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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.

Rodrigo de Azevedo
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Ivo Terek
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    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. – Neal Aug 23 '16 at 01:39
  • I'm not familiar with $QR$ decompositions, but I am with Lorentz transformations, though. I'll look into it, thanks! – Ivo Terek Aug 23 '16 at 01:44
  • Hi @IvoTerek, did u find how to prove that if the span is non-degenerate each $u_i$ is non degenerate? – Moza Apr 30 '22 at 19:12
  • See page 9 after Proposition 1.14 of https://www.asc.ohio-state.edu/terekcouto.1/texts/MAT6702_OSU-USP19_notes.pdf 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. [...] – Ivo Terek Apr 30 '22 at 19:52
  • [...] 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). – Ivo Terek Apr 30 '22 at 19:52
  • @IvoTerek Sorry! I wanted to say that if the span is non-degenerate it can't have lightlike vectors. And first of all, thanks for the fast reply, those texts helped me lots. What I'm trying to prove now is that we add a $v_{k+1]$ to our orthogonal base$ to mantain non-degeneracy the new orthogonal vector $\hat{v_{k+1}$ can´t be lightlike. I'm trying to prove that $g(\hat{v_{k+1},\hat{v_{k+1})=0$. But in the end I get : $g(\hat{v_{k+1},\hat{v_{k+1})= ag(v_{k+1},v_{k+1}+ sum(g(v_i , v_i)$. Don't now how to prove this is exactly zero to be lightlike. Any idea? – Moza Apr 30 '22 at 21:14
  • A non-degenerate subspace may absolutely contain lightlike vectors, this is not a problem. Non-degeneracy means that the only vector in the subspace which is orthogonal to everything in the subspace is the zero vector. With your notation, it is not true that $\hat{v_{k+1}}$ is lightlike. – Ivo Terek Apr 30 '22 at 21:17
  • Thanks again for your help. Sorry, bath attack got me and somehow I can't edit my post to fix all problems. I'm trying to prove that $g(\hat{v_{k+1},\hat{v_{k+1})=/=0$. – Moza Apr 30 '22 at 21:28
  • Let us continue this discussion in chat. – Moza Apr 30 '22 at 21:31
  • Again: if this were the case, $\hat{v_{k+1}}$ were lightlike, it would be orthogonal to itself and hence orthogonal to all vectors $\hat{v_1},...,\hat{v_{k+1}}$, against non-degeneracy of the subspace spanned by these $k+1$ vectors. – Ivo Terek Apr 30 '22 at 21:35
  • Got it! Thanks for the help – Moza Apr 30 '22 at 21:44

1 Answers1

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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)

Ivo Terek
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  • why would lightlike vectors interfere with non-degenerate spans? – Alex Bogatskiy Dec 17 '24 at 23:01
  • Try applying Gram-Schmidt to $v_1 = (1,0,0)$, $v_2 = (1,0,1)$, and $v_3 = (0,0,1)$, in this order, and see the disaster happen... :P – Ivo Terek Dec 18 '24 at 01:21
  • ok I see, because the only direction orthogonal to a null vector within a plane not tangent to the null cone and passing through that null vector is the original null direction, so you never end up producing a second basis element. – Alex Bogatskiy Dec 19 '24 at 17:21