How are rotation matrices considered linear mapping?

Question

This may sound like a dumb question, but its been bothering me for a while. We know that trigonometric functions are non linear functions. However, why are rotation matrices (and transformation matrices (such as affine)) considered linear mapping, since they are essentially made up of trigonometric entities. While from a higher level, the rotation matrices do satisfy the constraints that define what a linear mapping is (i.e. it preserves addition and scalar multiplication, additionally the rotation operation is invertible), however, when thinking of a perspective of building up from the individual trigonometric elements, I fail to understand how the non-linearity suddenly becomes a linear mapping, from an intuitive sense.

Note, I do understand that rotation by itself is a linear mapping, but I am more confused about the linearity of the rotation matrix while building up from trigonometric elements. Or, has my understanding been completely wrong and rotation cannot be considered linear mapping? Thanks

Answer 1

Given any four numbers $a,b,c,d\in\Bbb R$, the map$$\begin{array}{ccc}\Bbb R^2&\longrightarrow&\Bbb R^2\\(x,y)&\mapsto&(ax+by,cx+dy)\end{array}$$is linear. And, for every real number $\theta$, the numbers $\cos\theta$ and $\pm\sin\theta$ are real numbers. So, in particular, the map$$\begin{array}{ccc}\Bbb R^2&\longrightarrow&\Bbb R^2\\(x,y)&\mapsto&\bigl(\cos(\theta)x-\sin(\theta)y,\sin(\theta)x+\cos(\theta)y\bigr)\end{array}$$is linear. And it turns out that thse maps are precisely the rotations of $\Bbb R^2$ and that its matrix with respect to the standard basis is $\left[\begin{smallmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{smallmatrix}\right]$. The fact that the numbers $\cos\theta$ and $\pm\sin\theta$ were obtained from the trigonometric functions $\cos$ and $\sin$ doesn't change that.

Answer 2

To be somewhat explicit about it, a rotation matrix $\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$ is a linear map

from $\mathbb R^2$ to $\mathbb R^2$, which maps $\begin{bmatrix}x\\y\end{bmatrix}\mapsto\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}x\cos\theta-y\sin\theta\\x\sin\theta+y\cos\theta\end{bmatrix}$ linearly,

even though $\theta\mapsto \cos\theta$ is not a linear map from $\mathbb R$ into $\mathbb R$; think of $\theta$ as a parameter.

Answer 3

In fact, $\left[\begin{smallmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{smallmatrix}\right]$ isn't an actual rotation in case $\theta$ is a fixed number.

As an example, a window's wing may be rotated by $90^\circ$ to open it. Then the actual rotation, i.e., the act of opening it, isn't described by $\left[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right]$. Instead you have to imagine to look at the window, then close your eyes while someone opens the window and then open your eyes again to see the result.