How can I show that this matrix is a flip

Question

I am trying to show that

$$ F = \left(\begin{array}{cc} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{array} \right ) $$

is a flip about a line through the origin.

What I tried: Let $v \in \mathbb R^2$, $v = (r \cos \varphi, r \sin \varphi)$. Then

$$ vF = (r \cos (\varphi - \theta), -r \sin (\varphi - \theta) )$$

I notice that if $\theta = 2 \varphi$ then $v$ maps to itself. From this I conjecture that this is a flip about the line through the origin of angle $\theta / 2$.

But there I have two problems:

(1) I don't know how to prove this conjecture. Please can you help me?

(2) I computed $(1,0)F = (\cos \theta, \sin \theta)$. If it was a reflection around the line $\theta$ shouldn't $(1,0)F = (\cos 2\theta, \sin 2\theta)$?

@WillJagy How can I use eigenvectors to show that it is a flip? — learner, Dec 06 '14 at 01:50

Winther · Accepted Answer · 2014-12-06T01:48:26.173

1

To start with, a more convenient way to write $vF$ is by using $\cos(x)=\cos(-x)$ and $\sin(x)=-\sin(-x)$ to find $$vF = (r\cos(\theta-\varphi) , r \sin(\theta - \varphi))$$

Now, a line given by an angle $\beta$ reflected about a line with angle $\alpha$ gives a line with angle $\beta + 2(\alpha-\beta) = 2\alpha - \beta$ (draw a sketch). This gives the equation

$$2\alpha - \varphi = \theta - \varphi \to \alpha = \frac{\theta}{2}$$

edited Dec 06 '14 at 01:48

answered Dec 06 '14 at 01:42

Winther

24,478

What about (2)? – learner Dec 06 '14 at 01:51
Wait, I don't want to rotate lines, I want to reflect them. – learner Dec 06 '14 at 01:52
@learner I meant reflected. For 2) Note that $2\cdot(\theta/2) = \theta$ which is what we expect. Or another way to say it: it is not a reflection about a line with angle $\theta$, but rather $\theta/2$. – Winther Dec 06 '14 at 01:53
Wonderful, thank you. – learner Dec 06 '14 at 02:28

How can I show that this matrix is a flip

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