I'm currently working on an exercise which is about finding the base of an isotropic space $(V,\beta)$ The space $V$ is equipped with a symmetric bilinear form $\beta$ and together they form the so called metric space $(V,\beta)$.
I know that \begin{equation} V = V_\bot \oplus H \oplus W \end{equation} Where \begin{equation} H = \oplus^{m}_{i=1} H_i \end{equation} And the $H_i$ are the so called hyperbolic planes which have the proberty that their structure matrix - in regard to a basis containing of an isotropic vector $u_i$ and a proper corresponding vector $v_i$- take the form: \begin{bmatrix} 0&&1\\1&&0 \end{bmatrix}
So far so good. For my concrete example I have the matrix \begin{bmatrix}0&0&5&0&0\\0&0&3&1&0\\5&3&2&1&3\\0&1&1&0&0\\0&0&3&0&0\end{bmatrix} which corresponds to a bilinear form $\beta:V\times V \to K$ where $V=K^5$ and $K=\mathbb{Z}/{7\mathbb{Z}}$.
Step 1: Giving a base for $V_{\bot}=\operatorname{Rad}(V,\beta)$. That's easy, we immediately see, that $v=(5,0,0,0,1)$ fulfils the condition \begin{equation} \forall w \in V: \beta(v,w) = 0 \end{equation} So: \begin{equation} V_\bot = span\{v\} \end{equation} Step 2: Giving a base for the orthogonal complement of $V_\bot$. My attempt (which seems to be wrong) was just simply to take for example \begin{equation} B = (e_1,e_2,e_3,e_4) \end{equation} as a basis.
The corresponding structure matrix would then clearly be just the structure matrix before with cut off fifth column and row.
Step 3 & 4: Here is where it becomes tricky. I now want to determine the isotropic vectors $u_i$. I.e the vectors which fulfill the relation \begin{equation} \beta_1(u_i,u_i)=0. \end{equation} Where $\beta_1$ is the restriction I explained above. Clearly one sees that for example $e_1,e_4$ are both isotropic vectors. For I know that the space of isotropic vectors has dimension less than 3 (if I have two isotropic vectors the corresponding hyperbolic space has clearly dimension 4) I would have been done.
The only step left would be to find vectors $v_i$ which fulfill \begin{equation} (i): \beta(u_i,v_j)=\delta_{i,j} \end{equation} and \begin{equation} (ii): \beta(v_i,v_j)= 0 \end{equation} Of course the vectors $u_1,v_1,u_2,v_2$ have to be lineary indepent as they give us a base for the hyperbolic subspace. To visualice the corresponding structure matrix would be \begin{bmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}
I don't find vectors $v_i$ which fulfill this relations. Anyone out there who could tell me, what I was doing wrong.