Let matrix $A \in \mathbb{R}^{n\times n}$ be such that there is eigenbasis $e_1, \dots, e_n$ where eigenvector $e_i$ is associated with eigenvalue $\lambda_i$. Suppose that $|\lambda_1| > |\lambda_2| \geq \dots \geq |\lambda_n|$.
Since $e_1, \dots, e_n$ is eigenbasis every vector $q \in \mathbb{R}^{n}$ has unique representation $q = \sum_{i=1}^{n} q_i e_i$. Suppose that $q_1 \neq 0$ and consider the sequence $\frac{A^k q}{|| A^k q ||}$. I'm struggling to prove that this sequence is (not) convergent.
Let's denote $x_k = A^k q$. Vectors $e_1, \dots, e_n$ are eigenvectors so $Ae_i = \lambda_i e_i$. It's easy to see that $A^k e_i = \lambda_{i}^{k} e_i$ and so $$ A^k q = \sum_{i=1}^{n} q_i \lambda_{i}^{k} e_i = q_1 \lambda_{1}^{k} \left(e_1 + \sum_{i=1}^{n} \frac{q_i}{q_1} \left(\frac{\lambda_{i}}{\lambda_{1}}\right)^k \right)$$
$$ ||A^k q|| = |q_1| |\lambda_{1}^{k}| \left|\left|e_1 + \sum_{i=1}^{n} \frac{q_i}{q_1} \left(\frac{\lambda_{i}}{\lambda_{1}}\right)^k \right|\right|$$
$$ \frac{A^k q}{||A^k q||} = sign(q_1) sign(\lambda_{1}^{k}) \frac{e_1 + \sum_{i=1}^{n} \frac{q_i}{q_1} \left(\frac{\lambda_{i}}{\lambda_{1}}\right)^k}{\left|\left|e_1 + \sum_{i=1}^{n} \frac{q_i}{q_1} \left(\frac{\lambda_{i}}{\lambda_{1}}\right)^k \right|\right|}$$
Since $|\lambda_1| > |\lambda_i|$ for any $i \geq 2$ then $|\frac{\lambda_i}{\lambda_1}| < 1$ and thus we can conclude that $$ e_1 + \sum_{i=1}^{n} \frac{q_i}{q_1} \left(\frac{\lambda_{i}}{\lambda_{1}}\right)^k \longrightarrow e_1 $$ and so depedning of sign of $\lambda_1$ we have:
1) if $\lambda_1 > 0$ then $\frac{A^k q}{||A^k q||} \rightarrow sign(q_1) \frac{e_1}{||e_1||}$
2) if $\lambda_1 < 0$ then
2.1) $\frac{A^{2k} q}{||A^{2k} q||} \rightarrow sign(q_1) \frac{e_1}{||e_1||}$ 2.2) $\frac{A^{2k+1} q}{||A^{2k+1} q||} \rightarrow -sign(q_1) \frac{e_1}{||e_1||}$
So the sequence either converges to $v = sign(q_1) \frac{e_1}{||e_1||}$ or has exactly two limit points: $v$ and $-v$. Since $v \neq 0$ we can't state that the sequence converges in this case but still the larger k the closer we are to linear span of $e_1$.
Here's my question: is the above reasoning correct or am i missing something? Perhaps, we can modify somehow the sequence $\frac{A^k q}{||A^k q||}$ so that it's always convergent?
Seems right. Also note that $\lambda_1$ is real by default: a real matrix whose largest eigenvalue is complex has another eigenvalue of exactly the same modulus, which violates the assumptions of the power method. So you do not need to worry about the possibility of a complex eigenvalue (which such a method can never find).
As for guaranteeing convergence, one fix would be:
$$v^{(k+1)}=A u^{(k)};u^{(k)}=\frac{v^{(k)} \operatorname{sign}(v_{j_k})}{\| v^{(k)} \|}$$
where $j_k=\min \{ i : v^{(k)}_i \neq 0 \}$. (Of course for most matrices and most initial guesses, $j_k$ will just be $1$ forever.) After some number of iterations, the sequence $j_k$ will become constant, and after that you will have that $u^{(k)}_{j_k}$ keeps the same sign, which will force convergence. But this is a bit messy and unnecessary for practical purposes.