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How would you prove the following proposition...

The inverse of $r_h$ is $r_h$

where $r_h$ is a horizontal reflection and we are in the Euclidean plane.

I was thinking something like this, but I don't think I have it quite right...

If $r_h$ is the reflection about the x-axis on the Euclidean plane, then for any point $(x,y)$:

$r_h(x,y)=(x, y)$ and $r_h(x,y)^{-1}=(x, y)$

Now if we apply the composition to an arbitrary point in the plane then

$r_hr_h^{-1}(x,y)=r_h(r_h^{-1}(x,y))=r_h(x,y)=(x,y)$.

Therefore, the inverse of $r_h=r_h$.

user6259845
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1 Answers1

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Indeed, what is stated in the text of the question is not "quite right"; to wit:

Reflection about the $x$-axis reverses the sign of the $y$-coordinate, but leaves the $x$-coordinate fixed, viz.

$r_h(x, y) = (x, -y) \ne (x, y) \; \text{if} \; y \ne 0; \tag 1$

indeed, the assertion

$r_h(x, y) = (x, y) \tag 2$

is tantamount to saying that

$r_h = I, \tag 3$

the identity mapping of $\Bbb R^2$; certainly this is not the case for reflection about the $x$-axis in the $xy$-plane as we know it.

To see that $r_h$ as defined by (1) satisfies

$r_h = r_h^{-1}, \tag 4$

simply observe that

$r_h^2(x, y) = r_h(r_h(x, y)) = r_h(x, -y) = (x, -(-y)) = (x, y), \tag 5$

that is,

$r_h^2 = I; \tag 6$

this leads immediately to

$r_h^{-1} = r_h^{-1}I = r_h^{-1}(r_h^2) = r_h^{-1}(r_h r_h) = (r_h^{-1} r_h) r_h = I r_h = r_h, \tag 7$

the result we seek.

Robert Lewis
  • 71,180