As a part of my linear-algebra exam preparation, I am going through the surface equation and quadratic-bilinear form usage in my book which is a part we haven't really went through and left to explore on our own. I've found myself hitting the wall on the following type of exercise :
Given the equation $2x^2 + 2y^2 + 2z^2 + 2xy + 2xz + 2yz - 1=0$ show that the surface described by it has a unique center of symmetry $\big( O(0,0)\big)$ and after performing a coordinate system change, describe the type of the surface that this equations defines.
Now, what I have thought initially is that the $q_D(x,y,z)=2x^2 + 2y^2 + 2z^2 + 2xy + 2xz + 2yz $ is a quadratic form, which we can fully investigate and write its matrix down. After that, the question we need to answer is what $q_D(x,y,z) = 1$ describes. But, it seems like I cannot understand where the change of the coordinate system is needed and I also do not know how to proceed with this specific type of a problem. I've seen though another type of problem which was handled the way I tried to approach this one (it was a quadratic form-like equation again), but that doesn't seem to be the case here. Any help, hint or thorough solution will be really appreciated !
Put $v=(x,y,z)$ then
$$ v A v^{T} =1 $$
where $A=\begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{pmatrix} $
you can check that $A$ has eigenvalues $1,1,4$ with eigenvectors
$$ v_1=(1,0,-1)^{T} \;\;\; v_2=(0,1,-1)^{T} \;\;\; v_3=(1,1,1)^{T} $$
It's easy to see $v_1\cdot v_2\neq 0$ but any other pairs of vectors are orthogoanal, so apply Grams schmidt to these two to get an orthonormal basis $w_1,w_2,w_3$ where $w_3=\frac{1}{\sqrt{3}} v_3$. Set $P=[w_1,w_2,w_3]$, so
$$ P^{-1}AP=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4\end{pmatrix} $$
Now this is in canonnical form, and so surface is an ellipsoid