1

If a linear transformation swaps two axes is said to perform a reflection and the determinant will be negative. Testing for the sign of the determinant can tell me whether a reflection has happened.

What if we swap two pairs of axes instead? I understand that each swap will change the sign of our determinant. So, does this mean the given linear transformation is no longer performing a reflection? I'm confused…

Atcold
  • 249
  • 2
    The composition of two reflections with respect to hyperplanes (=$(n-1)$-dimensional subspace of $\Bbb{R}^n$) is a rotation in the plane of the normals of the two hyperplanes. If the angle between the two normals is $\theta$, then the rotation angle of the composition is $2\theta$. I once prepared an animation trying to demonstrate exactly that. In the animation the $(n-2)$-dimensional intersection of the two hyperplanes is squashed to a point. Anyway, Suzu Hirose got it right, too (+1 to all). – Jyrki Lahtonen Jul 07 '22 at 21:22
  • I fail to understand the animation… How are the two reflections composed? I see how the red vector is reflecting both across the green and blue direction, but I'm a little clueless otherwise. – Atcold Jul 08 '22 at 18:14
  • 1
    The orange vector is the original. The red vector is its reflected image with respect to the green line. The black vector is what you get when you reflect the red vector (in turn) with respect to the blue line. So the passage from the orange vector to the black vector is the composition of two reflections. The animation seeks to convery that the angle between the orange and the black vectors stays the same even when we vary the orange vector. In other words, the composition of those two reflections is a rotation. Sorry about not making all that clear. – Jyrki Lahtonen Jul 08 '22 at 18:19
  • Crystal clear now. Thanks. – Atcold Jul 11 '22 at 14:53

1 Answers1

2

If you swap two pairs of axes, what you end up with is a $180^\circ$ rotation around another axis. For example, if we reflect the $y$ value and the $z$ value as follows, it's the same as a rotation by $\pi$. $$ \begin{align} \left( \begin{matrix} 1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right) \times \left( \begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & -1 \\ \end{matrix} \right) &= \left( \begin{matrix} 1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end{matrix} \right)\\ &= \left( \begin{matrix} 1 & 0 & 0\\ 0 & \cos\pi & \sin\pi \\ 0 & -\sin\pi & \cos\pi \\ \end{matrix} \right) \end{align} $$ The general result is that the composition of two reflections is a rotation. You can find various results and demonstrations by searching for that.

Clearly the determinant of the above example is positive, and the general result is that the composition of two reflections is not a reflection.

Atcold
  • 249
Suzu Hirose
  • 11,949
  • 1
    I see, it makes sense. Thanks! Is there a $-$ missing in front of the top right $\sin \pi$? – Atcold Jul 08 '22 at 18:08
  • My original question comes from trying to understand how SVD may swap multiple left singular vectors in order to align them to the sorted list of singular values. I think I need to ponder a little longer, he he – Atcold Jul 08 '22 at 18:17
  • 1
    @atcold yes minus sign was missing! – Suzu Hirose Jul 08 '22 at 20:42
  • Ah, I see what's going on… When I think about ‘swapping a pair of axes’ I see it as flipping one and then rotating by $\frac{\pi}{2}$. And I understand that ‘swapping a pair of axes twice’ undoes the reflection. My confusion arises when (if it's even possible) one ‘swaps two distinct pairs of axes’, say, in a 4-dimensional space or larger. I believe I have to reason in terms of single-axis inversion rather than swaps. Does what I'm saying make any sense? I think my imagination is broken, LOL. – Atcold Jul 11 '22 at 18:18