rotation matrix in $\mathbb{R}^{2}$

Question

As a preparation for graduate entrance exam and mostly for fun, I am reading a linear algebra textbook.

It's been quite a while since I read that material last time.

So today, I got stuck at the part of deducing the 2 by 2 matrix which rotates a vector (x, y) to (x', y') by some angle $\theta$.

Thinking about it for some half an hour, I found out that polar coordinates do it.

My question is, then, is there any way to prove it not using polar coordinates or complex numbers?

That is, (x',y') is related to (x, y) by \[ x'= \cos \theta x - \sin \theta y, y' = \sin \theta x + \cos \theta y. \]

Is there any way to show this by drawing lines? (Or compass or ruler or whatever).

If you know that the rotation must be represented by a matrix, then it's enough to check the images of $(1,0)$ and $(0,1)$ under the rotation. These must be the columns of the matrix. The fact that any isometry $T$ fixing the origin is a linear mapping is actually geometrically obvious. It's clear that we must have $T(v + w) = T(v) + T(w)$ and $T(\lambda v) = \lambda T(v)$. — user49640, Jun 10 '17 at 06:08
Yes since the rank is 2 it is enough to check two linearly independent vectors map the way you want then every linear combination must do so to (because it is a linear map). — mathreadler, Jun 10 '17 at 06:15
There are a few geometric proofs in the answers to prove geometrically that $\cos(a + b) = \cos(a) \cos(b) − \sin(a) \sin(b)$, see also blue's answer to geometric proof of $2\cos A \cos B = \cos(A+B) + \cos(A−B)$. — Andrew D. Hwang, Jun 10 '17 at 09:41

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