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Ok so I have two matrices:

The reflection about the line y=x.

$$A=\pmatrix{0 & 1 \\ 1 & 0}$$

And the reflection about the line $y=0$.

$$B= \pmatrix{ 1 & 0 \\ 0 & -1} $$

I need to show that both $AB$ and $BA$ represent rotations of $\mathbb{R}^2$. I know that they are rotations of 90 degrees in opposite directions, but how can I show this? I don't think giving a few examples is enough.

Also, if I need to compute $ABABABAB$ and $BABABABA$, can I just bunch them up as $(AB)(AB)AB)(AB)$ and $(BA)(BA)(BA)(BA)$?

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    $\begingroup$ It might be a good idea to look around for tutorials on how to use MathJaX/LaTeX formatting so that we can understand your question better. $\endgroup$ Commented Feb 18, 2017 at 18:32
  • $\begingroup$ I've made some changes, thanks for the help @TheCount $\endgroup$ Commented Feb 18, 2017 at 18:36
  • $\begingroup$ Also thanks @pjs36 $\endgroup$ Commented Feb 18, 2017 at 18:37
  • $\begingroup$ Why don't you just work out AB and BA? They are rotations by 90 degrees... $\endgroup$ Commented Feb 18, 2017 at 18:40
  • $\begingroup$ ...so $(AB)^4=I=(BA)^4$ $\endgroup$ Commented Feb 18, 2017 at 18:48

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$$AB = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \ \ \textbf{and}\ \ BA=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$

Now take any vector $\textbf{x} = (u,v)$ then $AB(u,v) \cdot (u,v) = 0$ and $BA(u,v) \cdot (u,v) = 0$. Hence what can you say about the matrices since the vectors they give are always perpendicular? Once you see what their rotations correspond to, you can easily compute the matrix order.

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  • $\begingroup$ The scalar product establishes the angle, but how do we get the direction of the rotation in a rigorous way? $\endgroup$ Commented Feb 18, 2017 at 18:54
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    $\begingroup$ Easy. Consider the inclusion of the vector $AB(u,v)$ into 3-space i.e $(AB(u,v),0)$. Then take the determinant of the matrix: $[(AB(u,v),0) \ ((u,v),0) \ ((AB(u,v),0) \times ((u,v),0)]$. If positive then you have the right-handed system and so $[AB(u,v) \ (u,v)]$ would correspond to the counter-clockwise orientation i.e $AB(u,v)$ rotates in the counterclockwise direction. $\endgroup$ Commented Feb 18, 2017 at 19:12
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If you know about rotation matrices $R(\theta)$, showing that $M$ is a matrix boils down to finding an angle $\theta$ such that $M=R(\theta)$. Of course here $M$ stands for $AB$ or $BA$.

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