I already know how to find the matrix of reflection over the $x$-axis (in the plane) (ON-base). By looking at what happens to the standard basis vectors $\hat{e}_1 = (1,0)$ and $\hat{e}_2 = (0,1)$. The matrix for the transformation will be: $$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$

That's an easy example. But I wan't to find the matrix for the reflection over a line that goes through the origin and makes the angle $\pi/17$ with the x-axis.

How can I approach this problem?

  • 1
    $\begingroup$ Exactly the same way. $\endgroup$ – saulspatz May 3 '18 at 13:04

The same way. Look at the standard basis and find out the images of the standard basis under your transformation. These should involve sines and cosines of $\pi/17$...


OK, let's reflect $(1,0)$ over that line. Notice in a sense you would be just rotating the point by $2\pi/17$ radians around the origin, can you figure out the result?

Similarly, from $(0,1)$ to the line is $\pi/2 - \pi/17 = 15\pi/34$ radians, so the result after reflection is a negative rotation of $2 \cdot 15\pi/34$ radians from the $\pi/2$ angle, resulting in the final angle of$$\frac\pi2 - 2 \cdot \frac{15\pi}{34} = \frac{-13\pi}{34}.$$

Can you take it from here?

  • $\begingroup$ Could u maybe help me out more? The reflection over the x-axis is easy because the x - value will be the same but y will be -y. But the other one is harder .. $\endgroup$ – Fanny May 3 '18 at 13:14
  • $\begingroup$ @Fanny please see update $\endgroup$ – gt6989b May 3 '18 at 13:21
  • $\begingroup$ okej! I think I got it. thank you. $\endgroup$ – Fanny May 3 '18 at 13:28

Hint: make that reflection as a composition of the reflection you already have and a rotation. Alternatively, go at it the same way you already did: find the images of $\mathbf e_1$ and $\mathbf e_2$.


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