientation, sketch the surface $S$ with equation $x^TAx = 1$, and indicate on it those vectors $x$ at which $x^Tx$ attains its maximum and minimum values on $S$.
$x^TAx=y^TDy$ and $\lambda_{min}y^Ty\leq y^TDy\leq \lambda_{max}y^Ty$ but how to link it to $1$ and the surface?