Calculating isotropic basis of a bilinear form.

Question

Let $\phi:V\times V \longrightarrow R$ be a symmetric, bilinear form with signature. A not null vector $v\in V$ is isotropic if $\phi(v,v)=0$.

Suppose that the matrix of $\phi$ in one basis $B=(u_1,u_2,u_3,u_4)$ is:

$M = \begin{Bmatrix} 0 & -1 & 2 & 1 \\ -1 & -4 & 6 & 0 \\ 2 & 6 & 3 & -2 \\ 1 & 0 & -2 & 1 \\ \end{Bmatrix}$

I am asked to calculate a orthogonal basis to $\phi$ and calculate a isotropic basis to $\phi$.

I have already calculated an orthogonal basis: $(0,0,0,1),(0,1,0,0),(0,\frac{6}{4},1,2),(1,\frac{-3}{4},\frac{-5}{16},\frac{-26}{16})$ However I don't know how to calculate the isotropic basis, when do this basis exists, and how can I calculate it?

Answer 1

Hint:

If $\varphi(e,f)=0$, $\ \varphi(e,e)=1\ $ and $\ \varphi(f,f)=-1\ $ then $\varphi(e+f,\,e+f)=\varphi(e-f,\,e-f)=0$