Solution
= Solution
At $p$, choose local <normal coordinates> with $g_{\alpha\beta}(p)=\operatorname{diag}(-1,1)$. Let $e_0,e_1$ be their coordinate <basis>. The <vectors> $V=e_0+e_1$ and $W=e_0-e_1$ have zero <metric tensor> square, and their component <matrix> has <determinant> $-2\ne0$. Thus \b[two independent <null vectors> form a <basis> of the <tangent space>]. <Normal coordinates> are only needed at the chosen point, not on a whole neighbourhood with flat metric.