I am not very familiar with quadratic surfaces, but spontaneously I start to think about it. I thought that surfaces like $$ 2x^2+2xy+4yz+z^2=1 $$ can be written into the form $au^2+bv^2+cw^2=1$ on an orthonormal basis. To do this, I tried to write the surface into the form $$ (2x+y)^2-9y^2+(z-2y)^2=2, $$ but when I attempt to perform transform $u=2x+y, u=y$ and so on, I find that although the desired form is achieved, the resulting new basis is not orthonormal (because the matrix is not symmetric). I try to randomly guess other matrices to use, but that doesn't help.
My question is: is it at all possible to express the surface (actually a hyperboloid) in an orthonormal basis in the form $au^2+bv^2+cw^2=1$? If so, how can I find the desired real and symmetric matrix?
If you do not demand orthonormal change,
$$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 \\ 0 & - 4 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 9 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & - 4 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 1 \\ \end{array} \right) $$
leads to $$ 2 \left(x + \frac{y}{2} \right)^2 - \frac{1}{2} ( y - 4 z)^2 + 9 z^2 $$
Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 1 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$
$$ H = \left( \begin{array}{rrr} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 1 \\ \end{array} \right) $$
==============================================
$$ E_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 2 \\ 0 & 2 & 1 \\ \end{array} \right) $$
==============================================
$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - 2 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & - 4 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 9 \\ \end{array} \right) $$
==============================================
$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 1 & 0 \\ - 2 & 4 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - 2 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 9 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 \\ 0 & - 4 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 9 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & - 4 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 1 \\ \end{array} \right) $$