I am working on some practice problems for the convergence of power method for some given recursion relationship and I am trying to generalize/reflect on the question after having been stuck on the question as follows:
First, some background on the question as given:
I understood everything here but the following parts are what kind of stumped me:
So in theory, I tried out the following answer from part (b) where I used an arbitrarily large loop (5000000) to try to check the convergence with an initial vector of $$x_0 = 2v_2 + 3v_3 = \left[ \begin{matrix} -0.7060 + 0.3752i \\ -1.5471 - 0.4471i \\ 3.6868 + 0.0000i \end{matrix} \right]$$
where $v_2$ and $v_3$ are the eigenvectors corresponding to the complex eigenvalues of $A$. However, I managed to get that $$x_n = \left[ \begin{matrix} 0.8501 + 0.0216i \\ 0.4622 + 0.0117i \\ 0.2513 + 0.0064i \end{matrix}\right]$$
Which does not appear to be converging at all.
In theory, I understand that since $\left| \lambda_2 \right| < 1$ and $\left| \lambda_3 \right| < 1$, then it follows that $\lim_{n\rightarrow \infty} A^nv_2= \lambda_2^nv_2$ where $v_2$ is an eigenvector corresponding to $\lambda_2$. Since $\left| \lambda_2 \right| < 1$, then $\lambda^n = 0$ and so $A^nv_2 = 0$ at the limit.
Why is it in this case that I am not seeing anything near convergence towards 0 here ? Is it because of the fact that it is asymptotic and relatively, 5000000 is very small compared to infinity ?
Also as a second question: Wouldn't the eigenvector,$v_1$, which corresponds to the eigenvalue ($\lambda_1 = 1$) also be guaranteeing a real $x_0$ ?
Any advice is appreciated. Thank you
Think about what multiplying your matrix $A$ by a vector is doing. You are essentially applying your rule to a given input set $x_n,x_{n+1},x_{n+2}$ to obtain $x_{n+1},x_{n+2},x_{n+3}$. If you apply your rule with initial values as nonreal complex numbers, in general you are going to end up with nonreal complex outputs after many applications of your rule.
If you have a complex eigenvector $v$ with corresponding real eigenvalue $\lambda$ then $Av=\lambda v$ is again in general going to be nonreal. If you input a real eigenvector-eigenvalue pair you should be getting purely real outputs.