I need to pick a basis of elements $u$, $v$ and $w$ such that $u$ and $w$ are lightlike (i.e., $g(u,u) = 0 = g(w,w)$, where $g(x,y) = x_1y_1+x_2y_2-x_3y_3$), $g(v,v) = 1 = -g(u,w)$, $\langle u \rangle^\perp = \langle u,v\rangle$ and $u$ is a lightlike eigenvector with eigenvalue $\lambda$ such that $\lambda^2=1$.
My doubt comes by understanding how can I use that $u$ is a lightlike eigenvector with eigenvalue $\lambda=\pm1$. I don't know the expression of the matrix of isometries, and the property $$g(f(u),f(u))= g(u, u)$$ doesn't help much because $g(u,u)=0$, so $$0 = g(u,u) = g(f(u),f(u)) = g(\lambda u,\lambda u) = \lambda^2 g(u,u) = \lambda^2\cdot 0 = 0.$$
Is there any hint that I'm missing here? Thanks beforehand!