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I'm learning about computer vision and came across this point that my instructor made in the material regarding image transformations.

He claimed that rotation has 3 degrees of freedom, but I'm having trouble wrapping my head around how that is. As far as I know, the rotation matrix is:

$$\mathbf{R} = \begin{bmatrix}\cos{(\theta)} & -\sin{(\theta)} \\ \sin{(\theta)} & \phantom{-}\cos{(\theta)}\end{bmatrix}$$

Doesn't this mean that we only have 1 degree of freedom, namely $\theta$?

I understand that a rigid (Euclidean) transformation (which is basically rotation + translation) has 3 degrees of freedom, but how does rotation have that?

I've also checked the Wikipedia page for rotation but it seems that rotation has 3 degrees of freedom when in $\mathbb{R}^3$. I also checked the Wikipedia page for the rotation matrix but it doesn't mention anything about degrees of freedom.

Sean
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Rotation in 2d about a fixed point has only 1 degree of freedom, while in 3d has 3 degrees of freedom. In general, in n dimensions you have $\frac{1}{2}n(nāˆ’1)$ degrees of freedom. I think the important point here is that when usually talking, we assume the origin point is fixed. You should clear this up with your instructor.

kHarshit
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