Let $\theta>0$, and for all $k\in\mathbb{N}$, define $f_k\triangleq \begin{bmatrix}\cos k\theta\\\sin k\theta\end{bmatrix}\in\mathbb{R}^2$. Show that for all $k\in\mathbb{N}$, $${\rm spec}\big((I_2-f_kf_{k}^{\rm T})(I_2-f_{k+1}f_{k+1}^{\rm T})\big)=\{\lambda_1,0\}, \qquad {\rm where}\qquad |\lambda_1|<1$$
What I know sofar:
Note that $$f_{k+1}=Rf_k,\qquad {\rm where}\quad R\triangleq\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}. $$
Also note that for all $k\in\mathbb{N}$, $${\rm spec~}(I_2-f_kf_{k}^{\rm T})=\{1,0\}.$$
Brute force computation shows that $$(I_2-f_kf_k^T)(I_2-f_{k+1}f_{k+1}^T)=\left( \begin{array}{cc} \cos (\theta ) \sin (k \theta ) \sin ((k+1) \theta ) & -\cos (\theta ) \cos ((k+1) \theta ) \sin (k \theta ) \\ -\cos (\theta ) \cos (k \theta ) \sin ((k+1) \theta ) & \cos (\theta ) \cos (k \theta ) \cos ((k+1) \theta ) \\ \end{array} \right)$$
The characteristic equation is $$Det((I_2-f_kf_k^T)(I_2-f_{k+1}f_{k+1}^T)-\lambda I_2)=\frac{1}{2} \lambda (-\cos (2 \theta )+2 \lambda -1)$$ This gives $$\lambda =\frac{1}{2} (\cos (2 \theta k-2 \theta (k+1))+1)$$