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Suppose we have some real numbers $x_1, x_2, ..., x_n$. If each number represents an angle, and therefore a point on the unit circle with co-ordinates $(\sin x_i, \cos x_i)$, the centroid of all these points is \begin{equation} C = \left( \frac{1}{n} \sum_{i=1}^{n} \cos x_i , \frac{1}{n} \sum_{i=1}^{n} \sin x_i\right) \end{equation} Let $G$ be the angle made by $C$ with the x axis. Now suppose I rotate each of these points about the origin by some $\alpha$ such that my new centroid is \begin{equation} C' = \left( \frac{1}{n} \sum_{i=1}^{n} \cos (x_i - \alpha) , \frac{1}{n} \sum_{i=1}^{n} \sin (x_i - \alpha)\right) \end{equation} If $G'$ is the angle made by $C'$ with the x axis, how can I prove that $G' = G - \alpha$? That is, how can I prove that the rotation i've applied uniformly to all of my individual points is also applied to the centroid?

First of all, I'm aware that this makes perfect intuitive and geometric sense. If I treat all of my points as a figure, then rotate that figure by some value, then of course the centroid will rotate equivalently. I understand that it should be, and is true, but I don't really know how to show it mathematically (without using a visual proof).

Note also that this is part of a proof of a larger problem that i'm trying to understand. The proof in question involves using the rotation of this centroid to prove the result and offers the following as justification for this step: "$G'$ is the rotated position of $G$ (because rotation is a linear operation)." So perhaps there is a straightforward proof of this that involves using rotation matrices.

I've attempted to work this a few different ways but I keep getting stuck so I have a feeling that my approach is just generally off. Any help would be hugely appreciated.

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  • $\begingroup$ I'm not sure whether I understand the penultimate paragraph correctly. Is the proof that you're looking for allowed to use the fact that a rotation is a linear transformation? Would you be satisfied with a proof that shows more generally that the centroid of linearly transformed points is the linearly transformed centroid? Or are you looking concretely for an algebraic proof of $G'=G-\alpha$ based on your expression for $C'$? $\endgroup$ – joriki Sep 10 '18 at 19:14
  • $\begingroup$ @joriki I would absolutely be satisfied with a proof based on the fact that rotation is a linear transformation. I guess i'm just looking for something a bit more detailed than "rotation is a linear operation" -> QED. $\endgroup$ – zulu1310 Sep 10 '18 at 19:37
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I hope I'm interpreting your reply to my comment correctly in offering this proof:

The centroid of $n$ vectors $x_i$ is $\frac1n\sum_{i=1}^nx_i$. If we apply a linear transformation $f$ to the vectors $x_i$, the centroid of the transformed vectors is

$$ \frac1n\sum_{i=1}^nf(x_i)=\frac1nf\left(\sum_{i=1}^nx_i\right)=f\left(\frac1n\sum_{i=1}^nx_i\right)\;, $$

where both equalities use the linearity of $f$. Thus, the centroid of the transformed vectors is the transformed centroid.

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