The complex eigenvalues of a Rotation matrix are $e^{-i\theta}$ and $e^{i\theta}$. Corresponding to these we get complex eigenvectors.

We know that the eigenvector corresponding to the eigenvalue 1 is the axis of rotation. How to prove that the $\theta$ from complex eigenvalues is the angle of rotation.

Comparing it with some Rotation matrix for rotation about x,y or z axis is easy, but how to conclude this for a generic R belonging to SO(3).


Suppose $e^{i\theta}\neq e^{-i\theta}$. I leave to you the case $e^{i\theta}=e^{-i\theta}$.

Let $v=v_1+iv_2$ be a column vector of $\mathbb{C}^3$ such that $Rv=e^{i\theta}v$, where $v_1,v_2\in\mathbb{R^3}$ (Prove that $v_1\neq 0$ and $v_2\neq 0$). Let $v^t$ be its transpose.

Now, $R^tR=Id$ then $v=R^tRv=e^{i\theta}R^tv$. So $R^tv=e^{-i\theta}$.

Next, $e^{i\theta}v^tv=v^tRv=(v^tRv)^t=v^tR^tv=e^{-i\theta}v^tv$.

Since $e^{i\theta}\neq e^{-i\theta}$ then $0=v^tv=|v_1|^2-|v_2|^2+i2(v_2^tv_1)$. So $|v_1|=|v_2|$ and $v_1\perp v_2$. Without loss of generality assume $|v_1|=|v_2|=1$.

Now, $R(v_1)+iR(v_2)=R(v_1+iv_2)=(cos(\theta)+isin(\theta))(v_1+iv_2)=$ $(cos(\theta)v_1-sin(\theta) v_2)+i(sin(\theta) v_1+cos(\theta)v_2)$.

So $R(v_1)=(cos(\theta)v_1-sin(\theta) v_2)$ and R$(v_2)=(sin(\theta) v_1+cos(\theta)v_2)$.

Let $W$ be the plane generated by $v_1,v_2$.

Finally, the matrix of the restriction of R to $W$ relative to the orthonormal basis $v_1,v_2$ is $\left(\begin{array}{cc} cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta) \end{array}\right) $.


We know that a rotation matrix has real entries: $${ \bf R_\theta } = \left[\begin{array}{rr} \cos(\theta) &-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{array}\right] = /a = \cos(\theta), b = \sin(\theta)/ = \left[\begin{array}{rr} a &-b\\b&a\end{array}\right]$$

Now consider $$|{\bf R_\theta} - \lambda {\bf I}| = (a-\lambda)^2+b^2 = 0$$ Expand to find a polynomial in $\lambda$ with real coefficients. Such a polynomial must have pairs of complex conjugate roots. We could just factor directly using the conjugacy rule $x^2-y^2 = (x+y)(x-y)$:

$$(a+bi - \lambda)(a-bi-\lambda) = 0$$

And we see the roots are $\lambda = a \pm bi= \cos(\theta) \pm i \sin(\theta)$

Finally since sin is odd, this means we can put the signs inside, altering only the sign of sine:

$\lambda = \cos(\theta) \pm i \sin(\theta) = \cos(\theta) + i\sin(\mp \theta)$

So far is enough to show a 2D rotation matrix eigenvalues has the properties you sought. A 3D rotation is just a shift of basis to the plane of rotation and the axis of rotation. The vector corresponding to the axis is identity and the vectors in the plane are rotated. As this is a direct sum it has the eigenvalues of the elements of the direct sum, so it "inherits" the eigenvalues from it's 2D and 1D components.


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