# finding a plane of reflection of a matrix

• Dec 14th 2010, 09:15 AM
worc3247
finding a plane of reflection of a matrix
How would you going about finding the plane of reflection of the matrix:
$
\frac{1}{3}
(\begin{array}{ccc}
2 & 2 & -1 \\
2 & -1 & 2 \\
-1& 2 & 2 \end{array})]
$

Not looking for an answer, just looking for a method of finding the plane. Thanks!
• Dec 14th 2010, 09:21 AM
snowtea
If the matrix is a reflection matrix, what happens when it transforms vectors in the reflection plane?
If you know 2 vectors in the reflection plane, can you find the plane?
If you transformed v into v', is v + v' in the reflection plane?
• Dec 14th 2010, 03:35 PM
worc3247
Not really sure about the first bit. As for the second, you can't find the plane only knowing 2 vectors, you would need a third that is not collinear.
For the last bit, yes v + v' would be in the reflection plane, but I don't see how this helps, sorry. :(
• Dec 14th 2010, 05:36 PM
snowtea
2 non-parallel vectors is enough to determine the reflection plane (since the plane needs to contain the origin (0,0,0)).
The plane consists of all points that are the linear combinations of the 2 vectors.

Pick two different vectors v, w. Transform them into v', w'. If v + v' and w + w' are two non-parallel vectors in the plane, then you have your plane. If they happen to be parallel, you can also use this to determine the plane in another way (draw a diagram).
• Dec 14th 2010, 10:30 PM
snowtea
This may be an easier approach.

For a vector V not in the reflection plane, and it's reflected vector V'.
The difference V' - V is normal to the plane.
• Dec 15th 2010, 06:38 AM
worc3247
Awesome got it!
• Dec 15th 2010, 07:56 AM
TheGreenLaser
I believe you could also find an eigenvector associated with the eigenvalue -1. This would give you a normal to the plane, because if you reflect a vector that's normal to the plane about that same plane, the resulting vector will be -1 times the original vector. That's how I've been taught to solve this, but I think snowtea's way might be easier.