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Given any point x in R2, the point Qx is the point x rotated by some constant angle (what angle?) counter-clockwise about the origin. Such a matrix Q is called a rotation matrix for this reason.So the question just said to show that Q = (cos, -sin // sin cos) is orthogonal, and that x (in R2) and Qx are equidistant. Easy.
What does it mean here to say that Q acts as a rotation on R2?
I edited it in using complex numbers. It is clear though that it has to be some rotation for a given fixed x (at the very least a trivial one of angle 0) from the fact that the length is preserved, which means it's constrained to the circle it started on.Er... hmm how do I prove that there is a rotation then? I haven't been taught dot product yet (ignoring my 1 day spent on it during holidays).
I might quickly define it (the joys of second year maths in first year when they dump it on you)....Er... hmm how do I prove that there is a rotation then? I haven't been taught dot product yet (ignoring my 1 day spent on it during holidays).