Rotation of ellipsoid(quadric) Consider $$φ(x, y , z) = x^2 + 2y^2 + 4z^2 −xy −2xz −3yz$$
find the coordinate transformation (translation or rotation) to eliminate $xy$, $xz$ and $yz$.
In $\mathbb R²$, with conic sections, I would do this with 
$$\begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}\cos\alpha &-\sin\alpha\\\sin\alpha & \cos\alpha\end{pmatrix}.\begin{pmatrix}x' \\ y'\end{pmatrix}$$
to find the rotation angle. But how is it done with quadrics?
 A: Perhaps this will help:
\begin{align}
\phi(x,y,z) &= x^2 -x(y+2z)+{(y+2z)^2\over 4} -{(y+2z)^2\over 4}+ 2y^2 + 4z^2  −3yz\\
& = (x-{y+2z\over 2})^2+{7\over 4}y^2+3z^2-4yz\\
& = {1\over 4}(2x-y-2z)^2+{1\over 12}(36z^2-48yz+21y^2)\\
& = {1\over 4}(2x-y-2z)^2+{1\over 12}((6z)^2-2 \cdot4y\cdot 6z+16y^2+5y^2)\\
& = {1\over 4}(2x-y-2z)^2+{1\over 12}(6z-4y)^2+{5\over 12}y^2\\
& = {1\over 4}(2x-y-2z)^2+{1\over 3}(3z-2y)^2+{5\over 12}y^2
\end{align}
$$\begin{pmatrix}x'\\ y'\\z' \end{pmatrix}=
\begin{pmatrix}
1 &-1/2&-1\\
0 &\sqrt{5\over 12}&0\\
0 &-2\over \sqrt{3}& 3 
\end{pmatrix}
\begin{pmatrix}x \\ y\\z \end{pmatrix}$$
A: There is another way to 'see' it, given $\varphi(x,y,z)=x^2+2y^2+4z^2-xy-2xz-3yz$, you can also write this as $u^{\top}Qu$ for $u=(x,y,z)$ and 
\begin{equation}
Q = \begin{pmatrix}
1 & -0.5 & -1\\
-0.5 & 2 & -1.5\\
-1 & -1.5 & 4
\end{pmatrix}\succ 0
\end{equation}
Then diagonalize $Q$, i.e. $Q=T\Lambda T^{\top}$, where now $u^{\top}Q u = u^{\top}T\Lambda T^{\top}u$, for $\Lambda$ diagonal, such that $T^{\top}$ is your desired transformation.
