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Linear independence doesn't guarantee that you can rotate, for that you need to know how to exchange one dimension for another. Essentially you need to define
by zardo 5y ago
Linear independence doesn't guarantee that you can rotate, for that you need to know how to exchange one dimension for another.
Essentially you need to define a Pythagorean theorem for your space. You can do that for the car vector space, but there isn't a natural choice.
- javajosh 5y agoIsn't that the definition of rotation, the "exchanging one dimension for another"? In 2-D it keeps a point invariant, in 3-D a line, and in N-D it keeps the N-1 object invariant. (Side note: the more basic operation is reflection, since you can get rotation from two reflections)
- zardo 5y ago>Isn't that the definition of rotation, the "exchanging one dimension for another"? Maybe? I think it allows solutions not typically considered to be rotations, like the special relativity example.