: Conflict between reflection matrix and rotation matrix Consider the following matrix used for reflection. | 0 -1 | | -1 0 | This matrix produces the reflection across y=-x

@Cugini998

Not every reflection is possible to rotate. In this case you'll notice that a 90 degree rotation produces:

| 0 -1 |
| 1 0 |


There is not in fact any value for x that satisfies the equation*

-sin(x) = -1
sin(x) = -1


Any two (or multiples thereof 2,4,6 etc...) reflections can be expressed as one rotation tough. See Euclidean Spaces article. Mainly, because twice reflected is right way around which can all be represented as a rotation. So the space of x time mirrored objects is twice as big as the total space of rotations.

* This is easy to experiment, no matter how you turn your hand there's no way your left and right hand aligns. This is also easy to visualize. If we plot both sin(x) and -sin(x), then if their lowest point ever meets that's the solution.



Image 1: Plot of Sin(x) and its negative pair never meet at the lowest point. Because the functions are in fact in antiphase.

There is a rotation matrix that is different handed that can describe the mirrored image, it is however eqivalent of a mirror operation so it should not count.

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