Periodic Sequence - Radius of Trajectory

The general solution of this dynamic system is:

$\mathrm{v}(k)= A^k \mathrm{v}(0)$

If the solution is a circle and is periodic, then there must be a positive integer $k$ such that:

$\mathrm{v}(k)=\mathrm{v}(0) = A^k \mathrm{v}(0)$ $\implies A^k = I$

The eigenvalue decomposition of A is can be done as such:

$A = V D V^{-1}$

$A^k = VD^k V^{-1}$

For the above matrix to be the identity matrix, basically:

$D^k = I$

Here, $D$ is already a diagonal matrix with entries:

$D = \left[\begin{matrix} 0.9397+0.342i&0&0\\0&0.9397-0.342i&0\\0&0&1\end{matrix}\right]$

Let:

$\mathrm{e}^{i\theta} = 0.9397+0.342i$

$\implies D = = \left[\begin{matrix}\mathrm{e}^{i\theta} &0&0\\0&\mathrm{e}^{-i\theta} &0\\0&0&1\end{matrix}\right]$ $\implies D^k = \left[\begin{matrix}\mathrm{e}^{ik\theta} &0&0\\0&\mathrm{e}^{-ik\theta} &0\\0&0&1\end{matrix}\right]$

For the above matrix to be the identity matrix:

$k\theta = 2\pi$ $k =18$

So the 19th term which is $\mathrm{v}(18)$ is when the trajectory completes a complete circle. The term in between the first and 19th term is the 10th term which is $\mathrm{v}(9)$ . Therefore, the radius is:

$R = \frac{\lvert \mathrm{v}(9)- \mathrm{v}(0)\rvert}{2} =1.232$

$\mathrm{v}(9)$ corresponds to

$k\theta = \pi$

It is essentially the point diametrically opposite to the initial vector, on the trajectory.