[andrewla]

To a degree I think this is true, but it requires (at least in my experience) that you have an intrinsic grasp of trigonometry for it to make sense. If you have some complex function analysis and e^itheta then you can skirt the problem for a bit, but if you're like me and have to break out soh-cah-toa whenever you break down a triangle then this method ends up being pretty tedious too.

[Sharlin]

I’m not sure what you mean. Beyond rotation matrices, there’s really only trig involved in graphics if you actively want it.

[andrewla]

Maybe I was making unwarranted assumptions about the nature of your way to learn linear algebra. The approaches that I've seen invariably have to produce a sample matrix, and rotation is really the best example. The rotation matrix is going to have sines and cosines, and understanding what that means is not trivial; and even now if you asked me to write a rotation matrix I would have to work it out from scratch. Easy enough to do mechanically but I have no intuitions here even now.

[Sharlin]

Rotation matrices are somewhat mysterious to the uninitiated, but so is matrix multiplication until it "clicks". Whether it ever clicks is a function of the quality of the learning resource (I certainly do not recommend trying to learn linalg via 3D graphics by just dabbling without a good graphics-oriented textbook or tutorial – that usually doesn’t end well).

Anyway, I believe that it's perfectly possible to explain rotation matrices so that it "clicks" with a high probability, as long as you understand the basic fact that (cos a, sin a) is the point that you get when you rotate the point (1, 0) by angle a counter-clockwise about the origin (that's basically their definition!) Involving triangles at all is fully optional.

[srean]

In 2D there's an alternative. One can rotate purely synthetically, by that I mean with compass and straight edge. This avoids getting into transcendentals.

Of course I am not suggesting building synthetic graphics engines :) but the synthetic approach is sufficient to show that the operation is linear.