Where $f : \mathbb{R}^3 \times \mathbb{R}^3 \rightarrow \mathbb{R}$ corresponding to the quadratic form $q : \mathbb{R}^3 \rightarrow \mathbb{R}$,
$q(x,y,z) = x^2 + 2xy + y^2 + 2yz + z^2$
I found that if $u = (x,y,z)$ and $q(u,u) = u^tAu = q(x,y,z) = x^2 + 2xy + y^2 + 2yz + z^2$,
$A= \begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix} $
I've tried using the Gram-Schmidt Process however I am getting the problem of dividing by zero. What would be the best way for me to find the orthogonal basis in this case?
Edit:
Let $B = \{b_1, b_2,b_3\}$
where $b_1 = (1, 0, 0)$, $b_2 = (0, 1, 0)$, $b_3 = (0, 0, 1)$.
Now let $C = \{c_1, c_2,c_3\}$ be an the orthonormal basis generated iteratively using the Gram-Schmidt method.
Let $f(u,v) = u^tAv$ and $c_1 = b_1$
$c_2 = b_2 - \frac{f(b_2, c_1)}{f(c_1, c_1)}c_1$ $= (0,1,0) - \frac{1}{1}(1,0,0)$
$c_2 = (-1, 1, 0)$
$c_3 = b_3 - \frac{f(b_3, c_1)}{f(c_1, c_1)}c_1 - \frac{f(b_3, c_2)}{f(c_2, c_2)}c_2$
At this point $f(c_2, c_2) = 0$, would I maybe need to retrace and pick another $c_2$ or is this one ok?
Diagonalize orthogonally $\;A\;$ , which is possible since $\;A\;$ is symmetric:
$$|xI-A|=\begin{pmatrix}x-1&-1&0\\-1&x-1&-1\\0&-1&x-1\end{pmatrix}=(x-1)^3-2(x-1)=$$
$$(x-1)\left(x^2-2x-1\right)=(x-1)(x-1-\sqrt2)(x-1+\sqrt2)$$
Now just calculate eigenvectors of the above eigenvalues, which will be orthogonal (by symmetry, again). If you want the basis orthonormal just divide each of those eigenvectors by its length, and there you have an orthonormal basis.