Orthogonal 3x3 matrix with determinant -1

• Mar 22nd 2011, 08:28 PM
Pinkk
Orthogonal 3x3 matrix with determinant -1
What does such a matrix do geometrically? I'm having a hard time visualizing this; I used some basic examples but I am not seeing a general pattern. Any help would be appreciated.

Edit: From the examples I've seen it looks kinda like a glide reflection, but again, I'm having a hard time coming up with the general case.
• Mar 22nd 2011, 09:19 PM
Drexel28
Quote:

Originally Posted by Pinkk
What does such a matrix do geometrically? I'm having a hard time visualizing this; I used some basic examples but I am not seeing a general pattern. Any help would be appreciated.

Orthogonal matrices act as isometries...since you took a course in geometry I know you you know what this means. Now, what do you think the negative determinant corresponds to?
• Mar 22nd 2011, 09:35 PM
Pinkk
Well, it preserves length, so it's definitely a rigid motion. It obviously can't be a rotation since all rotations have determinant 1. A determinant of -1 means it reverses orientation, so it's gonna be a reflection, or a rotation and a reflection. I think it's the latter but I really don't know how to go about finding it and proving it.
• Mar 22nd 2011, 10:46 PM
FernandoRevilla
As a consequence of a well known theorem, if $\displaystyle A\in\mathbb{R}^{3\times 3}$ is orthogonal and $\displaystyle \det A=-1$ then, $\displaystyle A$ is orthogonally similar to a matrix of the form:

$\displaystyle B=\begin{bmatrix}{-1}&{0}&{\;\;\;0}\\{\;\;0}&{\cos \alpha}&{-\sin \alpha}\\{\;\;0}&{\sin \alpha}&{\;\;\;\cos \alpha}\end{bmatrix}\quad (\alpha\in\mathbb{R})$

Besides,

$\displaystyle B=\begin{bmatrix}{-1}&{0}&{0}\\{\;\;0}&{1}&{0}\\{\;\;0}&{0}&{1}\end{b matrix}\begin{bmatrix}{1}&{0}&{\;\;\;0}\\{0}&{\cos \alpha}&{-\sin \alpha}\\{0}&{\sin \alpha}&{\;\;\;\cos \alpha}\end{bmatrix}$

Now, you can conclude.