Axis and angle of rotation

{e,f,g} an orthonormal basis in E^3.

For the transition matrix P I have Tr P = 2cos(theta) + cos^2(theta) = 1 + 2cos(theta) so angle of rotation is pi.

If theta = pi/2 then Tr P = 0 = 1 + 2cos(theta) so angle is 2pi/3.

Solving Px = x for the axis seems tedious and I'm not sure about justifying the formula.

And are the above angles correct?
Thank you.

'Fraid I know nothing about this, so can't help much. From a bit of Googling.

If this is just matrix multiplication then I agree with your trace of P as
$2\cos(\theta) + \cos^2(\theta)$

If this now equals $1+2\cos\phi$ (formula for the trace of a 3D rotation matrix) where phi is the angle of rotation, then after rearranging into that form we have:

$\phi = \cos^{-1}(\cos\theta -\frac{1}{2}\sin^2\theta)$

As to how you show it's a rotation matrix, I don't know.

If $\theta = \frac{\pi}{2}$

then $\phi = \cos^{-1}(-\frac{1}{2})$

Which gives two possibiities, 2pi/3 and 4pi/3. Don't know the rationale for choosing one over the other.

And that's about my limit.

Going to need someone with experience of this.

Any idea about justifying the formula? Not sure where root2 comes from.

From the formula for a 3D rotation using the trace, we have:

$\cos\phi = \cos\theta -\frac{1}{2}\sin^2\theta$

Then series expansion as far as first non-constant term in each, yields the desired result.

Is there a nice way to find the axis of rotation of the composition?

I have an answer but I'm not confident.

From the formula for a 3D rotation using the trace, we have:

$\cos\phi = \cos\theta -\frac{1}{2}\sin^2\theta$

Then series expansion as far as first non-constant term in each, yields the desired result.

What argument are we suppose to give though?