I'm trying to prove a version of Gram-Schmidt orthogonalization process in Minkowski space $\Bbb L^n$ (for concreteness, I'll put the sign last). I am not interested in the existence of orthonormal bases, but instead in the algorithm.
Namely, suppose that $\{v_1,\cdots,v_k\}\subseteq\Bbb L^n$ is a linearly independent set which does not contain any lightlike vectors, and whose span is non-degenerate.
I'd try to mimic the usual proof by induction. If $k=1$, take $u_1 = v_1$, done. And if I assume $\{u_1,\cdots,u_k\}$ constructed, I'd define $$u_{k+1} = v_{k+1} - \sum_{j=1}^n\frac{\langle v_{k+1},u_j\rangle}{\langle u_j,u_j\rangle}u_j = v_{k+1} - \sum_{j=1}^n\epsilon_j\frac{\langle v_{k+1},u_j\rangle}{\|u_j\|^2}u_j,$$which is orthogonal to the previous $u_i$'s.
But:
- One of the $u_i$'s could be lightlike and the construction would stop there.
- I'm not using (as far as I can see) non-degenerability of the span of the initial vectors.
- Also, I tried applying the GS process to the plane $y=z$ in $\Bbb L^3$, starting with a basis with no lightlike vectors... it produced a freaking lightlike vector (and gave me an orthogonal basis, hooray!). I mean... it's no surprise a lightlike vector came up, assuming the GS process works here... but why should it?
I'm terribly lost. Can someone help me state the result correctly and maybe give me a little push on the proof? Thanks.
I got some help outside here, and I'll summarize the idea: we must require that for each $i=1,\cdots,k$, the span $[v_1,\cdots,v_i]$ is non-degenerate. This ensures that each $u_i$ is not lightlike. This solves the first and second bullets. About the last one, the issue is that we might lose existance and uniqueness of the projection, so as far as I have understood, it could or could have not worked.
(I'll leave the question open for awhile in case someone wants to add anything)