I am look for a way to determine the definiteness of a quadratic form on the orthogonal complement of a line. More specifically, I have the quadratic form $Q^-(x) = -x_0^2 + x_1^2 + \cdots + x_n^2$, which I know is indefinite. The text I am reading makes the claim that if you have a vector $x$ such that $Q^-(x) >0 $, then it must be true that $Q^-$ is indefinite on the orthogonal complement of $x$. I am not sure how one would prove this. Any suggestions on a general strategy to prove something like this would be appreciated as I also would like to then determine what would be true about the orthogonal complement when $Q^-(x) = 0$ or $Q^-(x) < 0$.
Note that the inner product being used here is $\langle x,y\rangle = -x_0y_0 + x_1y_1 + \cdots + x_ny_n$.
Hint The standard basis of $\Bbb R^{n, 1}$ is orthogonal with respect to $Q^-$, and we can read off that the signature of $Q^-$ is $(n, 1)$
On the other hand, if $Q^- \vert_{x^{\perp}}$ is definite, we can choose an orthogonal basis $(E_1, \ldots, E_n)$ for whichever of $\pm Q^-$ is positive definite, so that $(x, E_1, \ldots, E_n)$ is again an orthogonal basis. What, then, is the signature of $Q^-$?
Note that this approach makes plain the exception Will Jagy pointed out in the comments.