I've seen a few short proofs that if that some transformation [itex]\Lambda[/itex] preserves the spacetime interval, then(adsbygoogle = window.adsbygoogle || []).push({});

[itex]\Lambda^\top g \Lambda = g[/itex]

where g is the spacetime metric.

They have all relied on an argument using some simple algebra to show that

[itex](\Lambda^\top g \Lambda) x \cdot x = g x \cdot x[/itex]

and since this is true for *any* x, it must be true that

[itex]\Lambda^\top g \Lambda = g[/itex]

This confuses me. I don't see how this "since it's true for any x" step is justified.

For example,

[itex]\left( \begin{array}{ccc} 0 & -1 \\ 1 & 0 \end{array} \right) x \cdot x = \left( \begin{array}{ccc} 0 & 0 \\ 0 & 0 \end{array} \right) x \cdot x[/itex]

for any x, and by the same argument, since it is true for *any* x, it must be true that the arrays are equal and -1 = 0 = 1. Did I just break math?

Intuitively, I don't think I could do the same trick to produce a counterexample in 3+ dimensions, but this seems like kind of a subtle point to sweep under the rug in a proof.

Does anyone have a more detailed version of this argument that would (hopefully) make more sense to me?

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# Confused by proof of Lorentz properties from invariance of interval

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