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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:

$$ \cos \alpha = \frac{\cos(\theta)+\cos(\tilde \theta)}{\lambda(\theta,\tilde \theta)}, \sin \alpha = \frac{\sin(\theta)+\sin(\tilde \theta)}{\lambda(\theta,\tilde \theta)}$$

Asaf Shachar
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    “Scaled rotation” matrices are also known as conformal matrices. Under multiplication, they’re isomorphic to complex numbers. Perhaps this isomorphism carries over to addition as well? – amd Jan 02 '17 at 21:35
  • You are right of course. I think the isomorphism does carriy over to addition. – Asaf Shachar Jan 02 '17 at 21:40
  • For me the easiest way to visualize this is to consider what happens to the unit vector $(1,0)$, because its image determines how every other point transforms as well. Addition of transformations is then simply vector addition. (Of course, the image interpreted as a complex number is precisely the complex number the transformation is isomorphic to.) –  Jan 03 '17 at 01:14

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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.

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amd
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  • You are right, of course! The key point is that these matrices are isomorphic to $\mathbb{C}$ as $\mathbb{R}$-algebra. – Asaf Shachar Jan 03 '17 at 09:50
  • One more thing: Every scaled rotation is conformal, but not vice versa, take for example $\pmatrix{1&0\0&-1}$. Scaled rotations are exactly orientation preserving conformal matrices (i.e positive determinant). – Asaf Shachar Jan 04 '17 at 11:11
  • @AsafShachar Good point. The definition of conformal that I had learned was more restricted, but I see now that it includes all angle-preserving maps. I would say non-negative determinant, though. You can’t have a vector space without a zero, and we’re still using normal matrix addition for this space. – amd Jan 04 '17 at 17:18
  • You are right. Actually, after talking with a colleague, it turns out that perhaps the convention that conformal maps are orientation preserving is more popular than the other. Doesn't really matter anyway... Thanks again. – Asaf Shachar Jan 04 '17 at 18:00