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Consider a $2 \times 2$ matrix that reflects about a line through the origin that makes an angle $\theta$ with the x-axis: $$L_{\theta} = \begin{bmatrix} cos(2\theta) & sin(2\theta) \\ sin(2\theta) & -cos(2\theta) \end{bmatrix}$$

This can also be thought of as the composition of three matrices. First a rotation matrix by $-\theta$, then a reflection about the x-axis, then a rotation by $\theta$ back.

Give a non-geometric proof that shows the composition of $L_{\theta}$ and $L_{\theta'}$ is a just another rotation about the origin.

Of course, we can simply multiply the two matrices, and use trig identities to show this. And geometrically it makes perfect sense since we're reflecting about a line twice. But I need to give a non-geometric proof.

I see that the reflection matrix $$\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$$ in the middle of the composition can be "cancelled out", since two of these is just $I_2$, but I'm not 100% certain how to show this.

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  • $\begingroup$ Hint: a rotation matrix is orthogonal and orientation-preserving. $\endgroup$ – amd Jun 1 '18 at 23:58
  • $\begingroup$ It might be a translation sometimes $\endgroup$ – user106688 Jan 15 '20 at 2:57
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One possible proof could be;

Since reflections and rotations are all orthogonal, reflections have determinant $-1$, and rotations have determinant $1$, then the product of two reflections is the product of two orthogonal matrices, hence it is orthogonal, and since $$\det(AB) = \det(A)\det(B)$$ then the determinant will be $(-1)(-1) = 1$, so the product of two reflections is an orthogonal matrix with determinant $1$, therefore it is a rotation.

However I would say that your proof of multiplying the matrices together and using trig identites certainly counts as a "non-geometric" proof. It just requires algebraic manipulation, not any geometric arguments.

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  • $\begingroup$ well "reflections have determinant −1, and rotations have determinant 1" isn't exactly obvious. $\endgroup$ – B.Li Jun 2 '18 at 1:28
  • $\begingroup$ Since you know the form of a general rotation matrix and a general reflection matrix, try taking the determinant and see what you get. $\endgroup$ – Kaleb R. Jun 2 '18 at 1:38
  • $\begingroup$ Although that doesn't prove the other direction, i.e. that any orthogonal matrix with det 1 is a rotation which is maybe a little more involved. $\endgroup$ – Kaleb R. Jun 2 '18 at 1:46
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    $\begingroup$ So if you don't already know that any orthogonal matrix with determinant 1 gives a rotation, maybe this way is more work than just working out the trig identities. But you end up also knowing that orthogonal and det 1 $\implies$ rotation, which is a very useful fact itself. $\endgroup$ – David K Jun 2 '18 at 14:34

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