Quadratic sequence that converges to the closest eigenvalue

Question

I am trying to generate a sequence $[q(k)]^TAq(k)$ where $q(k) \in \mathbb{R}^n$ and $A \in \mathbb{R}^{nxn}$ is symmetric such that this sequence converges to the eigenvalue of A closest to 2. So, I want to ask, under which conditions I can generate such a sequence and what would it be as an iterative formula?

Answer 1

You can do the following:

Let $v(k):=(A-2I)^{-1}q(k)$, starting from a random $q(0)$.
$q(k+1):=v(k)/|v(k)|$.

Then $q(k)$ generically converges to that eigenvector $w$ whose eigenvalue $\lambda$ is closest to $2$; alternatively it can oscillate between $\pm w$, but this won't matter. (There are exceptional cases, e.g. if $A-2I$ is not invertible, or if there happen to be more than one eigenvalue equally distant from $2$.)

So $q(k)^TAq(k)\to w^TAw=\lambda$.