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I have this problem that says: Prove that in the plane, every rotation about the origin is composition of two reflections in axis on the origin.

First I have to say that this is a translation, off my own, about a problem written in spanish, second, this is the first time I write a geometry question in english.

I don't know how to prove this, so I made a few drawings, but I believe I got more confused. I put a point P in the plane and then rotate it $\theta$ from the X axis and got $P_\theta$, I assume that what the problem wants is to get from P to the same $P_\theta$ but with two reflections, this is what I don't understand, why do we need two? if we bisect the angle that P and $P_\theta$ formed then we get an axis that works as the axis of reflection, then we don't need two, but one to get the same point.

Please tell me what's going on here.

I know that we can see rotations and reflections as matrix, should I try to multiply two reflections with different angles and then see if I can rewrite the result as a rotation?

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  • $\begingroup$ Your angle-bisecting reflection only works for a specific vector. I think you want a pair of reflections that work for every vector. $\endgroup$
    – Evan
    Commented Sep 8, 2013 at 4:55
  • $\begingroup$ Rephrasing what Evan is saying: you need to compose two reflections to get a rotation of the entire plane. You cannot achieve a rotation of the plane via one reflection. $\endgroup$ Commented Sep 8, 2013 at 5:17
  • $\begingroup$ @proximal ok, maybe I didn't understood well the problem, I thought that if a had a random point P first I rotate it generating $P_\theta$, then I had to prove that I can get the same point $P_\theta$ with two reflections, so we had to look for them, but what you're telling me is that first I have to compose the two reflections, wich just means to reflect once and then the other right?, then prove that they originate all the points in the plane? $\endgroup$
    – Ana Galois
    Commented Sep 8, 2013 at 22:13
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    $\begingroup$ @AnaGalois Let $R_\theta$ be the rotation that rotates every point about the origin by the angle $\theta$. You are being asked to find two reflections $T$ and $S$ about the origin such that their composition is equal to $R_\theta$; that is, $T\circ S=R_\theta$. It's easy to find two reflections whose composition only takes $P$ to $P_\theta$, but a bit harder to find reflections whose composition rotates every point. $\endgroup$ Commented Sep 10, 2013 at 2:50

2 Answers 2

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Please see this diagram. Lines $m,n$ are normals to reflexive axes with the angle between them $\frac\theta2$. Then $v''$, which is reflected twice by $m,n$ is such a vector rotated $\theta$ from the original vector $v$.

And I think this has also an algebraic explanation in geometric algebra. The reflection of $v$ by the axis $n$ is represented as $v'=-nvn$. Then $v''=-mv'm=-m(-nvn)m=(mn)v(nm)=RvR^\dagger$, where $R=mn$ and $R^\dagger$ is reverse of $R$. We speak of $R$ is rotor of angle $\theta$ if $m\cdot n=\cos\frac\theta2$. $RvR^\dagger$ is exactly the expression of a rotation in geometric algebra.

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If you draw a circle around the origin, and then reflect a point in two straight lines at an angle $\theta$, the point rotates $2\theta$.

Multiplying the matrices will show this.

While one can produce a rotation by two mirrors, not every rotation implies the existence of two mirrors. Rotating things by 120 deg will produce three images, not six.

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  • $\begingroup$ I'm sorry, what do you mean by "mirrors"? the reflections? I tried to draw what you said, but I don't get it. $\endgroup$
    – Ana Galois
    Commented Sep 8, 2013 at 22:02
  • $\begingroup$ This is similar to what wikipedia says. Reflection matrix can be got using this process $\endgroup$
    – An5Drama
    Commented Jan 5 at 7:07
  • $\begingroup$ Saying the wikipedia process more specifically. Assume $L_2$ is at the counterclockwise direction of $L_1$. (The following allows angle to be negative which means the opposite direction of the positive one). Assume $\angle POL_1=\alpha>0$ (This means $L_1$ is at the counterclockwise direction of $P$) then $\angle L_2OP'=\alpha-\theta$ Then $\angle P''OL_1=\alpha-2\cdot \theta$. So $\angle POP''=2\cdot \theta$ $\endgroup$
    – An5Drama
    Commented Jan 5 at 7:15
  • $\begingroup$ Proof link in wikipedia which is a bit from this answer is similar to this QA $\endgroup$
    – An5Drama
    Commented Jan 5 at 7:56

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