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 '16 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 '16 at 1:44

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)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.