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Question:

Determine the matrix for reflection in the line 2 − = 0 in ℝ2

Hint: split this transformation into simpler ones, and combine the result using matrix multiplication.

I thought about it this way.

y = 2x.

If I scale all y values down by 1/2 with the matrix,

\begin{pmatrix} 1 & 0 \\ 0 & 1/2 \\ \end{pmatrix}

And do reflection as if y=x,

\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}

And scale the y values back up by 2,

\begin{pmatrix} 1 & 0 \\ 0 & 2 \\ \end{pmatrix}

If I multiply the matrices in the same order, I should get the reflection matrix for the line 2x-y=0 However, the result I had below was not correct.

\begin{pmatrix} 0 & 2 \\ 1/2 & 0 \\ \end{pmatrix}

What is wrong about my approach?

Emre
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  • See https://math.stackexchange.com/questions/525082/reflection-across-a-line – A. Goodier Oct 30 '17 at 19:52
  • Hans Lundmark’s answer identifies one of the mistakes you’re making. Try rotations instead. – amd Oct 30 '17 at 20:35
  • @amd What do you mean by trying rotation, is it relevant with this question? – Emre Oct 30 '17 at 20:49
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    You tried decomposing the reflection into $M^{-1}RM$, where $R$ is a simple reflection in a coordinate axis, with $M$ as a scaling transform. That fails for the reason given by Hand Lundmark. Rotations don’t affect relative angles, though, so try finding a suitable rotation to use as $M$ instead. – amd Oct 30 '17 at 20:55

2 Answers2

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Scaling one axis but not the other will distort angles.

(Also, your result is obviously wrong, since you didn't get an orthogonal matrix.)

Hans Lundmark
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I don’t know why you are making it more difficult. You can use the projection of the unit vectors $(1,0)$ and $(0,1)$ onto the $(1,2)$.

the image of $x$ onto $a$ is $a' = 2(a^Tx)a/(a^Ta) - x$. Using this in your problem, for the image of $x = (1,0)$, $a = (1,2)$. So we have $a^tx = 1$, $a^T a = 5$ which gives $a' = 2/5(1,2) - (1,0) = (-3/5, 2/5)$. This $a'$ is your first column of the reflection matrix. To find the second column, find $(0,1)'$.

Rócherz
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abel
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  • The OP isn’t the one making it “more difficult.” It’s whoever created the assignment :) – amd Oct 30 '17 at 20:57