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.


  • 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.

  • 2
    $\begingroup$ One though --- Graham-Schmidt is basically a way to find the $QR$ decomposition of a matrix whose columns are basis vectors. So maybe you'd find profit interpreting this as a $QR$ decomp where the $Q$ is actually a Lorentz transformation. $\endgroup$
    – Neal
    Aug 23, 2016 at 1:39
  • $\begingroup$ I'm not familiar with $QR$ decompositions, but I am with Lorentz transformations, though. I'll look into it, thanks! $\endgroup$
    – Ivo Terek
    Aug 23, 2016 at 1:44
  • $\begingroup$ Hi @IvoTerek, did u find how to prove that if the span is non-degenerate each $u_i$ is non degenerate? $\endgroup$
    – Moza
    Apr 30, 2022 at 19:12
  • $\begingroup$ See page 9 after Proposition 1.14 of asc.ohio-state.edu/terekcouto.1/texts/… or Theorem 1.2.23 on page 13 of the book expanded from these notes. The gist of it is that if ${\rm span}(u_1,\ldots, u_k) = {\rm span}(\widetilde{u_1},\ldots \widetilde{u_k})$ with the left side non-degenerate, the indicated vectors spanning the right side orthogonal, then $\widetilde{u_k}$ being lightlike would imply that it is orthogonal to all spanning vectors on the right, against non-degeneracy. [...] $\endgroup$
    – Ivo Terek
    Apr 30, 2022 at 19:52
  • $\begingroup$ [...] The statement "the span is non-degenerate if each $u_i$ is non-degenerate" is simply false (you can find easy examples in Minkowski $3$-space). $\endgroup$
    – Ivo Terek
    Apr 30, 2022 at 19:52

1 Answer 1


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)


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