Let P be a quadratic form with real coefficients in $\mathbb{R}^{n}$ such that $P^{-1}(1)$ is non-empty and compact (bounded). Show that there is an orthogonal transformation that maps $P^{-1}(1)$ onto an ellipsoid $\sum_{k =1}^{n} a_{k} x^{2} = 1$, $a_{k} > 0$.
Ideas: There exists a unique square symmetric matrix $A$ associated with $P$ such that $P(x) = x^{T}Ax$. Then by Spectral Theorems, we can find an orthogonal matrix $Q$ such that $QAQ^{T} = D$ - where $D = (d_{km})$ with $d_{km} = 0$ for $k$ different from $m$.If $q$ is the quadratic form associated with $D$, then for $x$, we set $y = Qx$, then we have $P(x) = y^{T}Dy = q(y) = \sum_{k=1}^{n} c_{k}y_{k}$, where $c_{k} = \sum_{m=1}^{n}d_{km}y_{m} = d_{kk}y_{k}$. It follows that $P(x) = \sum_{k=1}^{n}d_{kk}y_{k}^{2}$. Since $P^{-1}(1) \neq \emptyset$, there exists $x$ such that $P(x) =1$ from which we get $\sum_{k=1}^{n}d_{kk}y_{k}^{2} = 1$.
I would appreciate a clean-up to this or a better solution. I haven't used boundedness of the pre-image in question! This was a qual problem.
