Given a simple rotation matrix: $$\mathbf{R}(\theta) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \text{cos}(\theta) & \text{sin}(\theta) \\ 0 & -\text{sin}(\theta)& \text{cos}(\theta) \end{bmatrix} $$
Why does $\mathbf{R}$ preserve the norm (2-norm) and angle between two vectors in $\mathbb{R} ^3$?
i.e. for two vectors $u \in \mathbb{R}^3$ and $v \in \mathbb{R} ^3$
$$\angle(u,v) = \angle (\mathbf{R}u, \mathbf{R}v)$$
I know this to be true simply working with rotation matrices, but how do you prove it? Would it suffice to start with a generic vector and simply follow the definitions?
That is:
1) Prove $\mathbf{R}$ is unitary by first showing each column of $\mathbf{R}$ is pairwise orthogonal
2) Show $\mathbf{R}$ is orthonormal (unitary) with the definition of norm. Would the 2-norm be used here?
Now here is where I start having trouble:
The norm of a vector rotated by $\mathbf{R}$ should have the same norm post rotation correct? Wouldn't proving $\mathbf{R}$ is orthonormal already ensure this? Or am I missing something?
Also, the angle between $u$ and $v$ is simply defined as $$\angle(u,v) = \text{cos}^{-1}\bigg(\frac{<u,v>}{||u||||v||}\bigg)$$
which should be equivalent to $\angle(\mathbf{R}u, \mathbf{R}v)$ since the norm of $\mathbf{R}$ is 1?
Any help would be greatly appreciated!
Since $\mathbf R$ is an orthogonal matrix, the map $v\mapsto \mathbf R.v$ preserves the scalar product; in particular, it prserves the norm. So, if $u,v\in\mathbb R^3$,$$\frac{\langle u,v\rangle}{\lVert u\rVert.\lVert v\rVert}=\frac{\langle\mathbf R.u,\mathbf R.v\rangle}{\lVert\mathbf R.u\rVert.\lVert\mathbf R.v\rVert}.$$That is, the angles are preserved.