MHB Is the 3-D Rotation Matrix Defined by Euler Rotations or a General Angle?

[ognik]

The question mentions an orthogonal matrix describing a rotation in 3D ... where $\phi$ is the net angle of rotation about a fixed single axis. I know of the 3 Euler rotations, is this one of them, arbitrary, or is there a general 3-D rotation matrix in one angle?

If I build one, I would start with the direction cosines $ \begin{bmatrix}cos(x', x)&cos(y', x)&cos(z', x)\\cos(x', y)&cos(y', y)...\\...\end{bmatrix}$

Lets say we rotate a total of $\phi$, I think this means $\phi = \phi_x + \phi_y + \phi_z$? But around the z axis only (for example), $\phi = \phi_z$?

So I'm not sure how to apply this to the matrix above, is everything except w.r.t. z = $\delta_{ij}$?

 

[ognik]

Would appreciate corrections/confirmations to the above please - and if I put something confusingly I'll be happy to improve it, if I know what it is :-)

 

[ognik]

I'm now sure the question could use any of the 3 Euler (orthogonal) rotation matrices, the diagonals of each have 2 $Cos \phi$ terms and a 1. i.e. the sum of the 3 eigenvalues is $2Cos \phi + 1$

The question now is - given 1 eigenvalue = 1, show the other 2 = $e^{\pm i\phi}$

Choosing the rotation about the z axis, $R_z = \begin{bmatrix}Cos&-Sin&0\\Sin&Cos&0\\0&0&1\end{bmatrix}$

My Characteristic eqtn is $ (1-\lambda)(Cos^2\phi -2\lambda Cos\phi + Sin^2 \phi) $ = $ (1-\lambda)(1-2\lambda Cos\phi) $

The simplest (to me anyway) roots are $\lambda = 1$ (as expected) and $ \lambda = \frac{1}{2 Cos\phi}$ (But 2 complex roots expected?)

Now I could say that $ Cos\phi =Re\left\{ \frac{e^{i \phi}+e^{-i \phi}} {2} \right\} $ and dredge $\lambda = e^{\pm i\phi}$ out of this - but the question states that these 2 should be complex eigenvalues and I have to take the real parts to make this work?

 

[ognik]

Just revisiting this and noticed a silly mistake, off course $ Cos \phi = \frac{1}{2} \left( e^{i \phi} + e^{-i \phi} \right)$. Also my characteristic eqtn was wrong (Doh) ,

$ (1-\lambda)(Cos^2\phi -2\lambda Cos\phi + \lambda^2 + Sin^2 \phi) $ = $ (1 -\lambda)(\lambda^2-2\lambda Cos\phi +1) $ ... which indeed provides 2 complex roots - $e^{i \phi}, e^{-i \phi} $ - in addition to $\lambda = 1$