impossiblewizardry

so a transformation from Rⁿ to Rⁿ preserves Euclidean distances:

||v - w|| = ||v’ - w’||

From that you can prove it preserves Euclidean norms of vectors:

||v|| = ||v’||

And then you expand ||v’ - w’||² and see that it preserves dot products:

v’ . w’ = v . w

Geometrically, that means if you preserve all distances, you also have to preserve all angles, which makes sense. Then... OK, it has to be linear. You can find that

||(s v)’ - s v’||² = 0

just by expanding the left side and canceling. Then you show that

||v’ + w’||² = ||v + w||²

which you do just by expanding and unexpanding, and that allows you to prove

||(v + w)’ - (v’ + w’)||² = 0

again by just expanding and canceling. So that means this transformation is linear.

And then... well, it also turns out that it’s orthogonal. Because, ok let’s call the matrix of this transformation A. Then,

v^T v = ||v||² = ||v’||² = ||Av||² = v^T A^T A v

v^T (A^T A - I) v = 0

Alright? So, that last thing on the bottom, it’s a quadratic polynomial in n variables which puts some upper bound on how many zeros it can have, or like how those zeros can be shaped or something, but it’s actually zero everywhere, so it must be the zero polynomial, meaning all the coefficients are zero, and guess what’s the matrix containing the coefficients, it’s A^T A - I. So.

A^T A = I

A is orthogonal. Alright. Well, it turns out that its got to have a determinant of 1 or -1. Why? Well, because of some determinant rules that I don’t know how to prove.

1 = det I = det (A^T A) = det(A^T) det(A) = det(A) det(A)

So, det(A)² = 1, so it’s either 1 or -1.