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If you have an orthogonal matrix with a determinant of -1, how do you determine the plane of reflection?

Thanks

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find the -1 eigenvector. orthogonal to that –  Will Jagy Jan 15 at 23:13
    
@WillJagy mind putting that in a solution? –  rschwieb Jan 15 at 23:35
    
@rschwieb, sure. –  Will Jagy Jan 15 at 23:50

2 Answers 2

Any real eigenvalue is $\pm 1,$ and it is $-1$ as you are in dimension 3 with negative determinant. And there is a real eigenvalue because the dimension is odd, the degree pf the characteristic polynomial is odd.

So, there is a -1 eigenvector $v,$ if the matrix is called $M$ we have $Mv=-v.$

$M$ also preserves angles, easy enough to prove. So, the plane orthogonal to $v$ is setwise fixed. Note that $M$ may also cause a rotation within that plane.

For example, compare $$ M \; = \; \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right) $$

versus

$$ M \; = \; \left( \begin{array}{rrr} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right) $$

versus

$$ M \; = \; \left( \begin{array}{rrr} \frac{1}{\sqrt 2} & \frac{-1}{\sqrt 2} & 0 \\ \frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} & 0 \\ 0 & 0 & -1 \end{array} \right) . $$

Thought question, or called a "Gesundheit" experiment in German, what if $M=-I?$ Possibly Gedanken experiment, or Weltanschauung. Could be Schadenfreude.

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Pick your favorite vector, and run it through the transformation. The displacement from the original vector to the transformed vector is normal to the plane. Half the displacement from one to the other lies in the plane. Then it's straightforward to find the equation for the plane if you know the normal to the plane and a point in it.

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