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Given a reflection matrix, e.g.:

$$A = \left( \begin{array}{ccc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{\sqrt{3}}{2} \end{array} \right)$$

how can I find the reflection line for that matrix? And for a 3x3 matrix, how can I find the reflection surface?

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  • $\begingroup$ Do you know eigenvalues and eigenvectors? $\endgroup$
    – Emilio Novati
    Oct 5, 2015 at 9:31
  • $\begingroup$ Thanks for your comment. So If I get an eigenvector for A, that must be the direction of the line correct? Then I can simply take the origin in $\mathbb{R}^2$ and go in the direction of the eigenvector to obtain the line of reflection? $\endgroup$
    – eager2learn
    Oct 5, 2015 at 9:31
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    $\begingroup$ yes, that works $\endgroup$
    – jorst
    Oct 5, 2015 at 9:35
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    $\begingroup$ @eager2learn No, the eigenvalues of a reflection matrix are $\pm 1$; more or less by definition, the $+1$-eigenvectors are precisely the vectors contained inside the reflection line (or plane), and the $-1$ eigenvectors are precisely those orthogonal to it. $\endgroup$
    – Travis Willse
    Oct 5, 2015 at 9:37
  • $\begingroup$ oops I was implicitely assuming you look at the eigenvalue 1, thanks for the correction! $\endgroup$
    – jorst
    Oct 5, 2015 at 10:01

2 Answers 2

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A reflection has eigenvalues which are either $-1$ and $1$. The dimensions of symmetry of reflection are the ones which are $1$ and the ones which are reflected are $-1$ Basically you can write them in this way:

$A = V^{-1}DV$ where $D$ is diagonal and the columns $V_{:,k}$ are the corresponding vectors which are either left alone or reflected (depending on if $D_{kk}$ is 1 or -1).

In three dimensions we just have 2 times as many combinations, each of the three values could be either 1 or -1, but the same principle holds.

  1. If one $-1$, then there is a plane which the vectors are reflected in.
  2. If two $-1$ then there is a "thread" or "uncooked spaghetti" of reflection around.
  3. If three $-1$ then each dimension is flipped 180 degrees.
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Hint: a vector on the reflection line is not changed by the transform. That is, $Ax=x$.

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