Let $v_1, \dots, v_m$ be eigenbasis of real matrix $A$ where eigenvector $v_i$ is associated with eigenvalue $\lambda_i$ and $|\lambda_1| > |\lambda_2| \geq \dots \geq |\lambda_m|.$
I'm trying to understand the following proof:
My problem is the last statement. It's more-or-less obvious that $\frac{A^k x0}{\lambda_{1}^{k}} \rightarrow c_1 v_1$ but i don't understand how normalization helps us to eliminate the unknown eigenvalue $\lambda_1$.
I can prove the following. Since all the vectors $\frac{A^k x0}{||A^k x0||}$ are normalized then $|| \frac{A^k x0}{||A^k x0||} || \rightarrow 1$. Since $\frac{A^k x0}{\lambda_{1}^{k}} \rightarrow c_1 v_1$ and vectors $v_1, \dots , v_m$ are normalized then $|| \frac{A^k x0}{\lambda_{1}^{k}} || \rightarrow |c_1| ||v_1|| = |c_1|$. From these two facts and equality $|| \frac{A^k x0}{||A^k x0||} || = \frac{|\lambda_{1}^{k}|}{||A^k x0||} || \frac{A^k x0}{\lambda_{1}^{k}} ||$ we can deduce that exists the limit $\frac{|\lambda_{1}^{k}|}{||A^k x_0||} \rightarrow \frac{1}{|c_1|}$
I'm not sure whether this helps but it's the only result i've managed to get so far. If $\lambda_1 \geq 0$ then
$$ \frac{A^k x0}{||A^k x0||} = \frac{\lambda_{1}^{k}}{||A^k x0||} \frac{A^k x0}{\lambda_{1}^{k}} = \frac{|\lambda_{1}^{k}|}{||A^k x0||} \frac{A^k x0}{\lambda_{1}^{k}} \rightarrow \frac{1}{|c_1|} c_1 v_1 = \frac{c_1}{|c_1|} v_1$$ and that's the result we need. But what if $\lambda_1$ is negative? How to prove the convergence in this case?
Thanks in advance for any ideas.
Best Answer
I think the power in the denominator should be $k-1$, not $k$, at least according to my understanding of the algorithm. If that is correct, the result would follow from the derivation in the proof if we could show that
$$\frac{c_1 \lambda_1^{k-1}}{||A^{k-1} x_0||} \to 1,$$
would you agree? Well, for this it suffices to notice that
$$||A^k x_0|| = \sqrt{\sum_{i=1}^m c_i^2 \lambda_i^{2{k-2}} } \sim c_1 \lambda_{1}^{k-1}$$
by the fact that $|\lambda_j/\lambda_1|^k \to 0$ for $j \neq 1$. And I believe this all of it. By the way, the unspecified multiple of the eigenvector is just the corresponding eigenvalue of $A$.