Recently I have learned that scaled rotations in $2$ dimensions form a vector space.
One way to see this is to verfiy that a $2 \times 2$ matrix $A$ is a scaled rotation if and only if $\operatorname{cof} A=A$ where $\operatorname{cof} A$ is the cofactor matrix of $A$. (Note that $\operatorname{cof} A$ is linear in $A$).
However, I find it hard to visualise that a sum of rotations is again (a scaled) rotation. (I do not have a "feeling" to what the resulting angle will be as a function of the original angles).
If we denote the original angles by $\theta, \tilde \theta$, and by $\alpha$ the resulting angle, then (see calculation below) $$ \cos \alpha = \frac{\cos(\theta)+\cos(\tilde \theta)}{\sqrt{2\big(1+\cos(\theta-\tilde \theta)\big)}}, \sin \alpha = \frac{\sin(\theta)+\sin(\tilde \theta)}{\sqrt{2\big(1+\cos(\theta-\tilde \theta)\big)}}$$
Question: Is there a nice way to visualizse this phenomena?
I guess we can plot $\alpha(\theta,\tilde \theta) = \arctan(\frac{\sin(\theta)+\sin(\tilde \theta)}{\cos(\theta)+\cos(\tilde \theta)})$ in some domain, but this wil not give the whole behaviour since $\arctan$ has a domain which is limited by $\pi$.
Another relevant question is what is the dimension of the space of scaled rotations (which we denote by $W$)? How does a basis for $W$ looks like?
It is trivial to see that $\dim W \ge 2$ (there are two linearly independent rotations). Since not every matrix is a scaled rotation, we also get $\dim W \le 3$.
Calculations:
Consider the following: Let $ R_{\theta} = \begin{pmatrix} c & -s \\ s & c \end{pmatrix}, R_{\tilde \theta} = \begin{pmatrix} \tilde c & -\tilde s \\ \tilde s & \tilde c \\ \end{pmatrix}$ be two rotation matrices. We exrpess their sum as a scaled rotation:
$$ \begin{pmatrix} \tilde c + c & -(\tilde s + s) \\ \tilde s+s & \tilde c+c \\ \end{pmatrix} = R_{\theta} + R_{\tilde \theta} = \lambda R_{\alpha}= \lambda \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$$
Then $$ \lambda^2 = (\lambda a)^2+(\lambda b)^2= (\tilde c + c)^2 + (\tilde s + s)^2=2(1+c\tilde c+s \tilde s).$$
Noting that $c\tilde c+s \tilde s=\cos(\theta)\cos(\tilde \theta)+\sin(\theta)\sin(\tilde \theta)=\cos(\theta-\tilde \theta)$, we get that
$$ \lambda(\theta,\tilde \theta) = \sqrt{2\big(1+\cos(\theta-\tilde \theta)\big)}$$
This makes sens! The closer the angles are to each other, the behaviour is more like "constructive interference" so the scaling gets larger.
We can also obtain:
$$ \cos \alpha = \frac{\cos(\theta)+\cos(\tilde \theta)}{\lambda(\theta,\tilde \theta)}, \sin \alpha = \frac{\sin(\theta)+\sin(\tilde \theta)}{\lambda(\theta,\tilde \theta)}$$
The condition $A=\operatorname{cof}A$ means that we’re dealing with matrices of the form $\pmatrix{a&-b\\b&a}$, the so-called conformal matrices. These matrices can be identified with the complex numbers, and addition of these matrices is good old vector addition in $\mathbb R^2$ in disguise. If you carry through with the computations that you’ve started, you’ll have reconstructed polar formulas for vector/complex-number addition. (There’s a nice discussion of such formulas in this question.)
A natural basis for this space is $I=\pmatrix{1&0\\0&1}$, $J=\pmatrix{0&-1\\1&0}$, which correspond to $1$ and $i$, respectively.