Axis of rotation, plane of reflection

[JamesGoh]

For a 3x3 orthogonal matrix with determinant= -1 (which means rotation followed by simple reflection), is the axis of rotation the same as the plane of reflection ?

My reasoning is follows (see attachment)

Say you have two vectors with the same angle size (which i call A), same x-values, but one of the z-values (height in this case), is the negative of the other (n.b. y-value is zero in both vectors )

Vector 1 rotates around the x-axis by 2A to get to the same spot as vector 2. Because the angle size is the same and because the z-component of vector 2 is the negative of the z-component of vector 1, we get a reflection ?

Since the "reflection" happens about the x-axis, this is why the plane of reflection is the same as the axis of rotation in the case of a 3x3 orthogonal matrix having determinant 1 ?

 

[jostpuur]

if A\in \textrm{O}(3,\mathbb{R}) and \det(A)=-1, there exists a U\in \textrm{SU}(3) such that

<br /> UAU^{\dagger} = \left(\begin{array}{ccc}<br /> -1 &amp; 0 &amp; 0 \\<br /> 0 &amp; e^{i\theta} &amp; 0 \\<br /> 0 &amp; 0 &amp; e^{-i\theta} \\<br /> \end{array}\right)<br />

with some \theta\in\mathbb{R}.

Then there exists a V\in \textrm{SU}(2) such that

<br /> V\left(\begin{array}{cc}<br /> e^{i\theta} &amp; 0 \\<br /> 0 &amp; e^{-i\theta} \\<br /> \end{array}\right)V^{\dagger}<br /> = \left(\begin{array}{cc}<br /> \cos(\theta) &amp; -\sin(\theta) \\<br /> \sin(\theta) &amp; \cos(\theta) \\<br /> \end{array}\right)<br />

So if you define

<br /> W = \left(\begin{array}{cc}<br /> 1 &amp; 0 \\<br /> 0 &amp; V \\<br /> \end{array}\right)U<br />

then WAW^{\dagger} will be of such form that reflection and rotation are clearly carried out with respect to the same axis. Only problem is that W doesn't necessarily have only real entries. How to prove that W is necessarily proportional to a real matrix?