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?
Your eigenvalues are $(4,4,8)$
So you your biggest $x$ will be something that loads exculsively on 4 eigenvalues
and its magnitude will be $\frac 12$
$(\frac 1{2\sqrt 2}, -\frac 1{2\sqrt 2}, 0)$ will work.
So you your smallest $x$ will parallel to the eigenvector associated with the 8 eigenvalue.
and its magnitude will be $\frac {1}{2\sqrt 2}$
$(\frac {1}{4\sqrt 2},\frac {1}{4\sqrt 2}, \frac 14)$
As for the surface, it is an ellipsoid. It has 2 equal axes and one short axis. The big eigenvector is the minor axis.