I'm stuck on an exercise in Barret O'Neill's book on Semi-Riemannian Geometry(ex. 12 ch. 2).
"Let b be a symmetric bilinear form on V.[...] The null cone of b is the set $\Lambda$ of all null vectors in V. Let $A = \Lambda \cup 0 $, so $A \supset N$. Prove: (a) N is a subspace, but A is not unless $A = 0$ or $V$.[...]" (N is the nullspace of b)
It looks simple but I don't see the trick. I don't even know how to start:
Find a non-null vector that is the sum of two null vectors? (<-> find two non-orthogonal null vectors)
Let Q be the quadratic form associated with b. $A = Q^{-1}(0)$, Q is "nice" but not linear, what are the conditions on Q for A to be a subspace (forgive my English)...
If there is a null vector x ($\neq 0$), $\alpha x$ is a subspace, let B be a "nullsubspace" of maximal dimension. Find a null vector not in B,
It's just a mess...
If b is non-degenerate, we know there is an orthonormal basis $\{e_1,\ldots , e_r, e_{r+1}, \ldots , e_n \}$, with $b(e_i,e_j)=\begin{cases} 1 &if\quad i=j\le r \\ -1, &if\quad i=j>r \\ 0 &otherwise \end{cases}$
Vectors of the form $e_p\pm e_q$, with $1\le p \le r$ and $r+1 \le q\le n$ span V and are null.
For the degenerate case we can choose an orthogonal basis $\{e_1, \ldots ,e_r, f_1,\ldots ,f_s, g_1,\ldots ,g_{n-r-s}\} $ with $b(g_i,g_i)=0$...