Well, I managed to figure this out after a bit more pondering. It's necessary to use the properties of g on both sides of the equation.
(My thought that maybe it worked because of some nutty property of rotations in dimensions greater than 2 was crap, since e.g. \left( \begin{array}{ccc} 0 &...
Actually, thinking about this a bit, I think it's different than a statement that vector length or dot product is preserved.
For example, for any orthogonal matrices A and B, (Ax) \cdot (Ax) = x \cdot x = (Bx) \cdot (Bx), which is directly related to the length being preserved.
But (Ax)...
Here's an example of this from "Problem Book in Quantum Field Theory" by Voda Radovanovic...
He does specifically mention x is in M4, which rules out my counterexamples, since they only work in 2D. I presume this is a well-known and obvious argument in order to be the solution to the very...
If it's the zero matrix that is making you think of division by zero, it's also true for a 90 degree rotation in the opposite direction.
\left( \begin{array}{ccc} 0 & -1 \\ 1 & 0 \end{array} \right) x \cdot x = \left( \begin{array}{ccc} 0 & 1 \\ -1 & 0 \end{array} \right) x \cdot x
Those...
I've seen a few short proofs that if that some transformation \Lambda preserves the spacetime interval, then
\Lambda^\top g \Lambda = g
where g is the spacetime metric.
They have all relied on an argument using some simple algebra to show that
(\Lambda^\top g \Lambda) x \cdot x = g x...