Orthogonal basis with respect to a symmetric bilinear form $\phi$ I'm confused about orthogonal bases with respect to a symmetric bilinear form $\phi$. 
Consider the quadratic form
$Q: \mathbb{R^{4}} \rightarrow \mathbb{R}$
 defined as:
$Q(\vec{x})=-x_2^2+2x_1x_3+x_4^2$
Which is associated to a symmetric bilinear form $\phi$ defined as:
$\phi(\vec{x},\vec{y})=-x_2y_2+x_4y_4+x_1y_3+x_3y_1$
Then consider this subspace 
$W= \Big\{ {(x_1,x_2,x_3,x_4) \in  \mathbb{R^{4}} \mid x_4=2x_1-x_2+4x_3=0 \Big\} }= \mathscr{L}((1,2,0,0),(0,4,1,0))$
Determine a basis $\mathscr{C}=(\vec{c_1},\vec{c_2})$ of $W$ orthogonal with respect to $\phi$.
(The answer is $\mathscr{C}=((1,2,0,0),(-\frac{7}{4},\frac{1}{2},1,0))$)
I started considering $\vec{c_1}=(1,2,0,0)$ but then I don't know how to determine a second vector of $W$ which is orthogonal to it.
My idea was to apply Gram-Schmidt process using $\phi$ instead of the standard dot product but it doesn't work and I can't understand why!
In general what is the method to determine an orthogonal basis of a subspace with respect to $\phi$?
Thanks a lot for your help
 A: Just use the definitions. You are looking for a vector
$$ \vec{c}_2 = t(1,2,0,0) + s(0,4,1,0) $$
such that
$$ 0 = \phi(\vec{c}_1, \vec{c}_2) = \phi((1,2,0,0), t(1,2,0,0) + s(0,4,1,0)) = t\phi((1,2,0,0), (1,2,0,0)) + s\phi((1,2,0,0), (0,4,1,0)) = -4t -7s.  $$
Hence, $4t + 7s = 0$. Taking $s = 1$, we get $t = -\frac{7}{4}$ and
$$ \vec{c}_2 = \left( -\frac{7}{4}, -\frac{7}{2}, 0, 0 \right) + (0,4,1,0) = \left( -\frac{7}{4}, \frac{1}{2}, 1, 0 \right). $$
In general, there is an algorithm that given a bilinear form on $\mathbb{F}^n$ (with $\mathrm{char} \mathbb{F} \neq 2$), generates an orthogonal basis with respect to the form. It works by applying simultaneous row and column operations on the symmetric matrix representing the bilinear form and is, in some sense, a generalization of the regular Gram-Schmidt procedure. However, in your case since you are working with a two dimensional space, this is quite an overkill.
A: Gram-Schmidt using $\phi$ as the scalar product should have worked for you.  
Let $v_1=u_1=(1,2,0,0)^T$ and $v_2=(0,4,1,0)^T$, so that $\phi(u_1,u_1)=-4$ and $\phi(u_1,v_2)=-7$. Then we have $$\begin{align}
u_2 &= v_2-{\phi(u_1,v_2)\over\phi(u_1,u_1)}u_1 \\
&=(0,4,1,0)^T-{-7\over-4}(1,2,0,0)^T \\
&=\left(-\frac74,\frac12,1,0\right)^T.
\end{align}$$
