Let $l(α)$ be a line through the origin, such that the angle between $l(α)$ and the $x$-axis is $α$.
Let $s_{l(α)}$ be a reflection in the line $l(α)$. And let $s_x$ be a reflectioin in the $x$-axis.
Let $r_α$ be clockwise α rotation.

By studying the element of $D_4$, I think that the following is true:

$$s_{l(α)}=r_{α/2}\circ s_x$$

Is this true in general? I'm trying to think of why this would make intuitive sense (visually).

I guess you can prove this using matrices, are there other ways ?

  • $\begingroup$ If it does the right thing to your two favourite vectors, which you might adjust for the situation, then it does the right thing to everybody. $\endgroup$ Mar 16, 2013 at 18:59
  • $\begingroup$ I think that you should use $r_{2\alpha}$ instead of $r_{\alpha/2}$. Think $D_4$: the composition $s_x\circ s_{\ell(\pi/4)}$ is a 90 degree rotation, right? Anyway, this is true. Good that you thought about this in this generality! As an exercise application of $r_{\pi/2}=s_{\ell(\pi/4)}\circ s_x$ (that you by multiplying your equation from the right by $s_x$) I invite you to give a different proof of the fact that the product of the slopes of two perpendicular lines equals $-1$. $\endgroup$ Mar 16, 2013 at 20:03
  • $\begingroup$ Proof? Oh! Follow André's suggestion. Check what happens to the vectors pointing in the directions of the two lines of reflection. $\endgroup$ Mar 16, 2013 at 20:05

1 Answer 1


$s_{l(α)}$ and $s_x$ are both isometries, so is $r = s_{l(α)}∘s_x$.

$s_{l(α)}$ and $s_x$ are reflections, then $^\det(s_{l(α)}) = \det(s_{l(α)}) = -1$. Therefore $\det(s_{l(α)} ∘ s_{l(α)}) = 1$ and $r$ is a rotation.

If a is the point where $Ox$ and $l(α)$ intersect (that is it, the origin), then it invariant by both reflections and $r(a) = s_{l(α)}∘s_x(a) = s_{l(α)}(a)=a$. So $r$ is a rotation around $a$=$O$.

The know the angle of the rotation, look at the image of the invariants of the first reflection : $r(Ox) = s_{l(α)}∘s_x(Ox) = s_{l(α)}(Ox)=l(2α)$.

Conclude that $s_{l(α)}∘s_x = r_{2α}$, which is what you were looking for because $s_x^2=1$.

It is true in any dimension and the converse is true: every rotation is a product of reflections. [E. Cartan]

  • $\begingroup$ Thanks. There are years I did use LaTex. $\endgroup$
    – AlainD
    Mar 17, 2013 at 14:54

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